From hlm@math.lsa.umich.eduMon Aug 26 22:18:30 1996
Date: Mon, 26 Aug 1996 13:18:11 -0400
From: Hugh Montgomery <hlm@math.lsa.umich.edu>
To: NMBRTHRY@listserv.nodak.edu
Subject: small prime quadratic residues

On August 24, Wolfgang Ruppert asked the following question:

If D is a fundamental quadratic discriminant, is there a prime p such
that (D/p) = 1 and |D|^{1/2} < p < (1+\epsilon)|D|^{1/2}, for all
D with |D| > N(\epsilon).

It may be that a proof of this is out of reach at the present, but
one may make the following observations:

(1) It can be shown that if L(s, (D/.)) has all its non-trivial zeros
    on the line sigma = 1/2, then there is a prime p such that (D/p) = 1
    and x < p < (1+epsilon)x, provided that x > C(epsilon)(log D)^2.

(2) A. I. Vinogradov and Ju. V. Linnik used Burgess's estimates for character
    sums and Siegel's lower bound for L(1, chi) to show that there is a
    prime p with (D/p) = 1 and p < |D|^{1/4+epsilon} for all D with
    |D| > N(epsilon).  It seems, however, that their method does not
    yield primes in short intervals, as asked for in the proposed problem.
    The method has been used by P. D. T. A. Elliott and by D. Wolke to
    obtain similar results.

(3) The result mentioned above is ineffective, but two weeks ago (at the
    Riemann Hypothesis Symposium in Seattle), D. R. Heath-Brown (of Oxford
    University) said that he thought he could prove effectively that
    there is a prime p with (D/p) = 1 and p < |D|^{1+epsilon}.  Maybe
    his proof only works for D < 0; you'll have to ask him.

(4) I guess that one could use Heath-Brown's sharp form of Linnik's theorem
    to give an unconditional variant of (1), but only for x > |D|^{5/2+epsilon}.


     I hope this helps.

                        Hugh L. Montgomery
                        Department of Mathematics
                        University of Michigan
                        Ann Arbor, MI 48109--1109
                        hlm@umich.edu