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Project Gutenberg's Diophantine Analysis, by Robert Carmichael

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Title: Diophantine Analysis

Author: Robert Carmichael

Release Date: December 9, 2006 [EBook #20073]

Language: English

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\frontmatter
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{\LARGE MATHEMATICAL MONOGRAPHS}

\textsc{edited by}

\textsc{MANSFIELD MERRIMAN and ROBERT S. WOODWARD}

\rule{0pt}{72pt}

\begin{Huge}
No.~16

\rule{0pt}{24pt}

DIOPHANTINE
ANALYSIS
\end{Huge}

\rule{0pt}{72pt}

BY

\rule{0pt}{12pt}

ROBERT D. CARMICHAEL,

\textsc{Assistant Professor Of Mathematics In The University Of Illinois}

\rule{0pt}{72pt}

FIRST EDITION

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{\small FIRST THOUSAND}

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NEW YORK

\textsc{JOHN WILEY \& SONS, Inc.}

\textsc{London: CHAPMAN \& HALL, Limited}

1915

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\textsc{Copyright, 1915,}

\textsc{by}

ROBERT D. CARMICHAEL

\rule{0pt}{72pt}

\textsf{THE SCIENTIFIC PRESS}

\textsf{\small ROBERT DRUMMOND AND COMPANY}

\textsf{BROOKLYN, N.~Y.}

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% [File: 007.png Page: iii]
\newpage
\pagestyle{plain}
\chapter*{PREFACE}

The author's purpose in writing this book has been to supply
the reader with a convenient introduction to Diophantine
Analysis. The choice of material has been determined by
the end in view. No attempt has been made to include all
special results, but a large number of them are to be found
both in the text and in the exercises. The general theory
of quadratic forms has been omitted entirely, since that subject
would require a volume in itself. The reader will therefore
miss such an elegant theorem as the following: Every
positive integer may be represented as the sum of four squares.
Some methods of frequent use in the theory of quadratic forms,
in particular that of continued fractions, have been left out
of consideration even though they have some value for other
Diophantine questions. This is done for the sake of unity
and brevity. Probably these omissions will not be regretted,
since there are accessible sources through which one can make
acquaintance with the parts of the theory excluded.

For the range of matter actually covered by this text there
seems to be no consecutive exposition in existence at present
in any language. The task of the author has been to systematize,
as far as possible, a large number of isolated investigations
and to organize the fragmentary results into a connected
body of doctrine. The principal single organizing idea
here used and not previously developed systematically in the
literature is that connected with the notion of a multiplicative
domain introduced in Chapter II.

The table of contents affords an indication of the extent
and arrangement of the material embodied in the work.

% [File: 008.png Page: iv]
Concerning the exercises some special remarks should be
made. They are intended to serve three purposes: to afford
practice material for developing facility in the handling of
problems in Diophantine analysis; to give an indication of
what special results have already been obtained and what
special problems have been found amenable to attack; and to
point out unsolved problems which are interesting either from
their elegance or from their relation to other problems which
already have been treated.

Corresponding roughly to these three purposes the problems
have been divided into three classes. Those which have
no distinguishing mark are intended to serve mainly the purpose
first mentioned. Of these there are 133, of which 45 are
in the Miscellaneous Exercises at the end of the book. Many
of them are inserted at the end of individual sections with
the purpose of suggesting that a problem in such position is
readily amenable to the methods employed in the section to
which it is attached. The harder problems taken from the
literature of the subject are marked with an asterisk; they
are 53 in number. Some of them will serve a disciplinary
purpose; but they are intended primarily as a summary of
known results which are not otherwise included in the text or
exercises. In this way an attempt has been made to gather
up into the text and the exercises all results of essential or
considerable interest which fall within the province of an
elementary book on Diophantine analysis; but where the
special results are so numerous and so widely scattered it can
hardly be supposed that none of importance has escaped
attention. Finally those exercises which are marked with
a dagger (35 in number) are intended to suggest investigations
which have not yet been carried out so far as the author
is aware. Some of these are scarcely more than exercises,
while others call for investigations of considerable extent
or interest.

\begin{flushright}
\textsc{Robert D. Carmichael.}\qquad\null
\end{flushright}

% [File: 009.png Page: v]
\tableofcontents

%% Original table of contents:
%% CONTENTS
%% 
%% CHAPTER I. INTRODUCTION. RATIONAL TRIANGLES. METHOD
%% OF INFINITE DESCENT
%% 
%% § 1. INTRODUCTORY REMARKS     1
%% § 2. REMARKS RELATING TO RATIONAL TRIANGLES     8
%% § 3. PYTHAGOREAN TRIANGLES. EXERCISES 1-6     9
%% § 4. RATIONAL TRIANGLES. EXERCISES 1-3     11
%% § 5. IMPOSSIBILITY OF THE SYSTEM x^2+y^2=z^2, y^2+z^2=t^2 APPLICATIONS. EXERCISES 1-3     14
%% § 6. THE METHOD OF INFINITE DESCENT. EXERCISES 1-9     18
%% GENERAL EXERCISES 1-10     22
%% 
%% CHAPTER II. PROBLEMS INVOLVING A MULTIPLICATIVE DOMAIN
%% § 7. ON NUMBERS OF THE FORM x^2+axy+by^2. EXERCISES 1-7      24
%% § 8. ON THE EQUATION x^2-Dy^2=z^2. EXERCISES 1-8     26
%% § 9. GENERAL EQUATION OF THE SECOND DEGREE IN TWO VARIABLES     34
%% § 10. QUADRATIC EQUATIONS INVOLVING MORE THAN THREE VARIABLES. EXERCISES 1-6     35
%% § 11. CERTAIN EQUATIONS OF HIGHER DEGREE. EXERCISES 1-3 44
%% § 12. ON THE EXTENSION OF A SET OF NUMBERS SO AS TO FORM A MULTIPLICATIVE DOMAIN     48
%% GENERAL EXERCISES 1-22     52
%% 
%% CHAPTER III. EQUATIONS OF THE THIRD DEGREE
%% § 13. ON THE EQUATION kx^3+ax^2y+bxy^2+cy^3=t^2     55
%% § 14. ON THE EQUATION kx^3+ax^2y+bxy^2+cy^3=t^3        57
%% § 15. ON THE EQUATION   x^3+y^3+z^3-3xyz=u^3=v^3+w^3-3uvw         62
%% § 16. IMPOSSIBILITY OF THE EQUATION x^3+y^3=2^mz^3     67
%% GENERAL EXERCISES 1-26     72
%% 
%% CHAPTER IV. EQUATIONS OF THE FOURTH DEGREE
%% § 17. ON THE EQUATION ax^4=bx^3y+cx^2y^2+dxy^3+ey^4=mz^2. EXERCISES 1-4    74
%% § 18. ON THE EQUATION ax^4+by^4=cz^2.    EXERCISES 1-4     77
%% § 19. OTHER EQUATIONS OF THE FOURTH DEGREE    80
%% GENERAL EXERCISES 1-20     83
%% 
% [File: 010.png Page: vi]
%% CHAPTER V. EQUATIONS OF DEGREE HIGHER THAN THE FOURTH. THE FERMAT PROBLEM
%% § 20. REMARKS CONCERNING EQUATIONS OF HIGHER DEGREE 85
%% § 21. ELEMENTARY PROPERTIES OF THE EQUATION x^n+y^n=z^n, n>2 86
%% § 22. PRESENT STATE OF KNOWLEDGE CONCERNING THE EQUATION x^p+y^p+z^p=0 100
%% GENERAL EXERCISES 1-13    102
%% 
%% CHAPTER VI. THE METHOD OF FUNCTIONAL EQUATIONS
%% § 23. INTRODUCTION. RATIONAL SOLUTIONS OF A CERTAIN FUNCTIONAL EQUATION 104
%% § 24. SOLUTION OF A CERTAIN PROBLEM FROM DIOPHANTUS 106
%% § 25. SOLUTION OF A CERTAIN PROBLEM DUE TO FERMAT 108
%% GENERAL EXERCISES 1-6     111
%% MISCELLANEOUS EXERCISES 1-71    112
%% INDEX     117
%% 
\newpage

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% [File: 011.png Page: 1]
{\Huge DIOPHANTINE ANALYSIS}

\rule{2cm}{.2pt}
\end{center}

\mainmatter
\chapter{INTRODUCTION\@. RATIONAL TRIANGLES\@. METHOD OF INFINITE DESCENT}
\pagestyle{myheadings}
\markright{INTRODUCTION\@. RATIONAL TRIANGLES\@.}

%§ 1.
\section[\textsc{Introductory Remarks}]{Introductory Remarks}\index{Equations of Second Degree|(}\index{Quadratic Equations|(}

\hspace{\parindent}In the theory of Diophantine analysis two closely related
but somewhat different problems are treated. Both of them
have to do primarily with the solution, in a certain sense,
of an equation or a system of equations. They may be characterized
in the following manner: Let $f(x, y, z, \ldots)$ be a
given polynomial in the variables $x$, $y$, $z$, \ldots\ with rational
(usually integral) coefficients and form the equation
\[
f(x, y, z, \ldots)=0.
\]
\index{Diophantine Equation, Definition of}This is called a \textit{Diophantine equation} when we consider it
from the point of view of determining the \textit{rational} numbers
$x$, $y$, $z$, \ldots\ which satisfy it. We usually make a further
restriction on the problem by requiring that the solution
$x$, $y$, $z$, \ldots\ shall consist of \textit{integers}; and sometimes we say that
it shall consist of positive integers or of some other defined
class of integers. Connected with the above equation we
thus have two problems, namely: To find the rational numbers
$x$, $y$, $z$, \ldots\ which satisfy it; to find the integers (or
the positive integers) $x$, $y$, $z$, \ldots\ which satisfy it.

Similarly, if we have several such functions $f_i(x, y, z, \ldots)$,
in number less than the number of variables, then the set
of equations
\[
f_i(x, y, z, \ldots) = 0
\]
is said to be a \index{Diophantine System, Definition of}\textit{Diophantine system} of equations.

% [File: 012.png Page: 2]
\index{Solution of Diophantine Equation}
Any set of rational numbers $x$, $y$, $z$, \ldots, which satisfies
the equation [system], is said to be a \index{Rational Solution}\index{Solution, Rational}\textit{rational solution} of the
equation [system]. An \index{Integral Solution}\index{Solution, Integral}\textit{integral solution} is similarly defined. The
\textit{general rational} [\textit{integral}] \textit{solution} is a solution or set of solutions
containing all rational [integral] solutions. A \index{Primitive Solution}\index{Solution, Primitive}\textit{primitive
solution} is an integral solution in which the greatest common
divisor of the values of $x$, $y$, $z$, \ldots\ is unity.

A certain extension of the foregoing definition is possible.
One may replace the function $f(x, y, z, \ldots)$ by another which
is not necessarily a polynomial. Thus, for example, one may
ask what integers $x$ and $y$ can satisfy the relation
\[
x^y-y^x = 0.
\]
This more extended problem is all but untreated in the literature.
It seems to be of no particular importance and therefore
will be left almost entirely out of account in the following
pages.

We make one other general restriction in this book; we
leave linear equations out of consideration. This is because
their theory is different from that of non-linear equations
and is essentially contained in the theory of linear congruences.

That a Diophantine equation may have no solution at
all or only a finite number of solutions is shown by the
examples
\[
x^2+y^2+1=0, \quad x^2+y^2-1=0.
\]
Obviously the first of these equations has no rational solution
and the second only a finite number of integral solutions.
That the number of rational solutions of the second is infinite
will be seen below. Furthermore we shall see that the equation
$x^2+y^2=z^2$ has an infinite number of integral solutions.

In some cases the problem of finding rational solutions
and that of finding integral solutions are essentially equivalent.
This is obviously true in the case of the equation
$x^2+y^2 = z^2$. For, the set of all rational solutions contains
the set of all integral solutions, while from the set of all integral
solutions it is obvious that the set of all rational solutions is
obtained by dividing the numbers in each solution by an
% [File: 013.png Page: 3]
arbitrary positive integer. In a similar way it is easy to see
that the two problems are essentially equivalent in the case
of every homogeneous equation.

In certain other cases the two problems are essentially
different, as one may see readily from such an equation as
$x^2+y^2 = 1$. Obviously, the number of integral solutions is
finite; moreover, they are trivial. But the number of rational
solutions is infinite and they are not all trivial in character,
as we shall see below.

Sometimes integral solutions may be very readily found
by means of rational solutions which are easily obtained in
a direct way. Let us illustrate this remark with an example.
Consider the equation
\[
x^2+y^2=z^2.   \tag{1}
\]
The cases in which $x$ or $y$ is zero are trivial, and hence they are
excluded from consideration. Let us seek first those solutions
in which $z$ has the given value $z = 1$. Since $x \neq 0$ we may write
$y$ in the form $y = 1-mx$, where $m$ is \textit{rational}. Substituting
in (1) we have
\[
x^2+(1-mx)^2 = 1.
\]
This yields
\[
x=\frac{2m}{1+m^2};
\]
whence
\[
y=\frac{1-m^2}{1+m^2}.
\]
This, with $z = 1$, gives a \textit{rational} solution of Eq.~(1) for every
rational value of $m$. (Incidentally we have in the values of
$x$ and $y$ an infinite set of rational solutions of the equation
$x^2+y^2 = 1$.)

If we replace $m$ by $q/p$, where $q$ and $p$ are relatively prime
integers, and then multiply the above values of $x$, $y$, $z$ by $p^2+q^2$,
we have the new set of values
\[
x=2pq, \quad y=p^2-q^2, \quad z=p^2+q^2.
\]
This affords a two-parameter \textit{integral} solution of (1).

In § 3 we return to the theory of Eq.~(1), there deriving
the solution in a different way. The above exposition has
% [File: 014.png Page: 4]
been given for two reasons: It illustrates the way in which
rational solutions may often be employed to obtain integral
solutions (and this process is frequently one of considerable
importance); \index{Historical Remarks|(}again, the spirit of the method is essentially
that of the Greek mathematician \index{Diophantus}Diophantus, who flourished
probably about the middle of the third century of our era
and who wrote the first systematic exposition of what is now
known as Diophantine analysis. The reader is referred to
Heath's \textit{Diophantos of Alexandria} for an account of this work
and for an excellent abstract (in English) of the extant writings
of Diophantus.

The theory of Diophantine analysis has been cultivated
for many centuries. As we have just said, it takes its name
from the Greek mathematician Diophantus. The extent to
which the writings of Diophantus are original is unknown, and
it is probable now that no means will ever be discovered for
settling this question; but whether he drew much or little
from the work of his predecessors it is certain that his \textit{Arithmetica}
has exercised a profound influence on the development
of number theory.

The bulk of the work of Diophantus on the theory of
numbers consists of problems leading to indeterminate equations;
these are usually of the second degree, but a few indeterminate
equations of the third and fourth degrees appear and at least
one easy one of the sixth degree is to be found. The general
type of problem is to find a set of numbers, usually two or
three or four in number, such that different expressions involving
them in the first and second and third degrees are
squares or cubes or otherwise have a preassigned form.

As good examples of these problems we may mention the
following: To find three squares such that the product of any
two of them added to the sum of those two or to the remaining
one gives a square; to find three squares such that their continued
product added to any one of them gives a square;
to find two numbers such that their product plus or minus
their sum gives a cube. (See Chapter VI.)

Diophantus was always satisfied with a rational result
% [File: 015.png Page: 5]
even though it appeared in fractional form; that is, he did
not insist on having a solution in integers as is customary in
most of the recent work in Diophantine analysis.

It is through \index{Fermat}Fermat that the work of \index{Diophantus}Diophantus has
exercised the most pronounced influence on the development
of modern number theory. The germ of this remarkable growth
is contained in what is only a part of the original Diophantine
analysis, of which, without doubt, Fermat is the greatest master
who has yet appeared. The remarks, method and results of
the latter mathematician, especially those recorded on the
margin of his copy of Diophantus, have never ceased to be
the marvel of other workers in this fascinating field. Beyond
question they gave the fundamental initial impulse to the
brilliant work in the theory of numbers which has brought
that subject to its present state of advancement.

Many of the theorems announced without proof by Fermat
were demonstrated by \index{Euler}Euler, in whose work the spirit
of the method of Diophantus and Fermat is still vigorous.
In the \textit{Disquisitiones Arithmeticæ}, published in 1801, \index{Gauss}Gauss
introduced new methods, transforming the whole subject and
giving it a new tendency toward the use of analytical methods.
This was strengthened by the further discoveries of \index{Cauchy}Cauchy,
\index{Jacobi}Jacobi, \index{Eisenstein}Eisenstein, \index{Dirichlet}Dirichlet, and others.

The development in this direction has extended so rapidly
that by far the larger portion of the now existing body of
number theory has had its origin in this movement. The
science has thus departed widely from the point of view and
the methods of the two great pioneers Diophantus and Fermat.

Yet the methods of the older arithmeticians were fruitful
in a marked degree.\footnote
 {Cf.\ G.~B.~\index{Mathews}Mathews, \textit{Encyclopaedia Britannica}, 11th edition, Vol.~XIX, p.~863.}
They announced several theorems
which have not yet been proved or disproved and many others
the proofs of which have been obtained by means of such
difficulty as to make it almost certain that they possessed
other and simpler methods for their discovery. Moreover
they made a beginning of important theories which remain
to this day in a more or less rudimentary stage.

% [File: 016.png Page: 6]
\index{Development of Method|(}\index{Method , Development of@Method, Development of|(}
During all the intervening years, however, there has been
a feeble effort along the line of problems and methods in indeterminate
equations similar to those to be found in the works
of Diophantus and \index{Fermat}Fermat; but this has been disjointed and
fragmentary in character and has therefore not led to the
development of any considerable body of connected doctrine.
Into the history of this development we shall not go; it will
be sufficient to refer to general works of reference\footnote
 {See \textit{Encyclop\'edie des sciences math\'ematiques}, tome I, Vol.~III, pp.~27--38, 201--214;
 \textit{Royal Society Index}, Vol.~I, pp.~201--219.}
by means
of which the more important contributions can be found.

Notwithstanding the fact that the Diophantine method
has not yet proved itself particularly valuable, even in the
domain of Diophantine equations where it would seem to be
specially adapted, still one can hardly refuse to believe that
it is after all the method which is really germane to the subject.
It will of course need extension and addition in some
directions in order that it may be effective. There is hardly
room to doubt that \index{Fermat}Fermat was in possession of such extensions
if he did not indeed create new methods of a kindred
sort. More recently \index{Lucas}Lucas\footnote
 {\textit{American Journal of Mathematics}, Vol.~I (1878), pp.~184, 289.}
has revived something of the
old doctrine and has reached a considerable number of interesting
results.

The fragmentary character of the body of doctrine in
Diophantine \label{anlysis}analysis seems to be due to the fact that the history
of the subject has been primarily that of special problems.
At no time has the development of method been conspicuous,
and there has never been any considerable body of doctrine
worked out according to a method of general or even of fairly
general applicability. The earliest history of the subject has
been peculiarly adapted to bring about this state of things.
It was the plan of presentation of Diophantus to announce
a problem and then to give a solution of it in the most convenient
form for exposition, thus allowing the reader but small
opportunity to ascertain how the author was led either to the
problem or to its solution. The contributions of \index{Fermat}Fermat were
% [File: 017.png Page: 7]
mainly in the form of results stated without proof. Moreover,
through their correspondence with \index{Fermat}Fermat or their relation
to him in other ways, many of his contemporaries also
were led to announce a number of results without demonstration.
Naturally there was a desire to find proofs of interesting theorems
made known in this way. Thus it happened that much of the
earlier development of Diophantine analysis centered around
the solution of certain definite special problems or the demonstration
of particular theorems.

There is also something in the nature of the subject itself
which contributed to bring this about. If one begins to investigate
problems of the character of those solved by Diophantus
and \index{Fermat}Fermat he is soon led experimentally to observe certain
apparent laws, and this naturally excites his curiosity as to their
generality and possible means of demonstrating them. Thus
one is led again to consider special problems.

Now when we attack special problems, instead of devising
and employing general methods of investigation in a prescribed
domain, we fail to forge all the links of a chain of reasoning
necessary in order to build up a connected body of doctrine
of considerable extent and we are thus lost amid our difficulties,
because we have no means of arranging them in a natural
or logical order. We are very much in the situation of the
investigator who tries to make headway by considering only
those matters which have a practical bearing. We do not
make progress because we fail to direct our attention to essential
parts of our problems.

It is obvious that the theory of Diophantine analysis is
in need of general methods of investigation; and it is important
that these, when discovered, shall be developed to a wide
extent. In this book are gathered together the important
results so far developed and a number of new ones are added.
Many of the older ones are derived in a new way by means of
two general methods first systematically developed in the
present work. These are the method of the multiplicative
domain introduced in Chapter II and the method of functional
% [File: 018.png Page: 8]
equations employed in Chapter VI\@. Neither of these methods
is here used to the full extent of its capacity; this is especially
true of the latter. In a book such as the present it is natural
that one should undertake only an introductory account of
these methods.\index{Development of Method|)}\index{Method , Development of@Method, Development of|)}\index{Historical Remarks|)}

%§ 2.
\Needspace*{4\baselineskip}
\section[\textsc{Remarks Relating to Rational Triangles}]{Remarks Relating to Rational Triangles}

\hspace{\parindent}A triangle whose sides and area are rational numbers is
called a \index{Triangles, Rational|(}\textit{rational triangle}. If the sides of a rational triangle
are integers it is said to be \textit{integral}. If further these sides
have a greatest common divisor unity the triangle is said to
be \index{Triangles, Primitive}\textit{primitive}. If the triangle is right-angled it is said to be a
\textit{right-angled rational triangle} or a \index{Triangles, Pythagorean}\index{Pythagoras}Pythagoras\textit{Pythagorean triangle} or a
\index{Triangles, Numerical}\textit{numerical right triangle}.

It is convenient to speak, in the usual language of geometry,
of the hypotenuse and legs of the right triangle. If $x$ and $y$
are the legs and $z$ the hypotenuse of a Pythagorean triangle,
then
\[
x^2+y^2 = z^2.
\]
\index{Triangles, Pythagorean|(}Any rational solution of this equation affords a Pythagorean
triangle. If the triangle is primitive, it is obvious that no
two of the numbers $x$, $y$, $z$ have a common prime factor. Furthermore,
all rational solutions of this equation are obtained
by multiplying each primitive solution by an arbitrary rational
number.

From the cosine formula of trigonometry it follows immediately
that the cosine of each angle of a rational triangle
is itself rational. Hence a perpendicular let fall from any
angle upon the opposite side divides that side into two rational
segments. The length of this perpendicular is also a rational
number, since the sides and area of the given triangle are
rational. Hence every rational triangle is a sum of two Pythagorean
triangles which are formed by letting a perpendicular
fall upon the longest side from the opposite vertex. Thus
the theory of rational triangles may be based upon that of
Pythagorean triangles.

A more direct method is also available. Thus if $a$, $b$, $c$
% [File: 019.png Page: 9]
are the sides and $A$ the area of a rational triangle we have
from geometry
\[
(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = 16A^2.
\]
Putting
\[
a=\beta+\gamma, \quad b=\gamma+\alpha, \quad c=\alpha+\beta,
\]
we have
\[
(\alpha+\beta+\gamma)\alpha\beta\gamma=A^2.
\]
Every rational solution of the last equation affords a rational
triangle.

In the next two sections we shall take up the problem of
determining all Pythagorean triangles and all rational triangles.

\index{Historical Remarks}It is of interest to observe that Pythagorean triangles
have engaged the attention of mathematicians from remote
times. They take their name from the Greek philosopher
\index{Pythagoras}Pythagoras, who proved the existence of those triangles whose
legs and hypotenuse in modern notation would be denoted
by $2\alpha+1$, $2\alpha^2+2\alpha$, $2\alpha^2+2\alpha+1$, respectively, where $\alpha$ is a positive
integer. \index{Plato}Plato gave the triangles $2\alpha$, $\alpha^2-1$, $\alpha^2+1$. \index{Euclid}Euclid
gave a third set, while \index{Diophantus}Diophantus derived a formula essentially
equivalent to the general solution obtained in the following
section.

\index{Fermat}Fermat gave a great deal of attention to problems connected
with Pyth\-ag\-or\-ean triangles, and it is not too much to
say that the modern theory of numbers had its origin in the
meditations of Fermat concerning these and related problems.\index{Historical Remarks}

%§ 3.
\Needspace*{4\baselineskip}
\section[\textsc{Pythagorean Triangles. Exercises 1-6}]{Pythagorean Triangles}

\hspace{\parindent}We shall now determine the general form of the positive
integers $x$, $y$, $z$ which afford a primitive solution of the equation
\[
x^2+y^2=z^2.   \tag{1}
\]

The square of the odd number $2\mu+1$ is $4\mu^2+4\mu+1$. Hence
the sum of two odd squares is divisible by 2 but not by 4; and
therefore the sum of two odd squares cannot be a square.
Hence of the numbers $x$, $y$ in (1) one is even. If we suppose
that $x$ is even, then $y$ and $z$ are both odd.

% [File: 020.png Page: 10]
Let us write Eq.\ (1) in the form
\[
x^2 = (z+y)(z-y).  \tag{2}
\]
Every common divisor of $z+y$ and $z-y$ is a divisor of their
difference $2y$. Thence, since $z$ and $y$ are relatively prime odd
numbers, we conclude that 2 is the greatest common divisor
of $z+y$ and $z-y$. Then from (2) we see that each of these
numbers must be twice a square, so that we may write
\[
z+y = 2a^2, \quad z-y = 2b^2,
\]
where $a$ and $b$ are relatively prime integers. From these two
equations and Eq.~(2) we have
\[
x=2ab, \quad y=a^2-b^2, \quad z=a^2+b^2.  \tag{3}
\]
Since $x$ and $y$ are relatively prime, it follows that one of the
numbers $a$, $b$ is odd and the other even.

The forms of $x$, $y$, $z$ given in (3) are necessary in order that
(1) may be satisfied, while at the same time $x$, $y$, $z$ are relatively
prime and $x$ is even. A direct substitution in (1) shows
that this equation is indeed satisfied by these values. Hence
we have the following theorem:

\textit{The legs and hypotenuse of any primitive Pythagorean triangle
may be put in the form
\[
2ab, \quad a^2-b^2, \quad a^2+b^2.  \tag{4}
\]
respectively, where $a$ and $b$ are relatively prime positive integers
of which one is odd and the other even and $a$ is greater than
$b$; and every set of numbers \textup{(4)} forms a primitive Pythagorean
triangle.}

If we take $a=2$, $b=1$, we have $4^2+3^2 =5^2$; if $a=3$, $b=2$,
we have $12^2+5^2=13^2$; and so on.

\Needspace*{4\baselineskip}
\begin{center}EXERCISES\end{center}

\begin{small}
1. Prove that the legs and hypotenuse of all integral Pythagorean triangles
in which the hypotenuse differs from one leg by unity are given by $2\alpha+1$, $2\alpha^2+2\alpha$,
$2\alpha^2+2\alpha+1$, respectively, $\alpha$ being a positive integer.

2. Prove that the legs and hypotenuse of all primitive Pythagorean triangles
in which the hypotenuse differs from one leg by 2 are given by $2\alpha$, $\alpha^2-1$, $\alpha^2+1$,
respectively, $\alpha$ being a positive integer. In what non-primitive triangles does
the hypotenuse exceed one leg by 2?

% [File: 021.png Page: 11]
3. Show that the product of the three sides of a Pythagorean triangle is
divisible by 60.

4. Show that the general formulæ for the solution of the equation
\[
x^2+y^2=z^4
\]
in relatively prime positive integers $x$, $y$, $z$ are
\[
z=m^2+n^2, \quad x, y=4mn(m^2-n^2), \quad \pm(m^4-6m^2n^2+n^4), \ m>n
\]
$m$ and $n$ being relatively prime positive integers of which one is odd and the
other even.

5. Show that the general formulæ for the solution of the equation
\[
x^2+(2y)^4=z^2
\]
in relatively prime positive integers $x$, $y$, $z$ are
\[
z=4m^4+n^4, \quad x=\pm(4m^4-n^4), \quad y=mn,
\]
$m$ and $n$ being relatively prime positive integers.

6. Show that the general formulæ for the solution of the equation
\[
(2x)^2+y^4=z^2
\]
in relatively prime positive integers $x$, $y$, $z$ are
\[
z=m^4+6m^2n^2+n^4, \quad x=2mn(m^2+n^2), \quad y=m^2-n^2, \ m>n,
\]
$m$ and $n$ being relatively prime positive integers of which one is odd and the
other even.
\index{Triangles, Pythagorean|)}\par\end{small}

%§ 4.
\Needspace*{4\baselineskip}
\section[\textsc{Rational Triangle. Exercises 1-3}]{Rational Triangles}

\hspace{\parindent}We have seen that the length of the perpendicular from
any angle to the opposite side of a rational triangle is rational,
and that it divides that side into two parts each of which is
of rational length. If we denote the sides of the triangle by
$x$, $y$, $z$, the perpendicular from the opposite angle upon $z$ by
$h$ and the segments into which it divides $z$ by $z_1$ and $z_2$, $z_1$ being
adjacent to $x$ and $z_2$ adjacent to $y$, then we have
\[
h^2=x^2-z_1^2=y^2-z_2^2, \quad z_1+z_2=z.  \tag{1}
\]
These equations must be satisfied if $x$, $y$, $z$ are to be the sides
of a rational triangle. Moreover, if they are satisfied by positive
rational numbers $x$, $y$, $z$, $z_1$, $z_2$, $h$, then $x$, $y$, $z$, $h$ are in order
the sides and altitude upon $z$ of a rational triangle. Hence
the problem of determining all rational triangles is equivalent
to that of finding all positive rational solutions of system (1).

% [File: 022.png Page: 12]
From Eqs.\ (1) it follows readily that rational numbers
$m$ and $n$ exist such that
\[
\begin{alignedat}{2}
  x+z_1 &= m, &\quad x-z_1 &= \frac{h^2}{m};  \\
  y+z_2 &= n, &\quad y-z_2 &= \frac{h^2}{n}.
\end{alignedat}
\]
Hence $x$, $y$, and $z$, where $z=z_1+z_2$, have the form
\[
\begin{aligned}
x&=\frac{1}{2} \left( m   + \frac{h^2}{m} \right),\\
y&=\frac{1}{2} \left( n   + \frac{h^2}{n} \right),\\
z&=\frac{1}{2} \left( m+n - \frac{h^2}{m}-\frac{h^2}{n} \right),
\end{aligned}
\]
respectively. If we suppose that each side of the given triangle
is multiplied by $2mn$ and that $x$, $y$, $z$ are then used to
denote the sides of the resulting triangle, we have
\[
\left. \begin{aligned}
x&=n(m^2+h^2),\\
y&=m(n^2+h^2),\\
z&=(m+n)(mn-h^2).
\end{aligned}
\right\} \tag{2}
\]
It is obvious that the altitude upon the side $z$ is now $2hmn$,
so that the area of the triangle is
\[
hmn(m+n)(mn-h^2).  \tag{3}
\]

From this argument we conclude that the sides of any
rational triangle are proportional to the values of $x$, $y$, $z$ in
(2), the factor of proportionality being a rational number.
If we call this factor $\rho$, then a triangle having the sides $\rho x$,
$\rho y$, $\rho z$, where $x$, $y$, $z$ are defined in (2), has its area equal to
$\rho^2$ times the number in (3). Hence we conclude as follows:

\textit{A necessary and sufficient condition that rational numbers
$x$, $y$, $z$ shall represent the sides of a rational triangle is that they
shall be proportional to numbers of the form $n(m^2+h^2)$, $m(n^2+h^2)$,
$(m+n)(mn-h^2)$, where $m$, $n$, $h$ are positive rational numbers
and $mn>h^2$.}

% [File: 023.png Page: 13]
Let $d$ represent the greatest common denominator of the
rational fractions $m$, $n$, $h$, and write
\[
m=\frac{\mu}{d}, \quad n=\frac{\nu}{d}, \quad h =\frac{k}{d}.
\]
If we multiply the resulting values of $x$, $y$, $z$ in (2) by $d^3$ we
are led to the integral triangle of sides $\bar x$, $\bar y$, $\bar z$, where\label{missingbar}
\[
\begin{aligned}
\bar{x} &= \nu(\mu^2+k^2),\\
\bar{y} &= \mu(\nu^2+k^2),\\  %[** Bar missing in original]
\bar{z} &= (\mu+\nu)(\mu\nu-k^2).
\end{aligned}
\]
With a modified notation the result may be stated in the following
form:

\textit{Every rational integral triangle has its sides proportional
to numbers of the form $n(m^2+h^2)$, $m(n^2+h^2)$, $(m+n)(mn-h^2)$,
where $m$, $n$, $h$ are positive integers and $mn > h^2$.}

To obtain a special example we may put $m=4$, $n=3$, $h = 1$.
Then the sides of the triangle are 51, 40, 77 and the area is 924.

For further properties of rational triangles the reader may
consult an article by \index{Lehmer}Lehmer in \textit{Annals of Mathematics}, second
series, Volume~I, pp.~97--102.

\Needspace*{4\baselineskip}
\begin{center}EXERCISES\end{center}

\begin{small}
1. Obtain the general rational solution of the equation
\[
(x+y+z)xyz= u^2.
\]

\textsc{Suggestion}.---Recall the interpretation of this equation as given in §~2.

2. Show that the cosine of an angle of a rational triangle can be written in
one of the forms
\[
  \frac{\alpha^2-\beta^2}{\alpha^2+\beta^2}, \quad
  \frac{2 \alpha \beta  }{\alpha^2+\beta^2},
\]
where $\alpha$ and $\beta$ are relatively prime positive integers.

3. If $x$, $y$, $z$ are the sides of a rational triangle, show that positive numbers
$\alpha$ and $\beta$ exist such that one of the equations,
\[
  x^2 - 2xy \frac{\alpha^2-\beta^2}{\alpha^2+\beta^2}+y^2=z^2, \quad
  x^2 - 2xy \frac{2 \alpha \beta  }{\alpha^2+\beta^2}+y^2=z^2,
\]
is satisfied. Thence determine general expressions for $x$, $y$, $z$.
\index{Triangles, Rational|)}\par\end{small}

% [File: 024.png Page: 14]
%§ 5.
\Needspace*{4\baselineskip}
\section[\textsc{Impossibility of the System $x^2+y^2=z^2,\ y^2+z^2=t^2$. Applications. Exercises 1-3}]{Impossibility of the System $x^2+y^2=z^2,\ y^2+z^2=t^2$. \\
Applications}

\hspace{\parindent}By means of the result at the close of §~3 we shall now
prove the following theorem:

I\@. \textit{There do not exist integers $x$, $y$, $z$, $t$, all different from
zero, such that}
\[
x^2+y^2=z^2, \quad y^2+z^2=t^2.  \tag{1}
\]

It is obvious that an equivalent theorem is the following:

II\@. \textit{There do not exist integers $x$, $y$, $z$, $t$, all different from
zero, such that}
\[
t^2+x^2=2z^2, \quad t^2-x^2=2y^2.  \tag{2}
\]

It is obvious that there is no loss of generality if in the
proof we take $x$, $y$, $z$, $t$ to be positive; and this we do.

The method of proof is to assume the existence of integers
satisfying (1) and (2) and to show that we are thus led to a
contradiction. The argument we give is an illustration of
\index{Fermat}Fermat's famous method of \index{Infinite Descent, Method of}\index{Method of Infinite Descent}``infinite descent,'' of which we
give a general account in the next section.

