This is not duplicate As I wanted to check why my argument is not working,Please have a look
$\mathbb Z[\frac{1+\sqrt{-7}}{2}]$ is euclidean domain.
My attempt:
I had thought following proof for $\mathbb Z[\frac{1+\sqrt{-3}}{2}]$ but there is problem to use same method to show $\mathbb Z[\frac{1+\sqrt{-7}}{2}]$ is ED. So I thought is my attempt is wrong?
$a,b\in \mathbb Z[\frac{1+\sqrt{-3}}{2}]$
Consider $\frac{a}{b}=r+s(\frac{1+\sqrt{-3}}{2})$ where $r,s \in \mathbb Q$
So choose $m,n \in \mathbb Z$ such that $-1/2N(b)\leq r-m\leq 1/2N(b)$ and $|n-s|\leq 1/2N(b)$
So $a=mb+nb(\frac{1+\sqrt{-3}}{2})+r_1+r_2(\frac{1+\sqrt{-3}}{2})$ where $r_1+mb=r$ and $r_2+nb=s$
Now $N(r_1+r_2(\frac{1+\sqrt{-3}}{2})$=$r_1^2+r_1r_2+r_2^2\leq 3/4N(b)<N(b)$
Hence Done
Similar proof I tried for
$\mathbb Z[\frac{1+\sqrt{-7}}{2}]$ but last step I get $N(r_1+r_2(\frac{1+\sqrt{-7}}{2})\leq N(b)$
So I could not conclude.
I would thankful if someone help me out
