By Hua Loo Keng (auth.)

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For if 2 < ql < ... 1. If p > 2, then C/) ± 1, we have 0 = ( - l)t(P-1). In other words, - 1 is a quadratic residue or non-residue mod p, according to whether p == 1 or 3 (mod 4). It follows from this that the odd prime divisors of x 2 + 1 must be congruent to 1 (mod 4). 2 (Gauss's Lemma). Let p > 2, p,(n.

A",(m) be a reduced residue system, and suppose that (k, m) = 1. Then ka10 ka2, ... ,ka",(m) is also a reduced residue system. Proof Clearly we have (ka;,m) = 1, so that each kai represents a residue class coprime with m. If kai == kaj (modm), then, since (k, m) = 1, we have ai == aj (mod m). Therefore the members kai represent distinct residue classes. The theorem is proved. 2 (Euler). If(k,m) = 1, then k",(m) == 1 (modm). 1 we have n (ka ",(m) v=l Since (m,ai) = 1, it follows that na ",(m) v) == v (modm).

Then = = We take b = (n - a 2 )j21-l. Then a + 21- 2 b is a solution with respect to mod21. Therefore a solution to X2 n (mod 21) certainly exists. Let Xl be a solution, and let X2 be any solution. Then xi - x~ (Xl - X2)(Xl + X2) 0 (mod21), and since Xl - X2, Xl + X2 are both even it follows that 1(Xl - X2) ·1(Xl + X2) = 0 (mod 21- 2). But 1(Xl - X2) and 1(Xl + X2) must be of opposite parity, since otherwise their sum Xl cannot be odd. Therefore we have either Xl = X2 (mod21- l ) or Xl = - X2 (mod21- l ), and this means that X2 = ± Xl + k21- l (k = 0 or 1).