TITLE: When is $2$ a quadratic residue mod $p$?
QUESTION [7 upvotes]: For which prime numbers $p$ is $2$ a quadratic residue modulo $p$.

I know that $2$ is a quadratic residue iff 
$$2^{\frac{p-1}{2}} =1 \;  \bmod \;(p)
$$ so 
$$2^{p-1} =1 \;  \mod \; (p).
$$
But I don't know what to do.

REPLY [4 votes]: The other method is based on roots of unity $$\sqrt{2}e^{2i \pi /8}= 1+i$$
For $p$ odd
$$(\sqrt{2}e^{2i \pi /8})^p = \sqrt{2} 2^{(p-1)/2}e^{2i \pi /8} (e^{2i \pi /8})^{p-1}=(1+i)(\frac{2}{p})i^{(p-1)/2}$$
$$(1+i)^p \equiv 1^p+i^p \equiv 1+i (-1)^{(p-1)/2}\equiv (1+i) i^{(p-1)/2} (-1)^{(p^2-1)/8} \bmod p $$
Thus $$ (\frac{2}{p}) \equiv (-1)^{(p^2-1)/8}  \bmod p$$