CHAP. VI] SUM OF TWO SQUARES. 237 



in the spirit of their method. From this point of view, Jacobi proved that, 

 if a given odd number N is the sum of the squares of two integers b and c 

 having no common factor 4n 1, every prime factor p of N is of the form 

 4k + 1. When b, c are relatively prime, the proof shows that p and hence 

 every divisor of TV is a sum of two rational squares. The fact that every 

 divisor is a sum of two integral squares is established by an argument 

 perhaps not known to Diophantus and not necessary for his assertion. 



F. Arndt, 58 using continued fractions as had Legendre 39 for the case of a 

 prune, proved that the hih power of a prime 4n + 1 is a GO in 2 h ~ l ways. 



J. B. Kulik 59 gave the representation as a El of each prime ^ 10529. 



V. A. Lebesgue 60 noted that x z + y 2 = & + t 2 becomes pq = rs if we set 



2x = p + q + r-s, 2y = p + q r + s, 

 2z = p q + r + s, 2t = p q r s. 



C. Hermite 61 developed a/p into a continued fraction, where a 2 = 1 

 (mod p), and employed two consecutive convergents m/n, m'/n', such that 

 n < Vp, n r > Vp. Then 



- = --- 1 --- -., e < 1: (na rap) 2 = e 2 p 2 /n' 2 < p. 

 p n nn' 



Since (na rap) 2 + n 2 is a multiple of p and is < 2p, it equals p. 



J. A. Serret 62 employed q 2 = 1 (mod p), q < p, and developed p/q 

 into a continued fraction so that the number of quotients is even (replacing 

 if necessary the last quotient QbyQ 1 + 1). In the series of quotients 

 the terms equidistant from the extremes are shown to be equal. Let mfn 

 be the convergent which includes the quotients of the first half of the series, 

 and ra /n the preceding convergent. Then the continued fraction whose 

 quotients are those of the second half of the series has the value ra/ra . 

 If co is the common middle quotient, the convergent following m/n equals 



raco -f ra 



n 

 Replacing co by ra/ra , we get the entire continued fraction. Thus 



p ra 2 + m z 2 



p = m? + ml. 



q mn + 



L. Wantzel 63 stated that the use of complex integers affords the simplest 

 proof that every prime divisor of a C3 is a CH . He proved that no complex 

 prime a + bi divides a product without dividing one factor [due to Gauss]]. 



68 Jour, fur Math., 31, 1846, 343-358; extract of Diss., Sundiae, 1845. Arndt, 124 Ch. XII. 



69 Tafeln der Quadrat- und Kubik-Zahlen aller Zahlen bis Hundert Tausend . . . , Leipzig, 



1848, Table 2. 



60 Nouv. Ann. Math., 7, 1848, 37. 

 81 Jour, de Math., 13, 1848, 15; Oeuvres, I, 264; Nouv. Ann. Math., 12, 1853, 45; Socidte' 



philomatique de Paris, 1848, 13-14. 



62 Algebre Super., ed. 1, 1849, 331; Jour, de math., (1), 13, 1848, 12-14; Nouv. Ann. Math., 12, 



1853, 12; Societe philomatique de Paris, 1848, 12-13, 



63 Socie'te' philomatique de Paris, 1848, 19-22. 



