Let $d>0$ be a square free integer, $\mathbb{Z}[\sqrt {-d}]=\{a+b\sqrt{-d}:a,b \in \mathbb{Z}\}$ and $N(a+b\sqrt{-d})=a^2+b^2d$.
Is $2$ irreducible in $\mathbb{Z}[\sqrt {-d}] \ ?$
Appreciate if one could advise on my attempt below? Thank you very much.
Suppose $2=xy.$ Let $x=a+b\sqrt{-d}$
If $d=1,$ then $2=(1-\sqrt{-1})(1+\sqrt{-1})$ and since $N(1\pm\sqrt{-1}) \neq 1,$ $2$ is not irreducible.
If $d=2,$ since $N(2) = 4= N(x)N(y),$ we have $N(x) = 1$ or $2.$
If $N(x) = a^2+2b^2 = 2,$ then $a=0$ and $b=\pm 1,$ so $x=\pm\sqrt{-2}.$ We may argue similarly that $y = \pm \sqrt{-2}.$ Since $N(\pm \sqrt{-2}) \neq 1, 2$ is not irreducible.
If $d \geq 3,$ following previous argument, we have $a^2+b^2d=1$ or $2.$
If $a^2+b^2d=1,$ then $2$ is irreducible.
There is no solution to $a^2+b^2d=2,$ since $b^2d \geq 3$ and $d \neq 0 ($ otherwise $a \not\in \mathbb{Z})$
Hence, $2$ is irreducible in $\mathbb{Z}[\sqrt{-d}]$ for $d>2$