If any two of the numbers $x$, $y$, $z$, $t$ have a common prime
factor $p$, it follows at once from (1) and (2) that all four of
them have this factor. For, consider an equation in (1) or
in (2) in which the two numbers divisible by $p$ occur; this
equation contains a third number of the set $x$, $y$, $z$, $t$, and it
is readily seen that this third number is divisible by $p$. Then
from one of the equations containing the fourth number it follows
that this fourth number is divisible by $p$. Now let us
divide each equation of systems (1) and (2) by $p^2$; the resulting
systems are of the same forms as (1) and (2) respectively.
If any two numbers in these resulting systems have a common
prime factor $p_1$, we may divide each system through by $p_1^2$;
and so on. Hence if a pair of simultaneous equations (2)
exists then there exists a pair of equations of the same form
in which no two of the numbers $x$, $y$, $z$, $t$ have a common factor
other than unity. Let this system of equations be
\[
t_1^2+x_1^2=2z_1^2, \quad t_1^2-x_1^2=2y_1^2.  \tag{3}
\]

% [File: 025.png Page: 15]
From the first equation in (3) it follows that $t_1$ and $x_1$ are
both odd or both even; and, since they are relatively prime,
it follows that they are both odd. Evidently $t_1 > x_1$. Then
we may write
\[
t_1 = x_1 + 2\alpha,
\]
where $\alpha$ is a positive integer. If we substitute this value of
$t_1$ in the first equation in (3), the result may readily be put
in the form
\[
(x_1+\alpha)^2+\alpha^2=z_1^2.   \tag{4}
\]
Since $x_1$ and $z_1$ have no common prime factor it is easy to see
from this equation that $\alpha$ is prime to both $x_1$ and $z_1$, and hence
that no two of the numbers $x_1+\alpha$, $\alpha$, $z_1$ have a common factor
other than unity.

Then, from the general result at the close of §~3 it follows
that relatively prime positive integers $r$ and $s$ exist, where
$r > s$, such that
\begin{alignat*}{2}
x_1+\alpha&=2rs, &\alpha&=r^2-s^2,  \tag{5}\\
\intertext{or}
x_1+\alpha&=r^2-s^2, \quad &\alpha&=2rs.  \tag{6}
\end{alignat*}
In either case we have
\[
t_1^2-x_1^2=(t_1-x_1)(t_1+x_1)=2\alpha\cdot2(x_1+\alpha)=8rs(r^2-s^2).
\]
If we substitute in the second equation of (3) and divide by 2,
we have
\[
4rs(r^2-s^2)=y_1^2.
\]

From this equation and the fact that $r$ and $s$ are relatively
prime, it follows at once that $r$, $s$, $r^2-s^2$ are all square numbers;
say
\[
r=u^2, \quad s=v^2, \quad r^2-s^2=w^2.
\]
Now $r-s$ and $r+s$ can have no common factor other than
1 or 2; hence, from
\[
w^2=r^2-s^2=(r-s)(r+s)=(u^2-v^2)(u^2+v^2)
\]
we see that either
\begin{alignat*}{2}
u^2+v^2&=2w_1^2, \quad & u^2-v^2 &= 2w_2^2,  \tag{7}\\
\intertext{or}
u^2+v^2&=w_1^2, \quad & u^2-v^2 &= w_2^2.
\end{alignat*}
% [File: 026.png Page: 16]
And if it is the latter case which arises, then
\[
w_1^2+w_2^2=2u^2, \quad w_1^2-w_2^2=2v^2.     \tag{8}
\]
Hence, assuming equations of the form (2), we are led either
to Eqs.~(7) or to Eqs.~(8); that is, we are led to new equations
of the form with which we started. Let us write the equations
thus:
\[
t_2^2+x_2^2=2z_2^2, \quad t_2^2-x_2^2=2y_2^2;     \tag{9}
\]
that is, system (9) is identical with that one of systems (7),
(8) which actually arises.

Now from (5) and (6) and the relations $t_1=x_1+2\alpha$, $r>s$,
we see that
\[
  t_1 = 2rs+r^2-s^2 > 2s^2+r^2-s^2 = r^2+s^2 = u^4+v^4.
\]
Hence $u<t_1$. Also,
\[
w_1^2 \leqq w^2 \leqq r+s<r^2+s^2.
\]
Hence $w_1 < t_1$. Since $u$ and $w_1$ are both less than $t_1$, it follows
that $t_2$ is less than $t_1$. Hence, obviously, $t_2 < t$. Moreover,
it is clear that all the numbers $x_2$, $y_2$, $z_2$, $t_2$ are different from
zero.

From these results we have the following conclusion: If
we assume a system of the form (2) for given values of $x$, $y$,
$z$, $t$, we are led to a new system (9) of the same form; and
in the new system $t_2$ is less than $t$.

Now if we start with (9) and carry out a similar argument
we shall be led to a new system
\[
t_3^2+x_3^2=2z_3^2, \quad t_3^2-x_3^2=2y_3^2,
\]
with the relation $t_3 < t_2$; starting from this last system we shall
be led to a new one of the same form, with a similar relation
of inequality; and so on \textit{ad infinitum}. But, since there is
only a finite number of integers less than the given positive
integer $t$, this is impossible. We are thus led to a contradiction;
whence we conclude at once to the truth of II and likewise
of I.

By means of theorems I and II we may readily prove the
following theorem:

% [File: 027.png Page: 17]
III\@. \textit{The area of a \index{Triangles, Pythagorean}Pythagorean triangle is never equal to
a square number}.

Let the legs and hypotenuse of a Pythagorean triangle
be $u$, $v$, $w$, respectively. The area of this triangle is $\frac{1}{2}uv$. If
we assume this to be a square number $\rho^2$, we shall have the
following simultaneous Diophantine equations:
\[
u^2+v^2 = w^2, \quad uv = 2\rho^2.  \tag{10}
\]
We shall prove our theorem by showing that the assumption
of such a system for given values of $u$, $v$, $w$, $\rho$ leads to a contradiction.

From system (10) it is easy to show that if any two of
the numbers $u$, $v$, $w$ have the common prime factor $p$, then
the remaining one of these numbers and the number $\rho$ are
both divisible by $p$. Thence it is easy to show that if any
system of the form (10) exists there exists one in which $u$, $v$,
$w$ are prime each to each. We shall now suppose that (10)
itself is such a system.

Since $u$, $v$, $w$ are relatively prime it follows from the first
equation in (10) and the theorem in §~3 that relatively prime
integers $a$ and $b$ exist such that $u$, $v$ have the values $2ab$,
$a^2-b^2$ in some order. Hence from the second equation in
(10) we have
\[
\rho^2 = ab(a^2-b^2)=ab(a-b)(a+b).
\]
It is easy to see that no two of the numbers $a$, $b$, $a-b$, $a+b$,
have a common factor other than unity; for, if so, $u$ and $v$
would fail to satisfy the restriction of being relatively prime.
Hence from the last equation it follows that each of these
numbers is a square. That is, we have equations of the form
\[
a=m^2, \quad b=n^2, \quad a+b = p^2, \quad a-b = q^2;
\]
whence
\[
m^2-n^2=q^2, \quad m^2+n^2 = p^2.
\]
But, according to theorem I, 
no such system of equations
can exist. That is, the assumption of Eqs.~(10) leads to a
contradiction. Hence the theorem follows as stated above.

From the last theorem we have an almost immediate proof
of the following:

% [File: 028.png Page: 18]
IV\@. \textit{There are no integers $x$, $y$, $z$, all different from zero, such that}
\[
x^4-y^4=z^2.  \tag{11}
\]

If we assume an equation of the form (11), we have
\[
(x^4-y^4)x^2y^2 = x^2y^2z^2.  \tag{12}
\]
But, obviously,
\[
(2x^2y^2)^2+(x^4-y^4)^2 = (x^4+y^4)^2. \tag{13}
\]
Now, from (12), we see that the Pythagorean triangle determined
by (13) has its area $(x^4-y^4)x^2y^2$ equal to the square
number $x^2y^2z^2$. But this is impossible. Hence no equation
of the form (11) exists.

\textsc{Corollary}.---\textit{There exist no integers $x$, $y$, $z$, all different
from zero, such that}
\[
x^4+y^4=z^4.
\]

\Needspace*{4\baselineskip}
\begin{center}EXERCISES\end{center}

\begin{small}
1. The system $x^2-y^2 = ku^2$, $x^2+y^2=kv^2$ is impossible in integers $x$, $y$, $k$, $u$, $v$,
all of which are different from zero.

2. The equation $x^4+4y^4=z^2$ is impossible in integers $x$, $y$, $z$, all of which are
different from zero.

3. The equation $2x^4+2y^4=z^2$ is impossible in integers $x$, $y$, $z$, except for the
trivial solution $z= \pm 2x^2 = \pm 2y^2$.
\par\end{small}

%§ 6.
\Needspace*{4\baselineskip}
\section[\textsc{The Method of Infinite Descent. Exercises 1-9}]{The Method of Infinite Descent}
\index{Infinite Descent, Method of|(}\index{Method of Infinite Descent|(}

\hspace{\parindent}In the preceding section we have had an example of Fermat's
famous method of infinite descent. In its relation to
Diophantine equations this method may be broadly characterized
as follows:

Suppose that one desires to prove the impossibility of
the Diophantine equation
\[
f(x_1,x_2, \ldots, x_n)=0,  \tag{1}
\]
where $f$ is a given function of its arguments. One assumes
that the given equation is true for given values of $x_1$, $x_2$, \ldots,
$x_n$, and shows that this assumption leads to a contradiction
in the following particular manner. One proves the existence
% [File: 029.png Page: 19]
of a set of integers $u_1$, $u_2$, \ldots, $u_n$ and a function $g(u_1, u_2, \ldots, u_n)$
having only positive integral values such that
\[
f(u_1, u_2, \ldots u_n)=0,  \tag{2}
\]
while
\[
g(u_1, u_2, \ldots, u_n) < g(x_1, x_2, \ldots, x_n).
\]
The same process may then be applied to Eq.~(2) to prove
the existence of a set of integers $v_1$, $v_2$, \ldots, $v_n$, such that
\[
f(v_1, v_2, \ldots, v_n)=0, \quad g(v_1, v_2, \ldots, v_n) < g(u_1, u_2, \ldots, u_n).
\]
This process may evidently be repeated an indefinite number
of times. Hence there must be an indefinite number of different
positive integers less than \\ $g(x_1, x_2, \ldots, x_n)$. But this
is impossible. Hence the assumption of Eq.~(1) for a given
set of values $x_1$, \ldots, $x_n$ leads to a contradiction; and therefore
(1) is an impossible equation.

By a natural extension the method may also be employed
(but usually not so readily) to find all the solutions of certain
possible equations. It is also applicable, in an interesting way,
to the proof of a number of theorems; one of these is the
theorem that every prime number of the form $4n+1$ is a
sum of two squares of integers. See lemma~II of §~10.

We shall now apply this method to the proof of the following
theorem:

I\@. \textit{There are no integers $x$, $y$, $z$, all different from zero, satisfying
either of the equations}
\[
x^4-4y^4=\pm z^2.  \tag{3}
\]

Let us assume the existence of one of the equations (3)
for a given set of positive integers $x$, $y$, $z$. If any two of these
numbers have a common odd prime factor $p$, then all three
of them have this factor, and the equation may be divided
through by $p^4$. The new equation thus obtained is of the
same form as the original one. The process may be repeated
until an equation
\[
x_1^4-4y_1^4=\pm z_1^2
\]
is obtained, in which no two of the numbers $x_1$, $y_1$, $z_1$ have a
common odd prime factor.

% [File: 030.png Page: 20]
If $z_1$ is even, it is obvious that $x_1$ is also even, and therefore
the above equation may be divided through by 4; a result
of the form
\[
y_2^4 - 4x_2^4 = \mp z_2^2
\]
is obtained. The process may be continued until an equation
of one of the forms
\[
y_3^4 - 4x_3^4 = \pm z_3^2
\]
is obtained, in which $z_3$ is an odd number. Then $y_3$ is also odd.
Then if the second member in the last equation has the minus
sign we may write $y_3^4 + z_3^2 = 4x_3^4$. This equation is impossible,
since the sum of two odd squares is obviously divisible
by 2 but not by 4. Hence we must have
\[
4x_3^4 + z_3^2 = y_3^4.   \tag{4}
\]

Now it is clear that no two of the numbers $x_3$, $y_3$, $z_3$ have
a common factor other than unity and that all of them are
positive. Hence, from the last equation it follows (by means
of the result in §~3) that relatively prime positive integers
$r$ and $s$, $r>s$, exist such that
\[
x_3^2=rs, \quad z_3 = r^2 - s^2, \quad y_3^2 = r^2 + s^2.
\]
From the first of these equations it follows that $r$ and $s$ are
squares; say $r = \rho^2$, $s = \sigma^2$. Then from the last exposed equation
we have
\[
\rho^4 + \sigma^4 = y_3^2.
\]
It is easy to see that $\rho$, $\sigma$, $y_3$ are prime each to each.

The last equation leads to relations of the form
\begin{alignat*}{3}
y_3&= r_1^2 +s_1^2, \quad &  \rho^2 &= 2r_1s_1,       \quad & \sigma^2 &= r_1^2 - s_1^2,\\
\intertext{or of the form}
y_3&= r_1^2 +s_1^2, \quad &  \rho^2 &= r_1^2 - s_1^2, \quad & \sigma^2 &= 2r_1s_1.
\end{alignat*}
In either case we see that $2r_1s_1$ and $r_1^2 - s_1^2$ are squares, while
$r_1$ and $s_1$ are relatively prime and one of them is even. From
the relation $r_1^2 - s_1^2 =$ square, it follows that $r_1$ is odd, since
otherwise we should have the sum of two odd squares equal
to the even square $r_1^2$, which is impossible. Hence $s_1$ is even.
% [File: 031.png Page: 21]
But $2r_1s_1 =$ square. Hence positive integers $\rho_1$, $\sigma_1$, exist such
that $r_1 = \rho_1^2$, $s_1 = 2\sigma_1^2$. Hence, we have an equation of the
form $\rho_1^4 - 4\sigma_1^4 = w_1^2$, since $r_1^2 - s_1^2$ is a square; that is,
we have
\[
4\sigma_1^4 + w_1^2 = \rho_1^4.      \tag{5}
\]

Now the last equation has been obtained solely from Eq.\
(4). Moreover, it is obvious that all the numbers $\rho_1$, $\sigma_1$, $w_1$,
are positive. Also, we have
\[
  x_3^2=rs = \rho^2\sigma^2 = 2r_1s_1(r_1^2-s_1^2) 
= 4\rho_1^2\sigma_1^2(\rho_1^4-4\sigma_1^2) = 4\rho_1^2\sigma_1^2w_1^2
\]
Hence, $\sigma_1<x_3$. Similarly, starting from (5) we should be led
to an equation
\[
4\sigma_2^4+w_2^2=\rho_2^4,
\]
where $\sigma_2<\sigma_1$; and so on indefinitely. But such a recursion
is impossible. Hence, the theorem follows as stated above.

By means of this result we may readily prove the following
theorem:

II\@. \textit{The area of a \index{Triangles, Pythagorean}Pythagorean triangle is never equal to twice
a square number.}

For, if there exists a set of rational numbers $u$, $v$, $w$, $t$
such that
\[
u^2 + v^2 = w^2, \quad uv = t^2,
\]
then it is easy to see that
\[
(u+v)^2=w_2+2t^2, \quad (u-v)^2=w^2-2t^2;
\]
or,
\[
w^4-4t^4=(u^2-v^2)^2.
\]

Again, we have the following:

III\@. \textit{There are no integers $x$, $y$, $z$, all different from zero,
such that}
\[
x^4 + y^4 = z^2.
\]

For, if such an equation exists, we have a Pythagorean
triangle $(x^2)^2 + (y^2)^2 = z^2$, whose area $\frac{1}{2}x^2y^2$ is twice a square
number; but this is impossible.

% [File: 032.png Page: 22]
\Needspace*{4\baselineskip}
\begin{center}EXERCISES\end{center}

\begin{small}
1. In a \index{Triangles, Pythagorean}Pythagorean triangle $x^2+y^2=z^2$, prove that not more than one of
the sides $x$, $y$, $z$, is a square number. (Cf.\ Exs.~4, 5, 6 in §~3.)

2. Show that the number expressing the area of a Pythagorean triangle has
at least one odd prime factor entering into it to an odd power and thence show
that every number of the form $\rho^4-\sigma^4$, in which $\rho$ and $\sigma$ are different positive
integers, has always an odd prime factor entering into it to an odd power.

3. The equation $2x^4-2y^4=z^2$ is impossible in integers $x$, $y$, $z$, all of which
are different from zero.

4. The equation $x^4+2y^4=z^2$ is impossible in integers $x$, $y$, $z$, all of which
are different from zero.

\textsc{Suggestion}.---This may be proved by the method of infinite descent. (\index{Euler}Euler's
Algebra, $2_2$, §~210.) Begin by writing $z$ in the form
\[
z = x^2 + \frac{2py^2}{q},
\]
where $p$ and $q$ are relatively prime integers, and thence show that $x^2=q^2-2p^2$,
$y^2=2pq$, provided that $x$, $y$, $z$ are prime each to each.

5. By inspection or otherwise obtain several solutions of each of the equations
$x^4-2y^4=z^2$, $2x^4-y^4=z^2$, $x^4+8y^4=z^2$.

6. The equation $x^4-y^4=2z^2$ is impossible in integers $x$, $y$, $z$, all of which
are different from zero.

7. The equation $x^4+y^4=2z^2$ is impossible in integers $x$, $y$, $z$, except for the
trivial solution $z= \pm x^2= \pm y^2$.

8. The equation $8x^4-y^4=z^2$ is impossible in integers $x$, $y$, $z$, all of which are
different from zero.

9. The equation $x^4-8y^4=z^2$ is impossible in integers $x$, $y$, $z$, all of which are
different from zero.
\index{Infinite Descent, Method of|)}\index{Method of Infinite Descent|)}\par\end{small}

\Needspace*{4\baselineskip}
\begin{center}GENERAL EXERCISES\end{center}
\addtocontents{toc}{\protect\contentsline {section}{\numberline {}\textsc {General Exercises} 1-10}{\thepage}}

\begin{small}
1. Find the general rational solution of the equation $x^2+y^2=a^2$, where $a$
is a given rational number.

2. Find the general rational solution of the equation $x^2+y^2=a^2+b^2$, where
$a$ and $b$ are given rational numbers.

3. Determine all primitive \index{Triangles, Pythagorean}Pythagorean triangles of which the perimeter is
a square.

4. Find general formulæ for the sides of a primitive Pythagorean triangle
such that the sum of the hypotenuse and either leg is a cube.

5. Find general formulæ for the sides of a primitive \index{Triangles, Pythagorean}Pythagorean triangle
such that the hypotenuse shall differ from each side by a cube.

6. Observe that the equation $x^2+y^2=z^2$ has the three solutions
\[
x=2mn, \quad y=m^2-n^2, \quad z=m^2+n^2,
\]
where
\[
\begin{alignedat}{2}
m&=k^2+kl+l^2, \quad &n&=k^2-l^2;\\
m&=k^2+kl+l^2, &n&=2kl+l^2;\\
m&=k^2+2kl,    &n&=k^2+kl+l^2;
\end{alignedat}
\]
% [File: 033.png Page: 23]
and show that each of the three \index{Triangles, Pythagorean}Pythagorean triangles so determined has the
area
\[
\phantom{(Hillyer,\ 1902.)}
(k^2+kl+l^2)(k^2-l^2)(2k+l)(2l+k)kl. \tag{\index{Hillyer}Hillyer, 1902.}
\]

7.* Develop methods of finding an infinite number of positive integral solutions
of the Diophantine system
\[
x^2+y^2=u^2, \quad y^2+z^2=v^2, \quad z^2+x^2=w^2.
\]

(See \textit{Amer.\ Math.\ Monthly}, Vol.~XXI, p.~165, and \textit{Encyc\-lo\-p\'edie des sciences
math\'e\-ma\-ti\-ques}, Tome~I, Vol.~III, p.~31).

8.* Obtain \label{intergal}integral solutions of the Diophantine system.
\[
x^2+y^2=t^2=z^2+w^2, \quad x^2-w^2=u^2=z^2-y^2.
\]

9. Solve the Diophantine system $x^2+t=u^2$, $x^2-t=v^2$.

10. Find three squares in arithmetical progression.
\par\end{small}

% [File: 034.png Page: 24]
%CHAPTER II

\chapter{PROBLEMS INVOLVING A MULTIPLICATIVE DOMAIN}
\markright{PROBLEMS INVOLVING A MULTIPLICATIVE DOMAIN}

\index{Domain, Multiplicative|(}\index{Multiplicative Domain|(}\index{Method of Multiplicative Domain|(}
%§ 7.
\setcounter{section}{6}
\section[\textsc{On Numbers of the Form $x^2+axy+by^2$. Exercises 1-7}]{On Numbers of the Form $x^2+axy+by^2$}

\hspace{\parindent}Numbers of the form $m^2+n^2$ have a remarkable property
which is closely connected with the fact that the equation
$x^2+y^2=z^2$ has a simple and elegant theory. This property
is expressed by means of the identities
\[
\begin{aligned}
(m^2+n^2)(p^2+q^2) &= (mp+nq)^2+(mq-np)^2,\\
&= (mp-nq)^2+(mq+np)^2.
\end{aligned}  \tag{1}
\]
A part of what is contained in these relations may be expressed
in words as follows: the product of two numbers of the form
$m^2+n^2$ is itself of the same form and in general in two ways.

If in (1) we put $p=m$ and $q=n$, we have
\[
(m^2-n^2)^2+(2mn)^2 = (m^2+n^2)^2.
\]
Thus we are led to the fundamental solution
\[
x=m^2-n^2, \quad y=2mn, \quad z=m^2+n^2,
\]
of the Pythagorean equation $x^2+y^2=z^2$.

In a similar manner, from the relations
\[
\begin{aligned}
(m^2+n^2)^3 = (m^2+n^2)^2(m^2+n^2) &= [(m^2-n^2)^2+(2mn)^2](m^2+n^2)\\
&= (m^3+mn^2)^2+(m^{2}n+n^3)^2,\\
&= (m^3-3mn^2)^2 + (3m^{2}n-n^3)^2,
\end{aligned}
\]
we have for the equation $x^2+y^2=z^3$ the following two double-parameter
solutions
\[
\begin{alignedat}{3}
x&=m^3+mn^2,  \quad &y&=m^{2}n+n^3,  \quad &z&=m^2+n^2;\\
x&=m^3-3mn^2, \quad &y&=3m^{2}n-n^3, \quad &z&=m^2+n^2.
\end{alignedat}
\]
Thus, if we take $m=2$, $n=1$, we have $10^2+5^2=5^3$, and
$2^2+11^2=5^3$.

% [File: 035.png Page: 25]
It is obvious that we may in a similar way obtain two-parameter
solutions of the equation $x^2+y^2=z^k$ for every positive
integral value of $k$.

Again from (1) we see that the equation
\[
x^2+y^2=u^2+v^2
\]
has the four-parameter solution
\[
x=mp+nq, \quad y=mq-np, \quad u=mp-nq, \quad v=mq+np.
\]
Thus, if we put $m=3$, $n = 2$, $p = 2$, $q = 1$, we have in particular
$8^2+1^2=4^2+7^2$.

There are several kinds of forms which have the same
remarkable property as that pointed out above for the form
$m^2+n^2$. Thus we have, in particular,
\[
(m^2+amn+bn^2)(p^2+apq+bq^2)=r^2+ars+bs^2,    \tag{2}
\]
when
\[
r=mp-bnq, \quad s=np+mq+anq,
\]
as one may readily verify by actual multiplication. This is
a special case of a general formula which will be developed
in §~12 in such a way as to throw light on the reason for its
existence. A special case of it will be treated in detail in §~8.

\Needspace*{4\baselineskip}
\begin{center}EXERCISES\end{center}

\begin{small}
1. Find a two-parameter solution of the equation $x^2+axy+by^2=z^2$.

2. Find a two-parameter solution of the equation $x^2+axy+by^2=z^3$.

3. Describe a method for finding two-parameter solutions of the equation
$x^2+axy+by^2=z^k$ for any given positive integral value of $k$.

4. Show that $(m^2+amn+n^2)(p^2+apq+q^2)=r^2+ars+s^2$, where $r$, $s$ have
either of the two sets of values
\[
\begin{alignedat}{2}
r&=mp-nq, \quad &s&=np+mq+anq;\\
r&=mq-np, \quad &s&=nq+mp+anp.
\end{alignedat}
\]

5. Find a four-parameter solution of the equation
\[
x^2+axy+y^2=u^2+auv+v^2.
\]

6. Find a six-parameter solution of the system
\[
x^2+axy+y^2=u^2+auv+v^2=z^2+azt+t^2.
\]

7. Find a two-parameter integral solution of the equation $x^2+y^2=z^2+1$.
\par\end{small}

% [File: 036.png Page: 26]
%§ 8.
\Needspace*{4\baselineskip}
\section[\textsc{On the Equation $x^2 - D y^2 = z^2$. Exercises 1-8}]{On the Equation $x^2 - D y^2 = z^2$}
\index{Equation of Pell|(}\index{Pell Equation|(}

\hspace{\parindent}We shall now develop a general theory by means of which
the solutions of the equation
\[
x^2-Dy^2=z^2     \tag{1}
\]
may be found. Naturally $D$ is assumed to be an integer.
Without loss of generality it may be taken positive; for if it
were negative the equation might be written in the form
$z^2-(-D)y^2 = x^2$, where $-D$ is positive. If $D$ is the square
of an integer, say, $D = d^2$, the equation may be written
\[
x^2 = z^2 + (dy)^2,
\]
so that the theory becomes essentially that of the Pythagorean
equation $x^2+y^2 = z^2$. Accordingly, we shall suppose that $D$
is not a square.

By suitably specializing Eq.\ (2) of the preceding section
we readily obtain the following two-parameter solution of (1):
\[
x=m^2 + Dn^2, \quad y=2mn, \quad z=m^2-Dn^2.
\]
But there is no ready means for determining whether this is
the general solution. Consequently we shall approach from
another direction the problem of finding the solution of (1).

We shall first show that Eq.\ (1) possesses a non-trivial
solution for which $z = 1$; that is, we shall prove the existence
of a solution of the equation
\[
x^2 - Dy^2 = 1.     \tag{2}
\]
different from the trivial solutions $x = \pm 1$, $y = 0$.

For this purpose we shall first show that \textit{integers $u$, $v$ exist
such that the absolute value\footnote
 {By the absolute value of $A$ is meant $A$ itself when $A$ is positive and $-A$
 when $A$ is negative. We denote it by $|A|$.}
of the $($positive or negative$)$ real
quantity $u-v\sqrt{D}$ is less than $1/v$ and also less than any preassigned
positive constant $\epsilon$}. (By $\sqrt{D}$ we mean the positive
square root of $D$.) Let $t$ be an integer such that $t\epsilon > 1$. Now
give to $v$ successively the integral values from 0 to $t$ and in
each case choose for $u$ the least integral value greater than
% [File: 037.png Page: 27]
$v\sqrt{D}$. In each case the quantity $u-v\sqrt{D}$ lies between 0 and
1 and in no two cases are its values equal. If we divide the
interval from 0 to 1 into $t$ subintervals, each of length $1/t$,
then two of the above values of $u-v\sqrt{D}$, say $u'-v'\sqrt{D}$ and
$u'' - v''\sqrt{D}$, must lie in the same interval. Then the expression
\[
(u'- u'')-(v'-v'')\sqrt{D}
\]
is different from zero, is of the form $u - v\sqrt{D}$ and has an absolute
value less than $1/t$ and hence less than $\epsilon$. That this absolute
value is less than that of $1/(v' - v'')$ follows from the fact
that the difference of $v'$ and $v''$ is not greater than $t$. This
completes the proof of the above statement concerning the
existence of $u$, $v$ with the assigned properties.

From the existence of one such set of integers $u$, $v$ it follows
readily that there is an infinite number of such sets. For,
let $u$, $v$ be one such set. Let $\epsilon_1$ be a positive constant less
than $|u-v\sqrt{D}|$. Then integers $u_1$, $v_1$ can be determined
such that $u_1 - v_1 \sqrt{D}$ is in absolute value less than $1/v_1$, and
also less than $\epsilon_1$. It is then less than $\epsilon$. Thus we have a
second set $u_1$, $v_1$ satisfying the original conditions. Then,
letting $\epsilon_2$ be a positive constant less than $|u_1 - v_1\sqrt{D}|$, we
may proceed as before to find a third set $u_2$, $v_2$ with the required
properties. It is obvious that this process may be continued
indefinitely and that we are thus led to an infinite number
of sets of integers $u$, $v$ such that $u- v\sqrt{D}$ is in absolute value
less than $\epsilon$ and also less than the absolute value of $1/v$.

Now let $u$ and $v$ be a pair of integers determined as above.
Then we have
\[
|u+v\sqrt{D}| \leqq |u-v\sqrt{D}|+|2v\sqrt{D}|<\left|\frac{1}{v}\right|+|2v\sqrt{D}|.
\]
Hence
\[
|u^2-Dv^2|=|u+v\sqrt{D}|\cdot|u-v\sqrt{D}|<\left|\frac{1}{v}\right| \left\{ \left|\frac{1}{v}\right|+|2v\sqrt{D}| \right\},
\]
so that
\[
|u^2-Dv^2|<\frac{1}{v^2}+2\sqrt{D}<1+2\sqrt{D}.
\]
Since $|u^2-Dv^2|$ is less than $1+2\sqrt{D}$ for every one of the infinite
number of sets $u$, $v$ in consideration, and since its value
% [File: 038.png Page: 28]
is always integral, it follows that an integer $l$ exists such that
\[
u^2 - Dv^2 = l
\]
for an infinite number of sets of values $u$, $v$. It is then obvious
that there is an infinite number of these pairs $u_1$, $v_1$; $u_2$, $v_2$; $u_3$,
$v_3$; \ldots, such that $u_i - u_j$ and $v_i - v_j$ are both divisible by $l$ for
every $i$ and $j$. Let $u'$, $v'$; $u''$, $v''$ be two pairs belonging to
this last infinite subset and chosen so that $u'' \ne \pm u'$ and
$v'' \ne \pm v'$. It is obvious that this choice is possible. From
the equations
\[
u'^2 - Dv'^2 = l, \quad u''^2 - Dv''^2 = l,
\]
we have (by Formula (2) of §~7):
\[
(u'u'' - Dv'v'')^2 - D(u'v'' - u''v')^2 = l^2.
\]
Here we take $u'=m$, $u''=p$, $v'=n$, $v''=-q$, $D=-b$, in applying
the formula referred to.

Setting
\[
x = \frac{u' u'' - D v' v''}{l}, \quad y = \frac{u' v'' - u'' v'}{l},    \tag{3}
\]
we have
\[
x^2 - D y^2 = 1.         \tag{4}
\]
It remains to show that the values of $x$ and $y$ in (3) are integers.
On account of (4) it is obviously sufficient to show that $y$ is
an integer. That $y$ is an integer follows at once from the
equations $u' = u'' + \mu l$, $v'' = v' + \nu l$, by multiplication member
by member. We show further that $y \ne 0$. If we suppose
that $y = 0$, we have
\[
u' v'' - u'' v' = 0, \quad u' u'' - D v' v'' = \pm l.
\]
These equations are satisfied only if $u'' = \pm u'$, $v'' = \pm v'$, relations
which are contrary to the hypothesis concerning $u'$, $u''$, $v'$, $v''$.

We have thus established the fact that Eq.\ (2) has at least
one integral solution which is not trivial. Since we may associate
with any solution $x$, $y$ of (2) the other solutions $-x$, $y$;
$-x$, $-y$; $x$, $-y$; it is clear that there is at least one solution
of (2) in which $x$ and $y$ are positive.

% [File: 039.png Page: 29]
Let $x_1$, $y_1$, and $x_2$, $y_2$ be any solutions of (2), whether the
same or different. Then we have
\[
1 = (x_1^2 - D y_1^2) (x_2^2 - D y_2^2) = (x_1 x_2 + D y_1 y_2)^2 - D (x_1 y_2 + x_2 y_1)^2,
\]
so that $x_1 x_2 + D y_1 y_2$ and $x_1 y_2 + x_2 y_1$ afford a solution of (2).
Hence from the solution $x$, $y$, whose existence has already
been proved, we have a second solution $x^2 + D y^2$, $2xy$. It is
easy to show that this process may be continued and that it
will lead to an infinite number of solutions of (2). But this
problem is a special case of one to be treated presently; and
hence will not be further pursued now.

In order to come upon the more general problem let us
seek solutions of Eq.~(1) in which $z$ shall have the positive value
$\sigma$; that is, let us seek solutions of the equation
\[
x^2 - D y^2 = \sigma^2.   \tag{5}
\]
If $x = x_1$, $y = y_1$ is a positive solution of Eq.~(2) then it is clear
that $x = \sigma x_1$, $y = \sigma y_1$ is a positive solution of (5). Hence from
what precedes we have at least two positive solutions of (5).

Now let $x=t_1$, $y=u_1$; $x=t_2$, $y = u_2$ be any two solutions
of Eq.~(5) and write
\[
\frac{t_1 + u_1 \sqrt{D}}{\sigma} \cdot \frac{t_2 + u_2 \sqrt{D}}{\sigma} = \frac{t + u \sqrt{D}}{\sigma},      \tag{6}
\]
where $t$ and $u$ are rational numbers. Then
\[
\left.
\begin{aligned}
   t &= \frac{t_1 t_2 + D u_1 u_2}{\sigma}, \\
   u &= \frac{t_1 u_2 + t_2 u_1}{\sigma}.
\end{aligned}
\ \right\} \tag{7}
\]
From (6) we have
\[
   \frac{t_1 - u_1 \sqrt{D}}{\sigma} \cdot \frac{t_2 - u_2 \sqrt{D}}{\sigma} = \frac{t - u \sqrt{D}}{\sigma}.     \tag{8}
\]
Multiplying Eqs.\ (6) and (8) member by member and making
use of the relations
\[
   t_1^2 - D u_1^2 = \sigma^2, \quad t_2^2 - D u_2^2 = \sigma^2,   \tag{9}
\]
we have
\[
   t^2 - D u^2 = \sigma^2.    \tag{10}
\]
% [File: 040.png Page: 30]
Hence $x=t$, $y=u$ afford a rational solution of (5), $t$ and $u$
having the values given in (7).

We shall now point out two cases in which this solution
is integral.

Suppose that $\sigma^2$ is a factor of $D$. Then from (9) it follows
that $\sigma$ is a factor of both $t_1$ and $t_2$ and hence from (7) that $u$
is an integer. Then from (10) it follows that $t$ is an integer.

Suppose that\footnote
 {The symbol $\equiv$ is read \textit{is congruent to}. For the elementary properties of
congruences see the author's \textit{Theory of Numbers}, pp.~37--41.}\index{Carmichael}
\[
   4D \equiv \sigma^2 \bmod 4 \sigma^2;
\]
that is, that $\sigma^2$ is a remainder obtained on dividing $4D$ by
$4 \sigma^2$. Then $\sigma$ is evidently an even number. Write $\sigma = 2 \rho$.
Then we have $D \equiv \rho^2 \bmod 4 \rho^2$. Hence $D$ is divisible by $\rho^2$.
Then from (9) it follows that both $t_1$ and $t_2$ are divisible by
$\rho$, since $\sigma = 2 \rho$. Put
\[
D = d \rho^2, \quad t_1 = \theta_1 \rho, \quad t_2 = \theta_2 \rho.
\]
Then $d$ is odd. Moreover, the following relations exist, as we
see from (9) and (7):
\[
   \theta_1^2 - d u_1^2 = 4, \quad \theta_2^2 - d u_2^2 = 4;   \tag{11}
\]
\[
   u = \tfrac{1}{2} (\theta_1 u_2 + \theta_2 u_1).   \tag{12}
\]
From Eqs.~(11) we see that $\theta_1$ and $u_1$ are both odd or both
even, and also that $\theta_2$ and $u_2$ are both odd or both even. Then
from (12) it follows that $u$ is an integer and hence from (10)
that $t$ is an integer.

