Show that if p is a prime such that there is an integer $b$ with $p = b^2 + 4$, then $Z[ \sqrt{p}]$ is not a unique factorization domain.

Question

I feel like the trick is to find an appropriate integer $a$ and then get $(a + \sqrt{b^2 + 4})(a - \sqrt{b^2 + 4}) = a^2 - b^2 + 4$ such that $a + \sqrt{b^2 + 4}$ and $a - \sqrt{b^2 + 4}$ are irreducible in $Z[\sqrt{p}]$ but $a^2 - b^2 + 4$ has another irreducible factor, but I am having trouble coming up with such an example.

Answer 1

Notice that $b^2 = b\cdot b$ and $$b^2 = (\sqrt{p}-2)\cdot (\sqrt{p}+2)$$

We know that $\sqrt{p}\pm 2\neq b$ since $b$ is an integer and $p$ is a prime. If $p\not =5$, then $\sqrt{p}\pm2$ is not a unit and so our conclusion follows.

If $p=5$, then $-4 = -2\cdot 2 = (\sqrt{5}-3)\cdot (\sqrt{5}+3)$, and so the statement holds

Answer 2

So since $p = b^2 + 4$ we know $p \geq 4$ and in particular $p \neq 2$ and hence $p$ is odd. Therefore since $p = b^2$ mod $4$ we see that $p = 1$ mod $4$. Let $p = 1 + 4k$ then the polynomial $x^2 + x - k = x^2 + x \frac{1 - p}{4} = (x + \frac{1 + \sqrt{p}}{2})(x + \frac{1 - \sqrt{p}}{2})$. So $x^2 + x - k = x^2$ is reducible in $\mathbb{Q}(\sqrt{p})$ but not $\mathbb{Z}(\sqrt{p})$. Therefore $\mathbb{Z}(\sqrt{p})$ is not a UFD by Gauss's Lemma.