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For prime numbers $p$, does the Legendre symbol $(\frac {5}{p})$

depend only on the congruence class of $p$ modulo $5$?

So I know that $(\frac {5}{11}) = 5^5$mod $11$.

I don't think it does, and I know that I need to use Quadratic Reciprocity and the fact that $5$ is not congruent to $11$ (mod $4)$ to show it but I'm not sure how.

How should I approach this?

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    $\begingroup$ Yes, by Reciprocity and the fact $5$ is of the form $4k+1$. $\endgroup$ – André Nicolas Oct 25 '15 at 21:08
  • $\begingroup$ Why does being of the form $4k +1$ mean that it only depends on the congruence class of $p$ modulo $5$ ? $\endgroup$ – user274933 Oct 25 '15 at 21:10
  • $\begingroup$ Because if at least one of $q$ or $p$ is congruent to $1$ mod $4$, then $(q/p)=(p/q)$. $\endgroup$ – André Nicolas Oct 25 '15 at 21:13
  • $\begingroup$ I know that to be true, but I don't see how that implies that depends on the congruence class of $p$ modulo $5$. $\endgroup$ – user274933 Oct 25 '15 at 21:18
  • $\begingroup$ Let $q=5$, or any prime, and let $a$ be not divisible by $q$. Then by definition $a$ is a QR modulo $q$ if and only if there is an $x$ such that $x^2\equiv a\pmod{q}$. Now let $b\equiv a\pmod{q}$, There is an $x$ such that $x^2\equiv b\pmod{q}$ if and only if there is an $x$ such that $x^2\equiv a\pmod{q}$, for we use the same $x$. $\endgroup$ – André Nicolas Oct 25 '15 at 21:28
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As $5\not\equiv 4\pmod 4$, we have (for odd $p$) $$(\frac {5}{p})=(\frac {p}{5})=(\frac {p\bmod 5}{5})$$ which shows that the result depends only on $p\bmod 5$ for odd $p$. In order for $(\frac {5}{p})$ to depend only on $p\bmod 5$ for all primes $p$, however, we need to check additionally that e.g. $(\frac {5}{2})=(\frac {5}{7})$. Does it?

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  • $\begingroup$ Why does $(p/5)$ =$( pmod(5) / 5)$? $\endgroup$ – user274933 Oct 25 '15 at 21:15
  • $\begingroup$ By Euler's criteria, I get that it depends on $p^2$ mod 5 so I don't quite see why it's $p$mod$5$ $\endgroup$ – user274933 Oct 25 '15 at 21:17
  • $\begingroup$ I would take the point of view that the Legendre symbol $(a/p)$ is only defined for odd primes. $\endgroup$ – André Nicolas Oct 25 '15 at 21:24

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