We are now in position to prove readily the following
theorem:

\textit{Let $D$ be any positive non-square integer and let $\sigma$ be any
positive integer such that $D \equiv 0 \bmod \sigma^2$ or $D \equiv \sigma^2 \bmod 4 \sigma^2$. Let
$x = t_1$ and $y = u_1$ be the least positive integral solution of the equation}
\[
x^2 - Dy^2 = \sigma^2.   \tag{$5^\textrm{bis}$}
\]
\textit{Then all the positive integral solutions\footnote
 {It is obvious that all integral solutions are readily obtainable from all positive
 integral solutions.}
of this equation are
contained in the set}
\[
   x = t_n, \quad y = u_n, \quad n = 1, 2, 3, \ldots,
\]
% [File: 041.png Page: 31]
\textit{where}
\[
\begin{aligned}
  t_n &= \frac{1}{\sigma^{n-1}} \bigg[ t_1^n 
  \begin{aligned}[t]
    &+ \frac{n(n-1)}{2!} D t_1^{n-2} u_1^2 \\
    &+ \frac{n(n-1) (n-2) (n-3)}{4!} D^2 t_1^{n-4} u_1^4
     + \ldots \bigg],
  \end{aligned}
\\
  u_n &= \frac{1}{\sigma^{n-1}} \bigg[\frac{n}{1!} t_1^{n-1} u_1 
       + \frac{n(n-1)(n-2)}{3!} D t_1^{n-3} u_1^3 + \ldots\bigg].
\end{aligned}
\]

That all these values indeed afford solutions follows readily
from the fact that the quantities $t_n$ and $u_n$ so defined satisfy
the relation
\[
\left(\frac{t_1 + u_1 \sqrt{D}}{\sigma}\right)^n = \frac{t_n + u_n \sqrt{D}}{\sigma}.                \tag{13}
\]
For then we also have
\[
\left(\frac{t_1 - u_1 \sqrt{D}}{\sigma}\right)^n = \frac{t_n - u_n \sqrt{D}}{\sigma};
\]
whence
\[
t_n^2 - Du_n^2 = \sigma^2,
\]
as one easily shows by multiplying the preceding two equations
member by member and simplifying the result by means of the
relation $t_1^2 - Du_1^2 = \sigma^2$. That these solutions are positive is
obvious. That they are integral follows from the results
associated with Eqs.~(7) and (10).

It remains to be shown that there are no other positive
integral solutions than those defined in the above theorem.
Let $x = T$, $y = U$ be any positive integral solution of Eq.~($5^{\textrm{bis}}$).
Then, from the relation
\[
\frac{T+U\sqrt{D}}{\sigma} \cdot  \frac{T-U\sqrt{D}}{\sigma} = \frac{T^2-DU^2}{\sigma^2} = 1
\]
it follows readily that
\[
0 < \frac{T-U\sqrt{D}}{\sigma} < 1 < \frac{T+U\sqrt{D}}{\sigma}.
\]
Hence from (13) it follows that
\[
\frac{t_n+u_n\sqrt{D}}{\sigma}  < \frac{t_{n+1}+u_{n+1} \sqrt{D}}{\sigma}.
\]

% [File: 042.png Page: 32]
Now suppose that the solution $T$, $U$ does not coincide with
any solution given in the above theorem. Then for some
value of $n$ we have the relations:
\[
\frac{t_n+u_n\sqrt{D}}{\sigma} < \frac{T+U\sqrt{D}}{\sigma} < \frac{t_{n+1}+u_{n+1}\sqrt{D}}{\sigma},
\]
whence
\[
\frac{t_n+u_n\sqrt{D}}{\sigma} < \frac{T+U\sqrt{D}}{\sigma} < \frac{t_n+u_n\sqrt{D}}{\sigma} \cdot \frac{t_1+u_1\sqrt{D}}{\sigma},
\]
or
\[
1 < \frac{T+U\sqrt{D}}{\sigma} \cdot \frac{\sigma}{t_n+u_n\sqrt{D}} < \frac{t_1+u_1\sqrt{D}}{\sigma}.
\]
But
\[
\frac{\sigma}{t_n+u_n\sqrt{D}} = \frac{\sigma( t_n - u_n\sqrt{D})}{t_n^2-Du_n^2} = \frac{t_n-u_n\sqrt{D}}{\sigma}.
\]
Thence we have
\[
1 < \frac{T+U\sqrt{D}}{\sigma} \cdot \frac{t_n-u_n\sqrt{D}}{\sigma} < \frac{t_1+u_1\sqrt{D}}{\sigma}.
\]
Writing
\[
\frac{T+U\sqrt{D}}{\sigma} \cdot \frac{t_n-u_n\sqrt{D}}{\sigma} = \frac{T'+U'\sqrt{D}}{\sigma},
\]
where $T'$ and $U'$ are rational, we have $x = T'$, $y = U'$ as a solution
of ($5^\textrm{bis}$). It is integral, as we see from the results
associated with Eqs.\ (7) and (10). Moreover, the relations
\[
1 < \frac{T'+U'\sqrt{D}}{\sigma} < \frac{t_1+u_1\sqrt{D}}{\sigma}  \tag{14}
\]
are verified.

Since $(T' + U' \sqrt{D})(T' - U' \sqrt{D}) = \sigma^2$, it follows from the first
inequality in (14) that $T' - U' \sqrt{D}$ is positive and less than
$\sigma$, and hence that $T'$ and $U'$ are both positive. If we suppose
that $T' \geqq t_1$, it follows from the relations, $T'^2 - DU'^2 = \sigma^2$,
$t_1^2 - Du_1^2 = \sigma^2$, that $U' \geqq u_1$, a result in contradiction with
relation (14). Hence, $T' < t_1$ and $U' < u_1$. But this is contrary
to the hypothesis that $t_1$, $u_1$ is the least positive integral
solution of ($5^\textrm{bis}$). Hence the given positive solution $T$, $U$
must coincide with one of those given in the theorem.

This completes the demonstration of the theorem.

% [File: 043.png Page: 33]
It is clear that the value $\sigma = 1$ satisfies the requisite conditions
on $\sigma$ for every non-square integer $D$, so that the above theorem
is applicable in particular to every equation of the form
$x^2 - Dy^2 = 1$. In order to apply the theorem in a particular
case it is necessary first to find, by inspection or otherwise,\footnote
 {How this may be done, by developing the numerical value of $\sqrt{D}$ into a
 continued fraction, is explained by \index{Whitford}Whitford, in \textit{The Pell Equation} (New York,
 1912). When $D$ is 1620, the value of $x$ has three figures; when D is 1621, it has
 76 figures.}
the least positive integral solution.

As an illustrative example, let us consider the equation
$x^2 - 7y^2 = 1$. If we try successively the values 1, 2, 3, \ldots\
for $y$ we find that 3 is the least positive integral value of $y$ for
which there is a corresponding integer $x$ satisfying the given
equation. This value is $x = 8$ so that $x = 8$, $y = 3$ is the least
positive integral solution of the equation $x^2 - 7y^2 = 1$. Setting
$D = 7$, $\sigma = 1$, $t_1 = 8$, $u_1 = 3$, in the last two equations of the above
theorem, we have formulæ for the general positive integral
solution of the equation $x^2 - 7y^2 = 1$. Giving $n$ successively
the values 1, 2, 3, \ldots, the particular positive integral solutions
are obtained without repetition and in the order of increasing
magnitude. The first three of these solutions are
\[
8,\ 3; \quad 127,\ 48; \quad 2024,\ 765.
\]

\Needspace*{4\baselineskip}
\begin{center}EXERCISES\end{center}

\begin{small}
1. Show how all integral solutions of the equation $x^2 - Dy^2 = -1$ may be
obtained from one of them, $D$ being as usual a positive non-square integer.

\textsc{Suggestion}.---Observe that the relations $a^2-Db^2=-1$, $c^2-Dg^2=-1$ imply
the relation $(ac + Dbg)^2 - D(ag+bc)^2 = 1$.

2. Solve each of the Diophantine equations $x^2 + 1 = 2y^2$, $x^2 - 1 = 2y^2$.

3. Let $s_n$ represent the sum of the legs and $h_n$ the hypotenuse of an integral
\index{Triangles, Pythagorean}Pythagorean triangle in which the legs differ by unity. Show that every possible
pair of values $s_n$ and $h_n$ is determined by the relation
\[
(1 + \sqrt{2})(3 + 2 \sqrt{2})^n = s_n + h_n \sqrt{2},
\]
$s_n$ and $h_n$ being rational.

4. Find all integral \index{Triangles, Pythagorean}Pythagorean triangles in which the legs differ by 2.

5. Obtain the general integral solution of each of the equations $x^2 - 5y^2 = 4$,
$x^2 - 20y^2 = 4$.

6. Obtain a formula giving an infinite number of integral solutions of the
equation $x^2 - 19y^2 = 81$.

% [File: 044.png Page: 34]
7. Find the general rational solution of the equation $x^2 - D y^2 = 4$. By means
of this rational solution obtain an infinite number of integral solutions.

8. Find the smallest integral solutions of $x^2 - 1620 y^2 = 1$ and $x^2 - 1666 y^2 = 1$.
\par\end{small}\index{Equation of Pell|)}\index{Pell Equation|)}

%§ 9.
\Needspace*{4\baselineskip}
\section[\textsc{General Equation of the Second Degree in Two Variables}]{General Equation of the Second Degree in Two
Variables}

\hspace{\parindent}Let us consider the general Diophantine equation of the
second degree in two variables
\[
ax^2 + 2bxy + cy^2 + 2dx + 2ey + f = 0,     \tag{1}
\]
where $a$, $b$, $c$, $d$, $e$, $f$ are integers.

In case $ac - b^2 = 0$, the equation may be written in the form
\[
(ax+by)^2 + 2adx + 2aey + af = 0.
\]
In order to obtain rational solutions it is sufficient to put
\[
ax + by = t, \quad 2adx + 2aey + af = -t^2,    \tag{2}
\]
where $t$ is any rational number, and solve these equations for
$x$ and $y$. This gives, in general,
\[
\left. \begin{aligned} 
x &= \frac{-bt^2-2aet-abf}{2a(bd-ae)},  \\
y &= \frac{t^2+2dt+af}{2(bd-ae)}.
\end{aligned} \right\} \tag{3}
\]
If the solution is to be integral, then $t$ must be integral, as one
sees from the first equation in (2). Then from (3) it follows
that a necessary and sufficient condition on the integer $t$ is
that it shall satisfy the following congruences:
\[
\begin{aligned}
t^2 + 2dt + af &\equiv 0 \bmod 2(bd-ae),\\
bt^2 + 2aet + abf &\equiv 0 \bmod 2a(bd-ae).
\end{aligned}
\]
In any particular case the general solution of this system of
congruences may be determined by inspection.

In case $ac - b^2 \neq 0$ the solution is not so easily determined.
Multiplying Eq.~(1) through by $(ac-b^2)^2$ we have a result
which may be put in the form
\[
au^2 + 2buv + cv^2 = m,       \tag{4}
\]
% [File: 045.png Page: 35]
where
\[
\left. \begin{aligned}
u &= (ac-b^2)x - (be-cd),\\
v &= (ac-b^2)y - (bd-ae),\\
m &= (ac-b^2) (ae^2 + cd^2 + fb^2 - acf - 2bde).
\end{aligned} \right\}     \tag{5}
\]

Thus the problem of solving Eq.~(1) is reduced to that of
solving Eq.~(4) for $u$ and $v$ and choosing those values only of
$u$ and $v$ for which $x$ and $y$ have the desired characteristic.\footnote
 {If an \textit{integral} solution is desired we may choose those values only of $u$ and $v$
 for which $x$ and $y$ are integral. When $x$ and $y$ are restricted merely to be rational
 every solution of (4) leads through (5) to a solution of (1).}
But the problem of solving Eq.~(4) is identical with that of
the representation of a given integer by means of a binary
quadratic form. The plan of this book does not permit the
detailed development of this latter subject. (See the preface.)
Consequently the problem of solving Eq.~(1) will be dismissed
with this remark.

%§ 10.
\Needspace*{4\baselineskip}
\section[\textsc{Quadratic Equations Involving More than Three
Variables}]{Quadratic Equations Involving More than Three
Variables}

\hspace{\parindent}Having now developed the general theory of the equation
\[
x^2 + y^2 = t^2,
\]
and certain generalizations of it involving still a total of three
variables, it is natural to extend the problem in another direction,
namely, by increasing the number of variables. We
should thus be led next to consider the equation
\[
x^2 + y^2 + u^2 = t^2.        \tag{1}
\]

Now the classes of numbers which have been involved in
the larger part of our previous theory and which have given
rise to the most interesting results, namely, those defined by
forms such as $x^2 +y^2$ and $x^2-Dy^2$, have had the following remarkable
property: the product of any two numbers in one
of the classes is itself in that class. We shall express this fact
by saying that the numbers of the class form a domain with
respect to multiplication. The sets of numbers mentioned
% [File: 046.png Page: 36]
also have this further property: from the representation of
two numbers in the given form that of their product is readily
obtained by means of an algebraic formula.

Numbers of the form $x^2+y^2+u^2$ do not form a domain with
respect to multiplication. This may be shown by means of
an example. We have
\[
3 = 1^2 + 1^2 + 1^2, \quad  5 = 2^2 + 1^2 + 0^2, \quad  21 = 4^2 + 2^2 + 1^2,
\]
while neither 15 nor 63 can be expressed as a sum of three
integral squares.

But if we should enlarge the set of numbers $x^2+y^2+u^2$
so that the new set shall contain all numbers of the form
$x^2+y^2+u^2+v^2$ then the set so enlarged forms a domain with
respect to multiplication. This is a special case of a more
general result which we shall presently give. In view of the
existence of this domain with respect to multiplication we have
a direct means of treating the problem of solving the equation
\[
x^2 + y^2 + u^2 + v^2 = t^2.       \tag{2}
\]
Putting to zero the quantity representing $v$ in this solution and
restricting the values of $x$, $y$, $u$, $t$ accordingly, we should arrive
at a solution of Eq.~(1).

We proceed at once to a more general problem including
that concerning Eq.~(1). Let us consider the Diophantine
equation
\[
x^2 + ay^2 + bu^2 = t^2.        \tag{3}
\]
where $a$ and $b$ are given integers. When $a = b = 1$ the equation
is the same as (1). We shall first treat the more general equation
\[
x^2 + ay^2 + bu^2 + abv^2 = t^2,      \tag{4}
\]
because, as we shall now show, the form of the first member
defines a class of numbers which form a domain with respect
to multiplication.

Let us employ the notation
\[
g(x,y,u,v) = x^2 + ay^2 + bu^2 + abv^2.
\]
% [File: 047.png Page: 37]
Then it may be readily verified that\footnote
 {If in these relations we take $a = b = 1$ we shall have a set of formulæ to which
 one is led directly by means of quaternions. Thus if we write
 \[
 (x+iy+ju-kv)(x_1+iy_1+ju_1+kv_1)=x_2+iy_2+ju_2+kv_2,
 \]
 where $i$, $j$, $k$ are the quaternion units, we may readily determine $x_2$, $y_2$, $u_2$, $v_2$, by
 direct multiplication of the quaternions in the first member. We obtain the
 values gotten from (6) by putting $a = b = 1$. Taking the norm of each member
 of the equation in this footnote we have the special case of equation (5) for which
 $a = b = 1$.}
\[
g(x, y, u, v) \cdot g(x_1, y_1, u_1, v_1) =g(x_2, y_2, u_2, v_2),   \tag{5}
\]
where
\[
\left. \begin{aligned}
x_2 &= xx_1 - ayy_1 - buu_1 + abvv_1,\\
y_2 &= xy_1 +x_1y - buv_1 - bu_1v,\\
u_2 &= ux_1 + u_1x + avy_1 + av_1y,\\
v_2 &= vx_1 - v_1x - uy_1 + u_1y.
\end{aligned} \right\}   \tag{6}
\]
It is obvious that Eq.~(5) will also be satisfied by values $x_2$, $y_2$,
$u_2$, $v_2$ obtained from (6) by replacing any number of the quantities
$x$, $y$, $u$, $v$, $x_1$, $y_1$, $u_1$, $v_1$ by their negatives. In case
$a = b = 1$ still other values of $x_2$, $y_2$, $u_2$, $v_2$ may be obtained
by any interchange of the quantities $x$, $y$, $u$, $v$ or of the quantities
$x_1$, $y_1$, $u_1$, $v_1$ among themselves. In case $a = 1$ and $b \neq 1$
the elements of the following pairs may be similarly interchanged: $x$, $y$; $u$, $v$; $x_1$, $y_1$; $u_1$, $v_1$. Not all the resulting
values of $x_2$, $y_2$, $u_2$, $v_2$ will be distinct, though there will in
general be two or more independent sets.

Thus we see that the class of numbers defined by the form
$x^2+ay^2+bu^2+abv^2$ form a domain with respect to multiplication
and that the product of any two numbers of the class is readily
expressible in the given form, and frequently in several ways.

From Eqs.~(5) and (6) and the transformations of them
indicated above, we have the following relations:
\begin{small}
\[
\left. \begin{aligned}
\{g(x,y,u,v)\}^2
  &=g(x^2-ay^2-bu^2+abv^2,\ \ 2xy-2buv,\ \ 2ux+2avy,\ \ 0) \ \\
  &=g(x^2-ay^2+bu^2-abv^2,\ \ 2xy+2buv,\ \ 0,\ \ 2vx-2uy),\\
  &=g(x^2+ay^2-bu^2-abv^2,\ \ 0,\ \ 2ux-2avy,\ \ 2vx+2uy),\\
  &=g(x^2-ay^2+bu^2+abv^2,\ \ 2xy,\ \  2avy,\ \ 2uy),\\
  &=g(x^2+ay^2-bu^2+abv^2,\ \ 2buv,\ \ 2ux,\ \  2uy),\\
  &=g(x^2+ay^2+bu^2-abv^2,\ \ 2buv,\ \ 2avy,\ \ 2vx),\\
  &=g(x^2-ay^2-bu^2-abv^2,\ \ 2xy,\ \  2ux,\ \  2vx).
\end{aligned} \right\} \tag{7}
\]
\end{small}

% [File: 048.png Page: 38]
Let us return to the consideration of Eq.~(4), writing it
now in the form
\[
\alpha^2+a\beta^2+b\rho^2+ab\sigma^2=t^2,      \tag{8}
\]
where $\alpha$, $\beta$, $\rho$, $\sigma$, $t$ are the integers to be determined. It is
clear that we have a four-parameter solution of this equation
by taking $g(x, y, u, v)$ for the value of $t$ and the arguments
(in order) in any right member of (7) for the values of $\alpha$, $\beta$,
$\rho$, $\sigma$, respectively.

Similarly for the equation
\[
\alpha^2+a\beta^2+b\rho^2=t^2,        \tag{9}
\]
we have the following four-parameter solution:
\[
\begin{aligned}
t &= x^2+ay^2+bu^2+abv^2,\\
\alpha &= x^2-ay^2-bu^2+abv^2,\\
\beta &= 2xy-2buv,\\
\rho &= 2ux+2avy,
\end{aligned}
\]
where $x$, $y$, $u$, $v$ are arbitrary integers. It is also possible to
obtain three-parameter solutions of (9) in several ways. For
instance, by taking $x=0$ in next to the last equation in (7),
we have the following solution:
\[
t=ay^2+bu^2+abv^2, \quad \alpha=ay^2+bu^2-abv^2, \quad \beta= 2buv, \quad \rho= 2avy.
\]

Whether the above formulæ give the general integral solutions
of Eqs.~(8) and (9) for a given $a$ and $b$ when $x$, $y$, $u$, $v$
are restricted to be integers is a question which is not
answered in the preceding discussion. It appears to be difficult
of treatment so long as $a$ and $b$ are unrestricted. We
shall take it up only for the most interesting special case,
namely, that of the equation
\[
x^2+y^2+z^2 = t^2.        \tag{10}
\]

This is a special case of Eq.~(9). Modifying our notation,
we may write the first solution obtained above in the form
\[
\left. \begin{aligned}
t&=m^2+n^2+p^2+q^2,\\
x&=m^2-n^2-p^2+q^2,\\
y&= 2mn-2pq,\\
z&= 2mp+2nq.
\end{aligned} \right\} \tag{11}
\]
% [File: 049.png Page: 39]
For the case of Eq.~(10) the other solutions obtained for Eq.~%
(9) are special cases of that given in (11).

Taking $m = 3$, $n = 3$, $p = 1$, $q = 2$, we have the particular
instance $3^2+14^2+18^2 = 23^2$.

We shall prove that formulæ (11) afford the general integral
solution of Eq.~(10) if each of the second members is multiplied
by the arbitrary integral factor $d$. In this demonstration
we shall have use for certain lemmas. These will first be
proved. They constitute in themselves remarkable theorems.
They are due to \index{Fermat}Fermat.

\textsc{Lemma} I\@. \textit{If a number is expressible as a sum of two integral
squares $\alpha^2+\beta^2$ and if the quotient $(\alpha^2+\beta^2)/(a^2+b^2)$ is an integer
$m$, where $a$ and $b$ are integers and $a^2+b^2$ is a prime number,
then $m$ is also a sum of two integral squares}.

We have
\[
\begin{aligned}
m=\frac{\alpha^2+\beta^2}{a^2+b^2} 
= \frac{(\alpha^2+\beta^2)(a^2+b^2)}{(a^2+b^2)^2} &\equiv
\frac{(\alpha a \pm \beta b)^2
     +(\alpha b \mp \beta a)^2}{(a^2+b^2)^2}\\
&\equiv \left(\frac{\alpha a \pm \beta b}{a^2+b^2}\right)^2
       +\left(\frac{\alpha b \mp \beta a}{a^2+b^2}\right)^2.
\end{aligned}
\]
It is sufficient to show that one of the numbers $\alpha a±\beta b$ is a
multiple of $a^2+b^2$, and hence that their product is such a
multiple, since $a^2+b^2$ is a prime. But their product is
\[
\alpha^2a^2-\beta^2b^2
=a^2(\alpha^2+\beta^2)-\beta^2(a^2+b^2) 
=(ma^2-\beta^2)(a^2+b^2).
\]
Hence lemma I is established.

\textsc{Lemma} II\@. \textit{Every prime number of the form $4n+1$ can be
represented in one and in only one way as a sum of two integral
squares}.

We start from the theorem that $-1$ is a quadratic residue
of every prime number of the form $4n+1$ and a quadratic
non-residue of every prime number of the form $4n+3$. (See
the \index{Carmichael}author's \textit{Theory of Numbers}, p.~79.) This is equivalent
to saying that every prime number of the form $4n+1$ is a
factor of a number of the form $t^2+1$ where $t$ is a positive integer,
while no prime number of the form $4n+3$ is a factor of such
a number $t^2+1$. If we take for $t$ the least integer such that
% [File: 050.png Page: 40]
a prime number $p$ of the form $4n + 1$ is a factor of $t^2 + 1$, it is
clear that we have the following relations:
\[
t^2 + 1 = pk, \ \ k<p.
\]

Now consider the set of numbers
\[
1^2 +1, \quad 2^2 + 1, \quad 3^2 + 1, \quad 4^2 + 1, \ldots    \tag{12}
\]
It is obvious that no one of these numbers contains two prime
factors neither of which is a factor of a preceding number of
the set; for, if so, the smaller of these primes must be a factor
of a number $t^2 + 1$ with a complementary factor less than
itself and hence less than the other prime.

Arrange all prime numbers not of the form $4n + 3$ in the
order in which they occur as factors of numbers in the set
(12); thus:
\[
p_1 = 2, \quad  p_2 = 5, \quad  p_3 = 17, \quad p_4 = 13, \quad  p_5, \quad  p_6,\ \ldots. \tag{13}
\]
This set contains every prime of the form $4n + 1$.

Suppose that $p_m$ is a prime number of the set (13) which
is not expressible as a sum of two integral squares. Let $t^2 + 1$
be the first number of the set (12) of which $p_m$ is a factor and
by means of which $p_m$ was assigned its place in (13). Then
we have
\[
p_{m}k_{m} = t^2 + 1,         \tag{14}
\]
where $k_m$ is such that every prime factor of $k_m$ appears earlier
than $p_m$ in the set~(13). If every prime factor of $k_m$ is a sum
of two integral squares, then a repeated use of lemma~I in
connection with Eq.~(14) would lead to the conclusion that
$p_m$ is a sum of two integral squares. But this is contrary
to the hypothesis concerning $p_m$. Hence there is some prime
factor of $k_m$ which is not expressible as a sum of two integral
squares.

Thus we have proved that, if any prime of the set~(13) is
not expressible as a sum of two integral squares, then there is
an earlier prime in the same set which likewise is not expressible
as a sum of two integral squares. This is in evident
contradiction with the fact that the first primes of this set are
each expressible as a sum of two squares. Hence every prime
% [File: 051.png Page: 41]
in set (13), and hence every prime of the form $4n + 1$, is expressible
as a sum of two integral squares.

It remains to show that no prime $p$ can be represented in
two ways as a sum of two integral squares. This we shall do
by assuming an equation of the form
\[
p = a^2 + b^2 = c^2 +d^2,       \tag{15}
\]
where $a$ and $c$ are even and $b$ and $d$ are odd, and proving that
$a^2 = c^2$, $b^2 = d^2$. From (15) we have
\[
\begin{aligned}
p^2 &= (ac+bd)^2 + (ad-bc)^2 = (ac-bd)^2 + (ad+bc)^2,\\
p(a^2 - c^2) &=a^2(c^2 + d^2) - c^2(a^2 + b^2) = (ad+bc)(ad-bc).
\end{aligned}
\]
From the last equation it follows that $p$ is a factor of one of
the numbers $ad+bc$, $ad-bc$. Now, neither of the numbers
$ac+bd$ or $ac-bd$ is equal to zero, since both of them are odd.
Therefore, from next to the last equation we see that $ad-bc$
and $ad+bc$ are both less than $p$ in absolute value. But one
of them is divisible by $p$, and hence that one is equal to zero.
Therefore, $a^2/c^2 = b^2/d^2$. From this relation and (15) it follows
at once that $a^2 = c^2$, $b^2 = d^2$.

This completes the proof of lemma II.

It is obvious that no prime of the form $4n + 3$ can be represented
as a sum of two integral squares; for, if so, one square
must be even and the other odd, and in this case their sum
is of the form $4n +1$.

\textsc{Lemma} III\@. \textit{Let $p$ be any prime number of the form $4n + 1$
and write $p = a^2 + b^2$, where $a$ and $b$ are integers. Let $m$ be any
integer such that $pm ={\alpha}^2 + {\beta}^2$, where $\alpha$ and $\beta$ are integers. Then
there exists a representation of $m$ as a sum of two integral squares,
\[
m = \left( \frac{\alpha a \pm \beta b}{a^2+b^2} \right)^2 + \left( \frac{\alpha b \mp \beta a}{a^2+b^2} \right)^2,     \tag{16}
\]
such that the representation ${\alpha}^2 + {\beta}^2$ of $pm$ is obtained from the
above representations of $p$ and $m$ by multiplication according to
the formula}
\[
(a^2 + b^2)(c^2 + d^2) = (ac+bd)^2 + (ad-bc)^2.
\]

% [File: 052.png Page: 42]
That $m$ has the representation (16) was shown in the
proof of lemma~I\@.
That this representation has the further
property specified here may be verified by a direct computation.

\textsc{Corollary}. \textit{If $h$ is a composite number containing no
prime factor of the form $4n+3$ and if we write $h = h_1 h_2$, where
$h_1$ and $h_2$ are positive integers, then every representation of $h$ as
a sum of two integral squares is obtained by taking every representation
of $h_1$ and $h_2$ and multiplying these expressions in
accordance with the formula}
\[
(a^2+b^2)(c^2+d^2) = (ac \pm bd)^2+(ad \mp bc)^2.
\]

We are now ready to return to the consideration of Eq.~(10).
We shall suppose that the integers $x$, $y$, $z$, $t$ contain no common
factor other than unity. Then at least one of them is
odd. If $t$ is even one of the numbers $x$, $y$, $z<$ is even; suppose
it is $x$. Then $y^2+z^2$ is divisible by 4 and hence $y$ and $z$ are
even. Then $x$, $y$, $z$, $t$ have the common factor 2 contrary to
hypothesis. Hence $t$ is odd. Then, one of the numbers $x$,
$y$, $z$ is odd. Suppose it is $x$. Then $y^2+z^2$ is divisible by 4
and hence $y$ and $z$ are even.

Now write Eq.~(10) in the form
\[
(t-x)(t+x)=y^2+z^2.      \tag{17}
\]
Suppose that $t - x$ and $t + x$ have a common odd prime factor
$r$ in common, where $r$ is of the form $4n + 3$. Then $r$ is a factor
of $(t-x)+(t+x)$, and hence of $t$, and hence likewise of $x$, while
\[
y^2+z^2 \equiv 0 \bmod r.       \tag{18}
\]
If $z$ is not divisible by $r$ then there exists an integer $z_1$ such
that
\[
zz_1 \equiv 1 \bmod r.
\]
(See the \index{Carmichael}author's \textit{Theory of Numbers}, p.~43.) Hence
\[
(yz_1)^2 + 1 \equiv 0 \bmod r.
\]
But this is impossible, since $-1$ is a quadratic non-residue
of every prime number of the form $4n+3$. Hence, $z$, and
therefore $y$, is divisible by $r$. Then $x$, $y$, $z$, $t$ have the common
% [File: 053.png Page: 43]
factor $r$, contrary to hypothesis. Hence $t-x$ and $t+x$ have
no common prime factor of the form $4n+3$.

If congruence (18) is verified we have just seen that $y$
and $z$ are both divisible by $r$. Hence from (17) it follows
that if either $t-x$ or $t+x$ contains a factor $r$ of the form $4n+3$
it contains that factor to an even power. From this result
and the foregoing lemmas it follows at once that integers $m$,
$n$, $p$, $q$ exist such that
\[
t+x = 2(m^2+q^2), \quad t-x = 2(n^2+p^2),
\]
since both $t-x$ and $t+x$ are even. Hence $t$ and $x$ have the
form given in (11), while
\[
y^2+z^2 = 4(m^2+q^2)(n^2+p^2).
\]
Since $y$ and $z$ are even it now follows readily from the corollary
to lemma~III that for a given $t$, $x$, $y$, $z$ the integers $m$, $n$, $p$, $q$
may be so chosen that $y$ and $z$ are representable in the form
given in (11).

Hence we conclude that formulæ (11) afford the general
integral solution of Eq.~(10) if each of the second members
is multiplied by the arbitrary integral factor~$d$.

\Needspace*{4\baselineskip}
\begin{center}EXERCISES\end{center}

\begin{small}
1. By means of equations (5), (6), (7) find values of $\xi$, $\eta$, $\mu$, $\nu$ such that
\[
\left\{g(x, y, u, v)\right\}^3 = g(\xi, \eta, \mu, \nu).
\]
Apply this result to the solution of each of the equations
\[
\xi^2+a\eta^2+b\mu^2+ab\nu^2=t^3, \quad \xi^2+a\eta^2+b\mu^2=t^3, 
\]
finding a four-parameter solution of the former and a one-parameter solution
of the latter. Here $\xi$, $\eta$, $\mu$, $\nu$, $t$ are the integers to be determined.

2. Obtain an eight-parameter solution of the equation
\[
x^2+ay^2+bu^2+abv^2=x_1^2+ay_1^2+bu_1^2+abv_1^2,
\]
where $x$, $y$, $u$, $v$, $x_1$, $y_1$, $u_1$, $v_1$ are the integers to be determined.

3. Obtain a six-parameter solution of the equation
\[
x^2+ay^2+bz^2 = u^2+av^2+bw^2,
\]
where $x$, $y$, $z$, $u$, $v$, $w$ are the integers to be determined.

4. Find an integral solution of the equation
\[
x^2+y^2+z^2 = u^2+v^2+w^2 = r^2+s^2+t^2,
\]
involving at least four independent parameters.

% [File: 054.png Page: 44]
5. Find an integral solution of the equation
\[
x^2+2y^2=u^2+v^2+w^2,
\]
containing at least two arbitrary parameters.

6. Given a relation of the form
\[
A^2+B^2+C^2=k(a^2+b^2+c^2),
\]
where $A$, $B$, $C$, $a$, $b$, $c$, $k$ are integers and $aB \neq bA$; find a one-parameter solution
of the equation
\[
x^2+y^2+z^2=k(u^2+v^2+w^2).
\]
Find several values of $k$, less than 100, and the corresponding integers $A$, $B$, $C$,
$a$, $b$, $c$ such that the first equation stated in this problem is satisfied. In particular,
show that $k$ may have the values $k = 7, 19, 67$. (\index{Realis}Realis, 1882.)
\index{Equations of Second Degree|)}\index{Quadratic Equations|)}\par\end{small}

%§ 11.
\Needspace*{4\baselineskip}
\section[\textsc{Certain Equations of Higher Degree. Exercises 1-3}]{Certain Equations of Higher Degree}\index{Biquadratic Equations|(}\index{Equations of Fourth Degree|(}

\hspace{\parindent}Let us consider the equation
\[
\alpha^4+a\beta^4+b\gamma^4 = \mu^2,         \tag{1}
\]
where $\alpha$, $\beta$, $\gamma$, $\mu$ are the integers to be determined. As the form
of the first member of this equation defines a set of numbers
which do not form a domain with respect to multiplication,
we shall naturally seek to extend the set in such way that
the resulting class of integers do form a domain with respect
to multiplication. For this purpose let us replace $\alpha^2$, $\beta^2$, $\gamma^2$
by $x$, $y$, $u$ respectively and adjoin the new variable $v$ so as
to give rise to the class of numbers defined by the form
\[
x^2+ay^2+bu^2+abv^2.
\]
This class forms a domain with respect to multiplication, as
we have already seen.

Concerning this extension let us observe that the set of
numbers $\alpha^4+a\beta^4+b\gamma^4$ have been extended simultaneously in
two different ways. One way is by the generalization of variables
already present; the other is by the adjunction of a new
variable. We have here, then, an example of two methods
of extending a set of numbers. These methods we shall find
of frequent use and great importance in the theory of Diophantine
equations.

Let us return now to Eq.~(1) and consider it in connection
with the group~(7) of equations in the preceding section. By
% [File: 055.png Page: 45]
means of the first equation in that group (with $v$ taken equal
to zero) we see that (1) will be satisfied if
\[
\left . \begin{aligned}
\mu &= x^2 + ay^2 + bu^2,\    \\
\alpha^2 &= x^2 - ay^2 - bu^2,\\
\beta^2 &= 2xy,\\
\gamma^2 &= 2ux.
\end{aligned} \right\} \tag{2}
\]
Here $x$, $y$, $u$ are integers to be so restricted that (2) shall be
satisfied.

From the last equation in group (7), already referred to,
we see that the second equation in (2) will be satisfied if
\[
\left. \begin{aligned}
\alpha &= {x_1}^2 - a{y_1}^2 - b{u_1}^2,\\
x &= {x_1}^2 + a{y_1}^2 + b{u_1}^2,\\
y &= 2{x_1}{y_1},\\
u &= 2{u_1}{x_1}.
\end{aligned}\ \right\} \tag{3}
\]
Then the third and fourth equations in (2) become
\[
\left. \begin{aligned}
\beta^2 &= 4{x_1}{y_1}({x_1}^2 + a{y_1}^2 + b{u_1}^2),\\
\gamma^2 &= 4{u_1}{x_1}({x_1}^2 + a{y_1}^2 + b{u_1}^2).
\end{aligned} \right\} \tag{4}
\]
These can be satisfied if we have
\[
\left. \begin{aligned}
&x_1 = {x_2}^2, \quad y_1 = {y_2}^2, \quad u_1 = {u_2}^2,\\
&{x_2}^4 + a{y_2}^4 + b{u_2}^4 = \rho^2.
\end{aligned}\ \right\} \tag{5}
\]

The last equation in (5) is of the same form as (1). Hence
if we know a single solution of (1), we can by its means satisfy
the last equation in (5). Then from the remaining Eqs.~(5)
and Eqs.~(2), (3), (4), we can determine values of $\alpha$, $\beta$, $\gamma$, $\mu$.
Therefore, from a single solution of (1) we can find a second
solution; from this one a third can be obtained; from the
third a fourth can be gotten; and so on. Thus we have at
hand a means for determining in general an infinite number
of solutions of (1) as soon as a single solution is known.

