Legendre symbol and variable denominator

Asked Oct 25 '19 at 11:44

Active Oct 25 '19 at 12:25

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Suppose $z\in\mathbb{Z}$ and let $p$ be a prime, such that $p>>z$. Can I say something what happens to Legendre symbol $(\frac{z}{p})$ if I vary $p$?

One obvious answer comes when $x^2=z$ for $x\in \mathbb{Z}$, namely $(\frac{z}{p})=1$ for $p$ big enough. Suppose now that $z$ is prime. Are sets: \begin{equation*} S_+(z)=\{\,p\;|\;(\frac{z}{p})=1\} \end{equation*} \begin{equation*} S_-(z)=\{\,p\;|\;(\frac{z}{p})=-1\} \end{equation*} of infinite cardinality?

edited Oct 25 '19 at 12:25

asked Oct 25 '19 at 11:44

Aleksander Strzelczyk

Your sets depend on a given $z$? So $S_+(z)$ and $S_-(z)$? For $z=1$ the set $S_+(1)$ is infinite, but the set $S_-(1)$ is empty. – Dietrich Burde Oct 25 '19 at 12:06
You are right of course. I have corrected it. – Aleksander Strzelczyk Oct 25 '19 at 12:27
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For $z=2$ we know that $(2/p)=(-1)^{(p-1)/2}$, so both sets are infinite. This will also happen for all $z\ge 2$ because of this duplicate. – Dietrich Burde Oct 25 '19 at 12:42
I see, that it solves the case of $S_+(z)$, but how does it help with $S_-(z)$? – Aleksander Strzelczyk Oct 25 '19 at 13:34
It is the same idea for $S_-(z)$, see Bruno's answer here. – Dietrich Burde Oct 25 '19 at 13:48
I think, I have understood it. Thank you for your time and attention. – Aleksander Strzelczyk Oct 25 '19 at 16:20

Legendre symbol and variable denominator

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