For any positive prime $p$, we have $p=x^2-2y^2$ for some $x,y\in \mathbb{Z}$ iff $-p=x'^2-2y'^2$ for some $x',y'\in \mathbb{Z}$. Show that for any positive prime, $p=x^2-2y^2$ for some $x,y\in \mathbb{Z}$ iff $-p=x'^2-2y'^2$ for some $x',y'\in \mathbb{Z}$. 
The key here appears to be that there exists some element in $\mathbb{Z}[\sqrt{2}]$ with norm $-1$, namely $1+\sqrt{2}$. But it is not clear to me how I would use this. I presume there is some trick involving the multiplicativity of norm, but I am not able to make it work. I prefer a solution using this fact, and would like to avoid an ad-hoc method using some elementary trick, if one exists.
Context: I have shown that for a positive prime $p$, either $p$ or $-p$ can be written in the form $x^2-2y^2$ iff $2$ is a square mod $p$, and indeed the purpose of the above exercise is to show that we may freely remove the signs in this equivalence. 

Using the fact that $(1+\sqrt{2})$ has norm $-1$ and that norm is multiplicative, we have, the norm of $x+y\sqrt{2}$ is $p$ iff the norm of $(x+y\sqrt{2})(1+\sqrt{2})$ is $-p$, as required.
Note that we did not need the assumption that $p$ is a prime. 