As an illustration of this method let us consider the special
equation in which $a = b = 2$, namely, the equation,
\[
\alpha^4 + 2\beta^4 + 2\gamma^4 = \mu^2.
\]
% [File: 056.png Page: 46]
A particular solution is afforded by the relation
\[
3^4 + 2 \cdot 4^4 + 2 \cdot 2^4 = 25^2.
\]
Associating this with Eq.~(5), we see that we may take $x_2 = 3$,
$y_2 = 4$, $u_2 = 2$, $\rho = 25$. Then we have $x_1 = 9$,  $y_1 = 16$, $u_1 = 4$;
whence from (3) we have $\alpha = -463$, $x = 625$, $y = 288$, $u = 72$.
Thence from (2) we see that
\[
\alpha = 463, \quad \beta = 600, \quad \gamma = 300, \quad \mu = 566,881
\]
is a second solution of the equation. From this a third may
be obtained in a similar manner; and so on \textit{ad infinitum}.

Let us now consider the equation
\[
a^2 + a\beta^2 + b\gamma^2 = \mu^4.         \tag{6}
\]
Again making use of Eqs.~(7) of the preceding section we see
that (6) will be satisfied if
\[
\left. \begin{aligned}
\mu^2 &= x^2+ay^2+bu^2+abv^2,\\
\alpha &= x^2-ay^2-bu^2+abv^2,\\
\beta &= 2xy-2buv,\\
\gamma &= 2ux+2avy.
\end{aligned}\ \right\} \tag{7}
\]
It is only the first equation of this set which puts a condition
on the integers $x$, $y$, $u$, $v$. That equation will be satisfied if
\[
\left. \begin{aligned}
\mu &= {x_1}^2 + a{y_1}^2 + b{u_1}^2 + ab{v_1}^2,\\
x   &= {x_1}^2 - a{y_1}^2 - b{u_1}^2 + ab{v_1}^2,\\
y &= 2x_1y_1 - 2bu_1v_1,\\
u &= 2u_1x_1 + 2av_1y_1,\\
v &= 0.
\end{aligned}\ \right\} \tag{8}
\]
Here $x_1$, $y_1$, $u_1$, $v_1$ are arbitrary integers. Hence Eqs.~(7)
and (8) afford a four-parameter solution of Eq.~(6).

Finally, let us consider the equation
\[
\alpha^4 + a\beta^4 = \mu^2 + b\nu^2.         \tag{9}
\]
There is often a measure of choice at our disposal concerning
the set of numbers which we shall extend to form a domain
% [File: 057.png Page: 47]
with respect to multiplication. Here we may extend the set
determined by either of the following forms:
\[
\alpha^4 + a \beta^4 - b \nu^2, \quad \mu^2 + b \nu^2 -a \beta^4.
\]
We shall employ the latter alone.

Proceeding as in the previous cases, we may write
\[
\left. \begin{aligned}
\alpha^2 &= x^2 + by^2 - au^2,\\
\mu &= x^2 - by^2 + au^2,\\
\nu &= 2xy,\\
\beta^2 &= 2ux.
\end{aligned}\ \right\} \tag{10}
\]
To satisfy the first of these equations it is sufficient to take
\[
\left. \begin{aligned}
\alpha &= {x_1}^2 + b{y_1}^2 - a{u_1}^2,\\
x      &= {x_1}^2 - b{y_1}^2 + a{u_1}^2,\\
y &= 2x_1y_1,\\
u &= 2u_1 x_1.
\end{aligned}\ \right\} \tag{11}
\]
Then from the last equation in (10) we have the further restriction:
\[
\beta^2 = 4u_1 x_1({x_1}^2 - b{y_1}^2 + a{u_1}^2).
\]
This can be satisfied if we write
\[
\left. \begin{aligned}
u_1 = {u_2}^2, \quad x_1 = {x_2}^2;\\
{x_2}^4 + a{u_2}^4 - b{y_1}^2  = \rho^2.
\end{aligned}\ \right\} \tag{12}
\]
But this last equation is of the same form as (9). Hence
from a single solution of (9) we have integers satisfying the
last equation in (12). From these we can determine a second
solution of (9) by means of Eqs.~(10), (11), (12). From this
a third can be found; and so on.

As a special case of (9) consider the equation
\[
\alpha^4 + \beta^4 = \mu^2 + \nu^2,
\]
which has a solution afforded by the relation
\[
6^4 + 5^4 = 39^2 + 20^2.
\]
% [File: 058.png Page: 48]
Comparing with (12) and remembering that we now have
$a = b = 1$, we see that we may take $x_2 = 6$, $u_2 = 5$, $y_1 = 20$ (or
$y_1$ might be taken equal to 39 and thus another result be finally
obtained). Then $x_1 = 36$, $u_1 = 25$. Thence from (11) one may
determine $x$, $y$, $u$ and thence from (10) values of $\alpha$, $\beta$, $\mu$, $\nu$ may
be found which will satisfy the relation $\alpha^4 + \beta^4 = \mu^2 + \nu^2$. From
this second solution a third may be found; and so on \textit{ad
infinitum}.

Many other equations of the sort treated in this section
are amenable to the same methods. Some of these are indicated
in the general exercises at the close of this chapter.

\Needspace*{4\baselineskip}
\begin{center}EXERCISES\end{center}

\begin{small}
1. Show how to find an infinite number of integral solutions of the equation
\[
\alpha^4 + a\beta^4 = \mu^2,
\]
from a single given integral solution. Find several values of $a$ for which integral
solutions certainly exist.

2. Obtain a two-parameter integral solution of the equation
\[
\alpha^2 + a\beta^2 = \mu^4.
\]

3. Show how to find a two-parameter integral solution of the equation
\[
\alpha^2 + a\beta^2 = \mu^n,
\]
for any given positive integral value of $n$.\index{Biquadratic Equations|)}\index{Equations of Fourth Degree|)}
\par\end{small}

%§ 12.
\Needspace*{4\baselineskip}
\section[\textsc{On the Extension of a Set of Numbers so as to
Form a Multiplicative Domain}]{On the Extension of a Set of Numbers so as to
Form a Multiplicative Domain}

\hspace{\parindent}In the preceding sections of this chapter we have instances
illustrating a matter of great importance in Diophantine analysis.
It is intimately connected with the fact that certain
classes of integers form a domain with respect to multiplication.
We have seen how the properties of such a domain
may be employed to obtain solutions of a considerable class
of equations. Moreover, we have found it profitable to generalize
a set of integers introduced by a given equation so that
the resulting set shall form a domain, whereas the original set
did not. In the first place we extended the given set by the
% [File: 059.png Page: 49]
adjunction of a new variable; this method can evidently be
generalized, when convenient, by the introduction of two or
more variables. In the second place we extended the given
set by the generalization of variables already present. These
are two widely useful methods of extension which may be
employed either separately or simultaneously. They serve to
unify and connect many problems which otherwise would
appear unrelated.

If a given equation is put forward for consideration it is
usually a matter of ingenuity to determine a suitable method
of extension so as to give rise to a domain with respect to
multiplication; and it is not at all improbable that no method
will be apparent. In fact, it is easy to propose problems which
are not yet amenable to solution either by this or by other
means. There are, however, certain general classes of equations
to which the method will apply, and these will be indicated
as we proceed.

But the chief value of the method of extension does not
consist so much in its use for the solution of equations proposed
at random, as in its use for the arrangement of large and
interesting classes of problems in an order in which they are
amenable to attack and for suggesting a uniform procedure
by which they may be investigated. One cannot fail to see
the value of this in accomplishing the desirable end of building
up a considerable body of connected doctrine.

There is one general case in which the extension by the
adjunction of new variables is possible and to which we wish
to direct attention. We begin with an illustration.

Suppose that a problem gives rise to the set of numbers
$x^3+y^3$. They do not form a domain with respect to multiplication.
Let us consider the product
\[
P(x,y,z)=(x+y+z)(x+\omega y+\omega^2 z)(x+\omega^2 y+\omega z),  \tag{1}
\]
where $\omega$ and $\omega^2$ are the imaginary cube roots of unity. By
multiplication we find that
\[
P(x,y,z)=x^3+y^3+z^3-3xyz.    \tag{2}
\]
% [File: 060.png Page: 50]
Hence $P(x,y,0)=x^3+y^3$. Hence the set of numbers $P(x,y,z)$
is an extension of the set $x^3+y^3$.

Now
\[
(x+\omega y+\omega^2 z)(u+\omega v+\omega^2 w)=r+\omega s+\omega^2 t   \tag{3}
\]
where
\[
\left. \begin{aligned}
r&=xu+yw+zv,\\
s&=xv+yu+zw,\\
t&=xw+yv+zu.
\end{aligned}\ \right\} \tag{4}
\]
Hence
\[
P(x,y,z) \cdot P(u,v,w) = P(r,s,t).
\]
Therefore, the numbers $P(x,y,z)$ form a domain with respect
to multiplication.

By interchanging the role of $v$ and $w$ we have also
\[
P(x,y,z) \cdot P(u,v,w) = P(r_1,s_1,t_1),    \tag{5}
\]
where
\[
\left. \begin{aligned}
r_1&=xu+yv+zw,\\
s_1&=xw+yu+zv,\\
t_1&=xv+yw+zu.
\end{aligned}\ \right\} \tag{6}
\]

Let us consider more generally the set of numbers
\[
  x_1 + a_1{x_1}^{n-1} x^2 + a_2{x_1}^{n-2} {x_2}^2 + \ldots 
+ a_{n-1} x_1{x_2}^{n-1} + a_n {x_2}^n,     \tag{7}
\]
where $a_1$, $a_2$, \ldots, $a_n$ are given quantities. The method of
extending this set so that the resulting set shall form a domain
with respect to multiplication grows out of a remark due to
\index{Lagrange}Lagrange (\textit{\OE uvres}, Vol.~VII, pp.~164--179), though Lagrange
seems nowhere to have utilized it in connection with Diophantine
problems. A partial use of it has been made by
\index{Legendre}Legendre (\textit{Th\'eorie des nombres}, Vol.~II, 3d edition, pp.~134--141);
but its consequences seem nowhere to have been systematically
developed.

Let $\alpha_1$, $\alpha_2$, \ldots, $\alpha_n$ be the roots of the equation
\[
t^n - a_1 t^{n-1} + a_2 t^{n-2} - \ldots + (-1)^{n-1} a_{n-1}t + (-1)^n a_n = 0.    \tag{8}
\]
Form the product
\[
P(x) = \textstyle\prod\limits_{i=1}^{n}
       ( x_1 + \alpha_i x_2 + {\alpha_i}^2 x_3 + \ldots 
       + {\alpha_i}^{n-1}x_n).     \tag{9}
\]
% [File: 061.png Page: 51]
This is a symmetric function of the roots of Eq.~(8), whence
we see readily that it may be written as a homogeneous polynomial
of the $n$th degree in the quantities $x_1$, $x_2$, \ldots, $x_n$ with
coefficients which are polynomials in $a_1$, $a_2$, \ldots, $a_n$, the latter
having integral coefficients. For $x_3=x_4= \ldots =x_n=0$, it is
clear that $P(x)$ is identical with the expression in (7). Hence
the set of numbers (7) is extended into the set (9) by the adjunction
of the new variables $x_3$, $x_4$, \ldots, $x_n$.

It remains to show that the numbers $P(x)$ form a domain
with respect to multiplication. Let $P(x)$ and $P(y)$ be two
numbers of the form~(9). Then we have
\[
  P(x) \cdot P(y) 
= \textstyle\prod\limits_{i=1}^n
  (x_1 + \alpha_i x_2 + \ldots + {\alpha}_i^{n-1}x_n)
  (y_1 + \alpha_i y_2 + \ldots + {\alpha}_i^{n-1}y_n). \tag{10}
\]
Let us multiply together the two quantities
\[
  x_1 + \alpha_i x_2 + \ldots + {\alpha_i}^{n-1}x_n, \quad
  y_1 + \alpha_i y_2 + \ldots + {\alpha_i}^{n-1}y_n,
\]
and in the product repeatedly replace ${\alpha_i}^{n+k}$, $k\geqq 0$, by its value
\[
  {\alpha_i}^{n+k} 
= a_1{\alpha_i}^{n+k-1} 
- a_2{\alpha_i}^{n+k-2} 
+ a_3{\alpha_i}^{n+k-3} - \ldots, \ \ k\geq 0,
\]
until the result is reduced to the form
\[
z_1+\alpha_i z_2 + {\alpha_i}^2 z_3 + \ldots + {\alpha_i}^{n-1} z_n,
\]
where $z_1$, $z_2$, \ldots, $z_n$ are determinate polynomials in $x_1$, \ldots,
$x_n$, $y_1$, \ldots, $y_n$. Then it is obvious that we have
\[
P(x) \cdot P(y)=P(z);       \tag{11}
\]
that is, the product of two numbers of the given form is itself
of the same form. Hence the numbers $P(x)$ form a domain
with respect to multiplication.

It is obvious that one can obtain immediately an $n$-parameter
solution of the equation
\[
P(x)=t^k,
\]
where $k$ is any positive integer. For this purpose it is sufficient
to write $t=P(z)$ and to determine $x_1$, \ldots, $x_n$ from the
relation
\[
\{P(z)\}^k = P(x).
\]

% [File: 062.png Page: 52]
Again, from a single solution of the equation
\[
P(x)=0,
\]
an $n$-parameter solution may be obtained by means of Eq.~%
(11), $y_1$, \ldots, $y_n$ being taken as arbitrary and the quantities
$z_1$, \ldots, $z_n$ being the new values of $x_1$, \ldots, $x_n$.

In a similar manner, from a single solution of the equation
\[
P(x) = 1,         \tag{12}
\]
others may be found. From a solution of Eq.~(12) and a
solution of the equation
\[
P(x)=m,
\]
other solutions of the latter may obviously be obtained.

\Needspace*{4\baselineskip}
\begin{center}GENERAL EXERCISES\end{center}
\addtocontents{toc}{\protect\contentsline {section}{\numberline {}\textsc {General Exercises} 1-22}{\thepage}}

\begin{small}
1. If $p$, $q$, $r$ satisfy the relations
\[
p+q+r = 1, \quad \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0,
\]
show that
\[
a^2+b^2+c^2= (pa+qb+rc)^2+(qa+rb+pc)^2+(ra+pb+qc)^2.
\]
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(Davis, 1912.)\quad}\index{Davis}

2. Obtain solutions of the equation
\[
(x^2+y^2+z^2) ({x_1}^2+{y_1}^2+{z_1}^2)=u^2+v^2+w^2.
\]
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(Catalan, 1893.)\quad}\index{Catalan}

3. By means of the identity
\begin{multline*}
  ({a_1}^2 + {a_2}^2 + \ldots + {a_n}^2)^2 
\\
= ( {a_1}^2 + {a_2}^2 + \ldots + a_{n-1}^2-{a_n}^2 )^2
  + (2a_1a_n)^2 + (2a_2a_n)^2 + \ldots + (2a_{n-1}a_n)^2
\end{multline*}
show how to find any number of square numbers whose sum is a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Martin}Martin, 1896.)\quad}

4. Show how to write the square of a sum of $n$ squares as a sum of $n$ squares.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Moureaux}Moureaux, 1894.)\quad}

5. By means of the identities
\[
\begin{aligned}
(1+a+b+ab+a^2+b^2)^2 
\equiv (1+a)^2 (a+b)^2 + (1+b)^2 (a+b)^2 + (1+a+b-ab)^2&
\\
\equiv a^2(a+b+1)^2 + b^2(a+b+1)^2 + (a+b+1)^2 + (a+b+ab)^2,&
\end{aligned}
\]
show how to separate certain squares into a sum of three and of four squares.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(Avillez, 1897.)\quad}\index{Avillez}

6.* Show how to find other solutions of the system
\[
x^2-by^2=u^2, \quad x^2+by^2=v^2,
\]
when a single solution is known. Prove that a non-trivial solution does not exist
when $b$ is a square number. Show further that a necessary and sufficient condition
for the existence of an integral solution is that integers $m$, $n$, $p$ exist, such
that
\[
\phantom{(Lucas,\ 1876.)\quad}
bp^2=mn(m+n)(m-n).                \tag*{(\index{Lucas}Lucas, 1876.)\quad}
\]
%\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Lucas}Lucas, 1876.)\quad}

% [File: 063.png Page: 53]
7.* Discuss the solution of the Diophantine system
\[
\phantom{(Lucas,\ 1876.)\quad}
x^2+xy+2y^2=u^2, \quad x^2-xy-2y^2=v^2.                \tag*{(\index{Lucas}Lucas, 1876.)\quad}
\]
%\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Lucas}Lucas, 1876.)\quad}

8.* Show how to find a second solution of the system
\[
2v^2-u^2=w^2, \quad 2v^2+u^2=3z^2,
\]
when a single solution is known; thence find the general solution of the system.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Lucas}Lucas, 1877; \index{Pepin}Pepin, 1879.)\quad}

9.* Show that the only integral solution of the system
\[
2v^2-u^2=w^4, \quad 2v^2+u^2=3z^2,
\]
is the trivial one in which each of the numbers $u$, $v$, $w$, $z$ is equal to $\pm 1$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Lucas}Lucas, 1877.)\quad}

10. Let $p$, $r$, $s$ be three numbers satisfying the relation
\[
r^4+ar^2s^2+bs^4=p^2.
\]
Show that the equation
\[
x^4+ax^2y^2+by^4=z^2
\]
has the further solution
\[
x=r^4-bs^4, \quad y=2prs, \quad z=p^4-(a^2-4b)r^4s^4.
\]
Derive this result by a direct use of the method of extension by generalization
of variables.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Lebesgue}Lebesgue, 1853.)\quad}

11.* Prove that the equation
\[
x^4 - 2^m y^4=1
\]
is impossible in positive integers $x$, $y$, $m$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Thue}Thue, 1903.)\quad}

12.* Determine the values of $m$ for which the equation
\[
x^4+mx^2y^2+y^4=z^2
\]
has non-trivial solutions.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Werebr\"ussov}Werebr\"ussov, 1908.)\quad}

13.\dag\ Determine the integral values of $m$ and $n$ for which the equation
\[
x^4+mx^2y^2+ny^4=z^2
\]
has non-trivial solutions.

14.\dag\ Investigate further the problem of solving the equation
\[
x^2+ay^2+bz^2=t^k
\]
for integral values of $k$ greater than 2.

15.\dag\ Investigate the problem of solving the system
\[
x^2+y^2+z^2=k(u^2+v^2+w^2)=l(r^2+s^2+t^2)
\]
for particular values of $k$ and $l$. (Compare Exs.~4 and 6 in §~10.)

16.\dag\ Develop in further detail the theory of the equation
\[
x^4+ay^4+bz^4=t^2.
\]
(See special cases of this equation in Chapter IV.)

17.\dag\ Investigate the problem of solving the equation
\[
x^4+ay^4+bz^4=t^k
\]
for integral values of $k$ greater than 2. In particular, consider the case $k=4$.

% [File: 064.png Page: 54]
18.\dag\ Investigate the problem of solving the equation
\[
x^4+ay^4=u^2+av^2.
\]

19.\dag\ Develop in further detail the theory of the equation
\[
x^4+ay^4=u^2+bv^2.
\]

20.\dag\ Investigate the problem of solving the equation
\[
x^4+ay^4+bu^4+abv^4=t^2.
\]

21.\dag\ Let $a$, $b$, $k$ be given integers, $k$ being positive. Investigate the properties
of the integer $m$ such that the equation
\[
x^2+axy+by^2=mt^k
\]
shall possess integral solutions and find these solutions when they exist. In
particular, treat the cases $k=2$, $k=3$.

22.\dag\ Develop a similar theory for each of the following equations:
\[
\begin{aligned}
x^2+ay^2+bz^2&=mt^k,\\
x^4+ay^4+bz^4&= mt^k,\\
x^4+ay^4&=m(u^2+av^2),\\
x^4+ay^4&=m(u^2+bv^2).
\end{aligned}
\]\index{Domain, Multiplicative|)}\index{Multiplicative Domain|)}\index{Method of Multiplicative Domain|)}
\par\end{small}

% [File: 065.png Page: 55]
%CHAPTER III

\chapter{EQUATIONS OF THE THIRD DEGREE}\index{Cubic Equations|(}\index{Equations of Third Degree|(}
\markright{EQUATIONS OF THE THIRD DEGREE}

\setcounter{section}{12}
%§ 13.
\section[\textsc{On the Equation $kx^3+ax^2y+bxy^2+cy^3 = t^2$}]{On the Equation $kx^3+ax^2y+bxy^2+cy^3 = t^2$}

\hspace{\parindent}We shall now consider the problem of finding solutions of
the equation
\[
x^3+ax^2y+bxy^2+cy^3 = t^2,       \tag{1}
\]
where $a$, $b$, $c$ are rational numbers. By our general method
(in §~12) of extending the set of numbers defined by the first
member we are led to consider the function
\[
h(\alpha, \beta, \gamma) = \textstyle\prod\limits_{i=1}^3 (\alpha + \rho_i \beta + {\rho_i}^2 \gamma),  \tag{2}
\]
where $\rho_1$, $\rho_2$, $\rho_3$ are the roots of the equation
\[
\rho^3-a\rho^2+b\rho-c=0.        \tag{3}
\]
It is obvious that $h(x,y,0)=x^3+ax^2y+bxy^2+cy^3$.

Performing the requisite multiplications, we have
\[
\begin{aligned}
  h(\alpha,\beta,\gamma)
= \alpha^3 + a\alpha^2\beta + b\alpha\beta^2 + c\beta^3 
+ (a^2-2b)\alpha^2\gamma + (b^2-2ac)\alpha\gamma^2&
\\
+ ac\beta^2\gamma + bc\beta\gamma^2+c^2\gamma^3 
+ (ab-3c)\alpha\beta\gamma.&
\end{aligned} \tag{4}
\]
Since $\rho_i$ satisfies Eq.~(3), it is easy to see that we have a relation
of the form
\[
  (\alpha+\rho_i\beta+{\rho_i}^2\gamma) (u+\rho_iv+{\rho_i}^2w) 
= r+\rho_is+\rho_i^2t,
\]
where $r$, $s$, $t$ are rational. On computing the values of $r$, $s$, $t$,
we are led to the following result:

I\@. \textit{If $r$, $s$, $t$ have the values
\[
\left. \begin{aligned}
r&=\alpha u + c\gamma v + c\beta w + ac\gamma w,\\
s&=\beta u + \alpha v - b\gamma v - b\beta w + c\gamma w - ab\gamma w,\\
t&=\gamma u + \beta v + \alpha w + a\gamma v + a\beta w - b\gamma w + a^2\gamma w,
\end{aligned}\ \right\} \tag{5}
\]
then}%end\textit
\[
h(\alpha,\beta,\gamma) \cdot h(u,v,w)=h(r,s,t).      \tag{6}
\]

% [File: 066.png Page: 56]
From this formula we see readily that
\[
\begin{aligned}
\{ h(\alpha, \beta, \gamma) \}^2 
=h(\alpha^2 + 2c\beta\gamma + ac\gamma^2, \quad
   2\alpha\beta - &2b\beta\gamma + c\gamma^2 - ab\gamma^2,
\\
   2\alpha\gamma + \beta^2 &+ 2a\beta\gamma - b\gamma^2 + a^2\gamma^2).
\end{aligned} \tag{7}
\]

By means of the last relation we are able to find a two-parameter
solution of Eq.~(1). The latter we may write in
the form $h(x, y, 0) = t^2$.  Comparing this with Eq.~(7) we see
that a solution is afforded by the values
\[
\left. \begin{aligned}
t &= h(\alpha, \beta, \gamma),\\
x &= \alpha^2 + 2c\beta\gamma + ac\gamma^2,\\
y &= 2\alpha\beta - 2b\beta\gamma + c\gamma^2 - ab\gamma^2,
\end{aligned}\ \right\} \tag{8}
\]
provided that $\alpha$, $\beta$, $\gamma$ are connected by the relation
\[
2\alpha\gamma + \beta^2 + 2a\beta\gamma - b\gamma^2 - a^2\gamma^2 = 0.     \tag{9}
\]
Into the last relation $\alpha$ enters linearly. It is therefore a rational
function of $\beta$ and $\gamma$ with rational coefficients. Hence,

II\@.  \textit{A two-parameter rational solution of $(1)$ is afforded by
the set~$(8)$ where $\beta$ and $\gamma$ are arbitrary rational numbers $($except
that $\gamma$ must in general be different from zero$)$ and $\alpha$ is determined
by Eq.~$(9)$}.

If we set $\gamma = 2m$, $\beta = 2mn$, where $m$ and $n$ are integers, we
are led immediately to the following result:

III\@.  \textit{If $a$, $b$, $c$ are integers, then a two-parameter integral
solution of $(1)$ is afforded by the values
\[
\begin{aligned}
t &= h(\alpha, 2mn, 2m),\\
x &= \alpha^2 + 8cm^2 n + 4acm^2,\\
y &= 4\alpha m - 8bm^2 n + 4cm^2 - 4abm^2,
\end{aligned}
\]
where
\[
 \alpha = bm - a^2 m - 2amn - mn^2,
\]
$m$ and $n$ being arbitrary integers}.

It is obvious that these results may be applied readily to
the solution of the equation
\[
 kx^3 + ax^2 y + bxy^2 + cy^3 = t^2, \ \ k \neq 0;    \tag{10}
\]
for, if this equation is multiplied through by $k^2$ and $kx$ is replaced
by $x$, the resulting equation is of the form of (1).

% [File: 067.png Page: 57]
The reader may readily supply numerical illustrations of
these results.

A method of finding particular solutions of Eq.~(10) is
due to Fermat. It applies only when $k$ or $c$ is a square. Let
us suppose that $k$ is a square. Write the equation in the
form
\[
d^2 x^3 + ax^2 y + bxy^2 + cy^3 = t^2.      \tag{11}
\]
Take $x = 1$ and set
\[
t = d + \frac{a}{2d} y.
\]
Then we have
\[
  \left( d + \frac{a}{2d} y \right)^2 
+ \left( b - \frac{a^2}{4d^2} \right)y^2 + cy^3 
= \left( d + \frac{a}{2d} y \right)^2.
\]
This gives
\[
y = \frac{1}{c} \left( \frac{a^2}{4d^2} - b \right).
\]
From this value of $y$ we have a value of $t$, and hence a solution
of (11).

%§ 14.
\Needspace*{4\baselineskip}
\section[\textsc{On the Equation $kx^3 +ax^2 y + bxy^2 + cy^3 = t^3$}]{On the Equation $kx^3 +ax^2 y + bxy^2 + cy^3 = t^3$}

\hspace{\parindent}If we set
\[
\left. \begin{aligned}
u &= \alpha^2 + 2c\beta\gamma + ac\gamma^2,\\
v &= 2\alpha\beta - 2b\beta\gamma + c\gamma^2 - ab\gamma^2,\\
w &= 2\alpha\gamma + \beta^2 + 2a\beta\gamma - b\gamma^2 + a^2\gamma^2,
\end{aligned}\ \right\} \tag{1}
\]
then from Eqs.~(5), (6), (7) of the preceding section we see
that
\[
\{ h(\alpha, \beta, \gamma) \}^3 = h(\rho, \sigma, \tau),       \tag{2}
\]
where
\[
\left. \begin{aligned}
\rho &= \alpha u + c\gamma v + c\beta w + ac\gamma w,\\
\sigma &= \beta u + \alpha v - b\gamma v - b\beta w + c\gamma w - ab\gamma w,\\
\tau &= \gamma u + \beta v + \alpha w + a\gamma v + a\beta w - b\gamma w + a^2 \gamma w.
\end{aligned}\ \right\} \tag{3}
\]

For a solution of the equation
\[
x^3 + ax^2 y + bxy^2 + cy^3 = t^3,       \tag{4}
\]
it is obviously sufficient to take
\[
t = h(\alpha, \beta, \gamma), \quad x = \rho, \quad y = \sigma, \quad \tau = 0.     \tag{5}
\]
% [File: 068.png Page: 58]
Hence the equation $\tau = 0$ puts the necessary restriction on the
otherwise arbitrary quantities $\alpha$, $\beta$, $\gamma$.

Since $\tau$ is the crucial quantity, let us write out its value
in terms of $\alpha$, $\beta$, $\gamma$. We have
\begin{multline*} 
\tau = 3\alpha^2 \gamma + 3\alpha\beta^2 + 3(a^2 - b)\alpha\gamma^2 
+ 2(a^3 + 3a - ab)\alpha\beta\gamma + a\beta^3
\\
+ 3(a^2 - b)\beta^2 \gamma + (a^3 - 4ab + 3c)\beta\gamma^2 
+ (a^4 - 3a^2 b + 2ac + b^2)\gamma^3.   \tag{6}
\end{multline*}
From any solution of the equation $\tau = 0$ we have a solution
of (4).

In order that the equation $\tau = 0$ shall determine a rational
value of $\alpha$, it is obviously necessary and sufficient that the
expression
\[
\begin{aligned}
\{ 3\beta^2 &+ 3(a^2 - b)\gamma^2 + 2(a^3 + 3a - ab)\beta\gamma\}^2 
- 12\gamma\{a\beta^3
\\
&+ 3(a^2 - b)\beta^2 \gamma + (a^3 - 4ab + 3c)\beta\gamma^2 
+ (a^4 - 3a^2 b + 2ac + b^2)\gamma^3\},
\end{aligned}
\]
shall be equal to a square. If we call this $m^2$, we have
\[
\begin{aligned}[b]
 9\beta^4 &+ 12(a^3 + 2a - ab)\beta^3 \gamma\\
          &+ 2(2a^6 + 12a^4 - 4a^4 b + 9a^2 - 12a^2 b + 2a^2 b^2 + 9b)\beta^2 \gamma^2\\
          &+ 12(a^5 + 3a^4 - a^3 - 2a^3 b + ab + ab^2 - 3c)\beta\gamma^3\\
          &+ (-a^4 + 6a^2 b - b^2 - 8ac)\gamma^4 = m^2.
\end{aligned} \tag{7}
\]
A method of finding in general an infinite number of solutions
of this equation is given in §~17 of the following chapter. The
work done here consists essentially in reducing the problem
of solving Eq.~(4) to that of solving Eq.~(7).

A particular solution of Eq.~(7) is obvious. It is gotten
by setting $\gamma = 0$. This gives rise to the following solution of (4):
\[
\begin{aligned}
 t &= 2a^3 - 9ab + 27c,\\
 x &= 27c - a^3,\\
 y &= 9a^2 - 27b.
\end{aligned}
\]
But this particular solution is more readily obtained by the
method of Fermat described below.

The equation
\[
 k^3 x^3 + ax^2 y + bxy^2 + cy^3 = t^3,       \tag{8}
\]
can be readily reduced to the form of (4). It is sufficient to
multiply the equation through by $k^6$ and replace $x$ by $k^3 x$.
Hence the previously developed theory is applicable to Eq.~(8).

% [File: 069.png Page: 59]
Another method of effecting a reduction similar to that
above is the following (see \index{Schaewen}Schaewen, \textit{Jahresbericht der Deutschen
Mathematiker-Vereinigung}, Vol.\ XVIII, pp.~7--14): In the general
equation~(8) replace $x$ by
\[
  kx - \frac{a}{3k^2} y.
\]
Then we have a result which may be written
\[
  k^3 x^3 + pxy^2 + qy^3 = t^3.        
\tag{9}
\]
This equation may be put in the form
\[
  (t - kx)(t^2 + kxt + k^2 x^2) = y^2 (px + qy).
\]
The last equation will be satisfied if one writes
\begin{align*}
                 m(t-kx) &= ny,             \\
  n(t^2 + kxt + k^2 x^2) &= m(pxy + qy^2),
\end{align*}
where $m$ and $n$ are arbitrary quantities. If we put
\[
  t - kx = \frac{n}{m} y,
\]
the first of these equations is satisfied while the second becomes
\[
  3k^2 m^2 nx^2 - (pm^3 - 3kmn^2)xy - (qm^3 - n^3) y^2 = 0.
\]
If we denote the discriminant of this equation by $m^2 \rho^2$ we see
that it, and hence Eq.~(9), has a rational solution which may
be immediately derived as soon as one knows rational numbers
$m$, $n$, $\rho$, satisfying the equation
\[
  p^2 m^4 + 12k^2 qm^3 n - 6kpm^2 n^2 - 3k^2 n^4 = \rho^2.      
\tag{10}
\]
Thus we have reduced the problem of solving Eq.~(9) to that
of solving Eq.~(10).

There is an interesting method by which particular solutions
of Eq.~(4), and, in fact, of the more general equation
\[
  Ax^3 + Bx^2 y + Cxy^2 + Dy^3 = t^3,        
\tag{11}
\]
may often be obtained (see Schaewen, \textit{l.~c.}). It depends
essentially on writing Eq.~(11) in the form
\[
  P^3 + Q = t^3,       
\tag{11\textsuperscript{bis}}
\]
% [File: 070.png Page: 60]
where $P$ is a linear expression in $x$ and $y$ and $Q$ is a polynomial
which has at least one rational linear factor. A convenient
pair of values of $x$ and $y$ can be chosen so as to reduce such
a linear factor to zero. Then $t$ can readily be found. Thus
a particular solution of (11) is obtained. The applicability
of this method depends on whether or not Eq.~(11) may conveniently
be thrown into the form (11\textsuperscript{bis}).

As an example, we take the equation
\[
  x^3 - 5x^2 y - 6xy^2 + 8y^3 = t^3.        
\tag{12}
\]
It may be written in each of five forms as follows:
\begin{align*}
  8y^3 + x(x+y)(x-6y) &= t^3,
\\
  x^3 - (x+2y)(5x-4y)y &= t^3,
\\        
  \left( x - \frac{5}{3} y\right)^3 - \frac{y^2}{27} (387x - 341y) &= t^3,
\\
  (x + 2y)^3 - xy(11x + 18y) &= t^3,
\\        
  \left( 2y - \frac{x}{2} \right)^3 + \frac{x^2}{8} (9x - 52y) &= t^3.
\end{align*}
These give rise to the following pairs of values of $x$ and $y$, each
of which affords a solution of Eq.~(12), namely: 
$x =   1$,~$y =  -1$;
$x =   6$,~$y =   1$; 
$x =   2$,~$y =  -1$; 
$x =   4$,~$y =   5$; 
$x = 341$,~$y = 387$;
$x =  18$,~$y = -11$; 
$x =  52$,~$y =   9$.

A special case of this method is due to \index{Fermat}Fermat. Suppose
that $A = a^3$. Then for $P$ we may take
\[
  P = ax + \frac{B}{3a^2} y.
\]

Then $Q$ has the factor $y^2$ and hence a linear factor. Thus
one has a solution in all cases when $A$ is a cube. Similarly
there is a solution when $D$ is a cube. It must be observed that
for special equations this method may lead only to a trivial
solution of the given equation. This is true in the case of an
equation in the form $x^3 + cy^3 = t^3$.

% [File: 071.png Page: 61]
It may be remarked that the problem of finding integral
solutions of Eq.~(11) is obviously equivalent to that of finding
rational solutions of the equation
\[
  Ax^3 + Bx^2 + Cx + D = t^3;        
\tag{13}
\]
for (11) is reduced to (13) by dividing it through by $y^3$ and
in the resulting equation writing $x$, $t$ for $x/y$, $t/y$, respectively.
It is in this latter form that the problem was investigated by
Fermat and Euler.

