Let $p>2$ ,Existence of $a,b \in \mathbb{Z}$ such that $a^2+2b^2=p$.

Question

Let $p>2$ prime, such that $(-2)$ is a square in $\mathbb{F}_p$. I need to show that there exists $a,b\in \mathbb{Z}$ such that $a^2 +2b^2 = p$. How can I use the fact that $\mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain with the norm $$N(a+b\sqrt{-2}) = a^2 +2b^2$$?

Answer 1

First we show that $p$ is composite in $\mathbb{Z}[\sqrt{-2}]$.

Suppose that $p$ is not composite.

Since, $x^2\equiv -2\pmod{p}$ for some $x\in\mathbb{Z}$, we have $p\mid x^2+2$, so $p\mid (x+\sqrt{-2})(x-\sqrt{-2})$, so $p\mid (x+\sqrt{-2})$ and $p\mid (x-\sqrt{-2})$.

Now, $p(m\pm n\sqrt{-2})=x\pm\sqrt{-2}$ for some $m,n\in\mathbb{Z}$, so $pm=x$, so $x^2\equiv 0\pmod{p}$, contradiction as $p>2$.

So, there are $\alpha,\beta\in\mathbb{Z}[\sqrt{-2}]$ which are both non-units such that $p=\alpha\beta$. So, $p^2=N(\alpha)N(\beta)$, but $N(\alpha),N(\beta)\neq 1$, so $N(\alpha),N(\beta)=p$. But $\alpha=a+b\sqrt{-2}$ for some $a,b\in\mathbb{Z}$, so $a^2+2b^2=p$.

Edit: In fact the above is a small adaptation of a famous proof by Gauss that an odd prime number which is congruent to $1$ modulo $4$, can be written as a sum of two square integers.