# Given $N(\alpha) = pq$, $p$ ≡ $3$ mod $4$ and $q$ ≡ $3$ mod $4$, prove that $p = q$.

Suppose that $$p,q \in \mathbb{N}$$ are prime numbers and $$\alpha \in \mathbb{Z[i]}$$ satisfy $$N(\alpha) = pq$$, $$p$$$$3$$ mod $$4$$ and $$q$$$$3$$ mod $$4$$. Using the fact that $$p$$ and $$q$$ remain irreducible in $$\mathbb{Z[i]}$$, prove that $$p = q$$.

Proof attempt: Assume false, that is, $$p \neq q$$. As $$N(\alpha) = \alpha\alpha^* = pq$$, we have $$p \mid \alpha\alpha^*$$ in $$\mathbb{Z[i]}$$. Given that $$p$$ is congruent to $$3$$ modulo $$4$$, we have $$p$$ is irreducible in $$\mathbb{Z[i]}$$. Then $$p$$ is prime implying $$p \mid \alpha$$ or $$p \mid \alpha^*$$ in $$\mathbb{Z[i]}$$. Taking norms, we get in both cases that $$p^2 \mid pq$$ in $$\mathbb{Z}$$. As $$p$$ and $$q$$ are primes in $$\mathbb{Z}$$, we get $$p = q$$. Hence a contradiction so $$p = q$$.

This is my attempt. I'm not certain it's correct. The reason being I haven't used the fact that $$q$$ is congruent to $$3$$ modulo $$4$$.

I think a proof is okay and a more general fact holds: since the norm is multiplicative, then $$s | x$$ in $$\mathbb{Z}[i]$$ implies $$x = sy$$ for some $$y$$ and

$$N(x) = N(s)N(y),$$

so that $$N(x)| N(y)$$.

If $$p$$ is a Gauß prime, then $$p \mid N(x)= xx^\ast$$ implies $$p \mid x$$ and thus $$p^2 = N(p) \mid N(x)$$. In your case, we have $$p^2 \mid N(x) = pq$$, hence $$q = p$$ by uniqueness of prime factorization (in $$\mathbb{Z}$$).

In general, gaussian primes are of the form $$pi$$ and $$p$$ with $$p \in \mathbb{4\mathbb{N}_0}+3$$ or $$a+ib$$ such that $$ab \neq 0$$ and $$N(a+ib)$$ is prime in $$\mathbb{Z}$$. Thus, if $$x$$ has a certain prime factorization

$$x = p_1^{n_1} \cdots p_s^{n_s}f_1^{m_1} \cdots f_r^{m_r}$$

in $$\mathbb{Z}[i]$$, with each $$p_j \in 4\mathbb{Z}_0 +3$$ and $$q_i = f_i^2 \in \mathbb{N}$$ primes, taking norms we get

$$N(x) = N(p_1)^{n_1} \cdots N(p_s)^{n_s}N(q_1^{m_1})\cdots N(q_r^{m_r}) = p_1^{2m_1} \cdots p_s^{2n_s}f_1^{m_1} \cdots f_r^{m_r}.$$

This shows that gaussian primes appearing in the factorization of $$N(x)$$ for some $$x \in \mathbb{Z}[i]$$ always have even exponent.

Going back to the problem of the question, since $$N(x) = pq$$, either $$p = q$$ or no Gaussian primes divide $$x$$. But since $$p \equiv 3 \pmod{4}$$, the former is true. No extra hypotheses on $$q$$ are needed.