# Counting squares modulo $p$ that are also prime in an interval

What would be the best lower bound for the number of squares modulo $$p$$ in an interval $$[1,N]$$ with $$N that are prime?
Via the Burgess bound, I can find a lower bound for the number of squares modulo $$p$$ in $$[1,N]$$, but I would need a bound for the number of squares that are also prime. Since the size of $$N$$ matters, in my particular case I have $$N=\frac{\sqrt{p}}{2}.$$ Thank you very much!

• This seems like an odd phrasing — wouldn't it properly be 'the number of primes $\leq N$ that are squares modulo some $p\gt n$'? Jul 15 at 17:34

## 1 Answer

Here is the paper by P. Pollack on the distribution of non-residues and residues. Theorem 1.3 states that for any $$\varepsilon>0$$, $$A<\infty$$ and large enough $$m$$ there are at least $$(\ln m)^A$$ prime quadratic residues $$\mod m$$ below $$m^{1/4+\varepsilon}$$. I think, one can do a lot better under the assumption of some conjectures regarding distribution of zeros of $$L$$-functions. For example, if GRH is true, then there are $$\frac{x}{2\ln x}+O(\sqrt{x}\ln^2 p)$$ prime quadratic residues below $$x.

One can also probably derive some bounds similar to Pollack's result, but with $$(\ln m)^A$$ replaced with something like $$\exp(c\ln m\ln\ln\ln m/\ln\ln m),$$ if there are no Siegel zeros.

EDIT: The paper https://www.ams.org/journals/proc/2020-148-09/S0002-9939-2020-15011-3/, mentioned in the comment below, gives at least $$Cp^{9/160}$$ prime residues below $$\sqrt{p}/2$$.

EDIT 2: $$9/160$$ can actually be replaced by $$1/4-o(1)$$, see this article, Theorem 1.1.

• Pollack's lower bound of $(\log m)^A$ was improved for prime moduli to a small power of $m$ by Benli. See ams.org/journals/proc/2020-148-09/S0002-9939-2020-15011-3 Jun 1 at 22:07
• @so-called friend Don, thank you! Included your comment. The power turns out to be 9/160 in this case Jun 1 at 22:56