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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!

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  • $\begingroup$ 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
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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 $x<p$.

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/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.

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