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Daniele Tampieri
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counting Counting squares modulo p$p$ that are also prime in an interval

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

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<p that are prime? Via 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=sqrt(p)/2. Thank you very much!

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 $N<p$ 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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pali
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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<p that are prime? Via 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=sqrt(p)/2. Thank you very much!