Show that if $u,v,x,y$ are positive integers for which $u^2+2v^2=x^2+2y^2=p$ a prime number, then $u=x$ and $v=y$.
I get that if we had $\alpha=u+v\sqrt{-2} \in \mathbb{Z}[\sqrt{-2}]$, then the norm is $u^2+2v^2=p$. I also know that in $\mathbb{Z}[\sqrt{-2}]$, the only invertible elements are $\pm 1$
I'm not sure how to put these facts together to prove the required statement though.
From $N(u\pm v\sqrt{-2})=p$ prime you get that $u\pm v\sqrt{-2}$ are irreducible. Similarly for $x\pm y\sqrt{-2}$. But, as you were told in the comments by Daniel Fischer, $\mathbb Z[\sqrt{-2}]$ is a UFD, and therefore the irreducible elements are prime. From $(u-v\sqrt{-2})(u+v\sqrt{-2})=(x-y\sqrt{-2})(x+y\sqrt{-2})$ you can conclude.