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You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this eBook or online at www.gutenberg.net Title: Diophantine Analysis Author: Robert Carmichael Release Date: December 9, 2006 [EBook #20073] Language: English Character set encoding: TeX *** START OF THIS PROJECT GUTENBERG EBOOK DIOPHANTINE ANALYSIS *** Produced by Joshua Hutchinson, Keith Edkins and the Online Distributed Proofreading Team at http://www.pgdp.net (This file was produced from images from the Cornell University Library: Historical Mathematics Monographs collection.) \end{verbatim} \normalsize \pagestyle {empty} \newpage % [File: 001.png] \begin{center} \texttt{Production Note} \end{center} \noindent \texttt{Cornell University Library produced this volume to replace the \\ irreparably deteriorated original. It was scanned using Xerox \\ software and equipment at 600 dots per inch resolution and com- \\ pressed prior to storage using CCITT Group 4 compression. 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CARMICHAEL, \textsc{Assistant Professor Of Mathematics In The University Of Illinois} \rule{0pt}{72pt} FIRST EDITION \rule{0pt}{12pt} {\small FIRST THOUSAND} \rule{0pt}{72pt} NEW YORK \textsc{JOHN WILEY \& SONS, Inc.} \textsc{London: CHAPMAN \& HALL, Limited} 1915 \end{center} % [File: 006.png] \newpage \begin{center} \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.} \end{center} % [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 \begin{center} % [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 Arithmeticę}, 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 formulę 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 formulę 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 formulę 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 $us$, 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, $\sigma_1 1$. 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 formulę 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 formulę 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 formulę 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 formulę (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 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 formulę|(}We shall first derive the so-called Abelian formulę. 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 formulę (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 formulę 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 formulę|)} 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 formulę (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 Formulę, 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 %% Jonquičres, 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 % create TOC entry for Index for the NEXT page from the current \newcounter{nextpage} \setcounter{nextpage}{\thepage} \addtocounter{nextpage}{1} \addtocontents{toc}{\protect\contentsline {section}{\numberline {}\textsc {INDEX}}{\thenextpage}} \printindex \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 \pagestyle{empty} \pagenumbering{gobble} \begin{verbatim} End of Project Gutenberg's Diophantine Analysis, by Robert Carmichael *** END OF THIS PROJECT GUTENBERG EBOOK DIOPHANTINE ANALYSIS *** *** This file should be named 20073-t.tex or 20073-t.zip *** *** or 20073-pdf.pdf or 20073-pdf.pdf *** This and all associated files of various formats will be found in: http://www.gutenberg.org/2/0/0/7/20073/ Produced by Joshua Hutchinson, Keith Edkins and the Online Distributed Proofreading Team at http://www.pgdp.net (This file was produced from images from the Cornell University Library: Historical Mathematics Monographs collection.) Updated editions will replace the previous one--the old editions will be renamed. 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input line 3. %% %% ("C:\Program Files\MiKTeX 2.5\tex\latex\base\omscmr.fd" %% File: omscmr.fd 1999/05/25 v2.5h Standard LaTeX font definitions %% ) %% LaTeX Font Info: Font shape `OMS/cmr/m/n' in size <10> not available %% (Font) Font shape `OMS/cmsy/m/n' tried instead on input line 3. %% [5 %% %% ]) %% \tf@toc=\write4 %% [6] [7] %% Chapter I. %% LaTeX Font Info: Font shape `OMS/cmr/m/sc' in size <10> not available %% (Font) Font shape `OMS/cmsy/m/n' tried instead on input line 473. %% [1 %% %% %% ] [2] [3] [4] [5] [6] [7] [8] [9] %% LaTeX Font Info: Font shape `OMS/cmr/m/n' in size <9> not available %% (Font) Font shape `OMS/cmsy/m/n' tried instead on input line 1063. %% %% [10] [11] [12] [13] [14] [15] [16] [17] %% [18] %% Chapter II. %% [19 %% %% ] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] %% [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] %% Chapter III. %% [45 %% %% ] [46] [47] [48] [49] [50] [51] [52] [53] [54] [55] [56] [57] [58] [59] %% [60] 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