Prove that there exist infinitely many Gaussian integers with the following properties Prove that there exist infinitely many Gaussian integers $\alpha$ with the following properties:

*

*$N(\alpha)$ is the product of two prime numbers.

*At least one of the potential factors of $\alpha$, as discovered by the “norm method” for determining factors, is not actually a factor.

[Hint: you may assume without proof the fact that there are infinitely many prime numbers that are congruent to 1 modulo 4.]
This is an exercise I'm trying to do. I've listed my attempt below. I get stuck at the final case though I'm not even certain my proof is correct up until that point.
Let $\alpha \in \mathbb{Z[i]}$ and $\beta = a + bi$ a factor of $\alpha$ in $\mathbb{Z[i]}$. Then $\alpha = \beta\gamma$ for some $\gamma$ in $\mathbb{Z[i]}$.  Taking norms, $N(\alpha) = N(\beta)N(\gamma)$. Suppose that $N(\alpha) = pq$ for primes $p,q$ in $\mathbb{Z}$. Then $N(\beta)N(\gamma) = pq$. We have 4 cases to obtain potential factors of $\alpha$ in $\mathbb{Z[i]}$.
We ignore two cases, $N(\beta) = 1$ and $N(\beta) = pq$.  In both of these cases, $\beta$ and $\gamma$ are units or associates of $\alpha$. These are always factors of $\alpha$ so are not interesting to the proposition.
Then we are left with two cases to consider, $N(\beta) = p$ and $N(\beta) = q$.
When both $p,q$ are congruent to $3$ modulo $4$, $p,q$ cannot be written as a sum of squares.  So $\alpha$ is irreducible in $\mathbb{Z[i]}$ and there are no factors to consider.  There are infinitely many primes congruent to $3$ modulo $4$ (by the hint) so there are infinitely many such $\alpha$. We are done.
Now suppose only one of $p,q$ are congruent to $1$ modulo $4$. WLOG, let $p$ be congruent to $1$ modulo $4$. Then $q$ is congruent to $3$ mod $4$ so $q$ cannot be written as a sum of squares. We are done similar to the previous case.
Finally, consider when both $p,q$ are congruent to $1$ modulo $4$.
 A: You are only asked to demonstrate that there are infinitely many numbers with this property.  You don't have to exhaust the cases of $N(\alpha)$, just to find an infinite set that have this property.  For an extreme example, if you could show that the norm of all the Fermat numbers $2^{2^n}+1$ for $n \gt 1000$ was the product of two primes and one was not a factor, you would be done.
$N(\alpha)$ is always a natural and you are looking for natural number factors of it, not members of $\Bbb Z[i]$.  You then look for members of $\Bbb Z[i]$ that have that norm and ask if they are factors of $\alpha$
When $p,q$ are equivalent to $3 \bmod 4$ it is true that they cannot be written as the sum  of two squares, but neither can $N(\alpha)$.  This comes from the sum of two squares theorem.  As an example, $p=3, q=7$ gives a proposed $N(\alpha)$ of $21$.  As $21$ cannot be written as a sum of two squares, there are no numbers in $\Bbb Z[i]$ with norm $21$.
On the other hand, consider $65=5 \cdot 13$, which is the product of two primes equivalent to $1 \pmod 4$.  We know there are two ways to write it as the sum of two squares, to wit $65=8^2+1^2=7^2+4^2$.  We then know that $8+i$ and $7+4i$ have norm $65$.  Some factor of $8+i$ must not be a factor of $7+4i$, but this factor would be suggested by the norm method.  This applies to any product of two primes equivalent to $1 \pmod 4$
