BMO 2010 Shortlist

Write $x^3=(1+y\sqrt{-2})(1-y\sqrt{-2})$. The identity $$(1-y\sqrt{-2}-y^2)(1+y\sqrt{-2})-y^2(1-y\sqrt{-2})=1$$ shows that $1+y\sqrt{-2}$ and $1-y\sqrt{-2}$ are relatively prime in $\mathbb{Z}[\sqrt{-2}]$. Since $\mathbb{Z}[\sqrt{-2}]$ is a unique factorization domain, $$1+y\sqrt{-2}=(u+v\sqrt{-2}{)}^3=u(u^2-6v^2)+v(3u^2-2v^2)\sqrt{-2},$$ for some $u,v\in \mathbb{Z}$. Therefore, $u(u^2-6v^2)=1$, which forces $u=1$ and $v=0$. Consequently, $y=0$ and $x=1$, which are obviously solutions to the given equation. $\square$