For any odd prime $p$ , Gauss's lemma quickly yields \begin{eqnarray} \left( \frac{2}{p} \right) &=& 1 \text{ if }p\equiv\pm 1\pmod{8} \\ \left( \frac{2}{p} \right) &=& -1 \text{ if }p\equiv\pm 3\pmod{8} \end{eqnarray}But there is another way, which goes back to Euler, and is worth seeing, inasmuch as it is the prototype of certain more general arguments about character sums.

Let $\sigma$ be a primitive eighth root of unity in an algebraic closure of $\mathbb{Z}/p\mathbb{Z}$ , and write $\tau=\sigma+\sigma^{-1}$ . We have $\sigma^4=-1$ , whence $\sigma^2+\sigma^{-2}=0$ , whence $$\tau^2=2\;.$$ By the binomial formula, we have $$\tau^p=\sigma^p+\sigma^{-p}\;.$$ If $p\equiv\pm 1\pmod 8$ , this implies $\tau^p=\tau$ . If $p\equiv\pm 3\pmod 8$ , we get instead $\tau^p=\sigma^5+\sigma^{-5}= -\sigma^{-1}-\sigma=-\tau$ . In both cases, we get $\tau^{p-1}=\left( \frac{2}{p} \right)$ , proving (1) and (2).

A variation of the argument, closer to Euler's, goes as follows. Write $$\sigma=\exp(2\pi i/8)$$ $$\tau=\sigma+\sigma^{-1}$$ Both are algebraic integers. Arguing much as above, we end up with $$\tau^{p-1}\equiv\left( \frac{2}{p} \right)\pmod{p}$$ which is enough.