I need to prove that a prime $p$ is of the form $n^2-2m^2$ iff $p=2$ or $p=8k\pm 1$. So first I tried figuring out, for which $p$ does $n^2-2m^2=0$ has a solution if $\mathbb{F}_p$, or similarly all primes $p$ s.t there's a solution to the equation $(\frac{n}{m})^2=2$.
So now I know that if $\mathbb{F}_p$ has a square root for $2$, then $p$ divides a number of the form $n^2-2m^2$, but how can I prove that $p$ is equal to it? I tried looking at $\mathbb{Z}[\sqrt2]$ but that didn't get me so far.
Any help would be apperciated.
The number field $\Bbb Q[\sqrt 2]$ has class number $1$.
This means that the ring $\mathcal O = \Bbb Z[\sqrt 2]$ is a principal ideal domain.
If $p = 8k \pm 1$ is a prime, then we know that the ideal $p\mathcal O$ splits into a product $p\mathcal O = \mathfrak p \mathfrak q$, with both $\mathfrak p, \mathfrak q$ not equal to $\mathcal O$ (this is essentially the fact that $2$ has two different square roots in $\Bbb F_p$). This implies that the norms of $\mathfrak p, \mathfrak q$ are both equal to $p\Bbb Z$.
Since $\mathfrak p$ is principal, we may write $\mathfrak p = x\mathcal O$ for some $x \in \mathcal O$. We can write $x = n + m\sqrt 2$ with $n, m \in \Bbb Z$.
Since the norm of the ideal $x\mathcal O$ is equal to $p\Bbb Z$, we know that $N(x) = \pm p$, where $N$ denotes the norm map from $\mathcal O$ to $\Bbb Z$.
If $N(x)$ is equal to $p$, then we are done, as we get $n^2 - 2m^2 = p$.
If $N(x)$ is equal to $-p$, then we have $N(\epsilon x) = p$, where $\epsilon$ is the unit $1 + \sqrt 2$. By writing $\epsilon x = u + v\sqrt 2$, we get $u^2 - 2v^2 = p$.