When is $2$ a quadratic residue mod $p$? 
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.
 A: The second supplementary law of quadratic reciprocity says that:
$$\biggl(\frac{2}p\biggr)=\bigl(-1\bigr)^{\tfrac{p^2-1}8}$$
Namely, $2$ is a square modulo $p$ if and only if $p\equiv\pm 1\mod 8$.
A: The other method is based on roots of unity $$\sqrt{2}e^{2i \pi /8}= 1+i$$
For $p$ odd
$$(\sqrt{2}e^{2i \pi /8})^p = \sqrt{2} 2^{(p-1)/2}e^{2i \pi /8} (e^{2i \pi /8})^{p-1}=(1+i)\left(\frac{2}{p}\right)i^{(p-1)/2}$$
$$(1+i)^p \equiv 1^p+i^p \equiv 1+i (-1)^{(p-1)/2}\equiv (1+i) i^{(p-1)/2} (-1)^{(p^2-1)/8} \bmod p $$
Thus $$ \left(\frac{2}{p}\right) \equiv (-1)^{(p^2-1)/8}  \bmod p$$
A: Let $s = \frac{p-1}{2}$, and consider the $s$ equations
$$\begin{align}
1&= (-1)(-1)  \\
2&=2(-1)^2  \\
3&= (-3)(-1)^3 \\
4&= 4 (-1)^4 \\
 & \quad\quad \ldots\\
s&= (\pm s)(-1)^s
\end{align}$$
Where the sign is always chosen to have the correct resulting sign.
Now multiply the $s$ equations together. Clearly on the left we have $s!$. On the right, we have a $2,4,6,\dots$ and some negative odd numbers. But note that $2(s) \equiv -1 \mod p$, $2(s-1) \equiv - 3 \mod p$, and so on, so that the negative numbers are the rest of the even numbers mod $p$, but disguised. So the right side contains $s! (2^s)$ (where we intuit this to mean that one two goes to each of the terms of the factorial, to represent the even numbers $\mod p$).
We only have consideration of $(-1)^{1 + 2 + \ldots + s} = (-1)^{s(s+1)/2}$ left.
Putting this all together, we get that $2^s s! \equiv s! (-1)^{s(s+1)/2} \mod p$, or upon cancelling factorials that $2^s \equiv (-1)^{s(s+1)/2}$. And $s(s+1)/2 = (p^2 - 1)/8$, so we really have $2^{(p-1)/2} \equiv (-1)^{(p^2 - 1)/8}$.
So it depends on $p \pmod 8$. [This is probably the involved manipulation of factorial proof that André alludes to].
