Proving $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$

The standard method is:

Define a function $N\colon \mathbb{Z}[\sqrt{-5}]\to\mathbb{Z}$ by $N(a+b\sqrt{-5}) = (a+b\sqrt{-5})(a-b\sqrt{-5}) = a^2+5b^2$.

1. Prove that $N(\alpha\beta) = N(\alpha)N(\beta)$ for all $\alpha,\beta\in\mathbb{Z}[\sqrt{-5}]$.
2. Conclude that if $\alpha|\beta$ in $\mathbb{Z}[\sqrt{-5}]$, then $N(\alpha)|N(\beta)$ in $\mathbb{Z}$.
3. Prove that $\alpha\in\mathbb{Z}[\sqrt{-5}]$ is a unit if and only if $N(\alpha)=1$.
4. Show that there are no elements in $\mathbb{Z}[\sqrt{-5}]$ with $N(\alpha)=2$ or $N(\alpha)=3$.
5. Conclude that $2$, $3$, $1+\sqrt{-5}$, and $1-\sqrt{-5}$ are irreducible.

This is a common technique for dealing with rings of the form $\mathbb{Z}[\theta]$, where $\theta$ is an algebraic integer.

It can also be done directly, though it is a bit more laborious. Here's what I came up with on the fly:

Assume that $(a+b\sqrt{-5})(c+d\sqrt{-5}) = 2$. Then $ac+5bd = 2$ and $ad+bc=0$. If $a=0$, then we must have $c=0$ (since $bc=0$ but we cannot have $a=b=0$), but then $5bd=2$ is impossible. Thus, $a\neq0$ and $c\neq 0$. If $b=0$, then $d=0$, so the factorization occurs in $\mathbb{Z}$ and is trivial; symmetrically if $d=0$. So we may assume that all of $a,b,c,d$ are nonzero.

Then $ad=-bc$, so $2d=acd + 5bd^2 = -bc^2 + 5bd^2 = b(5d^2 - c^2)$. If $b$ is odd, then $b|d$, so writing $d=bk$ we get $abk + bc=0$, so $ak+c=0$, thus $c=-ak$. Hence $c+d\sqrt{-5} = -ak+bk\sqrt{-5} = k(-a+b\sqrt{-5})$. But this gives $$2 = (a+b\sqrt{-5})(c+d\sqrt{-5}) = k(a+b\sqrt{-5})(-a+b\sqrt{-5}) = -k(a^2+5b^2).$$ Since $a$ and $b$ are both nonzero, $a^2+5b^2$ is at least 6, which is impossible. So $b$ is even, $b=2b'$.

Then $b'(5d^2-c^2) = d$, so setting $5d^2-c^2 = k$ we have $$a+b\sqrt{-5} = a+2b'\sqrt{-5},\qquad c+d\sqrt{-5} = c+kb'\sqrt{-5}.$$ From $ad=-bc$ we get $ak=-2c$. If $a$ is even, then we have $a=2a'$, so $$2 = (2a'+2b'\sqrt{-5})(c+d\sqrt{-5}) = 2(a'+b'\sqrt{-5})(c+d\sqrt{-5}),$$ which yields that $c+d\sqrt{-5}$ is a unit (in fact, this is impossible with $c$ and $d$ both nonzero, but that doesn't matter). If $a$ is odd, then $k$ is even and $c=ak'$, with $2k'=k$. So now we have $$\begin{align*} 2 &= (a+b\sqrt{-5})(c+d\sqrt{-5})\\ &= (a + 2b'\sqrt{-5})(ak' + 2b'k'\sqrt{-5})\\ &= k'(a+2b'\sqrt{-5})(a+2b'\sqrt{-5})\\ &= k'(a^2 - 20b'^2) + 4ab'\sqrt{-5} \end{align*}$$ which implies $a=0$ or $b'=0$ (hence $b=0$), contradicting our hypotheses.

Thus, the only factorizations of $2$ in $\mathbb{Z}[\sqrt{-5}]$ are trivial.

(And you can probably see why the "standard method" is so much better....)

Let $A = \mathbb{Z}[\sqrt{-5}]$. For every element $\alpha$ of $A$ there exist unique $a, b \in \mathbb{Z}$ such that $\alpha = a + b \sqrt{-5}$. Consider the function $N \colon A \to \mathbb{N}$ defined by $N(a + b \sqrt{-5}) = a^2 + 5 b^2$. (If you know some field theory, $N$ is the restriction to $A$ of the norm of the field $\mathbb{Q}(\sqrt{-5})$ over $\mathbb{Q}$.) Prove the following facts:

1. $N$ is multiplicative, i.e. $N(\alpha \beta) = N(\alpha) N(\beta)$ for all $\alpha, \beta \in A$.
2. If $\alpha \in A$, then $\alpha$ is invertible in $A$ if and only if $N(\alpha) = 1$.

Do there exist element in $A$ with norm equal to $2$ or $3$? Now, $N(2) = 4$, $N(3) = 9$, $N(1 + \sqrt{-5}) = N(1- \sqrt{-5}) = 6$. As in Henning Makholm's comment, if they were irreducible then what could the norms of the factors be?