What would be the best lower bound for the number of squares modulo $p$ in an interval $[1,N]$ with $N<p$ 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!

$\begingroup$ This seems like an odd phrasing — wouldn't it properly be 'the number of primes $\leq N$ that are squares modulo some $p\gt n$'? $\endgroup$ – Steven Stadnicki Jul 15 at 17:34
Here is the paper by P. Pollack on the distribution of nonresidues 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 $x<p$.
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/202014809/S000299392020150113/, 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 $1/4o(1)$, see this article, Theorem 1.1.

3$\begingroup$ 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/202014809/S000299392020150113 $\endgroup$ – socalled friend Don Jun 1 at 22:07

2$\begingroup$ @socalled friend Don, thank you! Included your comment. The power turns out to be 9/160 in this case $\endgroup$ – Alexander Kalmynin Jun 1 at 22:56