CHAP. XIX] SQUARES, SUM OF ANY Two LESS THIRD A SQUARE. 507 

 by d/2, we have 



THREE SQUARES, SUM OF ANY TWO LESS THIRD A SQUARE. 

 L. Euler 46 gave four methods to solve 

 (1) ?/+z 2 -o; 2 = p 2 , x 2 +z*-y* = q z , x 2 +y--z* = r 2 . 



(i) Let s = rc 2 H-?/ 2 -{-2 2 . Since s = p~-\-2x~, etc., s must be expressible in 

 three ways in the form a 2 +2& 2 , whence s must have at least three prime 

 factors of that form. Take m = ac2bd, n = bc^ad, u = mf2ng, v = nf^Pmg. 

 Then 



Take u 2 -\-2v- = s. By using the four combinations of signs, we get four sets 

 of values of w, v. As we need only three sets, omit that given by both 

 lower signs. Set 



P, q=f(ac+2bd)2g(bc-ad), r=f(ac-2bd)+2g(bc+ad), 

 z> y=f(bc ad) = Fg(ac-}-2bd), z=f(bc+ad)g(ac 2bd), 



where the upper signs give p and x. Compute x 2 -\-y 2 -\-z~ and compare 

 with the earlier expression u 2 -}-2v 2 for s; we get 



222222 C= -(bc+ad)(ac-2bd). 



Taking F = 0, we get c:d = 2b a:b a or a 2b : 6+a, and also 

 / : gr= G : 2(7. The same solution results also from G = 0. 

 (ii) By (3), / : g = - (C F) : F, where 



Let Q be the square of c 2 4??icd+2d 2 . Then c : d = 2m 2 +l : 2m. 

 (iii) Use p, q, x, y given by (2), but take 



r=f(ac-2!3d}+2g(f3c+ad), z=f({3c+ad) -g(ac-2pd), 



where a, ft are such that a 2 +2jS 2 = a 2 +26 2 . Hence we now get new values 

 for F, G, C in (3). For F = 0, we get 



c : d= -a-2b : (3+a or -a+2b : p a. 



He deduced the following simple solution of the problem: Start with any 

 two integers m and n, m odd, and set 



s = m 2 +2n 2 , t = m 2 2n 2 , u = 2mn, 

 or take s, t, u such that s 2 = 2 +2w 2 ; we have the solution 



, q-stp+4t(s u)a, 

 where p = 3s+4w, o- = s-\-2u. 



46 Posth. paper, Comm. Arith., II, 603-16; Opera postuma, 1, 1862, 105-118. French transl. 

 in Sphinx-Oedipe, 1906-7, 163-83. 



