Show that $2$ is not prime in $\mathbb Z[\sqrt{-d}]$ for odd prime $d$

Question

I have to show that if $d > 2$ is a prime number, then $2$ is not prime in $\mathbb Z[\sqrt{-d}]$.

The case when $d$ is of the form $4k+1$ for integer $k$ is quite easy: $2\mid (1+\sqrt{-d})(1-\sqrt{-d})$, because the product is equal to $1+d$ and this is even of course. However $2 \nmid 1 \pm \sqrt{-d}$, because we can calculate the norm: $N(2) = 4$, $N(1 \pm \sqrt{-d}) =1+d = 4k+2$ and $4 \nmid 4k+2$.

This argument fails when the remainder after dividing by $4$ is $3$ as far as I can see. So how could I prove the statement in this case?

[EDIT]: The norm is given by $N(a+b\sqrt{-d}) = a^2+b^2d$.

Answer 1

I assume that by $\mathbb{Z}\left[\sqrt{-d}\right]$ you mean the numbers of the form $a+b\sqrt{-d}$, where $a$ and $b$ are integers. Note that $a+b\sqrt{-d}$ is divisible by $2$ in this ring if and only if both $a$ and $b$ are even.

Answer 2

I'm afraid your assertion is false: $2$ is inert in the ring of integers of $\mathbf Q(\sqrt{ -d})$ if $-d\equiv 5 \mod8$, which is the case for $d=3$.