Fermat observed that from a single solution of Eq.~(13)
others can usually be obtained. The method is as follows:
Let $x = m$ afford a rational solution of (13) and let $t = s$ be the
corresponding value of $t$. Then in (13) replace $x$ by $\xi + m$.
The equation takes the form
\[
  A\xi^3 + \overline{B}\xi^2 + \overline{C}\xi + s^3 = t^3.        
\tag{14}
\]
Now if we write
\[
  t = s + \frac{\overline{C}}{3s^2}\xi,
\]
we can readily determine a rational value of $\xi$, in general
different from zero, such that Eq.~(14) is satisfied. Adding
$m$ to this value of $\xi$, we have a new value of $x$ which affords
a solution of (13). Starting from this new value of $x$ we can
usually determine a third; and so on indefinitely. For certain
particular equations the method may fail to lead to new solutions.
This will be the case when the solution obtained for
(14) is $\xi = 0$, $t = s$.

As an illustrative example, let us consider the equation
\[
x^3 - 5x^2 + x + 4 = t^3.
\]
This has the solution $x = 1$, $t = 1$. Then write $x = \xi + 1$, whence
the equation takes the form
\[
\xi^3 - 2\xi^2 - 6\xi + 1 = t^3.
\]
Putting $t = 1 - 2\xi$ in the last equation and solving the resulting
equation for $\xi$ we have $\xi = 14/9$. Thence, as a second solution
of our given equation, we have $x = 23/9$, $t = -19/9$. From this
second solution a third may be found; and so on indefinitely.

% [File: 072.png Page: 62]
For an investigation concerning the existence of a solution
of Eq.~(13), see \index{Haentzschel}Haentzschel, \textit{Jahresbericht der Deutschen Mathematiker-Vereinigung},
\\ Vol.~XXII, pp.~319--329.

%§ 15.
\Needspace*{4\baselineskip}
\section[\textsc{On the Equation $x^3 + y^3 + z^3 - 3xyz = u^3 + v^3 + w^3 - 3uvw$}]{On the Equation \\
$x^3 + y^3 + z^3 - 3xyz = u^3 + v^3 + w^3 - 3uvw$}

\hspace{\parindent}Probably the most elegant particular equations of the
third degree are the following:
\begin{align*}
\tag{1}
x^3 + y^3 &= u^3,
\\
\tag{2}
x^3 + y^3 &= u^3 + v^3.
\end{align*}
That the former is impossible in integers different from zero
will be shown in the next section. The general solution of the
latter we shall derive in this section. It is convenient to consider
first the more general equation
\[
\tag{3}
x^3 + y^3 + z^3 - 3xyz = u^3 + v^3 + w^3 - 3uvw.
\]
This equation reduces to (2) on putting $z=w=0$. Moreover,
it is the equation to which one is led on extending each of the
two sets of integers $x^3 + y^3$, $u^3 + v^3$ so as to arrive at a multiplicative
domain, as one sees by reference to §~12. (It may
be observed that the problem of finding integral solutions
of (2) and (3) and that of finding rational solutions are essentially
the same.)

Let us denote the first member of (3) by $P(x, y, z)$ as in
§~12. Then from Eqs.~(3) to (6) in §~12 we see that
\[
P(r, s, t)\cdot P(m, n, p) = P(x, y, z) = P(u, v, w),
\]
where
\[
\tag{4}
\left.\begin{aligned}
x &= mr + nt + ps,\\
y &= ms + nr + pt,\\        
z &= mt + ns + pr,\end{aligned}\ \right\}
\]
and
\[
\tag{5}
\left.\begin{aligned}
u &= mr + ns + pt,\\
v &= mt + nr + ps,\\
w &= ms + nt + pr.\end{aligned}\ \right\}
\]
% [File: 073.png Page: 63]
Eqs.~(4) and (5) afford a six-parameter solution of (3). This
solution, so readily obtained, unfortunately lacks generality.

We proceed as follows to find a more general solution:
Write Eq.~(3) in the form
\begin{multline*}
(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\\
= (u + v + w)(u^2 + v^2 + w^2 - uv - vw - wu).
\end{multline*}
We may exclude as trivial the cases in which each member
is equal to zero; for all such solutions are obtained readily by
means of linear equations. (Compare the factorization of
$P(x, y, z)$ in §~12.) Then we may write
\[
\frac{x+y+z}{u+v+w} = \frac{u^2 + v^2 + w^2 - uv - vw - wu}
                           {x^2 + y^2 + z^2 - xy - yz - zx}.
\]
Now put
\[
\tag{6}
u = w + \alpha,\quad v = w + \beta,\quad x = z + \gamma,\quad y = z + \delta.
\]
Then we have
\[
\tag{7}
\frac{x+y+z}{u+v+w} = \frac{\alpha^2 - \alpha\beta + \beta^2}
                           {\gamma^2 - \gamma\delta + \delta^2}.
\]
Multiplying both numerator and denominator of the fraction
in the second member of (7) by the denominator we may
write the result in the form
\[
\frac{x+y+z}{u+v+w} 
= \frac{{\alpha_1}^2 - \alpha_1\beta_1 + {\beta_1}^2}
       {(\gamma^2 - \gamma\delta + \delta^2)^2} 
= {\alpha_2}^2 -\alpha_2 \beta_2 + {\beta_2}^2,
\]
where $\alpha_2$ and $\beta_2$ are rational numbers, when $u$, $v$, $w$, $x$, $y$, $z$
are rational numbers. Compare formula~(2) in §~7.

If we write
\[
  a = \frac{\alpha_2 + \beta_2}{2},\quad
  b = \frac{\alpha_2 - \beta_2}{2},
\]
then the preceding equation takes the form
\[
\tag{8}
x + y + z = (a^2 + 3b^2)(u + v + w).
\]
Hence, for every rational solution $x$, $y$, $z$, $u$, $v$, $w$ of Eq.~(3)
rational numbers $a$ and $b$ exist such that Eq.~(8) is satisfied.

From Eqs.\ (7) and (8) we have
\[
(a^2 + 3b^2)(\gamma^2 - \gamma\delta + \delta^2) = \alpha^2 - \alpha\beta + \beta^2.
\]
% [File: 074.png Page: 64]
This relation may be put in the form
\[
  (a^2 + 3b^2)
  \left[  \left( \frac{\gamma + \delta}{2} \right)^2
       + 3\left( \frac{\gamma - \delta}{2} \right)^2  \right] 
=         \left( \frac{\alpha + \beta }{2} \right)^2
       + 3\left( \frac{\alpha - \beta }{2} \right)^2 \!\!.   
\tag{9}
\]
By means of the relation
\[
  (a^2 + 3b^2) (c^2 + 3d^2) \equiv (ac + 3bd)^2 + 3(ad - bc)^2
\]
we see that Eq.\ (9) is satisfied if we have
\[
\left.
\begin{aligned}
  a(\gamma + \delta) + 3b(\gamma - \delta) &= \alpha + \beta,  \\
  a(\gamma - \delta) - \phantom{3}b(\gamma + \delta) &= \alpha - \beta.       
\end{aligned}
\right\}
\tag{10}
\]

From the homogeneous character of Eq.~(3) it follows that
if each of the numbers $x$, $y$, $z$, $u$, $v$, $w$ in a particular rational
solution is divided or multiplied by a given rational number,
the resulting numbers form a rational solution. Hence without
loss of generality we may take $u + v + w$ equal to unity. This
we do. Then, in view of (7), we have
\begin{align*}
  u + v + w &= 1,  \\
  x + y + z &= a^2 + 3b^2;
\end{align*}
and thence from (6),
\[
\left.
\begin{aligned}
  \alpha + \beta  &= 1 - 3w,          \\
  \gamma + \delta &= a^2 + 3b^2 - 3z.  
\end{aligned}
\right\}
\tag{11}
\]
Eqs.~(10) and (11) may now be solved for $\alpha$, $\beta$, $\gamma$, $\delta$. If the
results are put in Eq.~(6), we have
\[
\left.
\begin{aligned}
  x &= \frac{-(a-3b)(a^2 + 3b^2 - 3z) + 1 - 3w }{6b} + z,    \\
  y &= \frac{(a+3b)(a^2 + 3b^2 - 3z) - 1 + 3w }{6b} + z,    \\
  u &= \frac{-(a^2+3b^2) (a^2+3b^2-3z) + (a+3b) - 3w(a+3b)}{6b} + w,\\
  v &= \frac{(a^2+3b^2) (a^2+3b^2-3z) - (a-3b) + 3w(a-3b)}{6b} + w,  
\end{aligned}
\right\}
\tag{12}
\]
while $z$ and $w$ are arbitrary. This affords a rational solution
of (3) depending upon four rational parameters.

% [File: 075.png Page: 65]
Setting $z = w = 0$ and multiplying the resulting values of
$x$, $y$, $u$, $v$ by $6b$, we are led to the following solution of Eq.~(2):
\[
\left.
\begin{aligned}
  x &= -(a-3b)(a^2 + 3b^2) + 1,   \\
  y &=  (a+3b)(a^2 + 3b^2) - 1,   \\
  u &= -(a^2 + 3b^2)^2 + (a+3b),  \\
  v &=  (a^2 + 3b^2)^2 - (a-3b).       
\end{aligned}
\right\}
\tag{13}
\]
This solution is due to \index{Euler}Euler and \index{Binet}Binet; they obtained it by
a different method.

We shall now show that formulæ (13) afford the general
solution of Eq.~(2) (exclusive of trivial solutions) except for
an arbitrary constant $k$ multiplying the second member of
each equation in (13). In doing this it is convenient, first
of all, to transform the solution in the following manner: Put
\[
  a = \frac{\rho + \sigma}{2m}, \quad
  b = \frac{\sigma - \rho}{6m}, \quad
  n = \frac{\rho^2 + \rho\sigma + \sigma^2}{3m}.
\]
(This transformation is employed by \index{Fujiwara}Fujiwara in \textit{Arch.\ Math.\
Phys.}\ (3) 19 (1912): 369. See other papers referred to in
this note.) On replacing $x$, $y$, $u$, $v$ by $m^2$ times their former
values, we may write the resulting solution in the form:
\[
  x = -n\rho   + m^2, \quad
  y =  n\sigma - m^2, \quad
  u =  m\sigma - n^2, \quad
  v = -m\rho   + n^2.  
\tag{14}
\]
Here $\rho$, $\sigma$, $m$, $n$ are arbitrary parameters except for the relation
\[
  \rho^2 + \rho\sigma + \sigma^2 = 3mn.        
\tag{15}
\]

To prove that (14) affords the general solution of (2) it is
obviously sufficient to show how to determine $\rho$, $\sigma$, $m$, $n$ so
as to satisfy Eqs.~(14) and (15) when $x$, $y$, $u$, $v$, in order, are
replaced by a suitable multiple of any given set of values
satisfying (2). For this purpose we may proceed thus (see
\index{Schwering}Schwering, \textit{Arch.\ Math.\ Phys.}\ (3) 2 (1902): 281):

From (14) we have
\[
  x + y = n(\sigma - \rho),  \quad
  u + v = m(\sigma - \rho);
\]
whence
\[
  \frac{m}{n} = \frac{u+v}{x+y}.
\]
% [File: 076.png Page: 66]
Therefore, we may write
\[
m = \lambda (u + v), \quad n = \lambda (x + y).      \tag{16}
\]
Now again from (14) we have
\[
nv - mx = n^3 - m^3.
\]
If in this equation we substitute the values of $m$, $n$ from Eq.\
(16), we have
\[
\lambda^2 = \frac{v(x+y) - x(u+v)}{(x+y)^3 - (u+v)^3}.
\]
Since the numerator and denominator here are homogeneous,
the former of degree~2 and the latter of degree~3, it is obvious
that this equation will be satisfied by $\lambda = 1$, provided that
$x$, $y$, $u$, $v$ are replaced by $\tau x$, $\tau y$, $\tau u$, $\tau v$, respectively, and $\tau$ is
suitably determined. Replacing them by these same multiples
in Eqs.~(14) and (16) we see that (16) affords the values of
$m$ and $n$ and that (14) affords the values of $\rho$ and $\sigma$.

It remains to show that Eq.~(15) is satisfied. This we
do by substituting in Eq.~(2) the values of $x$, $y$, $u$, $v$ taken
from (14). Thus we have
\[
(-n\rho + m^2)^3 + (n\sigma - m^2)^3 + (-m\sigma + n^2)^3 + (m\rho - n^2)^3 = 0.
\]
Hence
\[
(m^3 - n^3)(\rho - \sigma)(\rho^2 + \rho\sigma + \sigma^2 - 3mn) = 0.
\]
Therefore Eq.~(15) or one of the relations $m = n$, $\rho = \sigma$ is satisfied.
If $m = n$ then $x = v$, $y = u$. If $\rho = \sigma$, then $x = -y$, $u = -v$.
Hence, except in the case of trivial solutions, Eq.~(15) is satisfied.
Therefore, Eqs.~(14), with the condition (15), afford
the general solution of (2). Likewise Eqs.~(13) afford the
general solution of (2). In each case the elements of the
solution given are all to be multiplied by an arbitrary constant
$k$.

If in (13) we set $a = 0$, $b = \tfrac{1}{2}$, and multiply the resulting
values of $x$, $y$, $u$, $v$ by 16, we are led to the relation
\[
34^3 + 2^3 = 15^3 + 33^3,
\]
as a particular case of the sum of two cubes equal to the sum
of two other cubes.

% [File: 077.png Page: 67]
%\§ 16. 
\Needspace*{4\baselineskip}
\section[\textsc{Impossibility of the Equation $x^3 + y^3 = 2^m z^3$}]{Impossibility of the Equation $x^3 + y^3 = 2^m z^3$}

\hspace{\parindent}We shall first consider the foregoing equation for the case
$m = 0$; it then has the form
\[
x^3 + y^3 = z^3.        \tag{1}
\]
In order to prove that this equation is impossible in integers
$x$, $y$, $z$, all of which are different from zero, we shall have need
of some lemmas similar to those in \S~10. We shall now state
them and indicate the means of proof.

%\begin{lemma}  
\textsc{Lemma I\@.} \textit{If a number is expressible in the form $\alpha^2 + 3\beta^2$,
and if the quotient $(\alpha^2 + 3\beta^2)/(a^2 + 3b^2)$ is an integer $m$, where
$\alpha$, $\beta$, $a$, $b$ are integers and $a^2 + 3b^2$ is a prime number, then $m$ is
expressible in the form $\gamma^2 + 3\delta^2$, where $\gamma$ and $\delta$ are integers.
}%\end{lemma}

%\begin{lemma}  
\textsc{Lemma II\@.} \textit{Every prime number of the form $6n + 1$ can be
represented in one and in only one way in the form $a^2 + 3b^2$ where
$a$ and $b$ are integers. No prime number of the form $6n - 1$ is
a divisor of a number of the form $a^2 + 3b^2$ where $a$ and $b$ are relatively
prime.
} %\end{lemma}

%\begin{lemma}  
\textsc{Lemma III\@.} \textit{Let $p$ be a prime number of the form $6n + 1$
and write $p = a^2 + 3b^2$, where $a$ and $b$ are integers. Let $m$ be any
integer such that $pm = \alpha^2 + 3\beta^2$, where $\alpha$ and $\beta$ are integers. Then
there exists a representation of $m$ as a sum of two integral squares,
\[
m=\left(\frac{a\alpha \pm 3b\beta}{a^2+3b^2}\right)^2 +3\left(\frac{\alpha b \mp a \beta}{a^2+3b^2}\right)^2, \tag{2}
\]
such that the representation $\alpha^2 + 3\beta^2$ of $pm$ is obtained from the
foregoing representations of $p$ and $m$ by multiplication in accordance
with the formula}
\[
(a^2 + 3b^2)(c^2 + 3d^2) = (ac+3bd)^2 + 3(ad - bc)^2.
\]
%\end{lemma}

%\begin{corollary}
\textsc{Corollary.} \textit{If $h$ is a composite number all of whose prime
factors are of the form $6n + 1$ and if we write $h = h_1 h_2$, where
$h_1$ and $h_2$ are positive integers, then every representation of $h$ in
the form $a^2 + 3b^2$ is obtained by taking every representation of
$h_1$ and $h_2$ and multiplying these expressions in accordance with
the formula}
\[
(a^2 + 3b^2)(c^2 + 3d^2) = (ac \pm 3bd)^2 + 3 (ad \mp bc)^2.
\]
%\end{corollary}

% [File: 078.png Page: 68]
The proof of lemma I is obtained by an obvious modification
of that of lemma~I in §~10. The proof of lemma~II is a close
parallel to that of lemma~II in §~10; in this case, however,
one starts from the fact that $-3$ is a quadratic residue of
primes of the form $6n + 1$ and of no other odd primes. The
proof of lemma~III is like that of lemma~III in §~10. The
reader can readily supply the necessary argumentation in all
cases.

Our immediate use of these lemmas is in finding the general
solution of the equation
\[
  p^2 + 3q^2 = s^3,        
\tag{3}
\]
where $p$ and $q$ are relatively prime integers and $s$ is odd. It
is clear that $s$ must have the form
\[
  s = t^2 + 3u^2,        
\tag{4}
\]
where $t$ and $u$ are integers. Writing $(t^2 + 3u^2)^3$ in the form
$p^2 + 3q^2$ by means of a repeated use of the last formula in the
foregoing corollary, we have
\[
  p = t^3 - 9tu^2, \quad  q = 3u(t^2 - u^2).        
\tag{5}
\]
Eqs.\ (4) and (5) give all the integral solutions of (3) subject
to the condition that $p$ and $q$ are relatively prime and $s$ is odd.

Let us now return to Eq.~(1). Our method of proof (see
\index{Euler}Euler, \textit{Opera Omnia}, series~1, Vol.~I, pp.\ 484--489) is to assume
that Eq.~(1) is satisfied by a set of numbers $x$, $y$, $z$, all of which
are different from zero, and to prove that we are then led
to a contradiction. Let $d$ be the greatest common divisor of
any two of these numbers. It is then a divisor of the third.
Eq.~(1) may then be divided through by $d^3$ and a new equation
of the same form obtained in which the resulting $x$, $y$, $z$ are
prime each to each. Hence, without loss of generality, we
may assume that Eq.~(1) itself is satisfied by integers $x$, $y$, $z$
which are prime each to each; and this we do. Then it is
easy to see that two of the numbers $x$, $y$, $z$ are odd and the
other one even. We shall assume that $x$ and $y$ are odd. This
assumption involves no loss of generality, since if one of these
numbers, say $x$, is even, then the other, in this case $y$, may be
% [File: 079.png Page: 69]
transposed to the second member so that we have $x^3 = z^3 + (-y)^3$,
an equation of the same form as (1), in which the two numbers
in the same member are both odd.

Let us now write
\[
  x + y = 2p, \quad  x - y = 2q.
\]
Then $x = p + q$, $y = p - q$, so that $p$ and $q$ are relatively prime
integers, one of them being odd and the other even. Substituting
in (1), we have
\[
  2p(p^2 + 3q^2) = z^3.        
\tag{6}
\]
From this equation it is clear that $p$ is even, since $p^2 + 3q^2$ is
odd; hence $q$ is odd. Since $p$ and $q$ are relatively prime it
follows that $p$ and $p^2 + 3q^2$ have the greatest common divisor
1 or 3. We treat these two cases separately.

In the first place suppose that $p$ and $p^2 + 3q^2$ have the
greatest common divisor 1. Then $p$ is not divisible by 3.
Then from Eq.~(6) we have
\[
  2p = r^3, \quad  p^2 + 3q^2 = s^3.        
\tag{7}
\]
We have seen above that the last equation can be satisfied
only when $p$ and $q$ have the values given in (5). Since $q$ is
odd it follows from the last equation in (5) that $u$ is odd and
$t$ is even. From the first equation in (5) we see that $t$ is not
divisible by 3. Moreover, $t$ and $u$ are relatively prime, since
the same is true of $p$ and $q$. From the two values of $p$ above
we have the relation
\[
  (2t)(t + 3u)(t - 3u) = r^3.
\]
It is clear that the three factors in the first member are prime
each to each. Hence each of them is a cube. Writing them
in order equal to the cubes $\mu^3$, $\rho^3$, $\sigma^3$, we have
\[
  \rho^3 + \sigma^3 = \mu^3.        
\tag{8}
\]
It is easy to verify that the numbers $\rho$, $\sigma$ are less in absolute
value than the numbers $x$, $y$ with which we set out. Moreover,
both of them are odd. Furthermore, $\mu$ is different from
zero.

Let us now consider the case in which $p$ and $p^2 + 3q^2$ have
% [File: 080.png Page: 70]
the greatest common divisor 3. Then write $p = 3\pi$. Then
from Eq.~(6) we have
\[
  6\pi = 9r^3, \quad  9\pi^2 + 3q^2 = 3s^3;
\]
or
\[
  2\pi = 3r^3, \quad  q^2 + 3\pi^2 = s^3.
\]
Here we see that $\pi$ is even. Hence $q$ is odd. In view of (5)
as the general solution of (3) we see that $q$ and $\pi$ have the
forms
\[
  q = t^3 - 9tu^2, \quad  \pi = 3u(t^2 - u^2).
\]
Since $\pi$ is even it follows from the last equation that $u$ is even
and $t$ is odd. From the two values of $\pi$ we have
\[
  (2u)(t + u)(t - u) = r^3.
\]
It is clear that the three factors in the first member of this
equation are prime each to each. From this we are led as
before to an equation of the form (8) with the same properties
as in the preceding case.

Therefore, starting with an equation $x^3 + y^3 = z^3$ in which
$x$, $y$, $z$ are all different from zero and are prime each to each
and $x$ and $y$ are odd, we are led to another equation
$x_1^3 + y_1^3 = z_1^3$ of the same sort, in which $x_1$, $y_1$ are less in
absolute value than $x$, $y$. Starting from the last equation,
we can proceed as before to an equation $x_2^3 + y_2^3 = z_2^3$ with
the same properties as in the preceding case, $x_2$, $y_2$ being less
in absolute value than $x_1$, $y_1$; and so on. But such a recursion
is evidently impossible. From this contradiction we conclude
that \emph{Eq.~$(1)$ cannot be satisfied by integers $x$, $y$, $z$, all of
which are different from zero.}

Let us now turn to the equation
\[
  x^3 + y^3 = 2^m z^3.        
\tag{9}
\]
In view of the result just attained, this is impossible in non-zero
integers $x$, $y$, $z$ if $m$ is a multiple of 3; hence we shall not
consider this case further. It is clear that $x$ and $y$ are both
odd or both even; if they are both even, it is clear that the
equation may be divided through by an appropriate power
of 2, so that in the resulting equation $x$ and $y$ shall be both
odd. This may necessitate a change in the value of $m$. After
% [File: 081.png Page: 71]
$x$ and $y$ are made odd, we suppose that $m$ is so chosen that
$z$ also is odd. We shall now show that \textit{the equation in this
reduced form is impossible in non-zero integers, except for the
trivial solution} $x = y = z$, \textit{which exists when} $m = 1$. It is obviously
sufficient to prove this for the case when $x$ and $y$ are relatively
prime.

Eq.\ (9) may be written in the form
\[
(x + y)(x^2 - xy + y^2) = 2^m z^3.
\]
The two factors of the first member have the greatest common
divisor 1 or 3 since $x$ and $y$ are to be taken relatively prime.
Hence the factor $x^2 - xy + y^2$ is of one of the forms $r^3, 3r^3$. For
the former case we may write
\[
x^2 - xy + y^2 = \left(\frac{x+y}{2}\right)^2 + 3\left(\frac{x-y}{2}\right)^2 = r^3, \quad x + y = 2^m s^3,        \tag{10}
\]
where $r$ and $s$ are odd integers. In the latter case we have
\[
x^2-xy+y^2 = \left(\frac{x+y}{2}\right)^2 + 3\left(\frac{x-y}{2}\right)^2 = 3r^3, \quad  x+y=2^m 3^2\cdot s^3.
\]
or
\[
\left(\frac{x-y}{2}\right)^2 + 3\left(\frac{x+y}{6}\right)^2 = r^3, \quad x+y=2^m \cdot 3^2 \cdot s^3. \tag{11}
\]

We shall merely outline the remainder of the proof. Consider
Eqs.~(10). From the first of these and the theory associated
with Eq.~(3) above, we have equations of the form
\[
r = t^2 + 3u^2, \quad \frac{x+y}{2} = t^3 - 9tu^2, \quad \frac{x-y}{2} = 3u(t^2 - u^2).
\]
Thence from the second equation in (10), we have
\[
t(t - 3u)(t + 3u) = 2^{m-1} s^3.
\]
Now $t^2 + 3u^2$ is odd, being equal to the odd integer $r$. Hence
$t^2 - 9u^2$ is odd. From the last equation it follows then that
$t$ has the factor $2^{m-1}$. Thence we have equations of the form
\[
t = 2^{m-1} \mu^3, \quad t - 3u = \alpha^3, \quad t + 3u = \beta^3,
\]
and hence the relation
\[
\alpha^3 + \beta^3 = 2^m \mu^3.
\]
% [File: 082.png Page: 72]
Similarly, by means of Eq.~(11), one is led again to an equation
of this last form. In each case it may be shown that the integers
involved in this deduced equation are smaller than
those in (9); and hence the conclusion announced above is
reached by the Fermatian method of infinite descent.

\Needspace*{4\baselineskip}
\begin{center}GENERAL EXERCISES\end{center}
\addtocontents{toc}{\protect\contentsline {section}{\numberline {}\textsc {General Exercises} 1-26}{\thepage}}

\begin{small}
1. Prove that the sum of the sixth powers of two integers cannot be the
square of an integer.

2. Prove that no integers $x$ and $y$ exist such that the difference of their sixth
powers is a square.

3. Prove that no relatively prime integers $x$ and $y$ exist such that the difference
of their fourth powers is a cube.

4. Show that the equation $x^2 + y^3 = z^6$ is impossible in non-zero integers.

5. By means of the identity
\[
(s^3 - t^3 + 6s^2 t + 3st^2)^3 + (t^3 - s^3 + 6t^2 s + 3ts^2)^3 = st\cdot(s + t)\cdot 3^3 (s^2 + st + t^2)^3
\]
show that the Diophantine equation $x^3 + y^3 = az^3$ has rational solutions (for which
$z\neq 0$) when and only when $a$ satisfies an equation of the form
\[
au^3 = st(s + t).
\]

6.* If $p$ and $q$ denote numbers of the form $18n + 5$ and $18n + 11$ respectively,
and if $a$ is a number of any one of the forms $p$, $p^2$, $q$, $q^2$, $9p$, $9p^2$, $9q$, $9q^2$, $2p$, $4p^2$,
$2q^2$, $4q$; then neither of the equations
\[
x^3 + y^3 = az^3, \quad xy(x + y) = az^3,
\]
has a non-trivial rational solution.

(For reference to several papers dealing with these equations see \textit{Encyclop\'edie
des sciences math\'ematiques}, Tome~I, Vol.~III, pp.~32--33.)

7. By means of a single solution of the equation $x^3 + y^3 = u^3 + v^3$ show directly
how to find a two-parameter solution.

8.* Obtain the general solution of the Diophantine system
\[
\phantom{(Werebrussov,\ 1909.)\quad}
x^3 + y^3 = u^3 + v^3 = s^3 + t^3.        \tag*{(\index{Werebr\"ussov}Werebr\"ussov, 1909.)\quad}
\]

9. Show that the equation $x^3 + y^3 + z^3 = 2u^3$ is satisfied by $x = u + v$, $y = u - v$,
$u = a^2 m^3$, $v = bn^3$, $z = -6mn^2$, $ab = 6$. By means of two solutions show how to find
a third. 
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Werebr\"ussov}Werebr\"ussov, 1908.)\quad}

10. By means of a single solution of the equation $x^3 + y^3 + u^3 + v^3 = t^3$ show
how to find a two-parameter solution. Generalize the result by increasing the
number of variables in the first member.

11. Consider the equation $1 + x + x^2 + x^3 = y^2$. Show that $x = 7$, $y = 20$ is the
only solution in which $x$ is a prime number. Show how to find other rational
solutions.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Gerono}Gerono, 1877.)\quad}

12. Show that no one of the following four equations has an integral solution:
\[
\phantom{(Delannoy,\ 1897.)\quad}
2x^3 \pm 1 = z^3,  \quad 2x^3 \pm 2 = z^3.        \tag*{(Delannoy, 1897.)\quad}\index{Delannoy}
\]

% [File: 083.png Page: 73]
13. Let $x = \alpha$, $y = \beta$, $z = \gamma$ be a solution of the equation $x^3 + y^3 = 9z^3$. Show
that another solution is obtained by means of the formulæ
\[
  x = \alpha(\alpha^3 + 2\beta^3), \quad
  y = -\beta(\beta^3 + 2\alpha^3), \quad
  z = \gamma(\alpha^3 - \beta^3).
\]
Obtain a similar result for the equation $x^3 + y^3 = 7z^3$. \hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad(\index{Realis}Realis, 1878.)\quad}

14. Show that the equation $x(x+1) = 2y^3$ has no integral solution except
$x = y = 0$, $x = y = 1$.

15. Show that the equation $8x^3 + 1 = y^2$ has no integral solution except $x = 0$,
$y = 1$; $x = 1$, $y = 3$.

16. By means of a single solution of the equation $x^3 + ay^3 = bz^3$ show how
to find a second; by means of this find a third; and so on. 
\hfil\penalty0\hfilneg\null\penalty5000\hfill \mbox{\qquad(\index{Legendre}Legendre.)\quad}

17. Find a two-parameter integral solution of the equation $x^3 + y^3 = z^2$.

18. Find a three-parameter integral solution of the equation $x^2 + 3y^2 = u^3 + v^3$.

\textsc{Suggestion}.---Observe that
\[
u^3 + v^3 = (u + v)\left\{ \left(\frac{u+v}{2}\right)^2 + 3\left(\frac{u-v}{2}\right)^2 \right\}.
\]

19. Obtain solutions of the equation $x^3 + y^3 = u^2 + v^2$.

20. Find three-parameter solutions of the equation $x^3 + y^3 + z^3 = u^3 + v^3$.

\textsc{Suggestion}.---Employ the identity
\[
(-a+b+c)^3 + (a-b+c)^3 + (a+b-c)^3 = (a+b+c)^3 - 24abc.
\]

21. Obtain a three-parameter solution of the equation
\[
x^3 + y^3 + z^3 - 3xyz = u^2 + 3v^2.
\]

22.\dag\; Find the general integral solution of the equation $t^3 = x^3 + y^3 + 1$.

23.\dag\; Construct another cubic equation for which large classes of solutions
may be found by the first method employed in \S~15. Develop the theory of
this equation.

24.\dag\; Investigate the properties of the integer $m$ such that the equation
\[
x^3 + y^3 + z^3 - 3xyz = mt^2
\]
shall have solutions and find solutions involving arbitrary parameters (when
$m$ is suitably determined). Treat the corresponding problems for the simpler
equation $x^3 + y^3 = mt^2$.

25.\dag\; Generalize the investigations called for in the preceding problem by
treating the more general equation
\[
h(x, y, z) = mt^2,
\]
the function $h$ being defined as in Eq.~(4) of \S~13.

26. From $3^3 + 4^3 + 5^3 = 6^3$ deduce $7^3 + 14^3 + 17^3 = 20^3$.
\index{Equations of Third Degree|)}\index{Cubic Equations|)}\par\end{small}

% [File: 084.png Page: 74]
%CHAPTER IV

\chapter[EQUATIONS OF THE FOURTH DEGREE]{EQUATIONS OF THE FOURTH DEGREE\protect\footnote
 {Other matter relating to equations of the fourth degree is to be found in
 \S~11 and in the exercises at the close of Chapter~II.}}
\markright{EQUATIONS OF THE FOURTH DEGREE}

\index{Biquadratic Equations|(}\index{Equations of Fourth Degree|(}
\setcounter{section}{16}
%\§ 17.
\section[\textsc{On the Equation $ax^4 + bx^3y + cx^2y^2 + dxy^3 + ey^4 = mz^2$. Exercises 1-4}]{On the Equation $ax^4 + bx^3y + cx^2y^2 + dxy^3 + ey^4 = mz^2$}

\hspace{\parindent}For the equation
\[
ax^4 + bx^3y + cx^2y^2 + dxy^3 + ey^4 = mz^2,     \tag{1}
\]
it is obvious that the problem of finding rational solutions and
that of finding integral solutions are essentially equivalent.
It will therefore be sufficient to consider only one of these
problems; it is convenient to treat the former rather than the
latter. If the equation is multiplied through by $m$ and $mz$ is
then replaced by $z$ we have a new equation of the form which
(1) takes on replacing $m$ by unity. We shall therefore consider
only the case when $m = 1$. Now it is clear that the problem
of finding rational solutions of (1) is equivalent to that of
finding rational solutions of the equation
\[
ax^4 + bx^3 + cx^2 + dx + e = z^2;       \tag{2}
\]
for if (1) is divided through by $y^4$ and in the result $x$ is put for
$x/y$ and $z$ for $z/y^2$, $m$ having been replaced by 1, we have
an equation of the form~(2). We shall confine our attention
to the reduced equation~(2).

The customary method of finding rational solutions of Eq.~%
(2) is due to \index{Fermat}Fermat. It takes different forms for different
cases. We shall examine these cases separately.

Suppose that $e$ is a square number, say $e = \epsilon^2$. Then write
\[
z = mx^2 + nx + \epsilon,
\]
where $m$ and $n$ are rational numbers subject to our choice.
% [File: 085.png Page: 75]
They are to be determined so that the value of $z^2$, namely,
\[
z^2=m^2x^4+2mnx^3+(n^2+2m\epsilon)x^2+2n\epsilon x+\epsilon^2,
\]
shall coincide with the first member of Eq.~(2) except for the
terms in $x^4$ and $x^3$. For this it is necessary and sufficient that
\[
n = \frac{d}{2\epsilon}, \quad m = \frac{c}{2\epsilon} - \frac{d^2}{8\epsilon^3}.
\]
Then we have
\[
ax^4+bx^3=m^2x^4+2mnx^3,
\]
an equation which is satisfied if
\[
  x = \frac{b - 2mn}{m^2 - a}.
\]
Thus we have in general a single rational solution of Eq.~(2).

As an illustration, it may thus be shown that the equation
\[
x^4+3x^3+5x^2+2x+1 = z^2,
\]
has the solution \hfill$x=-\frac{1}{3}$,\quad$z=\frac{8}{9}$.\hfill \phantom{has the solution}

In case $a$ is a square number, say $a=\alpha^2$, we may write
\[
z=\alpha x^2+mx+n,
\]
and proceed, in a manner similar to that employed in the preceding
paragraph. Here we choose $m$ and $n$ so as to make
the expression for $z^2$ coincide with the first member of (2)
except for the independent term and the term containing $x$,
and then determine $x$ as before. We have
\[
  m = \frac{b}{2\alpha}, \quad n = \frac{c}{2\alpha} - \frac{b^2}{8\alpha^3}, \quad x = \frac{n^2 - e}{d - 2mn}.
\]
Here again we have in general a single rational solution of
Eq.~(2).

If one applies this method to the particular equation
\[
x^4+3x^3+5x^2+2x+1=z^2,
\]
one finds that a solution is afforded by $x=-57/136$.

If both $a$ and $e$ are squares, say $a=\alpha^2$ and $e=\epsilon^2$, we may
write,
\[
z=\alpha x^2+mx+\epsilon,
\]
and then determine $m$ so that the expression for $z^2$ shall coincide
with the first member of Eq.~(2) except for the terms
% [File: 086.png Page: 76]
containing $x^3$ and $x^2$ or the terms containing $x^2$ and $x$. These
lead in order to the following pairs of values of $m$ and $x$:
\begin{alignat*}{2}
        m &= \frac{d}{2\epsilon},  &\quad
        x &= \frac{c- m^2-2\alpha\epsilon}{2\alpha m-b};
\\
        m &= \frac{b}{2\alpha},    &\quad   
        x &= \frac{d-2m\epsilon}{m^2+2\alpha\epsilon-c}.
\end{alignat*}
Here we have in general two solutions of Eq.~(2).

By means of the equation
\[
        x^4 + 3x^3 + 5x^2 + 2x + 1 = z^2
\]
the reader may readily supply a numerical illustration of this
method.

Gathering together the results thus obtained we may say
that we have a method which in general is effective for finding
a single rational solution of (2) when $e$ is a square or when
$a$ is a square and for finding two rational solutions when both
$a$ and $e$ are squares.

