Show that if $p \equiv 3$ mod $4$ is a prime in $\mathbb{Z}$, then $p$ is a prime in $\mathbb{Z}[i]$
I think I probably have to use something related to norm to solve this problem but I can't seem to figure out anything. Can someone suggest how I can solve this problem? Thanks
If $p=(a+bi)(c+di)$ then $$p^2=(a^2+b^2)(c^2+d^2)$$ Therefore $a^2+b^2$ is $1$, $p$ or $p^2$. If it is $1$ or $p^2$ it is easy to conclude that $p$ is a prime in $\Bbb Z[i]$. If it is $p$ we arrive at a contradiction taking mod $4$.
Hint: Use the fact that $\mathbf Z[\mathrm i]$ is a euclidean domain. Hence $p$ not being prime means it is reducible, say $$p=(a+b\mathrm i)(c+d\mathrm i)$$ Compute the norm of both sides and conclude that $p$ is the sum of two squares. Is that possible?
