<s>[INST] Irreducible elements of $\mathbb{Z}[i\sqrt{5}]$ Is there any neat way to show that $9$ and $3-3i\sqrt{5}$ are irreducible elements of $\mathbb{Z}[i\sqrt{5}]$, while $1+4i\sqrt{5}$ and $5-2i\sqrt{5}$ are reducible?
 [/INST] $9 = 3 \cdot 3$ is definitely reducible, and so is $3-3i\sqrt{5} = 3 \cdot (1-i\sqrt{5})$.
The norm of $a = 1+4i\sqrt{5}$ is $81$. There are no elements of norm $3$, so if it reducible, $a$ could only be the product of two elements of norm $9$, and these are $\pm 3, \pm 2 \pm i \sqrt{5}$. And in fact $a = (-2 + i \sqrt{5}) (2 - i \sqrt{5})$.
Finally, the norm of $b = 5-2i\sqrt{5}$ is $45$, and a similar argument leads to $b = -i \sqrt{5} \cdot (2 + i \sqrt{5})$.
So barring mistakes they all appear to be reducible, it seems.
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