$\Bbb Z[\sqrt{6}]$ is not a UFD

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!

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