Z[√-5] is not a UFD

Counterexample

The integral domain $\mathbb{Z}[\sqrt{-5}] = \{a + b\sqrt{-5} : a, b \in \mathbb{Z}\}$ is not a unique factorisation domain. In this ring, $$6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}).$$ Let $N(a + b\sqrt{-5}) = a^2 + 5b^2$. Then $N$ is multiplicative, and $N(2) = 4$, $N(3) = 9$, $N(1 \pm \sqrt{-5}) = 6$. No element has norm $2$ or $3$, because $a^2 + 5b^2 \notin \{2, 3\}$ for integers $a, b$. So the four factors are irreducible. The only units are $\pm 1$, so $2$ is not an associate of $1 + \sqrt{-5}$ or of $1 - \sqrt{-5}$. The two factorisations are different.