290 HISTORY OF THE THEORY OF NUMBERS. [CHAP, vm 



C. A. W. Berkhan 41 decomposed the integers < 360 into four rational 

 or integral squares, and into two or three squares if possible. 



G. L. Dirichlet 42 gave a simplification of Jacobi's 30 proof. According 

 as a factor a! of p' = a'q' has the form 4m + 1 or 4m + 3, set 5' = + 1 or 

 1. Then the number of positive solutions of 2p f w 2 + x 2 is 25'. 

 Hence each couple p', p" furnishes 25' -25" = 2r? solutions of (11), where 

 77 = -f 1 or 1 according as a' a" is or is not divisible by 4. Thus 

 H = 277, obtained by varying also p', p", so that there is a term 77 for each 

 set of odd solutions a', a" of 

 (13) a'q' + a"q" = 2p. 



Let 77' be a term obtained when a' = a", 77" one when a' > a". Then 

 M = 277' + 2277". From one set of odd solutions of (13), we obtain the 

 new odd solutions 



A' = q"(x + 1) + q'(x + 2), 



A" = q"x + q'(x + 1), 



Q' = - a 'x + a"(x + 1) = a" - (a r - a"}x, 



Q" = a'(x + 1) - - a"(x + 2) = (a' - a"}(x + 1) - a". 



Let a' > a". In order that Q' and Q" be positive, (a' a"}x must be the 

 least multiple of a' a" less than a". Then x is uniquely determined 

 and A' > A" > 0. If we repeat the process, starting with A', Q', A", Q", 

 we obtain merely the initial set a', q', a", q", since the preceding equations 

 hold after the interchange of a' with A', q' with Q', etc. Since 



a' - a" = Q' + Q", 



two such sets of solutions give values of 77" differing hi sign. Indeed, one 

 and but one of the even numbers a' a" and q' + q" is divisible by 4, 

 since a' = a", q' = T q" (mod 4) contradicts (13). Hence 277" = 0. 

 Thus = 277', with each 77' = + 1, so that M = N[a'(q f + ?")] = ff(p), 

 as above. Cf. Pepin. 72 



J. J. Sylvester 43 employed the lemma that, if 3M = p 2 + q 2 + r 2 + s 2 , 

 M is a sum of four squares. We may assume that p is divisible by 3 and, 

 by a proper choice of the signs of q, r, s, take q = r = s (mod 3). Then 

 M is the sum of the squares of the integers 



%(q + r + s), f (p + r - s), %(p - q + s), \(p + q - r). 



For N = 1 (mod 4), the function 3 2x+1 N - 2 of x is not rationally de- 

 composable and has no constant divisor; it is assumed to represent a prime 

 T for some integer x. Since T = 1 (mod 4), T is the sum of two squares. 

 Hence T + 2 = 3 2x+ W is the sum of four squares. The same is true of N 

 by the lemma. 



For N = 3 (mod 4), 3 2x N 2 is employed similarly. For N even, it 

 suffices to treat N = 2 (mod 4), by use of 3 X ./V 1, since the theorem is 

 true for 4N if true for N. 



41 Lehrbuch der Unbeatimmten Analytik, Halle, 2, 1856, 286. 



42 Jour, de Math., (2), 1, 1856, 210-214; Werke, 2, 1897, 201-8. 

 Quar. Jour. Math., 1, 1857, 196-7; Coll. Math. Papers, 2, 1908, 101-2. 



