From hlm@math.lsa.umich.eduMon Aug 26 22:18:30 1996
Date: Mon, 26 Aug 1996 13:18:11 -0400
From: Hugh Montgomery
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