We shall now show that Eq.~(2) may be reduced to one of
the special forms already considered provided that a single
rational solution is known.

Let $x=k$, $z=p$ be a rational solution of Eq.~(2). Then
replace $x$ by $t+k$. It is clear that a new equation is obtained
of the same form as (2) and that $p^2$ is the constant corresponding
to $e$. Since this constant is a square, the first method employed
above may be used to find a solution of the new equation.
Adding $k$ to the resulting value of $t$ we have a value of $x$ which
affords a rational solution of Eq.~(2). This in general is different
from the solution with which we started. By means
of this second solution a third can in general be obtained;
by means of this a fourth, and so on. Hence, if a single rational
solution of (2) is known it is possible in general to find as many
as may be desired.

As an illustrative example, consider the equation
\[
        2x^4 + 5x^3 + 7x^2 + 2 = z^2.
\]
It has the solution
\[
        x = 1, \quad  z = 4.
\]
% [File: 087.png Page: 77]
If we replace $x$ by $t+1$ the new equation takes the form
\[
        2t^4 + 13t^3 + 34t^2 + 37t + 16 = z^2,
\]
the constant term 16 being a square number. A solution of
the last equation may be found by the methods indicated
above. By means of this a second solution of the equation in
$x$ is readily obtained. With this in hand, the operation may
be repeated and thus a third solution of the equation in $x$
be obtained; and so on \textit{ad infinitum}.

\Needspace*{4\baselineskip}
\begin{center}EXERCISES\end{center}

\begin{small}
1. Determine integral \index{Triangles, Pythagorean}Pythagorean triangles such that the hypotenuse and
the sum of the legs shall be squares. (Compare Ex.~4 of \S~3.) 
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Fermat}Fermat.)\quad}

2. Determine integral \index{Triangles, Pythagorean}Pythagorean triangles such that the hypotenuse and
the difference of the legs shall be squares.

3. Find two or more pairs of integral \index{Triangles, Pythagorean}Pythagorean triangles such that in
each pair the difference of the legs of one shall be equal to the difference of the
legs of the other while the longer leg of one is equal to the hypotenuse of the
other.

4. Find two or more pairs of integral \index{Triangles, Pythagorean}Pythagorean triangles such that in
each pair the sum of the legs of one shall be equal to the sum of the legs of the
other, while the hypotenuse of one shall be equal to the greater leg of the other.
\par\end{small}

%§ 18.
\Needspace*{4\baselineskip}
\section[\textsc{On the Equation $ax^4 + by^4 = cz^2$. Exercises 1-4}]{On the Equation $ax^4 + by^4 = cz^2$}

\hspace{\parindent}The equation
\begin{equation}
        ax^4 + by^4 = cz^2,     \tag{1}
\end{equation}
which is a special case of that considered in the preceding
section, has been treated by several authors and detailed investigations
have been given of special cases, that is, of certain
special equations obtained by giving to $a$, $b$, $c$ particular values.
No attempt will be made to summarize these investigations.
On the other hand, an exposition will be given of a new method
of attack which leads to solutions in an important class of
cases. Convenient references to a representative part of the
literature concerning Eq.~(1) may be found in \textit{Encyclop\'edie
des sciences math\'ematiques}, Tome~I, Vol.~III, pp.~35--36, and in
\textit{Jahrbuch \"uber die Fortschritte der Mathematik}, at places indicated
by the following volume and page numbers: 10:~146,
148; 11:~136, 137; 12:~131, 136; 14:~133; 16:~154; 19:~187;
21:~181; 25:~295; 26:~211, 212.

% [File: 088.png Page: 78]
If Eq.~(1) is multiplied through by $a^3$ and $ax$ is then replaced
by $x$, an equation is obtained which is of the form
\[
  x^4 + my^4 = nz^2.  
\tag{2}
\]
It is in this latter form that we shall treat the equation. We
shall suppose $m$ to be any given integer and shall then restrict
$n$ to be an integer of the form $s^4 + mt^4$, where $s$ and $t$ are integers.
Then Eq.~(2) takes the form
\[
  x^4 + my^4 = (s^4 + mt^4)z^2.
\tag{3}
\]
Here $m$, $s$, $t$ are given integers and $x$, $y$, $z$ are unknown integers
suitable values of which are to be sought.

It should be observed that there is no real loss of generality
in confining $n$ to this form; for if Eq.~(2) has a solution, then
there is a square number $\rho^2$ such that $n\rho^2$ has the form $s^4 + mt^4$.

Since $x = s$, $y = t$, $z = 1$ is a solution of (3) we might proceed
to find another as in the work of the preceding section. We
shall, however, use a different method. Let us write $z$ in the
form
\[
  z = p^2 + mq^2, 
\tag{4}
\]
and seek to determine $p$ and $q$ and corresponding to them
values of $x$ and $y$ satisfying Eq.~(3). We have
\begin{align*}
  x^4 + my^4 &= (s^4 + mt^4)    (p^2 + mq^2)^2  \\
             &= (s^4 + mt^4) \{ (p^2 - mq^2)^2 + m(2pq)^2 \}  \\
             &=  \{ s^2(p^2 - mq^2) + 2mt^2 pq \}^2
              + m\{ t^2(p^2 - mq^2) - 2 s^2 pq \}^2.
\end{align*}
% [File: 089.png Page: 79]
This equation will be satisfied if $x^2$ and $y^2$ have the values
\[
\left. \begin{aligned}
  x^2 &= s^2(p^2 - mq^2) + 2mt^2 pq,  \\
  y^2 &= t^2(p^2 - mq^2) - 2 s^2 pq. 
\end{aligned} \right\}
\tag{5}
\]

The last two equations form a system of the type studied
by \index{Fermat}Fermat under the name \index{Equations , Double@Equations, Double}\index{Double Equations}\emph{double equations}, $x$, $y$, $p$, and $q$
being the unknown integers. We shall obtain solutions of
them by means similar to those employed by Fermat.

Multiplying the first equation in (5) by $t^2$ and the second
by $s^2$ and subtracting, we have
\[
  t^2 x^2 - s^2 y^2 = 2(s^4 + mt^4)pq.
\]
This equation will be satisfied if we put
\[
  tx + sy = 2stp, \quad  tx - sy = \frac{(s^4 + mt^4)q}{st}.
\]
(Here the coefficient of $p$ is chosen so as to be equal to twice
the square root of the coefficient of $p^2$ in the value of $t^2 x^2$
obtained by multiplying the first equation in (5) through by
$t^2$.) For $x$ and $y$ we have the values
\begin{align*}
  x &= sp + \frac{(s^4 + mt^4)q }{2st^2 }, \\
  y &= tp - \frac{(s^4 + mt^4)q }{2s^2t }.
\end{align*}
If we use these values of $x$ and $y$ and determine $p$ and $q$ so
that the first equation in (5) shall be satisfied, it is clear that
the second equation is also satisfied.

Substituting the foregoing value of $x$ in the first equation
in (5) and reducing, we have
\[
  4 s^2 t^2 (s^4 + mt^4)pq + (s^4 + mt^4)^2 q^2 
= 8ms^2 t^6 pq - 4ms^4 t^4 q^2.
\]
This equation will be satisfied if
\begin{align*}
  p &=           (s^4 + mt^4)^2 + 4ms^4 t^4,  \\
  q &= -4s^2 t^2 (s^4 - mt^4).
\end{align*}
Making use of these values of $p$ and $q$, we have for $x$, $y$, $z$ the
values:
\begin{align*}
  x &= s\{ (s^4 + mt^4)^2 + 4ms^4 t^4 - 2(s^8 - m^2 t^8) \},  \\
  y &= t\{ (s^4 + mt^4)^2 + 4ms^4 t^4 + 2(s^8 - m^2 t^8) \},  \\
  z &=  \{ (s^4 + mt^4)^2 + 4ms^4 t^4\}^2 + 16ms^4 t^4 (s^4 - mt^4)^2.
\end{align*}

These values afford a single integral solution of Eq.~(3).
Employing this solution and utilizing the methods of the preceding
section, one may obtain a second solution; from this
in turn a third may be found; and so on. The reader may
readily supply numerical illustrations.

% [File: 090.png Page: 80]
\Needspace*{4\baselineskip}
\begin{center}EXERCISES\end{center}

\begin{small}
1. Obtain solutions of the equation $7x^4 - 5y^4 = 2z^2$.
\hfill (\index{Pepin}Pepin, 1879.)\quad\null

2. Obtain solutions of each of the following equations:
\[
  x^4 - 140y^4 = z^2, \quad  4x^4 - 35y^4 = z^2. 
\tag*{(\index{Pepin}Pepin, 1879.)\quad}
\]

3. Obtain solutions of each of the following equations:
\[
  3x^4 - 2y^4 =  z^2,  \quad
   x^4 + 7y^4 = 8z^2,  \quad
  7x^4 - 2y^4 = 5z^2.
\tag*{(\index{Pepin}Pepin, 1879.)\quad}
\]

4. Show how to find solutions of the equation $ax^4 + by^4 = cz^2$ in all cases in
which $c = a + b$.
\par\end{small}

%§ 19. 
\Needspace*{4\baselineskip}
\section[\textsc{Other Equations of the Fourth Degree}]{Other Equations of the Fourth Degree}

\hspace{\parindent}We have seen (§~6) that the sum of two biquadrate numbers
cannot be a biquadrate number; in fact, such a sum cannot
be a square number. Furthermore, the sum of two biquadrates
cannot be the double of a square number (Ex.~7 of §~6).

That the sum of three biquadrates can be a square is readily
shown. Let $x$, $y$, $z$ be integers such that
\[
  x^2 + y^2 = z^2. 
\tag{1}
\]
Let us square each member of this equation, multiply through
by $z^4$, and then add $x^4 y^4 - 2z^4 x^2 y^2$ to each member of the resulting
equation; the relation thus obtained may be written
in the form
\[
  (xy)^4 + (yz)^4 + (zx)^4 = (z^4 - x^2 y^2)^2.        
\tag{2}
\]
Now Eq.\ (1) has the general primitive solution
\[
  x = m^2 - n^2, \quad y = 2mn, \quad z = m^2 + n^2.
\]
Substituting these values in (2) we have the identity
\begin{multline*}
  \{ 2mn(m^2 - n^2) \}^4 +\{ 2mn(m^2 + n^2) \}^4 + (m^4 - n^4)^4  \\
= (m^8 + 14m^4 n^4 + n^8)^2.
\end{multline*}
Thus we have a two-parameter solution of the problem of
finding three biquadrate numbers whose sum is a square.
If we put $m = 2$, $n = 1$, we have the particular illustrative relation
$12^4 + 20^4 + 15^4 = 481^2$.

Whether or not the sum of three biquadrate numbers can
be a biquadrate appears never to have been determined, though
\index{Euler}Euler seems to have been of the opinion that it is impossible.

% [File: 091.png Page: 81]
It is easy to find three biquadrate numbers whose sum is
equal to the double of a square. In fact, these are afforded
by the identity
\[
  x^4 + y^4 +(x + y)^4 = 2(x^2 + xy + y^2)^2. 
\tag{3}
\]
Now the relation
\[
  x^2 + xy + y^2 = (a^2 + ab + b^2)^2
\]
is satisfied if
\[
  x = a^2 - b^2, \quad y = 2ab + b^2.
\]
If we substitute these values of $x$ and $y$ in (3), we have the
identity
\[
  (a^2 - b^2)^4 + (2ab + b^2)^4 + (a^2 + 2ab)^4 = 2(a^2 + ab + b^2)^4.
\]
This affords a two-parameter solution of the problem of finding
three biquadrate numbers whose sum is equal to the double
of a biquadrate number. As a particular case we have
\[
  3^4 + 5^4 + 8^4 = 2\cdot 7^4.
\]

In a similar way one may obtain an identity affording a
two-parameter solution of the problem of finding three biquadrate
numbers whose sum is the double of a number of any
even power. For this purpose it is sufficient to determine
$x$ and $y$ so that
\[
  x^2 + xy + y^2 = (a^2 + ab + b^2)^k,
\]
$2k$ being the index of the even power under consideration, and
substitute the resulting values of $x$ and $y$ in (3). This determination
of $x$ and $y$ may readily be made by the method set
forth in §~7.

Whether the sum of four biquadrate numbers can be a
biquadrate number appears not yet to have been determined.
That the sum of five biquadrate numbers can be a biquadrate
number is readily shown. The smallest integers satisfying this
condition appear to be those involved in the relation
\[
  4^4 + 6^4 + 8^4 + 9^4 + 14^4 = 15^4.
\]
Several identities affording a solution of this problem have
been obtained by tentative methods. (See a paper by \index{Martin}Martin
% [File: 092.png Page: 82]
in the Proceedings of the International Congress of Mathematicians,
1912.) The following are two of them:
\[
\left. \begin{aligned}
(8s^2 + 40st - 24t^2)^4 &+ (6s^2 - 44st - 18t^2)^4\\
                        &+ (14s^2 - 4st - 42t^2)^4 + (9s^2 + 27t^2)^4\\
                        &+ (4s^2 + 12t^2)^4 = (15s^2 + 45t^2)^4; \\[1ex]
       (4m^2 - 12n^2)^4 &+ (2m^2 - 12mn - 6n^2)^4 + (4m^2 + 12n^2)^4\\
                        &+ (2m^2 + 12mn - 6n^2)^4 + (3m^2 + 9n^2)^4\\
                        &= (5m^2 + 15n^2)^4
\end{aligned} \right\} \tag{4}
\]

Another elegant problem concerning biquadrate numbers
is the following: To find two biquadrate numbers whose sum
is equal to the sum of two other biquadrate numbers; in other
words, to find solutions of the equation
\[
x^4 + y^4 = u^4 + v^4.        \tag{5}
\]
This problem we shall now consider. If we set
\[
x = a + b, \quad y = c - d, \quad u = a - b, \quad v = c + d,   \tag{6}
\]
and substitute in Eq.~(5), we have
\[
ab(a^2 + b^2) = cd(c^2 + d^2).
\]
\index{Euler}Euler has observed that this equation is identically satisfied if
\[
\begin{aligned}
a &= g(f^2 + g^2)(-f^4 + 18f^2g^2 - g^4),\\
b &= 2f(f^6 + 10f^4g^2 + f^2g^4 + 4g^6),\\
c &= 2g(4f^6 + f^4g^2 + 10f^2g^4 + g^6);\\
d &= f(f^2 + g^2)(-f^4 + 18f^2g^2 - g^4).
\end{aligned}
\]
With these values of $a$, $b$, $c$, $d$, Eqs.~(6) afford a two-parameter
solution of (5).

Formulas for resolving the equation
\[
x^4 + y^4 + z^4 = u^4 + v^4 + w^4
\]
have been given by several writers. See \textit{Interm\'ediaire des
Math\'ematiciens}, 19:~254; 20:~105. A two-parameter solution
of the equation
\[
w^4 + x^4 + y^4 + z^4 = s^4 + t^4 + u^4 + v^4
\]
% [File: 093.png Page: 83]
may be found from Eqs.~(4) above by putting $n = s$ and $m = 3t$
and equating the first members, this being legitimate since
the second members are identical.

\Needspace*{4\baselineskip}
\begin{center}GENERAL EXERCISES\end{center}
\addtocontents{toc}{\protect\contentsline {section}{\numberline {}\textsc {General Exercises} 1-20}{\thepage}}

\begin{small}
1. Find a two-parameter solution of each of the following equations:
\begin{alignat*}{2}
  x^4 +{}&&  y^4 + 4z^4 &= t^4,\\
  x^4 +{}&& 2y^4 + 2z^4 &= t^4,\\
  x^4 +{}&& 8y^4 + 8z^4 &= t^4,\\
  x^4 +{}&&  y^4 + 2z^4 &= 2t^4,\\
\phantom{(Carmichael,\ 1913.)\quad}\index{Carmichael}
  x^4 +{}&&  y^4 + 8z^4 &= 8t^4.  \tag*{(Carmichael, 1913.)\quad}
\end{alignat*}

2. Obtain solutions of the equation
\[
x^4 + y^4 + z^4 = u^4 + v^4.
\]

3. Find six biquadrate numbers whose sum is a biquadrate.

4. Find $n$ biquadrate numbers whose sum is a biquadrate in case $n = 7$, $8$,
$9$, $10$.

5.* Show that $x = 3$, $y = 1$, $z = 2$ are the only relatively prime integers which
satisfy the equation
\[
\phantom{(Fauquembergue,\ 1912.)\quad}
x^4 - y^4 = 5z^4.
\tag*{(\index{Fauquembergue}Fauquembergue, 1912.)\quad}
\]

6. The equation $x(x + 1) = 2y^4$ has no integral solution other than $x = y = 0$,
$x = y = 1$.

7. Starting from the equation $x^2 + ay^2 = z^2$ generalize the first result of §~19.

8.* Show how to find all the solutions of the equation $x^4 + 35y^4 = z^2$ which
lie under a given limit.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Pepin}Pepin, 1895.)\quad}

9.* Show that the equation $px^4 - 41y^4 = z^2$ has no rational solution when
the prime $p$ has any one of the values 5, 37, 73, 113, 337, 349, 353, \ldots, represented
by the form $5m^2 + 4mn + 9n^2$. (This and several similar results are
given by \index{Pepin}Pepin in the Comptes Rendus of the Paris Academy, Vol.~LXXXVIII,
pp.~1255--1257 and Vol.~XCIV, pp.~122--124.)

10.* Obtain the general solution of the equation
\[
\phantom{(Pepin,\ 1898.)\quad}
x^4 - 8x^2y^2 + 8y^4 = z^2.
\tag*{(\index{Pepin}Pepin, 1898.)\quad}
\]

11.* Obtain formulæ affording solutions of the equation
\[
\phantom{(Aubry,\ 1911.)\quad}
ax^4 + bx^2y^2 + cy^4 = dz^2.
\tag*{(\index{Aubry}Aubry, 1911.)\quad}
\]

12.* Solve the equation
\[
x^4 + 4hx^2y^2 + (2h - 1)^2y^4 = z^2,
\]
where $h$ is such that $4h - 1$ and $2h - 1$ are primes.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Pietrocola}Pietrocola, 1898.)\quad}

13. Obtain solutions of the equation
\[
\phantom{(Moret-Blanc,\ 1881.)\quad}
x^4 + x^3y + x^2y^2 + xy^3 + y^4 = z^2.
\tag*{(\index{Moret-Blanc}Moret-Blanc, 1881.)\quad}
\]

14. Obtain solutions of the equation
\[
\phantom{(Moret-Blanc,\ 1881.)\quad}
x^4 - 5x^2y^2 + 5y^4 = z^2.
\tag*{(Moret-Blanc, 1881.)\quad}
\]

% [File: 094.png Page: 84]
15.* Obtain the general solution of the equation $x^4 - 4x^2y^2 + y^4 = z^4$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Paraira}Paraira, 1913.)\quad}

16. Obtain solutions of the equation
\[
\phantom{(Aubry,\ 1913.)\quad}
(x^2 - y^2 - 2xy)^2 - 8x^2y^2 = z^2.
\tag*{(\index{Aubry}Aubry, 1913.)\quad}
\]

17.\dag\ Determine whether the sum of three biquadrate numbers can be equal
to a biquadrate number.

18.\dag\ Determine whether the sum of four biquadrate numbers can be equal
to a biquadrate number.

19.\dag\ Discuss the values of $a$ for which the equation
\[
x^4 + y^4 + a^2z^4 = u^4
\]
has non-zero integral solutions.

20.\dag\ Discuss the values of $a$ (if any exist) for which the equation
\[
x^4 + a^2y^4 + a^2z^4 = u^4
\]
has non-zero integral solutions.\index{Biquadratic Equations|)}\index{Equations of Fourth Degree|)}
\par\end{small}

% [File: 095.png Page: 85]
%CHAPTER V

\chapter{EQUATIONS OF DEGREE HIGHER THAN THE FOURTH\@.
THE FERMAT PROBLEM}
\markright{EQUATIONS OF HIGHER DEGREE}

\index{Equations of Higher Degree|(}\setcounter{section}{19}
%§ 20.
\section[\textsc{Remarks Concerning Equations Of Higher Degree}]{Remarks Concerning Equations Of Higher Degree}

\hspace{\parindent}When we pass to equations of degree higher than the
fourth we find that but little effective progress has been made.
Often it is a matter of great difficulty to determine whether
a given solution is the general solution of a given equation.
Indeed this is true, to a large extent, of equations of the third
and fourth degrees. Even here there are but few equations
for which a general solution is known or for which it is known
that no solution at all exists. As the degree of the equation
increases, the generality of the known results decreases in a
rapid ratio. Only the most special equations of degree higher
than the fourth have been at all treated and for only a few of
these is one able to answer the questions naturally propounded as
to the existence or generality of solutions. (See Ex.~71, p.~116.)

Several papers have been written on the problem of determining
a sum of different $n$th powers equal to an $n$th power;
but the methods employed are largely tentative in character
and the results are far from complete. Reference may be
made to \index{Barbette}Barbette's monograph of 154 quarto pages, entitled,
``\textit{Les sommes de $p^{i\grave{e}mes}$ puissances distinctes \'egales \`a une $p^{i\grave{e}me}$
puissance},'' and to a paper by A.~\index{Martin}Martin in the Proceedings
of the International Congress of Mathematicians, 1912. See
also the references in the latter paper.

It is an easy matter to construct equations of any degree
desired in such a way that formulæ are readily obtained yielding
rational solutions involving one or more parameters; but this
is a trivial exercise. What is more profitable is a satisfactory
treatment of those equations which first come to mind when
% [File: 096.png Page: 86]
one considers the question as to what are the simplest equations
of various given degrees. Probably the most elegant equation
of degree $n$ is the following:
\[
  x^n + y^n = z^n. 
\tag{1}
\]
Concerning equations of higher degree the most desirable thing
to do first is to ascertain methods by which one may treat
completely the special cases, such, for example, as Eq.~(1)
above.

For a long time Eq.~(1) has held an interesting place in
the history and literature of the theory of numbers. This
chapter is devoted principally to a study of that equation.
It was first introduced to notice by \index{Fermat}Fermat in the seventeenth
century. Fermat stated, without proof, the following theorem,
commonly known as \index{Fermat's Last Theorem}Fermat's Last Theorem:

\textit{If $n$ is an integer greater than $2$ there do not exist integers
$x$, $y$, $z$, all different from zero, such that $x^n + y^n = z^n$.}

This theorem was written down by Fermat on the margin
of his copy of Diophantus. He added that he had discovered
a truly remarkable proof of the theorem, but that the margin
of the book was too narrow to contain it. No one has rediscovered
Fermat's proof, if indeed he had one (and there seems
to be no sufficient reason for doubting his statement). In
fact, no general proof of the theorem has yet been found. For
various special values of $n$ proofs have been given; in particular
for every value of $n$ not greater than 100.

In the next section we shall develop the more elementary
properties of Eq.~(1); and in the following section we shall
give a brief general account of the present state of knowledge
concerning this equation.


%§ 21. 
\Needspace*{4\baselineskip}
\section[\textsc{Elementary Properties of the Equation 
$x^n + y^n = z^n$, $n > 2$}]{Elementary Properties of the Equation \\
$x^n + y^n = z^n$, $n > 2$}
\index{Fermat Problem|(}

\hspace{\parindent}In the study of the equation
\[
  x^n + y^n = z^n,\quad n > 2, 
\tag{1}
\]
it is convenient to make some preliminary reductions. If
there exists any particular solution of (1), there exists also
% [File: 097.png Page: 87]
a solution in which $x$, $y$, $z$ are prime each to each. This may
be readily proved as follows: if any two of the numbers $x$, $y$,
$z$ have the greatest common factor $d$, then from (1) itself it
follows that the third number of the set has also this same
factor. Hence the equation may be divided through by $d^n$.
The resulting equation is of the same form as (1). It is clear
that $x$, $y$, $z$ in this resulting equation are prime each to each.
Hence, in proving the impossibility of (1), it is sufficient to
treat only the case in which $x$, $y$, $z$ are prime each to each.

Again, since $n$ is greater than 2, it must contain the factor
4 or an odd prime factor $p$. If $n$ contains the factor 4, we
may write $n = 4m$, whence we have
\[
  (x^m)^4 + (y^m)^4 = (z^m)^4.
\]
From the corollary to theorem~IV in §~5 it follows that this
equation is impossible. Hence, if Eq.~(1) is satisfied, $n$ does
not contain the factor 4. Now, if $n$ contains the odd prime
factor $p$, we may write $n = pm$, whence we have
\[
  (x^m)^p + (y^m)^p = (z^m)^p.
\]
Therefore, in order to prove the impossibility of Eq.~(1) it is
sufficient to show that it is impossible when $n$ is equal to an
odd prime number $p$; that is, it is sufficient to prove the impossibility
of the equation $x^p + y^p = z^p$, where $p$ is an odd prime.
By changing $z$ to $-z$ this may be written in the more symmetric
form
\[
  x^p + y^p + z^p = 0. 
\tag{2}
\]
We shall take $x$, $y$, $z$ to be prime each to each.

Special proofs of the impossibility of Eq.~(2) for particular
values of $p$ are known. One of these for the case $p = 3$ has
been reproduced in §~16 above. The remainder of this section
is devoted to the derivation, by elementary means, of certain
properties of $x$, $y$, $z$, $p$, which are necessary if Eq.~(2) is to be
satisfied.

\index{Abelian formulæ|(}We shall first derive the so-called Abelian formulæ. Let
us write Eq.~(1) in the form
\[
  (x + y)
  (x^{p-1} - x^{p-2} y + x^{p-3} y^2 + \ldots - xy^{p-2} + y^p-1) 
= (-z)^p. 
\tag{3}
\]
% [File: 098.png Page: 88]
The second factor of the first member of this equation may be
written in the form
\begin{align*}
   \frac{x^p + y^p}{x+y} 
&= \frac{\{(x+y) - y \}^p + y^p }{x+y}
\\
&= \frac{(x+y)^p - p(x+y)^{p-1}y + \ldots + p(x+y)y^{p-1} }{x+y}  
\\
&= (x+y)Q(x, y) + py^{p-1}, 
\tag{4}
\end{align*}
where $Q(x, y)$ is a polynomial in $x$ and $y$ with integral coefficients.
From this and the fact that $x$ and $y$ are relatively
prime, it follows readily that the numbers represented by the
two factors in the first member of (3) have the greatest common
divisor 1 or $p$.

If $z$ is prime to $p$, this greatest common divisor must be 1.
In this case the two factors of the first member of (3) are
relatively prime. Then, since their product is a $p$th power,
it is clear that each of them is a $p$th power. Hence we may
write
\[
  x + y = u^p, \quad \frac{x^p + y^p}{x+y} = v^p, \quad z = -uv. 
\tag{5}
\]

Let us next consider the case in which $z$ is divisible by $p$.
Then $x^p +y^p \equiv 0 \bmod p$; thence, from the theorem of \index{Fermat}FermatFermat
(see \index{Carmichael}Carmichael's \textit{Theory of Numbers}, p.~48), it follows that
$x + y \equiv 0 \bmod p$. That is, $x + y$ has the factor $p$. Then, by
means of Eq.~(4), we see that the second factor in the first member
of (3) also has the factor $p$. But the greatest common
divisor of the two factors in the first member of (3) is 1 or
$p$; therefore, in this case, it is $p$. Hence one of these two
factors contains $p$ to only the first power, while the other contains
it to the $(kp - 1)$th power, where $k$ is a suitably determined
positive integer. We shall show that it is the factor $x + y$
which contains $p$ to the higher power. Suppose that $x + y$
contains the factor $p$ to the $\nu$th power, and let us write
\[
  x + y = p^\nu t,
\]
where $t$ is prime to $p$. From this we have
\[
  x^p = (-y + p^\nu t)^p = -y^p + p^{\nu+1}ty^{p-1} 
      - \frac{p-1}{2} p^{2\nu+1} t^2 y^{p-2} + \ldots;
\]
% [File: 099.png Page: 89]
whence
\[
x^p + y^p = p^{\nu+1} ty^{p-1} + p^{\nu + 2} I,
\]
where $I$ is an integer. Now, in the case under consideration,
$y$ is prime to $p$, since by hypothesis, $y$ is prime to $z$ and $z$ is
divisible by $p$. Hence, $x^p + y^p$ is divisible by $p^{\nu + 1}$, but by
no higher power of $p$. Therefore $(x^p + y^p)/( x + y)$ is divisible
by $p$ but not by $p^2$. Thence it follows that the first and second
factors in the first member of (3) contain $p^{kp - 1}$ and $p$ respectively. 
Therefore, we have relations of the form
\[
  x + y = p^{kp - 1} u^p, \quad \frac{x^p + y^p}{x + y} = pv^p, \quad
  z = -p^k uv,  \tag{6}
\]
where $k$ is a suitably chosen positive integer.

We can now readily prove the following theorems:

I\@. \textit{If an equation of the form
\[
x^p + y^p + z^p = 0, \tag{$2^\textrm{bis}$}
\]
in which $p$ is an odd prime, is satisfied by integers $x$, $y$, $z$, which
are prime each to each and to $p$ and are all different from zero,
then integers $u_1$, $u_2$, $u_3$, $v_1$, $v_2$, $v_3$, prime to $p$, exist such that}
\[
\left. \begin{alignedat}{3}
x + y =& {u_1}^p,& \quad \frac{x^p + y^p}{x + y} =& v_1^p, \quad &z=& -u_1v_1,\\
y + z =& {u_2}^p,&&  %[**note: no equation here]
                                                             &x=& -u_2v_2,\\
z + x =& {u_3}^p,&&  %[**note: no equation here]
                                                             &y=& -u_3v_3;
\end{alignedat}\ \right\} \tag{7}
\]
% [File: 100.png Page: 90]
\textit{whence it follows that}
\[
\left. \begin{aligned}
x &= \tfrac{1}{2}( {u_1}^p - {u_2}^p + {u_3}^p),\\
y &= \tfrac{1}{2}( {u_1}^p + {u_2}^p - {u_3}^p),\\
z &= \tfrac{1}{2}(-{u_1}^p + {u_2}^p + {u_3}^p).
\end{aligned}\ \right\} \tag{8}
\]

II\@. \textit{If an equation of the form
\[
x^p + y^p + z^p = 0,    \tag{$2^\textrm{bis}$}
\]
in which $p$ is an odd prime, is satisfied by integers $x$, $y$, $z$, which
are prime to each, and if $z$ is divisible by $p$, then integers
$u_1$, $u_2$, $u_3$, $v_1$ $v_2$, $v_3$, prime to $p$, and a positive integer $k$, exist
such that}
\[
\left. \begin{alignedat}{3}
x + y &= p^{kp-1}{u_1}^p,& \quad \frac{x^p +y^p}{x + y} =& p{v_1}^p, \quad  &z =& -p^k u_1 v_1,\\
y + z &= {u_2}^p,        &&                             &x =& -u_2v_2,\\
z + x &= {u_3}^p,        &&                             &y =& -u_3v_3;
\end{alignedat}\ \right\} \tag{9}
\]
\textit{whence it follows that}
\[
\left. \begin{aligned}
x &= \tfrac{1}{2} ( p^{kp-1} {u_1}^p - {u_2}^p + {u_3}^p),\\
y &= \tfrac{1}{2} ( p^{kp-1} {u_1}^p + {u_2}^p - {u_3}^p),\\
z &= \tfrac{1}{2} (-p^{kp-1} {u_1}^p + {u_2}^p + {u_3}^p).
\end{aligned}\ \right\} \tag{10}
\]

To prove these theorems it is sufficient to show that formulæ
(7) and (9) are true. The equations in the first line in (9)
are equivalent to those in (6). The equations in the other
two lines in (9) and in all three lines in (7) are essentially
equivalent to those in (5), the only difference being in the
interchange of the roles of $x$, $y$, $z$. This interchange is legitimate,
since $x$, $y$, $z$ enter symmetrically into Eq.~($2^\textrm{bis}$).

The formulæ contained in these theorems were given by
\index{Legendre}Legendre (\textit{M\'em.\ Acad.\ d.~Sciences, Institut de France}, 1823
[1827], p.~1). They are also to be found in a letter of \index{Abel}Abel's
to \index{Holmboe}Holmboe and published in Abel's \textit{\OE uvres}, Vol.~II, pp.~254--255.\index{Abelian formulæ|)}

In the second theorem above we have said nothing concerning
the character of the integer $k$ except that it is positive.
We shall now show that it is greater than 1, whence it will
follow that $z$ is divisible by $p^2$. From formulæ (9) we have
\[
  {u_2}^p + {u_3}^p = x + y + 2z  \equiv  0 \bmod p.
\]
From this it follows that $u_2 + u_3$ is divisible by $p$. Let us write
\[
u_2 = -u_3 + p\alpha.
\]
Then
\[
  {u_2}^p \equiv -{u_3}^p    \bmod p^2 \quad \textrm{or} \quad
  {u_2}^p + {u_3}^p \equiv 0 \bmod p^2.
\]
Thence, by aid of the last formula in (10), we see that $z$ is
divisible by $p^2$, and hence that $k > 1$.

We shall show next that the prime factors of $v_1$ in both
theorems are of the form $2hp^2 + 1$, where $h$ is a positive integer.
Let $q$ be a prime factor of $v_1$. Then in either case we have
\[
  v_1 \equiv 0, \quad z \equiv 0, \quad
  y \equiv {u_2}^p, \quad x \equiv {u_3}^p, \quad
  x^p + y^p \equiv {u_2}^{p^2} + {u_3}^{p^2} \equiv 0 \bmod q    \tag{11}
\]
Let $\alpha$ be an integer such that $u_3 \alpha \equiv 1 \bmod q$. Then the last
congruence in (11) gives rise to the following:
\[
(u_2 \alpha)^{p^2} + 1 \equiv 0 \bmod q.
\]
% [File: 101.png Page: 91]
From this relation we see that
\[
(u_2\alpha)^{2p^2} \equiv 1, \quad u_2\alpha \not\equiv 1,\quad (u_2\alpha)^p \not\equiv 1
\bmod q. \tag{12}
\]

Let $m$ be the exponent to which $u_2\alpha$ belongs modulo $q$.
(See \index{Carmichael}Carmichael, \textit{l.~c.}, pp.~61--63.) From relations (12) it follows
that $m$ is a divisor of $2p^2$, but is different from 1 and $p$.
Hence $m$ must have one of the values, $2$, $2p$, $p^2$, $2p^2$. We shall
show that it cannot have the value 2 or the value $2p$, and hence
that $m = p^2$ or $m = 2p^2$.

Suppose that $m = 2$. Then $u_2 \alpha \equiv -1 \bmod q$. This, together 
with the relation $u_3 \alpha \equiv 1 \bmod q$, yields the
congruence $u_2 + u_3 \equiv 0 \bmod q$; 
whence $x + y \equiv 0 \bmod q$. But this is impossible, since
$v_1$ is prime to $x+y$ and $q$ is a factor of $v_1$. Hence $m \neq 2$.

Next suppose that $m = 2p$. Then, in view of the last relation
in (12), we see that $(u_2 \alpha)^p \equiv -1 \bmod q$. But
$(u_3 \alpha)^p \equiv 1 \bmod q$.
Hence ${u_2}^p + {u_3}^p \equiv 0 \bmod q$; whence $x+y\equiv 0 \bmod q$.
Since the last relation is impossible, it follows that $m\neq 2p$.

Then $m = p^2$ or $m = 2p^2$. In either case $q-1$ is divisible
by $p^2$, and hence by $2p^2$ since $q$ obviously is odd. Therefore
$q$ is of the form $2hp^2 + 1$, as was to be proved.

In the case of theorem~I we see by symmetry that the
prime factors of $v_2$ and $v_3$ are also of the form $2hp^2 + 1$.

In the case of theorem~II it may be shown similarly that
the prime factors of $v_2$ and $v_3$ are of the form $2hp + 1$. It is
sufficient to treat one of the numbers, say $v_2$. Then we have
\[
y^p + z^p \equiv 0 \bmod q,
\]
where $q$ is a prime factor of $v_2$ and is hence prime to $y$ and $z$.
This relation may be treated in the same manner as the last
relation in (11) was treated above, and with the result already
stated. The reader can readily supply the argument.

