# Show that $R = \mathbb{Z}[\sqrt{-17}]$ is not a Euclidean ring.

Show that $R = \mathbb{Z}[\sqrt{-17}]$ is not a Euclidean ring. To do this I tried showing that the ring is not a principal ideal domain. I wonder if this is enough and how to actually show that it is not a PID. Thanks!

Hint. Notice that: $$18=2\times 3\times 3=(1+i\sqrt{17})(1-i\sqrt{17}).$$ If you can prove that $2,3,1-i\sqrt{17}$ and $1+i\sqrt{17}$ are irreducible, you are done!
Remark. If you can show that $\mathbb{Z}[i\sqrt{17}]$ is not a PID, you are also done. Try to consider the ideal generated by $2$ and $1+i\sqrt{17}$ and proceed by contradiction. What if there exists $z=a+ib\sqrt{17}$ such that: $$(z)=(2,1+i\sqrt{17}).$$ What will happen if you take the norm of $z$ which is $a^2+17b^2$?