GCD with Gaussian integers Let $\alpha,\beta\in\mathbb{Z[i]}$, show that if
  $gcd(N(\alpha),N(\beta))=p$ where $p$ is a prime $\in\mathbb{N}$ then
  $p\not\equiv 3\pmod 4$.


I suppose $gcd(N(\alpha),N(\beta))=p$ where $p\in\mathbb{N}$ is prime.

$N(gcd(\alpha,\beta))|gcd(N(\alpha),N(\beta))=p$ and $N(gcd(\alpha,\beta))=a^2+b^2$ for some $a,b\in\mathbb{Z}$

So $p=d*(a^2+b^2)$ for some $d\in\mathbb{Z}$

But $p$ is prime, so either $d=1$ and $a^2+b^2=p$ or $d=p$ and $a^2+b^2=1$

Suppose $d=1\implies a^2+b^2=p$, then $p$ is the sum of two squares and p must be $\not\equiv 3\pmod 4$.

How can I get $p\not\equiv 3\pmod 4$ in the case where $a^2+b^2=1$ and $d=p$? I know that $\alpha$ and $\beta$ are coprime, should that help me?
Assume that $p|a^2+b^2$. If $p\not |a$ then for some $c$ $$ac\equiv 1(\mod p)$$

This implies that $$(bc)^2\equiv -1(\mod p)$$ this is impossible since $p\equiv 3(\mod 4)$. Thus $p|a$ and so $p|b$ thus $p^2|a^2+b^2$.
Assume that $p|a^2+b^2$. If $p\not |a$ then for some $c$ $$ac\equiv 1(\mod p)$$

This implies that $$(bc)^2\equiv -1(\mod p)$$ this is impossible since $p\equiv 3(\mod 4)$. Thus $p|a$ and so $p|b$ thus $p^2|a^2+b^2$.