# 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 $$4$$.

• I've done that now. Please undo your downvote. Dec 21, 2020 at 3:23
• Thanks. I have undone the downvote as well given an upvote Dec 21, 2020 at 5:19

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 $$1 \pmod 4$$
• I assume that your comments from the second paragraph onwards are in regards to my proof. I know that $N(\alpha)$ is natural. I'm assuming that I have $\alpha$ satisfying the first property, that is, $N(\alpha) = N(\beta)N(\gamma) = pq$ and then I look at what $N(\beta)$ can be. We know that $N(\beta) = a^2 + b^2$. When $p,q$ are congruent to $3$ mod $4$, $N(\beta)$ and $N(\gamma)$ cannot be written as a sum of squares so these cases are impossible. Thus $\alpha$ is irreducible and satisfies the properties. There are infinitely many such $\alpha$. Dec 21, 2020 at 4:00
• But neither are there any $\alpha$ with $N(\alpha)=pq$ when $p,q \equiv 3 \bmod 4$, so there are certainly not infinitely many such $\alpha$. Write $\alpha=a+bi$, then $N(\alpha)=a^2+b^2$. This shows the norm is expressible as the sum of two squares, but a number of the form $pq$ with $p,q \equiv 3 \bmod 4$ is not expressible this way. There are no $\alpha$ with $N(\alpha)=21, 33, 77, 57$ or so on Dec 21, 2020 at 4:02
• I think you should be looking at the sum of two squares theorem and the conditions that make there be more than one way to express $pq$ as a sum of two squares. For example, $65=5\cdot 13=8^2+1^1=7^2+4^2$ so both $8+i$ and $7+4i$ have norm $65$ Dec 21, 2020 at 4:22