I've been trying to prove that $\mathbb{Z}[\sqrt{10}]$ is not factorial. I did this by defining the norm $N(a+b\sqrt{10})=a^2-10b^2$.
I was able to show for myself that $N(z)=\pm 2$ and $N(z)=\pm 5$ have no solutions, and my idea is to show that $2,5,\sqrt{10}$ are irreducible, and then $2\cdot 5=10=\sqrt{10}\sqrt{10}$ gives two factorizations of $10$ into nonassociate irreducibles.
The only hitch is I need to show $N(z)=\pm 1$ implies $z$ is a unit, from which it will follow that $2,5,\sqrt{10}$ are irreducibles since $N(2)=4$, $N(5)=25$, and $N(\sqrt{10})=-10$, but norms never take values $\pm 2,\pm 5$. I know that if $z$ is a unit, then $N(z)=\pm 1$, but I don't know how to show the converse. If there is a better solution, I'd be happy to see that instead. Thanks.