232 HISTORY OF THE THEORY OF NUMBERS. [CHAP, vi 



to 6. Since A is prime to a and &, we may set p = mA fa,q = nA gb. 

 Thus / 2 a 2 + g 2 b 2 is divisible by A. The error in the conclusion that g = f 

 was pointed out in a marginal note by means of the case p = 17, q = 6, 

 a = 7, b = 4. However, (0 2 / 2 )6 2 and hence # 2 / 2 is divisible by A. 

 If we assume that A is a prime, we see that g / is divisible by A, so that 

 g = vA dz/6. Hence 



P/A = (/ =fc wo "&) 2 + ( ?a =F m&) 2 . 



Thus Euler's proof of the first step in the Lemma is valid if A is a prime. 

 He gave (p. 570) another proof by setting p = ma nb, q = na + mb + s. 

 Then P = A(m 2 + n 2 ) + sk, k = 2(na + w&) -f s. Since A is a prime, 

 either s = tA or k = 2A. In either case, 



PI A = (m + bt) 2 + (n + at) 2 . 



J. L. Lagrange 31 deduced from Wilson's theorem the fact that the 

 prime 4n + 1 divides (1-2 2ri) 2 + 1. He 32 proved the Lemma in con- 

 nection with the general problem to find the form of the divisors of numbers 

 represented by Bt 2 + Ctu + Du 2 . He 33 deduced Girard's theorem from 

 the fact that a prime p of the form 4n + 1 divides x 2n + 1 for 2n integral 

 values of x numerically < %p (it being a factor of x p ~ l 1). 



Beguelin 75 of Ch. I failed in his attempt to prove Girard's theorem. 



P. S. Laplace 34 remarked that every prime 4n + 1 will be a El if proved 

 to divide a El, in view of Lagrange. 32 But 4n + 1 divides (a 271 + 1) (a 2n 1) 

 and not the last factor for every a, since 



(2ri)l = {(2n + 1) 2 - 1} - 2n{(2n) 2n -!} + -, 



by the formula for the 2nth order of differences of x 2n 1 for x = 1 

 [Euler 25 ]. 



J. Leslie 340 solved x 2 + y 2 = a 2 + b 2 by setting 



x + a = (b y)m, x a = (b + y)/tn- 



C. F. Kausler 35 gave tentative numerical methods of expressing a given 

 number A as a sum of 2, 3 or 4 squares. 



Let A = 4C + 1 = (2P) 2 + (2Q + I) 2 . Then C = P 2 + Q(Q + 1). 

 If C = 2D + 1, then P = 2T + 1 and D - Q(Q + 1) = 2T(T -f 1). 

 Hence we subtract from D in turn the halves of the pronic numbers 

 Q(Q + 1), given by a table (extending to Q = 225), and note if any re- 

 mainder is double a pronic number. If C = 2D, then P = 2T and we 

 use D - Q(Q + 1) = 2T 2 . 



A number A = 4B + 2 can only be the sum of two odd squares, whence 



B = P(P + 1) + Q(Q + 1). 

 Thus B = 2(7. Set P = Q + R. Solving the quadratic for Q, we see 



Nouv. M<Sm. Acad. Berlin, ann<e 1771 (1773), 125; Oeuvres, III, 431. 



* 2 Ibid., ann<5e 1773, 275; Oeuvres, III, 707. 



"Ibid., ann6e 1775, 351; Oeuvres, III, 789-790. 



M Th6orie abrdgde des nombres premiers, 1776, p. 24. 



* Trans. Roy. Soc. Edinburgh, 2, 1790, 193. 



Nova Acta Acad. Petrop., 11, ad annum 1793 (1798), Histoire, 125-156. 



