Question about irreducible Gaussian integers 
Let $\pi \in \mathbb{Z}[i]$ irreducible and $\pi$ divides $p$ ($p\in\Bbb Z$ prime). Then show that $N(\pi)=p$.

($N(\alpha)= a^2 +b^2$ where $\alpha = a+ib$.)
 A: Let us write your element as $\alpha=a+ib$ with $a,b\in\mathbb{Z}$ and $N(a)=q=(a+ib)(a-ib)=\alpha\overline{\alpha}$.
You assume that $\alpha$ divides $p$, hence $p=\alpha\beta$ for some $\beta\in\mathbb{Z}[i].$ Observing that the norm is multiplicative (follows from its definition and straight forward calculation), you get $$p^2=N(p)=N(\alpha\beta)=N(\alpha)N(\beta)= qN(\beta).$$
Hence $q$ divides $p^2$. Since $q$ is a positive integer and $p$ is a prime number you have either $q=1$, $q=p$ or $q=p^2$. 
$a)$ $q=1$ implies that $(a,b)\in \{(0,\pm 1),(1,\pm 1)\}$ and thus that $\alpha$ is unitary, which is not true by hypothesis (you assume that $\alpha$ is irreducible).
$b)$ $q=p^2$ implies that $N(\beta)=1$, so $\beta$ is unitary. Since $\alpha$ is irreducible, then so is $p=\alpha\beta$. But you assumed that $p=1\pmod{4}$, so by Fermat's theorem on sum of squares you can write $p=c^2+d^2=(c+id)(c-id)$, which implies that $p$ is not irreducible.
The only remaining case is $q=p$, which gives $N(\alpha)=p$ as you wanted.
A: Hint: Definite the norm by $N(\alpha) = N(a + ib) = a^2 + b^2$, where $\alpha \in \mathbb{Z}[i]$.  Suppose a prime $p$ factors nontrivially as $p = (a + ib)(c + id)$. Then $N(p) = p^2 = (a^2 + b^2)(c^2 + d^2)$ because $N$ is multiplicative. Since $p$ factors nontrivially, neither $a^2 + b^2$ or $c^2 + d^2$ is a unit. Hence $p = a^2 + b^2 = c^2 + d^2$.
