280 HISTORY OF THE THEORY OF NUMBERS. [CHAP, viil 



(ii) If 7 2 + S 2 is divisible by m 2 + n 2 , the quotient t is a sum of two 

 squares. 



Let I be the g.c.d. of 7 = Ip, 5 = lq, m = Ir, n = Is. Then p 2 + q 2 is 

 divisible by r 2 + s 2 = p. Hence, by (i), t = (p 2 + q 2 )fp is a sum of two 

 squares. 



(iii) If P = p- + q 2 + r 2 + s 2 is divisible by a prime A > VP, then A 

 is a sum of four squares. 



Set P = Aa. Then a < A. A common divisor d of p, g, r, s is < A, 

 so that d 2 divides a and may be deleted from a, p 2 , , s 2 . Let therefore 

 d = 1. 



Let p be the g.c.d. of a = 6p and p 2 + g 2 = tp. Then (r 2 + s 2 )/p is an 

 integer u. By (i), t = m 2 + n 2 , u = h 2 -}- I 2 . Thus 



w = x 2 + 2/ 2 , # = ink -}- nl, y = ml nh. 

 From P = Aa follows 



Ab = t + u, AU = t 2 + x 2 + y 2 . 



Since 6 is prime to t, there exist integers a, , 6 such that 



x = at + 76, # = # + 56, | a | < !&, | /3 < |&, 



(3) 46* = to 2 + 2a7*6 + 205*6 + (y 2 + 5 2 )6 2 , & = 1 + a 2 + /3 2 . 



Hence Atf 2 is divisible by 6. Thus /c = ajb, where GI < #/2 + 1/6. Then 



At = aj 2 + 2ayt + 206* + (y 2 + 5 2 )6, 



a^At = (ai + 7 + /35) 2 + 7 2 (i& - - 2 ) + 5 2 (ai6 - /5 2 ) - 2a/?7. 

 Replacing o 16 by 1 + a 2 + /3 2 , we get 



a,At = (ai< + 7 + /35) 2 + (fly - ad) 2 + 7 2 + S 2 . 

 (3), 7 2 + 5 2 is divisible by t = m 2 + ft 2 - By the last equation and (ii), 

 2 + S 2 = t(p\ + q\], (ait 



s. 



If a = 6p is > 1, Oi < 6/2 + 1/6 < . Similarly, if ai > 1, a^A is the 

 sum of four squares, where cig < i, etc. But each a t ^ 1. Thus a certain 

 o* = 1, and a*A = A is the sum of four squares. 



(iv) Any prime which divides the sum of four or fewer squares which 

 have no common factor is itself the sum of four or fewer squares. 



If the prime A divides p 2 + q 2 + r 2 + s 2 , it divides the sum obtained by 

 replacing p by db (p mA), where m is such that =i | p mA \ < %A, 

 etc. The sum of the four new squares is < A 2 and is divisible by A. 

 Then (iii) may be applied, even if some of the four squares are zero. 



(v) If B and C are integers not divisible by the odd prime A, there 

 exist integers p and q such that p 2 Bq 2 C is divisible by A. 



Suppose that there is no integer q which makes 6 = Bq 2 + C divisible 

 by A (since otherwise we may take p = 0). For 



P = /-' + 6^- 5 + 6 2 /- 7 + 

 (p 2 - 6)P = p x -' - 1 - (6 ( ^ 1)/2 - 1). 