Among the methods which conceivably might be used
separately for the proof of the Fermat theorem are the following:
Assume that Eq.~(2) is satisfied and find a set of contradictory
properties of $x$, $y$, $z$; assume that Eq.~(2) is satisfied
and find a set of contradictory properties of $p$. (These
two methods might clearly be included in the more general
one in which contradictory properties of $x$, $y$, $z$, $p$ would be
% [File: 102.png Page: 92]
obtained.) Theorems~I and II above give properties of $x$,
$y$, $z$; they may be thought of in connection with the first
method of proof just mentioned. We shall now derive some
properties of $p$ which are necessary if (2) is to be satisfied;
the results so obtained may be thought of in connection with
the second method of proof just mentioned.

Let us suppose that Eq.~(2) has solutions in integers $x$,
$y$, $z$ which are prime to each other and to $p$. Let $q$ be a prime
number of the form $2hp+1$. Suppose first that $q$ is not a
factor of any one of the numbers $x$, $y$, $z$. Then from Eq.~(2) we
have
\[
x^p + y^p + z^p \equiv 0 \bmod q.      \tag{13}
\]
Let $z_1$ be a number such that $zz_1 \equiv 1 \bmod q$. Then we have
\[
(xz_1)^p + 1 \equiv (-yz_1)^p \bmod q,
\]
a relation which we shall write in the form
\[
s^p + 1 \equiv t^p \bmod q.        \tag{14}
\]
Raising each member of this congruence to the $2h$th power
and simplifying by means of the relations
\[
s^{2hp} \equiv 1, \quad t^{2hp} \equiv 1 \bmod q,      \tag{15}
\]
we have
\[
\binom{2h}{1} s^{(2h-1)p} + \binom{2h}{2} s^{(2h-2)p} + \ldots + \binom{2h}{2h-1} s^p + 1 \equiv 0 \bmod q,          \tag{$16a$}
\]
the parentheses quantities being binomial coefficients. Multiplying
this congruence through repeatedly by $s^p$ and simplifying
by means of the first relation in (15), we have
\[
\left. \begin{matrix} \begin{aligned}
& \binom{2h}{2} s^{(2h-1)p} + \binom{2h}{3} s^{(2h-2)p} + \ldots + 1 \cdot s^p
+ \binom{2h}{1} \equiv 0 \bmod q,
\\
& \binom{2h}{3} s^{(2h-1)p} + \binom{2h}{4} s^{(2h-2)p} + \ldots + \binom{2h}{1} s^p
+ \binom{2h}{2} \equiv 0 \bmod q,
\\
&\hdotsfor[20]{1}\quad
\\
& 1 \cdot s^{(2h-1)p} + \binom{2h}{1} s^{(2h-2)p} + \ldots + \binom{2h}{2h-2} s^p
+ \binom{2h}{2h-1} \equiv 0 \bmod q
\end{aligned} \end{matrix} \right\} \tag{$16b$}
\]
It is clear that the existence of congruences (16) implies that
\[
D_{2h} \equiv 0 \bmod q,       \tag{17}
\]
% [File: 103.png Page: 93]
where $D_{2h}$ denotes the determinant
\[
D_{2h}=
\begin{vmatrix}
  \displaystyle \binom{2h}{1}  
& \displaystyle \binom{2h}{2}  &  \cdots  
& \displaystyle \binom{2h}{2h-1} &  1 
\\[2ex]
  \displaystyle \binom{2h}{2} 
& \displaystyle \binom{2h}{3} & \cdots & 1 
& \displaystyle \binom{2h}{1} 
\\
\hdotsfor[15]{5}\\[1ex]
1 & \displaystyle \binom{2h}{1} & \cdots 
& \displaystyle \binom{2h}{2h-2} 
& \displaystyle \binom{2h}{2h-1}
\end{vmatrix}
\]
We may look on (14) and (17) as giving necessary properties
of $q$ when no one of the numbers $x$, $y$, $z$ is divisible by $q$.

Let us next suppose that $q$ is a factor of one of the numbers
$x$, $y$, $z$; say that it is a factor of $z$. Then from Eq.~(8) we
have
\[
(-u_{1})^{p}+{u_{2}}^{p}+{u_{3}}^{p}\equiv 0\mod q. \tag{18}
\]
Now, $q$ is a factor of $u_{1}$ or of $v_{1}$, since it is a factor of $z$. Suppose
that $q$ is a factor of $v_{1}$. Then it is not a factor of $-u_{1},
u_{2}$, or $u_{3}$. Therefore congruence~(18) may be used just as (13)
was employed above to derive a necessary relation of the form
(14) and thence the necessary relation~(17). Suppose next that
$q$ is a factor of $u_{1}$. Then (18) becomes
\[
{u_{2}}^{p} + {u_{3}}^p\equiv 0 \bmod q;
\]
whence
\[
{u_{2}}^{2p} \equiv {u_{3}}^{2p} \bmod q. \tag{19}
\]

Now, by means of Eqs.~(2) and (8) and the polynomial
theorem we see that
\begin{multline*}
  ( {u_{1}}^{p}+{u_{2}}^{p}+{u_{3}}^{p} )^{p} 
= ( {u_{1}}^{p}+{u_{2}}^{p}+{u_{3}}^{p} )^{p} 
- ( {u_{1}}^{p}+{u_{2}}^{p}-{u_{3}}^{p} )^{p}
\\
\shoveright{- ( {u_{1}}^{p}-{u_{2}}^{p}+{u_{3}}^{p} )^{p} 
- (-{u_{1}}^{p}+{u_{2}}^{p}+{u_{3}}^{p} )^{p} }
\\
= \sum{}\frac{p!}{\alpha! \beta! \gamma!}
  {u_{1}}^{\alpha p} {u_{2}}^{\beta p} {u_{3}}^{\gamma p}
  (1-(-1)^{\alpha} - (-1)^{\beta} - (-1)^{\gamma}),  \qquad \tag*{\llap{(20)}}
\end{multline*}
where the summation is taken over all non-negative numbers
$\alpha$, $\beta$, $\gamma$ for which $\alpha + \beta + \gamma = p$. Of the numbers $\alpha$, $\beta$, $\gamma$ in a given
set, either one is odd or three are odd, since their sum is the
odd number $p$. For a set in which only one of them is odd
% [File: 104.png Page: 94]
the parenthesis expression in (20) vanishes. Hence from (20)
we have the relation
\begin{multline*}
({u_1}^p + {u_2}^p + {u_3}^p)^p
\\
= 4p{u_1}^p {u_2}^p {u_3}^p  \sum \frac{(p-1)!}{(2\lambda+1)!(2\mu+1)!(2\nu+1)!}
  {u_1}^{2\lambda p} {u_2}^{2\mu p} {u_3}^{2\nu p},        \tag{21}
\end{multline*}
where the summation is taken over all non-negative numbers $\lambda$,
$\mu$, $\nu$ for which $\lambda + \mu + \nu = \frac{1}{2}(p - 3)$.
Hence we may write
\[
  {u_1}^p + {u_2}^p + {u_3}^p = 2p u_1 u_2 u_3 P,
\]
\[
\sum \frac{(p-1)!}{(2\lambda+1)!(2\mu+1)!(2\nu+1)!} 
     {u_1}^{2\lambda p} {u_2}^{2\mu p} {u_3}^{2\nu p} =
 2^{p-2} p^{p-1} P^p.        \tag{22}
\]

For the case under consideration $u_1$ is a multiple of $q$ while
from (19) we see that
\[
  {u_2}^{2\mu p} {u_3}^{2\nu p} \equiv {u_2}^{2(\mu + \nu)p} \bmod q.
\]
Hence from (22) we have
\[
 {u_2}^{(p-3)p}  \sum \frac{(p-1)!}{(2\mu+1)!(2\nu+1)!}
\equiv  2^{p-2} p^{p-1} P^p \bmod q,
\]
since $q$ is a factor of every term in the first member of (21) for
which $\lambda \neq 0$. Here the summation is for all non-negative
values of $\mu$ and $\nu$ for which $\mu + \nu = \frac{1}{2}(p -
3)$. Now the sum indicated by $\Sigma$ in the last congruence has
the value $2^{p-2}$, as one sees readily by expanding $(1+1)^{p-1}$
and $(1-1)^{p-1}$ by the binomial theorem and taking half their
difference. Hence
\[
  {u_2}^{(p-3)p} \equiv p^{p-1} P^p \bmod q.
\]
From this it follows that $P$ is not divisible by $q$. Moreover,
taking the $2h$th power of each member of the last congruence
we have
\begin{align*}
& 1 \equiv p^{2h(p-1)} \bmod q;\\
\intertext{or,}
& p^{2h} \equiv 1 \bmod q.
\end{align*}
This is a necessary condition on $q$ in the case now under consideration.

Gathering together the last result and those associated
with relations (14) and (17) we have the following theorem:

% [File: 105.png Page: 95]
III\@. \textit{If $p$ is an odd prime number having the property that a prime
number $q$ exists, $q = 2hp + 1$, such that $p^{2h} - 1$ is not
divisible by $q$ and either}
\[
D_{2h} \not\equiv 0 \bmod q,
\]
\textit{where $D_{2h}$ denotes the determinant introduced in Eq.~(17), or
the congruence}
\[
s^p + 1 \equiv t^p \bmod q,
\]
\textit{has no solution in integers $s$ and $t$ which are prime to $q$; then
the equation $x^p + y^p + z^p = 0$ cannot be satisfied by integers
$x$, $y$, $z$, which are prime to $p$.}

Next let us suppose that Eq.~(2) has a solution in integers
$x$, $y$, $z$ which are prime each to each, one of them being divisible
by $p$. We shall suppose that it is $z$ which is divisible by $p$.
In this case we shall need to employ the relations in theorem~II\@.
Replacing Eq.~(21), we shall now have another, obtained
in a similar manner; namely:
\begin{multline*}
  (p^{kp-1} {u_1}^p + {u_2}^p + {u_3}^p)^p
\\
= 4p^{kp} {u_1}^p {u_2}^p {u_3}^p
  \sum \frac{(p-1)!}{(2\lambda+1)!(2\mu+1)!(2\nu+1)!}
  p^{2\lambda(kp-1)} {u_1}^{2\lambda p} {u_2}^{2\mu p} {u_3}^{2\nu p};
\end{multline*}
whence we may write
\[
p^{kp-1} {u_1}^p + {u_2}^p + {u_3}^p = 2p^k u_1 u_2 u_3 \cdot P,
\]
\[
\sum\frac{(p-1)!}{(2\lambda+1)!(2\mu+1)!(2\nu+1)!}  p^{2\lambda(kp-1)}
   \cdot {u_1}^{2\lambda p} {u_2}^{2\mu p} {u_3}^{2\nu p} = 2^{p-2} P^p.  \tag{23}
\]
We shall presently have need for the last relation.

Again, let $q$ be a prime number of the form $2hp + 1$. If
we suppose that $q$ is not a factor of any one of the numbers
$x$, $y$, $z$, we shall be led as before to relations (14) and (17). 
We shall therefore direct our attention to the other case, namely,
that in which $q$ is a factor of one of the numbers $x$, $y$, $z$.

Suppose that $q$ is a factor of $x$. Then from (10) we have
\[
p^{kp-1} {u_1}^p - {u_2}^p + {u_3}^p \equiv 0 \bmod q.        \tag{24}
\]
Now, $q$ is a factor of $u_2$ or of $v_2$, since it is a factor of $x$. First
suppose that it is a factor of $v_2$. Then it is prime to $u_1$, $u_2$,
$u_3$. Let $u$ be chosen so that $p^{k-1} u_1 u \equiv 1 \bmod q$ and
multiply
% [File: 106.png Page: 96]
both members of congruence~(24) by $u^p$. A relation is
obtained which may be written in the form
\begin{equation}
\tag{25}
p^{p-1} \equiv s^p + t^p \bmod q,
\end{equation}
where $s$ and $t$ are prime to $q$. This congruence may now be
employed in the way in which (14) was used in the preceding
argument. A set of congruences modulo $q$, $2h$ in number,
may be found in the following manner: The first arises from
(25) by raising each member to the $2h$th power and simplifying
by means of relations~(15). The others come from
this one by repeated multiplication by $s^p t^{(2h-1)p}$ and reduction
by means of (15). The existence of these $2h$ congruences
implies that
\begin{equation}
\tag{26}
\Delta_{2h} \equiv 0 \bmod q,
\end{equation}
where
\[
\Delta_{2h} =
\left|
\begin{array}{lllll}
  \displaystyle \binom{2h}{1} 
& \displaystyle \binom{2h}{2} & \cdots 
& \displaystyle \binom{2h}{2h-1} & 2-p^{2h(p-1)} 
\\[2ex]
  \displaystyle \binom{2h}{2} 
& \displaystyle \binom{2h}{3} & \cdots & 2-p^{2h(p-1)} 
& \displaystyle \binom{2h}{1} 
\\[1ex]
\hdotsfor[20]{5}
\\[1ex]
  \displaystyle 2-p^{2h(p-1)} 
& \displaystyle \binom{2h}{1} & \cdots 
& \displaystyle \binom{2h}{2h-2} 
& \displaystyle \binom{2h}{2h-1}
\end{array}
\right|.
\]

Next let us suppose that $q$ is a factor of $u_2$. Then from
(24) we have readily
\[
u_3^{2p} \equiv p^{2(kp-1)} {u_1}^{2p} \bmod q.
\]
Employing (23) now as we did (22) in the previous case, we
have
\begin{equation}
\tag{27}
p^{(p-3)(kp-1)} {u_1}^{p(p-3)} \equiv P^p \bmod q.
\end{equation}
Raising each member of this congruence to the $2h$th power
and simplifying, we have
\[
p^{6h} \equiv 1 \bmod q.
\]
This, in connection with the relation $p^{2hp} \equiv 1 \mod q$, leads to
the congruence
\[
p^{2h} \equiv 1 \bmod q,
\]
provided that $p$ is greater than 3.

% [File: 107.png Page: 97]
It is obvious that in the case in which $q$ is a factor of $y$,
we should be led to the same relations as those just obtained
when $q$ is a factor of $x$.

Finally, let us suppose that $q$ is a factor of $z$. Then from
(10) we have
\[
 -p^{kp-1} u_1^p + u_2^p + u_3^p \equiv 0 \bmod q.
\]
Now, $q$ is a factor of $u_1$ or of $v_1$, since it is a factor of $z$. If
we suppose that $q$ is a factor of $v_1$, we shall be led as before to
relation (26). If we suppose that $q$ is a factor of $u_1$, then from
(9) we have $x + y \equiv 0 \mod q$.

Gathering together the results just deduced and those in
theorem~III, we have the following theorem:

IV. \emph{If there exists a prime number $q$, $q = 2hp + 1$, which is
not a factor of any of the numbers
\[
  D_{2h}, \quad \Delta_{2h}, \quad p^{2h} - 1,
\]
and if the equation
\[
  x^p + y^p + z^p = 0
\]
is satisfied by integers $x$, $y$, $z$, which are prime each to each, then
one of these integers $($say $z)$ and the sum of the other two $($say
$x + y)$ are both divisible by $q$.}

We shall give one other theorem which may be demonstrated
by elementary means; namely, the following:

V. \emph{If p is an odd prime and the equation
\[
  x^p + y^p + z^p = 0,       
\tag{$2^\text{bis}$}
\]
has a solution in integers $x$, $y$, $z$, each of which is prime to $p$,
then there exists a positive integer $s$, less than $\frac{1}{2}(p-1)$, such that
\[
  (s + 1)^p \equiv s^p + 1 \bmod p^3.  
\tag{28}
\]
}

From Eq.~(7) we have
\[
  (x + y)^{p-1} \equiv u_1^{p(p - 1)} \equiv 1 \bmod p^2.
\]
This relation and two similar ones lead to the following:
\[
  (x + y)^p \equiv x + y, \quad
  (y + z)^p \equiv y + z, \quad
  (z + x)^p \equiv z + x \bmod p^2.  
\tag{29}
\]
Now,
\[
  x + y \equiv -z \bmod p;
\]
whence
\[
  (x + y)^p \equiv -z^p \equiv x^p + y^p \bmod p^2.  
\tag{30}
\]
% [File: 108.png Page: 98]
Similarly,
\[
        (y+z)^p \equiv y^p+z^p, \quad (z+x)^p \equiv z^p+x^p \bmod p^2.      \tag{31}
\]
From relations (29), (30), (31) and Eq.~($2^{bis}$), we have
\[
        x+y+z \equiv 0 \bmod p^2.
\]
From this we have
\[
        (x+y)^p \equiv -z^p \equiv x^p+y^p \bmod p^3.
\]
Let $u$ be an integer such that $yu \equiv 1 \bmod p^3$. Then we have
\[
        (xu+1)^p \equiv (xu)^p+1 \bmod p^3.
\]
Hence we have the congruence
\[
        (\sigma+1)^p \equiv \sigma^p+1 \bmod p^3,      \tag{32}
\]
where $\sigma$ is a positive integer less than $p^3$.

We shall next show that congruence (32) implies and is implied by
the congruence
\[
        (\sigma+1)^{p^2} \equiv \sigma^{p^2}+1 \bmod p^3.      \tag{33}
\]
Let us define integers $\lambda$ and $\mu$ by means of the relations
\begin{align*}
        (\sigma + 1)^p &= \sigma + 1 + \lambda p,\quad \sigma^p = \sigma + \mu p.
\\
\intertext{Then }
        (\sigma + 1)^p &= \sigma^p + 1 + (\lambda - \mu)p.     \tag{34}
\end{align*}
We have also
\begin{align*}
(\sigma+1)^p &\equiv (\sigma+1)^p + \lambda p^2(\sigma+1)^{p-1} \bmod p^3,
\\
    &\equiv \sigma +1+\lambda p +\lambda p^2 \bmod p^3.
\\
\intertext{Likewise }
   \sigma^{p^2} &\equiv \sigma+\mu p+\mu p^2 \bmod p^3.
\\
\intertext{From the last two congruences we have }
        (\sigma +1)^{p^2} &\equiv \sigma^{p^2} +1+(\lambda-\mu)(p+p^2) \bmod p^3.  \tag{35}
\end{align*}
From (34) and (35) we see that a necessary and sufficient condition
for either (32) or (33) is that $\lambda-\mu \equiv 0 \bmod p^2$. Therefore,
(32) and (33) are equivalent; that is, if one of these congruences
is satisfied for a given value of $\sigma$, so is the other.

In view of this result we shall have proved relation~(28)
as soon as we have shown the existence of an integer $s$ less than
$\frac{1}{2}(p-1)$ and such that
\[
        (s-1)^{p^2} \equiv s^{p^2}+1 \bmod p^3.      \tag{36}
\]
% [File: 109.png Page: 99]
Let $t$ be that residue of $\sigma$ modulo $p$ for which the absolute
value is a minimum. Then from (33) we have
\[
(t + 1)^{p^2} \equiv t^{p^2} + 1 \bmod p^3.
\]
Three cases are possible. (\textit{a}) We may have $t = \frac{1}{2} (p-1)$. Then
\begin{align*}
(p + 1)^{p^2} &\equiv (p - 1)^{p^2} + 2^{p^2} \bmod p^{3},
\\
\intertext{or }
2^{p^2} &\equiv 1^p + 1 \bmod p^3,
\end{align*}
so that for this case we may take $s = 1$. (\textit{b})~We may have $t$
positive and \label{extrabrac}less than $ \frac{1}{2}(p - 1)$. 
In this case we may take
$s = t$. (\textit{c})~We may have $t$ negative and greater than $-\frac{1}{2}(p - 1)$.
In this case we write $s + 1 = -t$; then
\[
(-s)^{p^2} \equiv (-s - 1)^{p^2} + 1 \bmod p^3;
\]
whence (36) follows readily and then (28).

This completes the proof of the theorem.

By means of theorem III, \index{Legendre}Legendre has shown that the
equation $x^p + y^p + z^p = 0$ cannot be satisfied by integers $x$, $y$, $z$,
each of which is prime to $p$ if $p < 197$. \index{Maillet}Maillet has shown that
the same is true if $p < 223$. \index{Mirimanoff}Mirimanoff has extended the
result to every $p$ less than 257. By a further penetrating discussion
\index{Dickson}Dickson (\textit{Quart.\ Journ.\ Math.}, vol.~40) has proved
that the equation is without a solution in integers prime to $p$
if $p < 6857$.

We shall illustrate the means of obtaining these results
by proving that \textit{the equation $x^p + y^p + z^p = 0$ has no solution in
integers $x$, $y$, $z$, prime to $p$ if $2p + 1$ or $4p + 1$ is a prime.} For
this purpose we employ theorem~III.

If $2p + 1$ is a prime, we may take $q = 2p + 1$ and $h = 1$. Then
\[
D_2 = \begin{array}{|cc|}
2&1\\
1&2
\end{array} = 3.
\]
Now $2p + 1$ is not a factor either of $3$ or of $p^2 - 1 = (p - 1)(p + 1)$.
If $4p + 1$ is a prime, we may take $h = 2$. Then
\[
D_4 = \begin{array}{|cccc|}
4&6&4&1\\
6&4&1&4\\
4&1&4&6\\
1&4&6&4
\end{array} =  -3\cdot 5^3. 
\]
% [File: 110.png Page: 100]
Now $4p + 1$ is not a factor of $3\cdot 5^3$
or of
\[
p^4 - 1 = (p - 1)(p + 1)(p^2 + 1).
\]
From these results and theorem III, we conclude that the
equation $x^p + y^p + z^p = 0$ has no solution in integers $x$, $y$, $z$,
prime to $p$ if either $2p + 1$ or $4p + 1$ is a prime; in particular,
therefore, if
\[
p = 3,\: 5,\: 7,\: 11,\: 13,\: 23,\: 29,\: 41, \ldots
\]

%§ 22.
\Needspace*{4\baselineskip}
\section[\textsc{Present State of Knowledge Concerning the
Equation $x^p + y^p + z^p = 0$}]{Present State of Knowledge Concerning the
Equation \\ $x^p + y^p + z^p = 0$}

\hspace{\parindent}In the present section we shall give a brief statement of
the more important known facts about the equation
\[
x^p + y^p + z^p = 0, \ \ p = \text{odd prime}, \tag{1}
\]
over and above those which we have already mentioned. These
have not yet been demonstrated by elementary means; and
therefore a proof of them would be out of place in this introductory
book.

\index{Cauchy}Cauchy (\textit{Comptes Rendus} of Paris, Vol.~XXV, p.~181; \textit{{\OE}uvres},
(1) 10:~364) states without proof the remarkable theorem
that if Eq.~(1) is satisfied by integers $x$, $y$, $z$ which are prime
to $p$, then
\[
1^{p-4} + 2^{p-4} + 3^{p-4} + \cdots + \{\tfrac{1}{2}(p-1)\}^{p-4} \equiv 0 \bmod p.
\]

It is to \index{Kummer}Kummer that we owe the most important development
of the theory of Eq.~(1). (See references to Kummer's
work in H.~J.~S.\ \index{Smith}Smith's Report on the Theory of Numbers
in Smith's \textit{Collected Mathematical Papers}, Vol.~I, p.~97.) Kummer
makes use of complex numbers and by aid of them proves
the following general theorem:

If $p$ is a prime number which is not a factor of the numerator
of one of the first $\frac{1}{2}(p-3)$ Bernoulli numbers, then Eq.~(1)
has no solution in integers $x$, $y$, $z$, all of which are different
from zero.

In case $p$ is a factor of one of the first $\frac{1}{2}(p-3)$ Bernoulli
numerators, Kummer finds other properties which it must
possess. These we shall not state.

% [File: 111.png Page: 101]
By means of his general theorem \index{Kummer}Kummer shows in particular
that Eq.~(1) is impossible if $p\leqq 100$.

Starting from results due to Kummer, \index{Wieferich}Wieferich (Crelle's
Journal, Vol.\ CXXXVI) has shown that if Eq.~(1) is satisfied
by integers prime to $p$ then\footnote
  {The smallest prime $p$ for which this relation is satisfied is $p=1093$. There
  is no other $p$ less than 2000 satisfying this relation.}
\[
2^{p-1} \equiv 1 \bmod p^2.
\]
\index{Mirimanoff}Mirimanoff (Crelle's Journal, Vol.\ CXXXIX) has shown that
$p$ must in this case also satisfy the relation
\[
3^{p-1} \equiv 1 \bmod p^2.
\]
He also derives other relations which are less simple in form.

Later \index{Furtw\"angler}Furtw\"angler has proved two theorems from which
the above criteria of Wieferich and \index{Mirimanoff}Mirimanoff may be deduced.
These theorems are as follows:

If $x_1$, $x_2$, $x_3$ are three integers, different from zero and without
common divisor, among which subsists the equation
\[
        {x_1}^p + {x_2}^p + {x_3}^p = 0,
\]
where $p$ is an odd prime, then

I\@. Every factor $r$ of $x_i\ (i = 1, 2, 3)$ satisfies the congruence
\[
r^{p-1} \equiv 1 \bmod p^2,
\]
if $x_i$ is prime to $p$;

II\@. Every factor $r$ of $x_i\pm x_k\ (i, k = 1, 2, 3)$ satisfies the
congruence
\[
r^{p-1} \equiv 1 \bmod p^2,
\]
if $x_i + x_k$ and $x_i - x_k$ are prime to $p$.

By means of relations due to \index{Mirimanoff}Mirimanoff and \index{Furtw\"angler}Furtw\"angler,
\index{Vandiver}Vandiver (Trans.\ Amer.\ Math.\ Soc., Vol.~XV) has shown that
if Eq.~(1) has a solution in integers $x$, $y$, $z$, all of which are
prime to $p$, then $p$ has the following property:

(1) If $p$ is of the form $3n + 1$, then either
\[
  2^{p-1} \equiv 1 \bmod p^4  \quad\text{or}\quad
  5^{p-1} \equiv 1 \bmod p^2;
\]

(2) If $p$ is of the form $3n + 2$, then either
\[
  2^{p-1} \equiv 1 \bmod p^4  \quad\text{or}\quad
  5^{p-1} \equiv 7^{p-1} \equiv 1 \bmod p^2;
\]
% [File: 112.png Page: 102]
\index{Vandiver}Vandiver has also recently announced (Bull.\ Amer.\ Math.\ 
Soc., March, 1915) that he has found a relation which implies
several of those previously obtained, in particular, those in
the theorems of \index{Furtw\"angler}Furtw\"angler above.

Reference should also be made to two papers by \index{Bernstein}Bernstein,
one by Furt\-w\"ang\-ler, and one by \index{Hecke}Hecke in the \textit{G\"ottinger
Nachrichten} for 1910.

In conclusion it is to be remarked that the Academy of
Sciences of G\"ottingen holds a sum of 100,000 marks which is
to be awarded as a prize to the person who first presents a
rigorous proof of Fermat's Last Theorem. The existence of
this prize money has called forth a large number of pseudo-solutions
of the problem. Unfortunately, the number of
untrained workers attacking the problem seems to be increasing.

\Needspace*{4\baselineskip}
\begin{center}GENERAL EXERCISES\end{center}
\addtocontents{toc}{\protect\contentsline {section}{\numberline {}\textsc {General Exercises} 1-13}{\thepage}}

\begin{small}
1.* There do not exist three binary forms which afford a solution of the
equation $x^n + y^n = z^n$, $n > 2$, for every pair of values of
the variables in those
forms.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Carlini, 1911.)\quad}\index{Carlini}

2. If the equation $x^n + y^n = z^n$ is impossible, so is each of the equations
$u^{2n} - 4v^n = t^2$ and $s(2s + 1) = t^{2n}$. 
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Lind}Lind, 1910.)\quad}

3. If the equation $x^n + y^n = z^n$ is impossible, so is each of the equations
$u^{2n} + v^{2n} = t^2$ and $u^{2n} - v^{2n} = 2t^n$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Liouville}Liouville, 1840.)\quad}

4. If the equation $x^k + y^k = z^k$ is impossible for every $k$ greater than 2, then
is the equation $u^m v^n + v^m w^n + w^m u^n = 0$ impossible for every pair of integers $m$
and $n$, except for the trivial solutions 1,\:0,\:0; 0,\:1,\:0; 0,\:0,\:1.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Hurwitz}Hurwitz, 1908.)\quad}

5.\dag\ Determine systematically a large number of simple equations which
are impossible when $x^n + y^n = z^n$ has no solution.

6.* If we write
\[
s_1 = x + y + z, \quad s_2 = xy + yz + zx, \quad s_3 =xyz,
\]
then the condition
\[
x^p + y^p + z^p = 0,            \tag{l}
\]
can be written in the form
\[
\phi_p(s_1, s_2, s_3) = 0,              \tag{2}
\]
while $x, y, z$ are roots of the cubic equation
\[
t^3 - s_1 t^2 + s_2 t - s_3 = 0.            \tag{3}
\]
Then Eq.~(1) can have a rational solution only when all the roots of (3) are
rational, its coefficients being subject to the condition (2). By aid of this remark
show that (1) is impossible when $p=17$. 
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Mirimanoff}Mirimanoff, 1909.)\quad}

% [File: 113.png Page: 103]
7.* If the equation $x^p + y^p + z^p = 0$ has the primitive solution $x$, $y$, $z$, $p$ being
an odd prime, and $G$ is the greatest common divisor of $x + y + z$ and $x^2 + xy + y^2$,
then integers $I$, $K$, $L$ exist such that
\[
  y^2 + yz + z^2 = GI, \quad
  z^2 + zx + x^2 = GK, \quad
  x^2 + xy + y^2 = GL,
\]
and all the factors of the numbers $I$, $K$, $L$ are of the form $6\mu p + 1$. Show that
to demonstrate the impossibility of the equation $x^p + y^p + z^p = 0$ it is sufficient
to prove that two of the numbers $I$, $K$, $L$ are equal or that one of them is unity.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Fleck}Fleck, 1909.)\quad}

8. Show that the equation $3u(4v^3 - u^3) = t^2$ is impossible in integers $u$, $v$, $t$,
all of which are different from zero.

9.\dag\ Note that when $p$ is an odd prime the equation
\[
x^{2p} + y^{2p} = z^{2p},
\]
with the condition that $x$, $y$, $z$ are prime to $p$, implies the three \index{Triangles, Pythagorean}Pythagorean
equations
\[
  x^2 + y^2 = {z_1}^2, \quad
  {x_1}^2 + y^2 = z^2, \quad
  x^2 + {y_1}^2 = z^2.
\]
What numbers $x$, $y$, $z$ can satisfy these three equations?

10. Show that the equation $x^{2p} + y^{2p} = z^{2p}$, in which $p$ is a prime number,
implies the coexistence of two equations of the form
\[
a^p + b^p = c^p, \qquad b^p + c^p = d^p.
\]

11.* Investigate the problem of solving the equation $x^p + y^p = pz^p$, where
$p$ is an odd prime.    
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Hayashi}Hayashi, 1911.)\quad}

12.* Investigate the problem of solving the equation $x^p + y^p = cz^p$, where
$p$ is an odd prime.        
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Maillet}Maillet, 1901.)\quad}

13.* Prove that neither of the equations $t^2 = (z^2 + y^2)^2 - (zy)^2$, $t^2=(z^2-y^2)^2-(zy)^2$
possesses an integral solution. By means of this result prove the impossibility
of each of the equations
\[
\phantom{(Kapferer,\ 1913.)\quad}
u^6 + v^6 = w^6,\quad  u^{10} + v^{10} = w^{10}.   \tag*{(\index{Kapferer}Kapferer, 1913.)\quad}
\]
\index{Fermat Problem|)}\index{Equations of Higher Degree|)}\par\end{small}

% [File: 114.png Page: 104]
%CHAPTER VI

\chapter{THE METHOD OF FUNCTIONAL EQUATIONS}
\markright{THE METHOD OF FUNCTIONAL EQUATIONS}

\index{Functional Equations, Method of|(}\index{Equations , Functional@Equations, Functional|(}\index{Method of Functional Equations|(}
\setcounter{section}{22}
%§ 23.
\section[\textsc{Introduction. Rational Solutions of a Certain
Functional Equation}]{Introduction. Rational Solutions of a Certain
Functional Equation}

\hspace{\parindent}There is a method which will sometimes be found useful
in the theory of Diophantine analysis and which we have
had no occasion to employ in the preceding pages. It may
conveniently be described as the method of functional equations.
It consists essentially in the use of rational solutions
of functional equations as an aid in solving Diophantine problems
of a certain type, a type in fact which plays an important role
in the work of \index{Diophantus}Diophantus. In this chapter we shall give a
brief illustration of the method by employing it in the solution
of certain problems first treated by Diophantus and \index{Fermat}Fermat.

It should be said that the principal value of this method
lies not so much in its use for the solution of given problems
as in the fact that it renders possible an arrangement of certain
problems in an order in which they may profitably be
investigated. A treatment of these problems from this point
of view seems not to exist in the literature. The primary
purpose of this chapter is to direct attention to the possibilities
of this general method of functional equations and to
give an indication of how it may be employed. A general
systematic development of the method is not attempted.

Diophantus more than once makes use of the identity
\[
a^2(a+1)^2 + a^2 + (a+1)^2 = (a^2+a+1)^2
\]
in the solution of problems. This identity may be looked
upon as affording a solution of the functional equation
\[
a^2{u_a}^2 + a^2 + {u_a}^2 = {v_a}^2,       \tag{1}
\]
% [File: 115.png Page: 105]
in which $u_a$ and $v_a$ are to be determined as rational functions
of $a$. It is clear that this equation may be written in the form
\[
(a^2 + 1)({u_a}^2 + 1) = {v_a}^2 + 1.     \tag{2}
\]
The first member may be written as a sum of two squares,
thus
\[
(a^2 + 1)({u_a}^2 + 1) = (au_a \pm 1)^2 +(u_a \mp a)^2.    \tag{3}
\]
The second member may be written as a sum of two squares
in a great variety of ways. Thus, if we write
\[
  {v_a}^2 + 1 = (v_a + x_a)^2 + (1 + m_a x_a)^2,           \tag{4}
\]
we have
\[
  2v_ax_a + {x_a}^2 + 2m_ax_a + {m_a}^2 {x_a}^2 = 0.
\]
Besides the solution $x_a = 0$ of this equation, we have
\[
  x_a = -\frac{2(m_a + v_a)}{{m_a}^2 + 1}.
\]
Thence we see that
\[
  {v_a}^2 + 1 
= \left\{ v_a - \frac{2   (m_a + v_a)}{{m_a}^2 + 1} \right\}^2 \!
+ \left\{ 1   - \frac{2m_a(m_a + v_a)}{{m_a}^2 + 1} \right\}^2. \tag{5}
\]
From (3) and (5) we see that (2) will be satisfied if
\[
\left. \begin{aligned}
  au_a \pm 1 &= v_a - \frac{2   (m_a + v_a)}{{m_a}^2 + 1},\\
   u_a \mp a &= 1   - \frac{2m_a(m_a + v_a)}{{m_a}^2 + 1}.
\end{aligned}\ \right\} \tag{6}      
\]
It is obvious that these two equations may be solved rationally
for $u_a$ and $v_a$ in terms of $m_a$, so that we have a rational solution
of (2), and hence of (1), for every rational function $m_a$.
This solution may be written in the following form:
\[
\left. \begin{aligned}
  u_a
&= \pm a - 1 + \frac{2}{{m_a}^2 + 1} 
 - \frac{2m_a}{{m_a}^2 + 1}
   \left\{ -a \pm \frac{(a^2 + 1)(m_a \pm 1)^2}{{m_a}^2 + 2am_a - 1} \right\},
\\
  v_a &=   -a \pm \frac{(a^2 + 1)(m_a \pm 1)^2}{{m_a}^2 + 2am_a - 1}
\end{aligned}\ \right\} \tag{7}
\]
To the solution of (1) afforded by (7) we should adjoin those
gotten by taking $x_a = 0$ in (4), namely:
\[
u_a = \pm a + 1, \quad v_a = a \pm (a^2 + 1), \quad u_a = \frac{2}{a}, \quad v_a = a + \frac{2}{a}.
\]

% [File: 116.png Page: 106]
We write down some simple particular solutions for which
we shall have use in §~25.
\[
\left. \begin{alignedat}{3}
u_a &= a+1,         &\quad &\frac{2}{a},     \quad & &2a^2,\\
v_a &= a^2 + a + 1, &\quad &a + \frac{2}{a}, \quad & &a(2a^2 + 1),\\
u_a &\multispan{5}{${}= 4a^3 + 4a^2 + 3a + 1$,\hfill}\\
u_a &\multispan{5}{${}= 4a^4 + 4a^3 + 5a^2 + 3a + 1$.\hfill}
\end{alignedat}\ \right\} \tag{8}
\]

%§ 24.
\Needspace*{4\baselineskip}
\section[\textsc{Solution of a Certain Problem from Diophantus}]{Solution of a Certain Problem from Diophantus}

\hspace{\parindent}In Book V of his \textit{Arithmetica} \index{Diophantus}Diophantus proposes and shows
how to solve the following problem:

\textit{To find three squares such that the product of any two of them,
added to the sum of those two or to the remaining one, gives a square.}

If we denote one of these squares by $a^2$ it will be convenient
to take ${u_a}^2$ for a second one, where $u_a$ is one of the functions
denoted by this symbol in the preceding section. For this
purpose Diophantus uses $u_a = a+1$. He then observes that
the three numbers,
\[
a^2, \quad (a+1)^2, \quad 4a^2+4a+4,      \tag{1}
\]
have the property that the product of any two of them, added
to the sum of those two or to the remaining one, gives a square.
This may readily be verified by the reader. Then to complete
the solution of the problem it is sufficient to render $4a^2+4a+4$,
and hence $a^2+a+1$, equal to a square. Setting, as usual,
\[
a^2+a+1 = (m-a)^2,
\]
we have
\[
a = \frac{m^2 - 1}{2m + 1},        \tag{2}
\]
where $m$ is any rational number whatever. If this value of
$a$ is set in expressions~(1) we have the three squares sought.

