374 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xii 



where T, U is a fundamental solution, i. e., one for which N(T-\-U^D) is 

 the minimum of all the N(t j s-u-^)>l. If D is real and positive, t, u are 

 both real or both pure imaginaries. Thus if the fundamental solution is 

 real, all solutions are real. But if it be imaginary, only even values of n 

 give real solutions. Since pure imaginary solutions give real solutions of 

 PDu 2 =l, the fundamental solution is imaginary or real according as 

 the latter equation has real solutions or not, and the least positive solutions 

 are T/i, U/i in the former case. 



Du Hays 121 derived, for the case 6 = 0, Euler's 65 recursion formulae be- 

 tween consecutive sets of solutions of ao; 2 +c= D, and gave the nth set. 



Chabert 122 treated ny 2 -\-py-\-q = D by equating it to n(y ^)(y /3 > ) and 

 setting (y{f)n/f=(yp f )f. Use an irrational / if /?, /3' are irrational. 

 While we cannot always get rational x, y, the process is said to be far simpler 

 than Legendre's. 



G. Eisenstein 123 proposed the problem to find a criterion to decide a 

 priori if p 2 Dg 2 = 4 is solvable in odd integers, given that D is a positive 

 integer 8n+5, i. e., if the number of improperly primitive classes of quad- 

 ratic forms of determinant D equals the number of properly primitive 

 classes of determinant D or is three times the latter number. 



F. Arndt 124 extended the work of Legendre 88 on p 2 Aq* = l, who treated 

 only the cases in which A is a prime or a product of two primes. Let p, 

 q be the least positive solutions. First, let A be odd. Let 0i be the g.c.d. 

 of p+1, q, and 2 that of p 1, q. Then 



p+l = i0ip!, p-l = 6 2 2 p 2 , 6 i e 2 = 2q, p lPz = A, l = Q0i) 2 pi-(ip 2 (podd); 



P+l = 0i(Ti, p 1 = 02<T 2 , 0!0 2 = q, ffi<T2=A, 2 = 0J O-i 0* <r 2 (p CVCn) . 



If A=4m+l, only the first system of relations holds and pi, p 2 are both 

 = 1 (mod 4) if ^0i is odd, while both are =3 (mod 4) if \Qi is even. If 

 A = 4w+3, either system may hold; if the first holds, pi= 1, p 2 = 3 (mod 4). 

 If A is an odd power of a prime 4m+l, then pi = A, P2 = l ? and 



-i=(WO'-(JW^, 



whence 1 is a quadratic residue of A, and the number k of terms in the 

 period of the continued fraction for VZ is odd. Let (VZ+/ n )/-Sn be any 

 complete quotient; then if k is odd A=B 2 ,-\-I 2 s , s = (&+l)/2. If, for 

 A = 4m+l, the number k of terms in the period is even, the denominator 

 of the middle complete quotient is odd. If A = 2 n , where n is odd and 

 >3, it is proved that 



If po, q be the least solutions of the latter, then p, q give the least solutions 

 of p 2 2 n g 2 =l. Since Po = 3, g = l when A = 8, we can find the solutions 

 step by step. Finally there is treated the case A = 2 H A', A' odd. 



121 Jour, de Math., 7, 1842, 325-30. 



122 Nouv. Ann. Math., 3, 1844, 250-3. 

 m Jour, fiir Math., 27, 1844, 86. 



124 Disquisitiones nonnullac de fractionibus continuis, Diss. Sundiae, 1845, 32 pp. Extract 

 in Jour, fur Math., 31, 1846, 343-358. 



