For which primes is $-2$ a quadratic residue? For which primes is $-2$ a quadratic residue?
We are trying to find primes that have solution for $x^2 \equiv -2 \mod p.$  Using the Lagrange symbol I know that $2$ is a quadratic residue when $p \equiv 1$ or $7 \mod 8,$ but when is $-2$ a quadratic residue?
Work so far:
$\left(\dfrac{-2}{p} \right) = \left(\dfrac{-1}{p} \right)\left(\dfrac{2}{p} \right).$
We have that
$\left(\dfrac{-1}{p}\right) = \begin{cases} 1 & : p \equiv 1 \pmod{8} \text{ or } p \equiv 5 \pmod{8} \\ -1 & : p \equiv 3 \pmod{8} \text{ or } p \equiv 7 \pmod{8} \end{cases}$
and 
$\left(\dfrac{2}{p} \right) = \left\{
     \begin{array}{lr}
       1 & : p \equiv 1 \pmod 8 \text{ or } p \equiv 7 \pmod 8 \\
      -1 & : p \equiv 3 \pmod 8 \text{ or } p \equiv 5 \pmod 8
     \end{array}
   \right.$
If $p \equiv 5 \pmod 8,$ then $\left(\dfrac{-2}{p} \right) = \left(\dfrac{-1}{p} \right)\left(\dfrac{2}{p} \right) = 1 \cdot -1 = -1,$ so $-2$ is not a quadratic residue for this particular case. If $p \equiv 7 \mod 8,$ then $\left(\dfrac{2}{p} \right) = 1$ and $\left(\dfrac{-1}{p} \right) = -1 \Longrightarrow \left(\dfrac{-2}{p} \right) = -1 \cdot 1 = -1.$ Hence $-2$ is not a quadratic residue in this particular case either. But if $p \equiv 1 \mod 8$ or $p \equiv 3 \mod 8,$ then $\left(\dfrac{-2}{p} \right) = 1 \cdot 1 = 1$ or $\left(\dfrac{-2}{p} \right) = -1 \cdot -1 = 1.$ Thus $-2$ is a quadratic residue for primes congruent to $1$ or $3$ modulo $8.$ 

Hint:
$$
\left(\frac{-1}{p}\right) = \begin{cases} 1 & : p \equiv 1 \pmod{8} \text{ or } p \equiv 5 \pmod{8} \\ -1 & : p \equiv 3 \pmod{8} \text{ or } p \equiv 7 \pmod{8} \end{cases}
$$
