The set of odd primes p such that -5 is a quadratic residue mod p

Question

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)

Answer 1

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.

Answer 2

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.