422 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xm 



If (5) be solvable, also fx 2 -}-gy 2 = hiZ 2 is solvable when hi is a certain 

 integer <h. For, k=t 2 -}~fg is divisible by h for some integer t<hj2; call 

 the quotient hi. Then 



so that fX 2 +gY 2 = hiZ 2 is solvable. It is not shown that hi<h. In case 

 a set of decreasing values h, hi, h z , eventually contains / or g, we can 

 determine x, y (p. 569, 62). 



A. M. Legendre 114 proved [Legendre 87 of Ch. XII] that, if each of the 

 positive integers a, b, c has no square factor and if no two of them have a 

 common factor, then ax 2 -\-by 2 = cz 2 has integral solutions not all zero if and 

 only if there exist three integers X, /*, v such that 



d\ 2 -\-b C[j? b cv 2 a 



~T~' a ' ~b~ 



are all integers. 



Legendre 115 explained the method of Lagrange 110 to solve (2), modified 

 by use of a principle employed elsewhere by Lagrange 76 of Ch. XII. The 

 present method is essentially due to Lagrange. 1150 We may take A and B 

 positive, since otherwise 



x 2 -Ay 2 =-Bz 2 or x 2 +Ay 2 =Bz 2 (A>0, 5>0). 



In the second write Bz = z', AB = A', whence z' 2 A'y 2 = Bx 2 . The second 

 is obtained from the first by the transpositions of two terms. Consider 

 therefore x 2 By 2 = Az 2 , A>B>0, where y is prime to A and x, while A 

 and B have no square factors. Set x = ay Ay'. Then 



The first coefficient must be an integer, say A 'k 2 , where A' has no square 

 factor. By changing a by a multiple of A, we may take a between A/2 

 and A/2. Multiply the resulting equation by A 'k 2 and set kz = z', 

 A f k 2 y-ay' = x r ; we get x f2 -By' 2 = A'z' z , A'<A. If A'>B, we repeat the 

 process. Finally we get a similar equation with one coefficient unity and 

 hence easily solved. While this method is not the simplest one for solving 

 the proposed equation, it is a very luminous one. 



C. F. Gauss 116 proved by use of ternary quadratic forms the theorem of 

 Legendre 114 that, if no two of a, b, c have a common factor and if each is 

 neither zero nor divisible by a square, then ax 2 -\-by 2 -\-cz 2 = has integral 

 solutions not all zero if and only if be, ac, ab are quadratic residues of 

 a, b, c, respectively, and a, b, c are not all of the same sign. If a, b, c are 



1M M6m. Acad. Sc. Paris, 1785, 512-3; The"orie des nombres, 1798, 49; ed. 2, 1808, 41; 



ed. 3, 1830, I, 47; German transl. by Maser, I, 49. 

 116 The"oric des nombres, 1798, 36-41; ed. 2, 1808, 28-32; ed. 3, 1830, I, 33-39. (Maser, I, 



36-39.) For his remark on x 2 +by 2 = z 2 see Legendre. 39 

 1160 Addition V to Euler's Algebra, 2, 1774, 538-55; Euler's Opera Omnia, (1), I, 586-94; 



Oeuvres de Lagrange, VII, 102-14. 

 116 Disquisitiones Arith., arts. 294-8; Werke, I, 1863, 349. German transl. by Maser, pp. 



335-343. 



