I am working on the following exercise:
Show that if $p \equiv 1 \mod 4$ then $p$ is not prime in $\mathbb{Z}[i]$, but instead splits as the product of two distinct prime. [Hint: Show that $p|(a^2+1)$ for some $a$; if $p$ remained prime in $\mathbb{Z}[i]$ show that $p|a\pm i$ and obtain a contradiction.]
I have been able to prove that $p|(a^2+1)$ for some $a$. This implies that $a^2 + 1 = pk $ for some integer $k$. In $\mathbb{Z}[i]$, we then have $$a^2 +1 = (a+i)(a-i) =pk$$ If $p$ is prime in $\mathbb{Z}[i]$, it must divide either $a-i$ or $a+i$. This follows from the fact that $\mathbb{Z}[i]$ is a unique factorization domain. Without loss of generality, suppose that $p|(a+i)$. This implies that $N(p)|N(a+i)$. That is, $p^2|(a^2+1)$. How does this yield a contradiction?
