$\textbf Z[\sqrt{pq}]$ is not a UFD if $\left( \frac{q}p \right) = -1$ and $p \equiv 1 \pmod 4$. 
Let $p$ and $q$ be primes such that $p \equiv 1 \pmod 4$ and $\left( \frac q p \right) = -1$. Show that $\textbf Z[\sqrt {pq}]$ is not a UFD.

I tried some examples like $p=5$ and $q = 2$. But I have no clue about the general case. Any hint?

If $$\left(\frac{q}{p}\right) = -1,$$ that means that $q$ is not a quadratic residue modulo $p$. That much is obvious, right? It also means that $q$ is not a quadratic residue modulo $pq$ either. Thus, in your example, since 2 is not a quadratic residue modulo 5, it can't be a residue modulo 10 either, and is therefore irreducible.
Since $q$ is irreducible, it's not divisible by $\sqrt{pq}$. So we conclude that $pq$ has two distinct factorizations in $\mathbb Z[\sqrt{pq}]$: $$(\sqrt{pq})^2 = pq.$$
Well, that might slightly wrong if $p$ can be broken down, but $q$ can't. In order to present this answer first, I have not actually pondered the significance of quadratic reciprocity to your question.