\index{Fermat}Fermat has shown that this result\footnote
 {It may be remarked that the result of the next section may also be used
 for the same purpose.}
of Diophantus may be
employed in the solution of the following problem:

% [File: 117.png Page: 107]
\textit{To find four numbers such that the product of any two of them
added to the sum of those two gives a square.}

For three of the numbers sought take a set of numbers (1)
where $a$ has the form given in (2). Following \index{Fermat}Fermat, we
shall take the particular set given by \index{Diophantus}Diophantus, namely:
\[
\frac{25}{9}, \quad \frac{64}{9}, \quad \frac{196}{9}.
\]
This is obtained by taking $m = -2$ in equation~(4). Let $x$ be
the fourth number sought. Then it is necessary and sufficient
that $x$ satisfy the conditions\footnote
 {The symbol $\square$ stands for a square number with whose value we are not
 concerned. It may differ from one equation to another.}
\[
  \frac{34}{9}  x + \frac{25}{9}  = \square, \quad
  \frac{73}{9}  x + \frac{64}{9}  = \square, \quad
  \frac{205}{9} x + \frac{196}{9} = \square;
\]
or more simply the conditions
\[
34x+25=\square, \quad 73x+64=\square, \quad 205x+196=\square.   \tag{3}
\]
This is an example of the so-called \index{Equations , Triple@Equations, Triple}\index{Triple equation}\textit{triple equation} of Fermat.
We shall find a solution by the method originated by Fermat.
Replace $x$ by a function of $t$ in such way that the first equation
in (3) shall be satisfied. For this purpose it is sufficient to
take
\[
x = 34t^2 + 10t.
\]
Then the other two equations in (3) become
\[
2482t^2 + 730t + 64 = \square, \quad 14,\!965t^2 + 2050t + 196 = \square.  \tag{4}
\]
We have to determine $t$ so as to satisfy these equations, an
example of the so-called \index{Equations , Double@Equations, Double}\index{Double Equations}\textit{double equation} of Fermat.

The interesting method of Fermat enables one to find an
indefinitely great number of solutions of system~(4). Multiplying
the first equation through by 196 and the second by 64,
we have two new equations of the same form as (4) with the
further condition that the independent terms in the first members
are equal. These equations are
\[
\left. \begin{aligned}
486,\!472t^2 + 143,\!080t + 12,\!544 &= \square,\\
957,\!760t^2 + 131,\!200t + 12,\!544 &= \square,
\end{aligned}\ \right\} \tag{5}
\]
% [File: 118.png Page: 108]
The difference of the two first members is $471,\!288t^2 - 11,\!880t$.
We may separate this into two factors, thus:
\[
\frac{1425}{28} t \cdot \left( \frac{4,\!398,\!688}{475} t - 224 \right).  \tag{6}
\]
The separation is effected in such way that the independent
term in the second factor is twice the square root of the independent
term 12,544 in Eqs.~(5). Now, if half the sum of
the two factors in (6) is squared and the result equated to the
first member of the second equation in (5), it is obvious that a
rational value of $t$ will be obtained satisfying that equation.
It is clear that this value will then also satisfy the other equation
in (5). This value of $t$ affords a value of $x$, the fourth
number in the set to be determined.

Eqs.~(4) have not merely a single solution, but an infinite
number. These may be found one after the other as follows:
Let $t_1$ be a value of $t$ satisfying Eqs.~(4) and write $t=u+t_1$.
Putting this value of $t$ in (4), we obtain a pair of equations in
$u$ of the same form as (4). These can be solved by the method
just given for solving (4). We thus obtain a single solution
$u=u_1$ of these equations. Then $t=u_1+t_1$ is a solution of (4).
By the aid of this solution of (4) another may be obtained;
and so on indefinitely.

By means of each solution of (4) we obtain a new value of
$x$ affording a solution of the problem proposed.

It should be observed that the method of solving Eqs.~(4),
and hence that of solving Eqs.~(3), is general, being applicable
to all equations of the types (3) and (4).

%§ 25.
\Needspace*{4\baselineskip}
\section[\textsc{Solution of a Certain Problem Due to Fermat}]{Solution of a Certain Problem Due to Fermat}

\hspace{\parindent}Fermat\index{Fermat} has given attention to the following problem:

\textit{To find three squares such that the product of any two of
them, added to the sum of those two, gives a square}.

He has indicated that this problem is capable of a solution
different from that which is incidental to the solution
given by \index{Diophantus}Diophantus for the first problem treated in the preceding
% [File: 119.png Page: 109]
section; but he gives no hint as to the method which
he employs. He says, however, that it leads to an indefinitely
great number of solutions. Making use of the solutions of
the functional equation treated in §~23, we shall now give
two methods for solving this problem with such result.

Let $u_a$ and $w_a$ be two rational functions of the rational
number $a$ such that
\[
\left. \begin{aligned}
  a^2{u_a}^2 + a^2 + {u_a}^2 &= \square,\\
  a_2{w_a}^2 + a^2 + {w_a}^2 &= \square.
\end{aligned}\ \right\} \tag{1}
\]
Then if we take $a_2$, ${u_a}^2$, ${w_a}^2$ for the three squares sought, we
have to determine $a$ so as to satisfy the single equation
\[
  {u_a}^2 {w_a}^2 + {u_a}^2 + {w_a}^2 = \square.         \tag{2}
\]
For determining appropriate functions $u_a$ and $w_a$ we have the
results of §~23.

Let us take
\[
u_a=a+1, \quad  w_a=\frac{2}{a}.
\]
Then (2) becomes
\[
  (a+1)^2   \left(\frac{2}{a}\right)^2 \! 
+ (a+1)^2 + \left(\frac{2}{a}\right)^2 \! = \square,
\]
or
\[
a^4 + 2a^3 + 5a^2 + 8a + 8 = \square.
\]
By means of the general method of §~17 in Chapter~IV, it is
possible to find an infinite number of values of $a$ satisfying
this equation. For every such value of $a$ the three numbers
$a^2$, $(a+1)^2$, $4/a^2$ furnish a solution of our problem.

We may also proceed as follows: Denoting the square
in the second member of (2) by ${t_a}^2$, we may write that equation
in the form
\[
  ({u_a}^2+1)({w_a}^2+1) = {t_a}^2+1.
\]
The second member may be separated into a sum of two squares
as in Eq.~(5) of §~23. Thus we have an equation of the form
\[
  ({u_a}^2 + 1)({w_a}^2 + 1) 
= \left\{ t_a - \frac{2   (n_a + t_a)}{n_a^2 + 1} \right\}^2  \!
+ \left\{ 1   - \frac{2n_a(n_a + t_a)}{n_a^2 + 1} \right\}^2,
\]
% [File: 120.png Page: 110]
where $n_a$ is an arbitrary rational function of $a$. This equation
will be satisfied if
\[
\begin{aligned}
  u_a w_a + 1 &= t_a - \frac{2   (n_a + t_a)}{{n_a}^2 + 1},\\
  u_a - w_a   &= 1   - \frac{2n_a(n_a + t_a)}{{n_a}^2 + 1}.
\end{aligned}
\]

These equations may be solved rationally for $w_a$ and $t_a$
in terms of $n_a$ and $u_a$. Thus, we have for $w_a$ the value
\[
  w_a = \frac{ (u_a-1)({n_a}^4-1) + 2n_a({n_a}^3+{n_a}^2+n_a+1) } 
             { ({n_a}^2+1) ({n_a}^2-2n_au_a-1) }.     \tag{3}
\]

With this value of $w_a$, Eq.~(2) will be satisfied whatever
rational functions $u_a$ and $n_a$ may be. If $u_a$ is given any value
such as those in Eqs.~(7) and (8) of §~23, the first equation
in (1) is satisfied. It is then sufficient to determine $a$ so that
the second equation in (1) is satisfied. Then for this value
of $a$ the squares $a^2$, ${u_a}^2$, ${w_a}^2$ furnish a solution of our problem.

As an illustration of this result let us take
\[
u_a = a+1, \quad n_a = 1.
\]
Then
\[
w_a = - \frac{2}{a+1},
\]
so that the condition on $a$ may be written
\[
\frac{4a^2}{(a+1)^2} + a^2 + \frac{4}{(a+1)^2} = \square;
\]
or
\[
a^4 + 2a^3 + 5a^2 + 4 = \square.
\]
An unlimited number of values of $a$ may be found satisfying
this equation (see §~17). We may get one of them by taking
for the square in the second member the quantity
\[
  \left( 2 + \frac{5}{4} a^2 \right)^2,
\]
and proceeding according to the methods of §~17 in Chapter
IV\@. Thus, we have $a = 32/9$. Then our three squares are
\[
\frac{1024}{81}, \quad \frac{1681}{81}, \quad \frac{324}{1681}.
\]\index{Functional Equations, Method of|)}\index{Equations , Functional@Equations, Functional|)}\index{Method of Functional Equations|)}

% [File: 121.png Page: 111]
\Needspace*{4\baselineskip}
\begin{center}GENERAL EXERCISES\end{center}
\addtocontents{toc}{\protect\contentsline {section}{\numberline {}\textsc {General Exercises} 1-6}{\thepage}}

\begin{small}
1.$\dag$ Determine all the polynomial solutions of the functional equation
\[
  a^2 {u_a}^2 + a^2 + {u_a}^2 = {v_a}^2.
\]
Apply the result to the solution of a group of Diophantine problems.

2.$\dag$ Investigate the problem of finding three squares such that the product
of any two of them exceeds the sum of those two by a square.

3.$\dag$ Obtain a solution of the system of equations
\[
u_x v_x - 1 = \square, \quad v_x w_x - 1 = \square, \quad w_x u_x - 1 = \square,
\]
in which $u_x$, $v_x$, $w_x$ are unknown rational functions of $x$. Apply the result to the
solution of problems in Diophantine analysis. (Cf.~\index{Diophantus}\textit{Diophantus}, Book~IV,
Problem~24.)

\textsc{Suggestion}.--The given equations may be written in the form
\[
  u_x v_x = {\rho_x}^2 + 1, \quad
  v_x w_x = {\sigma_x}^2 + 1, \quad
  w_x u_x = {\tau_x}^2 + 1.      \tag{1}
\]
Then if equations of the form
\[
  u_x = {a_x}^2 + {b_x}^2,  \quad
  v_x = {c_x}^2 + {d_x}^2, \qquad
  w_x = {e_x}^2 + {f_x}^2      \tag{2}
\]
are assumed and substitution is made in system (1), certain of the functions
introduced in Eq.~(2) may be determined in terms of the others. A rational
solution of the given system of equations is thus obtained. This process is also
capable of generalization in accordance with the suggestion afforded by Eq.~(5)
of \S~23.

4.$\dag$ Treat the corresponding problems for the system of functional equations
\[
u_x v_x + 1 = \square, \quad v_x w_x + 1 = \square, \quad w_x u_x + 1 = \square.
\]
(Cf.~\textit{Diophantus}, Book~IV, Problem~23.)

5.$\dag$ Find rational functions $u_x$, $v_x$, $w_x$ such that the continued product of their
squares increased by the square of each one of them separately shall be the
square of a rational function of $x$. Apply the result to problems in Diophantine
analysis. (Cf.~\textit{Diophantus}, Book~V, Problem~24.)

6.$\dag$ Find rational functions $u_x$, $v_x$, $w_x$ such that the continued product of
their squares decreased by the square of each one of them separately shall be
the square of a rational function of $x$. Apply the result to problems in Diophantine
analysis. (Cf.~\textit{Diophantus}, Book~V, Problem~25.)
\par\end{small}

% [File: 122.png Page: 112]
\Needspace*{5\baselineskip}
\begin{center}MISCELLANEOUS EXERCISES\end{center}
\addtocontents{toc}{\protect\contentsline {section}{\numberline {}\textsc {MISCELLANEOUS EXERCISES} 1-71}{\thepage}}

\begin{small}
1. Show how to find four numbers such that if one takes the square of their
sum \emph{plus} or \emph{minus} any one singly, then all the eight resulting numbers are squares.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Diophantus}Diophantus.)\quad}

2. Show how to find three numbers whose sum is a square, such that the
sum of the square of each and the succeeding number is a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.)\quad}

3. Show how to find two numbers such that their product \textit{plus} or \textit{minus}
their sum is a cube.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.)\quad}

4. Show how to find three numbers such that the square of any one of them
\textit{plus} or \textit{minus} the sum of the three is a square.   
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus; \index{Hart}Hart, 1876.)\quad}

5. Show how to find three numbers such that the product of any two of
them \textit{plus} or \textit{minus} the sum of the three is a square.      
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.)\quad}

6. Show how to find four numbers such that the product of each two of them
increased by unity shall be the same square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus; \index{Lucas}Lucas, 1880.)\quad}

7.* Show how to find five numbers such that the product of each two of
them increased by unity shall be a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Euler}Euler, \index{Legendre}Legendre.)\quad}

8.* Obtain the general solution of the Diophantine system $y = x^2 + (x + 1)^2$,
$y^2 = z^2 + (z + 1)^2$. Generalize the results by treating also the system $y = x^2 + t(x + \alpha)^2$,
$y^2 = z^2 + t(z + \beta)^2$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Jonqui\`eres}Jonqui\`eres, 1878.)\quad}

9.* Develop a theory of the Diophantine system $x = 4y^2 + 1$, $x^2 = z^2 + (z + 1)^2$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Gerono}Gerono, 1878.)\quad}

10. Obtain a single-parameter solution of the system $x^2 + y^2 - 1 = u^2$,
$x^2 - y^2 - 1 = v^2$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Arch.\ Math.\ Phys., 1854.)\quad}

11.* Obtain the general solution of the Diophantine equation
\[
\phantom{(Pepin,\ 1879.)\quad}
y^2 = x(x + 1)(2x + 1).      \tag*{(\index{Pepin}Pepin, 1879.)\quad}
\]

12. Apply the identity
\[
  (s^2 - 2st - t^2)^4 + (2s + t)s^2 t(2t + 2s)^4 
= (s^4 + t^4 + 10t^2 s^2 + 4st^3 + 12s^3 t)^2
\]
to the resolution of certain Diophantine equations.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Desboves, 1878.)\quad}\index{Desboves}

13. Find all the integral solutions of the equation $(x+1)^y = x^{y+1} + 1$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Meyl}Meyl, 1876.)\quad}

14. Develop methods for finding solutions of the Diophantine equation
\[
\phantom{(Valroff,\ 1912.)\quad)}
2x^2 y^2 + 1 = x^2 + y^2 + z^2.      \tag*{(\index{Valroff}Valroff, 1912.)\quad}
\]

15. Develop methods for obtaining solutions of the Diophantine system
\[
\phantom{(Gerono,\ 1878.)\quad}
x = u^2, \quad  x + 1 = 2v^2, \quad 2x + 1 = 3w^2.   \tag*{(Gerono, 1878.)\quad}
\]

16. Determine those \index{Triangles, Pythagorean}Pythagorean triangles for each of which the sum of
the area and either of the legs is a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.\footnotemark[1])\quad}

17. Determine those \index{Triangles, Pythagorean}Pythagorean triangles for each of which the area exceeds
either leg by a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.\footnotemark[1])\quad}

% [File: 123.png Page: 113]
18. Determine those \index{Triangles, Pythagorean}Pythagorean triangles for each of which the area exceeds
the hypotenuse or one leg by a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.\footnotemark[1])\quad}

19. Determine those Pythagorean triangles for each of which the sum of
the area and either the hypotenuse or one leg is a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.\footnotemark[1])\quad}

20. Determine those Pythagorean triangles for each of which the line bisecting
an acute angle is rational.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.\footnotemark[1])\quad}

21. Determine those Pythagorean triangles for each of which the sum of
the area and the hypotenuse is a square and the perimeter is a cube.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.\footnotemark[1])\quad}

22. Determine those Pythagorean triangles for each of which the sum of
the area and the hypotenuse is a cube and the perimeter is a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.\footnotemark[1])\quad}

23. Determine those Pythagorean triangles for each of which the sum of
the area and one side is a square and the perimeter is a cube.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.\footnotemark[1])\quad}

24. Determine those Pythagorean triangles for each of which the sum of
the area and one side is a cube and the perimeter is a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.\footnotemark[1])\quad}

25. Determine those Pythagorean triangles for each of which the perimeter
is a square and the area is a cube.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Diophantus.\footnotemark[1])\quad}

26. Determine those \index{Triangles, Pythagorean}Pythagorean triangles for each of which the perimeter
is a cube and the sum of the perimeter and area is a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Diophantus}Diophantus.\footnotemark[1])\quad}

\footnotetext[1]{In the case of Problems 16 to 26 Diophantus shows merely how to find
particular rational solutions. It is doubtless difficult to find general solutions
of some of these problems; but particular solutions may be found without great
difficulty.}

27. Give a method of finding an infinite number of solutions of each of the
equations $x^3 + y^3 = u^2 + v^2$;\quad $x^{2m+1} + y^{2m+1} = u^2 + v^2$,
$m > 1$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Aubry}Aubry, \index{Miot}Miot, 1912.)\quad}

28. If a Diophantine equation can be separated into two members each of
which is homogeneous and the numbers representing the degrees of the two
members are relatively prime, show how solutions may always be obtained in
an easy manner. By means of special examples show that this method may often
be used to obtain results which are not trivial in character.

29. Find three squares in arithmetical progression such that the square root
of each of them is less than a square by unity.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Evans}Evans and \index{Martin}Martin, 1873.)\quad}

30. Obtain all the integral solutions of the equation $x^y = y^x$.

31. Determine all the positive integral solutions of the equation
\[
\phantom{(Swinden,\ 1912)\quad}
4x^3 - y^3 = 3x^2 yz^2.     \tag*{(\index{Swinden}Swinden, 1912.)\quad}
\]

32. Show that the system $xy + x + y = a^2$, $xy - x - y = b^2$ is impossible in
integers $x$, $y$, $a$, $b$ all of which are different from zero. 
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Aubry}Aubry, 1911.)\quad}

33.* Obtain the general rational solution of the system of equations
\[
\phantom{(Welmin,\ 1912.)\quad}
ax^2 + b = u^2, \quad cx^2 + d = t^2.    \tag*{(\index{Welmin}Welmin, 1912.)\quad}
\]

34.* Show that the equation $u^m + v^n = w^k$ in which $m$, $n$, $k$ are positive integers
possesses an algebraic solution $u$, $v$, $w$ each function of which is expressible as
% [File: 124.png Page: 114]
a polynomial in the single variable $t$ in each of the following cases and only in
these:
\begin{flalign*}
\indent& (1)  &  u^m + v^2 &= w^2,  && \\
\indent& (2)  &  u^2 + v^2 &= w^k,  && \\
\indent& (3)  &  u^3 + v^3 &= w^2,  && \\
\indent& (4)  &  u^4 + v^3 &= w^2,  && \\
\indent& (5)  &  u^4 + v^2 &= w^3,  && \\
\indent& (6)\phantom{Welmin}  
              &  u^5 + v^3 &= w^2.   &\text{(\index{Welmin}Welmin,~1904.)}&\quad
\end{flalign*}

35.* Obtain all the solutions of the equation
\[
m \arctan \frac{1}{x} + n \arctan \frac{1}{y} = k \frac{\pi}{4}
\]
in integers $k$, $m$, $n$, $x$, $y$, showing that there are but the following four sets:
1, 1, 1, 2, 3; 1, 2, $-1$, 2, 7; 1, 2, 1, 3, 7; 1, 4, $-1$, 5, 239.  \hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{St\"ormer}St\"ormer,~1899.)\quad}

36.* Develop the theory of the equation $ax^{p^{t}} + by^{p^{t}} = cz^{p^{t}}$, $p$ being a prime
number.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Maillet}Maillet,~1898.)\quad}

37.* Develop the theory of the equation $ax^m + by^m = cz^n$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Desboves,~1879.)\quad}\index{Desboves}

38. Of each of the following equations find a solution involving two or more
parameters:
\begin{align*}
x^3 + y^3 + z^3 & = 2t^3, \\
x^3 + y^3 + z^3 & = 2t^{9k}, \\
x^3 + y^3 + z^3 + u^3 & = 3t^3, \\
x^3 + y^3 + z^3 + u^3 & = kt^m, \\
x^3 + 2y^3 + 3z^3 & = t^3, \\
x^3 + 2y^{3m} + 3z^{3n} & = t^3.
\end{align*}
\null\hfill {(Carmichael,~1913.)\quad}\index{Carmichael}

39. Show how to find $r$ rational numbers such that if a given number is added
to their sum or to the sum of any $r - 1$ of them the results shall all be squares.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Holm}Holm,~Cunningham,\index{Wallis}~Wallis,~1906.)\quad}\index{Cunningham}

40. Find several cubes such that the sum of the divisors of each is a square.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Fermat}Fermat.)\quad}

41. Find several squares such that the sum of the divisors of each is a cube.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Fermat}Fermat.)\quad}

42. Prove that 25 is the only square which is 2 less than a cube. Prove that
4 and 121 are the only squares each of which is four less than a cube. \hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Fermat.)\quad}

43. Show how to determine an unlimited number of Pythagorean triangles
having the same area.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Fermat}Fermat.)\quad}

44. Prove that the number $2(x^2 + xy + y^2)$ cannot be a square when $x$ and
$y$ are rational.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Fermat}Fermat.)\quad}

45. Prove that the equation $x^2 - 2 = m(y^2 + 2)$ has no solution in positive
integers $m$, $x$, $y$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Fermat}Fermat.)\quad}

46.* Prove that the equation $2x^2 - 1 = (2y^2 - 1)^2$ has the unique solution
$x = 5$, $y = 2$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Fermat}Fermat;~\index{Pepin}Pepin,~1884.)\quad}

% [File: 125.png Page: 115]
47. Obtain solutions of the system $x + y = u^2$, $x^2 + y^2 = v^4$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Euler.)\quad}

48. Find six fifth-power numbers whose sum is a fifth power.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Martin}Martin,~1898.)\quad}

49. Find a sum of sixth-power numbers whose sum is a sixth power.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Martin}Martin,~1912.)\quad}

50.\dag\ Find a set of powers of higher degree than the sixth such that their
sum is a power of the same degree. (Compare papers referred to in \S~20.)

51. Solve each of the Diophantine equations $xy = z(x + y)$, $z^2(x^2 + y^2) = (xy)^2$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Mathesis~(4)~3:~119.)\quad}

52. Obtain a solution of the Diophantine system $x^2 + y^2 + z^2 = u^2 $, $x^3 + y^3 + z^3 = v^3$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Martin}Martin~and~Davis,~1898.)\quad}\index{Davis}

53. Apply the following identities to the solution of Diophantine equations:
\[
(a^2 - b^2)^8 + (a^2 + b^2)^8 + (2ab)^8 = 2(a^8 + 14a^4 b^4 + b^8)^2, 
\]
\[
x^8 + y^8 + (x^2 \pm y^2)^4 = 2(x^4 \pm x^2y^2 + y^4)^2.
\]
\null\hfill {(\index{Barisien}Barisien,\index{Visschers}~Visschers,~1911.)\quad}

54.* Obtain the general solution of the equation
\[
\phantom{\index{Hurwitz}Hurwitz,\ 1907.)\quad}
  {x_1}^2 + {x_2}^2 + \dots + {x_n}^2 = x x_1 x_2 \dots x_n.
\tag*{(Hurwitz,~1907.)\quad}
\]

55. Show how to obtain solutions of the system
\[
%\phantom{(Legendre.)\quad}
   x^2 +  y^2 + 2z^2 = \square, \quad
   x^2 + 2y^2 +  z^2 = \square, \quad
  2x^2 +  y^2 +  z^2 = \square.
\tag*{(\index{Legendre}Legendre.)\quad}
\]

56. Show how to obtain solutions of the system $x^2 + y^2 - z^2 = \square$,
$x^2 - y^2 + z^2 = \square$,  $-x^2 + y^2 + z^2 = \square$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Legendre}Legendre.)\quad}

57.* Develop a theory of the Diophantine system
\begin{align*}
  a &= x^2 + y^2 + u^2 + v^2,  \\
  b &= x + y + u + v.
\end{align*}
Apply the results to several problems in the theory of numbers.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (Cauchy,~\index{Legendre}Legendre.)\quad}\index{Cauchy}

58.* Investigate the solutions of the equation
\[
x^3 + (x + r)^3 + (x + 2r)^3 + \dots + [x + (n - 1)r]^3 = y^3.
\]
\null\hfill {(\index{Genocchi}Genocchi,~1865.)\quad}

59. Determine systems of four numbers such that the sum of every two in
a system shall be a cube.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Fermat}Fermat.)\quad}

60.* Develop a general theory of the equation $(n + 4)x^2 - ny^2 = 4$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Realis}Realis,~1883.)\quad}

61.* Determine properties of the integers $a$, $b$, $c$, $d$ such that the equation
$ax^2 + by^2 + cz^2 + du^2 = 0$ shall have integral solutions.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Meyer}Meyer,~1884.)\quad}

62.* Show how to write the product of two sums of eight squares as a sum
of eight squares.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Thomson}Thomson, 1877.)\quad}

63.\dag\ Develop the theory of the Diophantine equation
\[
\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n} = 1,
\]
where $x_1$, $x_2$, $\dots$, $x_n$ are restricted to be positive integers. In particular
show that the maximum value of an $x$ which can occur in a solution is $u_n$ where
$u_{k+1} = u_k(u_k + 1)$ and $u_1 = 1$ and find the other integers which go to make up a
% [File: 126.png Page: 116]
solution containing this number $u_n$. (Problem and result communicated to the
author by O.~D. \index{Kellogg}Kellogg.)

64.\dag\ Develop a theory of the system
\begin{align*}
   x_1 + x_2 + \dots + x_n 
&= y_1 + y_2 + \dots + y_n,
\\
   {x_1}^2 + {x_2}^2 + \dots + {x_n}^2 
&= {y_1}^2 + {y_2}^2 + \dots + {y_n}^2.
\end{align*}
Generalize the results by adding further similar equations with exponents
3, 4, $\dots$
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Longchamps}Longchamps,~1889;~\index{Frolov}Frolov,~1889;~\index{Aubry}Aubry,~1914.)\quad}

65.\dag\ Observe that
\[
4 \frac{x^3 + y^3}{x + y}  = (x + y)^2 + 3(x - y)^2
\]
and thence show how to extend the set of numbers of the form $(x^3 + y^3)/(x + y)$
by generalization of variables so as to form a domain with respect to multiplication.
Treat likewise the forms $(x^5 + y^5)/(x + y)$, $(x^7 + y^7)/(x + y)$. Generalize
to the form $(x^p + y^p)/(x + y)$, where $p$ is any odd prime. (Compare
\index{Bachmann}Bachmann's \textit{Zahlentheorie}, III, p.~206.)

66.\dag\ Apply the results obtained in Exercise 65 to the solution of problems
in Diophantine analysis.

67.\dag\ Develop the theory of the equation $x^4 + y^4 = mz^2$ for given values of $m$.
(See examples of solutions in \textit{Interm\'ed.\ d.~Math.}, Vol.~XVIII, p.~45.)

68.\dag\ Develop the theory of the equation $x^n + y^n + z^n = u^n + v^n$ for various
values of the positive integral exponent $n$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Gerardin@G\'erardin}G\'erardin,~1910.)\quad}

69.\dag\ Equations of the form
\begin{equation}
\tag{1}
x^m = y^n + c,
\end{equation}
where $c$ is a given number, have been investigated by several writers. In particular,
the case $c = 1$ has been treated in several papers, the only known solution
for the latter case (in which $m$ and $n$ are greater than unity) being $x = 3$,
$m = 2$, $y = 2$, $n = 3$. Investigate the general theory of Eq.~(1), summarizing the
results in the literature and adding to them. In particular, determine whether
other consecutive integers than 8 and 9 can be perfect powers. 
(See Proc.\ Lond.\ Math.\ Soc.\ (2) 13 (1914): 60--80.)

70.\dag\ Determine whether the sum either of $n$ $n$th powers or of $n - 1$ $n$th powers
can itself be an $n$th power when $n$ is greater than 3.

71.* The equation
\[
q^r F\left(\frac{p}{q}\right) = c,
\]
in which $F(x)$ denotes an irreducible polynomial in $x$ of degree $r$ $(r > 2)$ with integral
coefficients and $c$ is an integer, has only a finite number of solutions in integers $p$
and $q$.
\hfil\penalty0\hfilneg\null\penalty5000\hfill\mbox{\qquad (\index{Thue}Thue,~1908.)\quad}
\par\end{small} 

% aliases for index
\index{Descent, Infinite|see{Infinite Descent}}
\index{Pythagorean Triangles|see{Triangles}}
\index{Rational Triangles|see{Triangles}}

% [File: 127.png Page: 117]
%% ORIGINAL INDEX
%% Abel, 90
%% Abelian Formulæ, 87-90
%% Aubry, 83, 84, 113, 116
%% Avillez, 52
%% Bachmann, 116
%% Barbette, 85
%% Barisien, 115
%% Bernstein, 102
%% Binet, 65
%% Biquadratic Equations, 44-48, 74-84
%% Carlini, 102
%% Carmichael, 30, 39, 42, 83, 88, 91, 114
%% Catalan, 52
%% Cauchy, 5, 100, 115
%% Cubic Equations, 55-73
%% Cunningham, 114
%% Davis, 52, 115
%% Delannoy, 72
%% Desboves, 112, 114
%% Descent, Infinite; see Infinite Descent
%% Development of Method, 6-8
%% Dickson, 99
%% Diophantine Equation, Definition of, 1
%% Diophantine System, Definition of, 1
%% Diophantus, 4, 5, 9, 104, 106, 107, 108, 111, 112, 113
%% Dirichlet, 5
%% Domain, Multiplicative, 24-54
%% Double Equations, 78, 107
%% Eisenstein, 5
%% Equation of Pell, 26-34
%% Equations, Double, 78, 107
%% Equations, Functional, 104-111
%% Equations, Triple, 107
%% Equations of Fourth Degree, 44-48, 74-84
%% Equations of Higher Degree, 85-103
%% Equations of Second Degree, 1-44
%% Equations of Third Degree, 55-73
%% Euclid, 9
%% Euler, 5, 22, 65, 68, 80, 82, 112
%% Evans, 113
%% Fauquembergue, 83
%% Fermat, 5, 6, 7, 9, 14, 39, 60, 74, 77, 78, 86, 88, 104, 106, 107, 108, 114, 115
%% Fermat Problem, 86-103
%% Fermat's Last Theorem, 86
%% Fleck, 103
%% Frolov, 116
%% Fujiwara, 65
%% Functional Equations, Method of, 104-111
%% Furtwängler, 101, 102
%% Gauss, 5
%% Genocchi, 115
%% Gérardin, 116
%% Gerono, 72, 112
%% Haentzschel, 62
%% Hart, 112
%% Hayashi, 103
%% Hecke, 102
%% Hillyer, 23
%% Historical Remarks, 4-8, 9
% [File: 128.png Page: 118]
%% Holm, 114
%% Holmboe, 90
%% Hurwitz, 102, 115
%% Infinite Descent, Method of, 14, 18-22
%% Integral Solution, 2
%% Jacobi, 5
%% Jonquières, 112
%% Kapferer, 103
%% Kellogg, 115, 116
%% Kummer, 100, 101
%% Lagrange, 50
%% Lebesgue, 52
%% Legendre, 50, 73, 90, 99, 112, 115
%% Lehmer, 13
%% Lind, 102
%% Liouville, 102
%% Longchamps, 116
%% Lucas, 6, 52, 53, 112
%% Maillet, 99, 103, 114
%% Martin, 52, 81, 85, 113, 114, 115
%% Mathews, 5
%% Method, Development of, 6-8
%% Method of Functional Equations, 104-111
%% Method of Infinite Descent, 14, 18-22
%% Method of Multiplicative Domain, 24-54
%% Meyer, 115
%% Meyl, 112
%% Miot, 113
%% Mirimanoff, 99, 101, 102
%% Moret-Blanc, 83
%% Moureaux, 52
%% Multiplicative Domain, 24-54
%% Paraira, 84
%% Pell Equation, 26-34
%% Pepin, 53, 80, 83, 112, 114
%% Pietrocola, 83
%% Plato, 9
%% Primitive Solution, 2
%% Pythagoras, 8, 9
%% Pythagorean Triangles; see Triangles
%% Quadratic Equations, 1-44
%% Rational Solution, 2
%% Rational Triangles; see Triangles
%% Realis, 44, 73, 115
%% Schaewen, 59
%% Schwering, 65
%% Smith, 100
%% Solution of Diophantine Equation, 2
%% Solution, Integral, 2
%% Solution, Primitive, 2
%% Solution, Rational, 2
%% Störmer, 114
%% Swinden, 113
%% Thomson, 115
%% Thue, 53, 116
%% Triangles, Numerical, 8
%% Triangles, Primitive, 8
%% Triangles, Pythagorean, 8, 9-11, 17, 21, 22, 33, 77, 80, 103, 112, 113
%% Triangles, Rational, 8-13
%% Triple Equation, 107
%% Valroff, 112
%% Vandiver, 101, 102
%% Visschers, 115
%% Wallis, 114
%% Welmin, 113, 114
%% Werebrüssov, 53, 72
%% Whitford, 33
%% Wieferich, 101

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\begin{center}\textsc{Typographical Errors corrected in Project Gutenberg edition}\end{center}

p. \pageref{anlysis} ``the body of doctrine in
Diophantine \underline{anlysis}'' in original, amended to ``\underline{analysis}''

p. \pageref{missingbar} the second equation under ``If we multiply the resulting values'', $y$
on lhs amended to $\bar{y}$

p. \pageref{intergal} ``\S 8. Obtain \underline{intergal} solutions'' in original, amended to ``\underline{integral}''

p. \pageref{extrabrac} ``less than $ (\frac{1}{2}(p - 1)$'' amended to ``less than $ \frac{1}{2}(p - 1)$''

\newpage
\small
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