Exercise 2.1.59 (Proof of the converse)

Proof. If $p=2$, then the congruence $x^2\equiv -2\equiv 0(\mathit{mod}2)$ has the solution $x=0$. Now we assume that $p$ is an odd prime.

Suppose that $p\mid a^2+2b^2$ and that $p\nmid a$ or $p\nmid b$.

If $p\nmid a$, then $p\nmid b$, otherwise $p\nmid b$ and $p\mid a^2+2b^2$ imply $p\mid a^2$, where $p$ is prime, thus $p\mid a$.

If $p\nmid b$, then $p\nmid a$, otherwise $p\nmid a$, and $p\mid a^2+2b^2$ imply $p\mid 2b^2$, where $p$ is an odd prime, thus $p\mid b^2$, and $p\mid b$.

This shows that in both cases,

Since $p\nmid b$, $b$ has an inverse modulo $p$, i.e. there exists an integer $b^{\prime }$ such that $b{b}^{\prime }\equiv 1(\mathit{mod}p)$. Then $a^2+2b^2\equiv 0(\mathit{mod}p)$ implies ${(a{b}^{\prime })}^2+2{(b{b}^{\prime })}^2\equiv 0(\mathit{mod}p)$, thus

The congruence $x^2\equiv -2(\mathit{mod}p)$ has a solution $a{b}^{\prime }$. □