Let $ n \geq 3 $. By factorising $ n $ or $n + 1 $ (as appropriate), show that $ \mathbb{Z}[\sqrt{-n}] $ is not a UFD.
My thoughts so far:
Define $ N(a + b \sqrt{-n}) = a^2 + n b^2 $.
Suppose $ n $ is odd. Then $ n + 1 $ is even, say $ n + 1 = 2k $. Now $ N(2) = 4 $, and the norm of an element in this ring can never be 2, so we have that 2 is irreducible. Now note that $ 1 + n = (1 + \sqrt{-n})(1 - \sqrt{-n}) $. Is $ 1 + \sqrt{-n} $ irreducible? Well, if $ 1 + \sqrt{-n} = z_1 z_2 $, then $ N(z_1)N(z_2) = 1 + n $. So $ N(z_i) \leq \frac{n+1}{2} < n $. But this means both $ z_i$ must be purely real, which clearly can't be the case. Similarly, $ 1 - \sqrt{-n} $ is irreducible. Neither of these factors are equal to 2, and so 2 appears in one factorisation but not another. Hence for $ n $ odd, we don't have a UFD.
What about $n$ even? How can I factorise $ n $ other than as $ 2k $ for some $k $?
Thanks
EDIT: I overlooked $ n = \sqrt{-n} ( -\sqrt{n}) $!