2
$\begingroup$

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 $4$.

$\endgroup$
3
  • $\begingroup$ Please post the contents of your linked question here. It is advisable not to use links for crucial parts of your post. $\endgroup$
    – Paramanand Singh
    Commented Dec 21, 2020 at 3:19
  • $\begingroup$ I've done that now. Please undo your downvote. $\endgroup$
    – someone1
    Commented Dec 21, 2020 at 3:23
  • $\begingroup$ Thanks. I have undone the downvote as well given an upvote $\endgroup$
    – Paramanand Singh
    Commented Dec 21, 2020 at 5:19

1 Answer 1

2
$\begingroup$

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 $1 \pmod 4$

$\endgroup$
4
  • $\begingroup$ 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 $\alpha$. $\endgroup$
    – someone1
    Commented Dec 21, 2020 at 4:00
  • $\begingroup$ 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 $\endgroup$ Commented Dec 21, 2020 at 4:02
  • $\begingroup$ Ah, I see. Thank you. I'll have to look at my argument again or try something else. I appreciate your Fermat numbers idea but I believe the question is intended to be answered with norm properties and irreducibility in the Gaussian integers. And possibly the sum of squares theorem you mention which was taught in the course. $\endgroup$
    – someone1
    Commented Dec 21, 2020 at 4:10
  • $\begingroup$ 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 $65$ $\endgroup$ Commented Dec 21, 2020 at 4:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .