Every element of the ring $\Z[\sqrt{5}]$ can be written as $a+b\sqrt{5}$ for some integers $a, b$.
The (field) norm $N$ of an element $a+b\sqrt{5}$ is defined by
\[N(a+b\sqrt{5})=(a+b\sqrt{5})(a-b\sqrt{5})=a^2-5b^2.\]

Consider the case when $a=3, b=1$.
Then we have
\[(3+\sqrt{5})(3-\sqrt{5})=4=2\cdot 2. \tag{*}\]

We prove that elements $2, 3\pm \sqrt{5}$ are irreducible in $\Z[\sqrt{5}]$.
Note that the norms of these elements are $4$. We claim that each element $\alpha \in \Z[\sqrt{5}]$ of norm $4$ is irreducible.

Suppose that $\alpha=\beta \gamma$ for some $\beta, \gamma \in \Z[\sqrt{5}]$.
Our objective is to show that either $\beta$ or $\gamma$ is a unit.

Since we have
\[4=N(\alpha)=N(\beta \gamma)=N(\beta) N(\gamma)\]
and the norms are integers, the value of $N(\beta)$ is one of $\pm 1, \pm 2, \pm 4$.

If $N(\beta)=\pm 1$, then $\beta$ is a unit.
If $N(\beta)=\pm 4$, then $N(\gamma)=\pm 1$ and hence $\gamma$ is a unit.

Let us consider the case $N(\beta)=\pm 2$.
We show that this case does not happen.
Write $\beta=a+b\sqrt{5}$ for some integers $a, b$.
Then we have
\[\pm 2 =N(\beta)=a^2-5b^2.\]
Considering the above equality modulo $5$ yields that
\[\pm 2 \equiv a^2 \pmod{5}.\]
However note that any square of an integer modulo $5$ is one of $0, 1, 4$.
So this shows that there is no such $a$.

Therefore, we have proved that either $\beta$ or $\gamma$ is a unit, hence $\alpha$ is irreducible.
The claim is proved.

It follows from (*) that the element $4 \in \Z[\sqrt{5}]$ has two different decompositions into irreducible elements.
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).

This problem only differs from the current problem by the sign.
($-5$ is used instead of $5$.)

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