Any element of the ring $\Z[\sqrt{-5}]$ is of the form $a+b\sqrt{-5}$ for some integers $a, b$.
The associated (field) norm $N$ is given by
\[N(a+b\sqrt{-5})=(a+b\sqrt{-5})(a-b\sqrt{-5})=a^2+5b^2.\]

Consider the case when $a=2, b=1$.
Then we have
\begin{align*}
(2+\sqrt{-5})(2-\sqrt{-5})=9=3\cdot 3. \tag{*}
\end{align*}

We claim that the numbers $3, 2\pm \sqrt{-5}$ are irreducible elements in the ring $\Z[\sqrt{-5}]$.

To prove the claim at once, we show that any element in $\Z[\sqrt{-5}]$ of norm $9$ is irreducible.

Let $\alpha$ be an element in $\Z[\sqrt{-5}]$ such that $N(\alpha)=9$.
Suppose that $\alpha=\beta \gamma$ for some $\beta, \gamma \in \Z[\sqrt{-5}]$.
Out goal is to show that either $\beta$ or $\gamma$ is a unit.

We have
\begin{align*}
9&=N(\alpha)=N(\beta)N(\gamma).
\end{align*}
Since the norms are nonnegative integers, $N(\beta)$ is one of $1, 3, 9$.

If $N(\beta)=1$, then it yields that $\beta$ is a unit.

If $N(\beta)=3$, then we write $\beta=a+b\sqrt{-5}$ for some integers $a, b$, and we obtain
\[3=N(\beta)=a^2+5b^2.\]
A quick inspection yields that there are no integers $a, b$ satisfying this equality.
Thus $N(\beta)=3$ is impossible.

If $N(\beta)=9$, then $N(\gamma)=1$ and thus $\gamma$ is a unit.

Therefore, we have shown that either $\beta$ or $\gamma$ is a unit.

Note that the elements $3, 2\pm \sqrt{-5}$ have norm $9$, and hence they are irreducible by what we have just proved.

It follows from the equalities in (*) that the factorization of the element $9$ into irreducible elements are not unique.
Thus, the ring $\Z[\sqrt{-5}]$ is not a UFD.

Related Question.

Problem.
Prove that the quadratic integer ring $\Z[\sqrt{5}]$ is not a Unique Factorization Domain (UFD).

The Quadratic Integer Ring $\Z[\sqrt{5}]$ is not a Unique Factorization Domain (UFD)
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Let $R$ be a principal ideal domain. Let $a\in R$ be a nonzero, non-unit element. Show that the following are equivalent.
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