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)