Proving $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$ Could anyone help me prove that $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$?
As $6=2*3=(1+\sqrt{-5})(1-\sqrt{-5})$ so $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. Therefore is not a PID or euclidean domain

Since $\sqrt{-5} \cdot \sqrt{-5}=-5 \in \mathbb{Z}$ holds, we have $\mathbb{Z}[\sqrt{-5}]=\{a+b \sqrt{-5} \mid a, b \in \mathbb{Z}\}.$
We use
$$
(a+b \sqrt{-5})(c+d \sqrt{-5})=a c-5 b d+\left(a d+bc\right) \sqrt{5}
$$
and $|a+b \sqrt{-5}|^2=a^2+5 b^2$.
$1+\sqrt{-5}$ is not a unit: it is valid that
$$
(1+\sqrt{-5})(a+b \sqrt{-5})=a-5 b+(a+b) \sqrt{5} \neq 1,
$$
since otherwise $a=-b$ and $-6b=0$.
If $(1+\sqrt{-5})=(a+b \sqrt{-5})(c+d \sqrt{-5})$, then
$$
6=1+5=|1+\sqrt{-5}|^2=\left(a^2+5 b^2\right)\left(c^2+5 d^2\right)
$$
holds.
There is no $a, b \in \mathbb{Z}$ with $a^2+5 b^2=2$ and $a^2+5 b^2=3$, since $a^2 \in\overline{0}, \overline{1}, \overline{4}\rangle \pmod{5}$
and if $a^2+5 b^2=1$, then $a+b \sqrt{-5}$ would be invertible with
$$
(a-b \sqrt{-5})^{-1}=(a-b \sqrt{-5}),
$$
since
$$
(a+b \sqrt{-5})(a-b \sqrt{-5})=a^2+5 b^2. \quad\square
$$
