Abstract

We will give a short and elementary proof of the existence of infinitely many primes p such that a given positive integer a congruent 3 modulo 4 is a quadratic non-residue modulo p. Let p be an odd prime and let a be an integer with (a, p) = 1. Recall that the quadratic residue symbol (ap ) is defined to be 1 or −1 according as the congruence x ≡ a (mod p) is solvable or not. It is not difficult to see that for a given a there exist infinitely many primes p such that (ap ) = 1. Indeed, it is rather straightforward to prove that if f(X) ∈ Z[X] is any polynomial with integral coefficients of positive degree, then there exist infinitely many primes p and positive integers n such that p|f(n) [2, chap. II, sect. 1, Satz 1.2]. Note that the analytic proof of Dirichlet’s theorem on primes in arithmetic progressions in fact shows that if a is not a perfect square, then the number of p such that (ap ) = 1 has density 1 2 [3, part II, chap. VI, sect. 4, Propos. 14]. If a is not a perfect square, then there exist infinitely many primes p with (ap ) = −1 as well, but the proof is more difficult and uses deeper tools. For example, one can proceed as follows. Say a > 0 and write a = a0p1 · · · pr with p1, . . . , pr different primes. Assume first that all the pν (ν = 1, . . . , r) are odd. If p ≡ 1 (mod 4), we have ( a p ) = ( p a ) = ( p1 a ) · · · (r a ) (“Jacobi symbols”) by quadratic reciprocity. Now by Dirichlet’s prime number theorem, in conjunction with the Chinese remainder theorem, we can find infinitely many primes p such that p ≡ 1 (mod 4), p ≡ 1 (mod pν) (ν = 1, . . . , r − 1), p ≡ α0 (mod pr), where α0 is a fixed quadratic non-residue modulo pr. For those p we then have (ap ) = −1. Received March 4, 2007. 2000 Mathematics Subject Classification. 11A07, 11A041.

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