I want to show that $\Bbb Z[\sqrt{6}]$ is not a UFD. We know that this is a Dedekind domain. (Hint: For the ideals $\mathfrak{p}=(2,4+\sqrt{6})$, $\mathfrak{q}=(5,4+\sqrt{6})$, compute $\mathfrak{p}^2$, $\mathfrak{q}\bar{\mathfrak{q}}$, $\mathfrak{p}\mathfrak{q}$ and $\mathfrak{p}\bar{\mathfrak{q}}$.)
Note that $\mathfrak{p}^2=(4,8+2\sqrt{6},22+8\sqrt{6})$, $\mathfrak{q}\bar{\mathfrak{q}}=(5, 20-5\sqrt{6},20+5\sqrt{6})$, $\mathfrak{p}\mathfrak{q}=(10,8+2\sqrt{6},20+5\sqrt{6},22+8\sqrt{6})$, and $\mathfrak{p}\bar{\mathfrak{q}}=(10,8-2\sqrt{6},20+5\sqrt{6})$. Also, we observe that $10=(4+\sqrt{6})(4-\sqrt{6})=(-1+\sqrt{6})(2+\sqrt{6})(1+\sqrt{6})(-2+\sqrt{6})$ and $10=2.5=(2+\sqrt{6})(-2+\sqrt{6})(-1+\sqrt{6})(1+\sqrt{6})$. But, this doesn't imply that $\Bbb Z[\sqrt{6}]$ is not a UFD.
I am not sure how to use the hint. Thanks!
$\mathbb Z[\sqrt 6]$ is indeed a unique factorization domain. Notice, for one thing, that $\langle 2 \rangle$ and $\langle 4 + \sqrt 6 \rangle$ are both contained in $\langle 2 + \sqrt 6 \rangle$. As for $\langle 5, 4 + \sqrt 6 \rangle$, that turns out to be the whole ring.
Maybe you meant $\mathbb Z[\sqrt{-6}]$ is not UFD? In which case the ideal $\langle 2, \sqrt{-6} \rangle$ would be far more pertinent than either $\langle 2, 4 + \sqrt{-6} \rangle$ or $\langle 5, 4 + \sqrt{-6} \rangle$.