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For any odd prime p -5 is a quadratic residue mod p $\Leftrightarrow$ $(\frac{-5}{p})$=1

By multiplicaticity of the legendre symbol $(\frac{-5}{p})$=$(\frac{-1}{p})$*$(\frac{5}{p})$

What i know:

$(\frac{-1}{p})$=1, if p $\equiv$1 (mod 4)

$(\frac{-1}{p})$=-1, if p $\equiv$3 (mod 4)

$(\frac{5}{p})$=1, if p $\equiv\pm$1 (mod 5)

$(\frac{5}{p})$=-1, if p $\equiv\pm$2 (mod 5)

I also know that -5 is a quadratic recidue mod p if and only mod p if and only if p$\equiv$1, 3, 7, 9 (mod 20)

I have tried back and forth to solve this in a nice way.. I know that quadratic reciprocity can be helpful to this solution... I just cant understand how I should show it. Can anyone with the information that I have stated above help to show that: -5 is a quadratic recidue mod p if and only mod p if and only if p$\equiv$1, 3, 7, 9 (mod 20)

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  • $\begingroup$ That should be $\left(\frac{5}{p}\right)=-1,\text{ if }p\equiv \pm 2\bmod 5$, no? Actually those $\bmod 5$ congruences already use quadratic reciprocity. What you need is the Chinese Remainder theorem to join appropriate combinations of those congruences together $\bmod 20$ $\endgroup$
    – sharding4
    Aug 20, 2017 at 19:35
  • $\begingroup$ of cource $\pm2$ :) How should i use Chinese remainder theorem to combine appropiate combinations? @sharding4 $\endgroup$
    – PerkinsN
    Aug 20, 2017 at 19:40
  • $\begingroup$ You need either both $\left(\frac{5}{p}\right)$ and $\left(\frac{-1}{p}\right)$ to be $1$ or both to be $-1$. You have congruence conditions for each case, e.g. $p\equiv 1\bmod 4$ and $p\equiv 4\bmod 5$. That says $p\equiv 9\bmod 20$. $\endgroup$
    – sharding4
    Aug 20, 2017 at 19:43
  • $\begingroup$ And $(\frac{5}{p}) = (\frac{p}{5})$ essentially follows from $\sqrt{5} = \sum_{n=0}^4 \zeta_5^{n^2}$. Not sure how to adapt it for $(\frac{-1}{p})$ and $(\frac{-5}{p})$ though. $\endgroup$
    – reuns
    Aug 20, 2017 at 20:02

2 Answers 2

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Just go through every case in turn. E.g., if $p\equiv 7\pmod{20}$ then $p\equiv3\pmod 4$ and $p\equiv 2\pmod 5$ so $\left(\frac{-1}p\right)=-1$ and $\left(\frac5p\right)=-1$, etc. Don't forget to do the cases where we expect $\left(\frac{-5}p\right)=-1$, for example if $p\equiv17\pmod{20}$ then then $p\equiv1\pmod 4$ and $p\equiv 2\pmod 5$ so $\left(\frac{-1}p\right)=1$ and $\left(\frac5p\right)=-1$, etc.

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if it helps, other than $2,$ the primes are those that can be written as either $$ p = x^2 + 5 y^2 $$ or $$ p = 2 u^2 + 2uv + 3 v^2 $$ in integers, not necessarily positive.

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