# If $p \equiv 3 (\bmod{4})$ is a rational prime, then p is a Gaussian prime

I understand the proof that my professor gave in class for the most part, but there's one snag that I'm having. Here's the proof to begin with [Side note: $N(x)=a^2+b^2$, where $a+bi$ is a complex number]:

Assume not, then $p =rs$ over $\mathbb{Z}[i]$, where $r, s$ are not units.

$N(p) = N(r)N(s) \rightarrow p^2 = N(r)N(s) \rightarrow N(r) = N(s) = p$

If $r = a + bi$, then $N(r) = p \rightarrow a^2 + b^2 = p \rightarrow a^2 + b^2 \equiv 3 (\bmod{4})$

My question is this (hopefully it isnt a silly): How does the conclusion imply (or show) that p must be a gaussian prime? Thank you for reading this and any help at all would be appreciated.

• I recommend the treatment in the first chapter of Neukirch's "Algebraic number theory", it is very clear. For this specific part, he proves (as first thing in the book) that a prime which is a sum of two squares must be congruent to $1$ mod $4$, hence here you'd get a contradiction – 57Jimmy Apr 12 '18 at 15:05
• See also similar questions at this site, e.g., here and here. – Dietrich Burde Apr 12 '18 at 15:09
• $a^2+b^2$ cannot be $3 \pmod 4$ for (rational) integers $a, b$. – metamorphy Apr 12 '18 at 16:11