Is $\{n: n = 3k + 1\}$ a unique factorization domain? (We are factoring into numbers of form $3k + 1$)

Question

Call a number of form $3k + 1$ a prime if and only it cannot be factored into smaller numbers all of form $3k + 1$. The set of all numbers of this form is obviously closed, and every number of this form clearly has at least one factorization. Is it unique? Prove or give counterexample.

Answer 1

I can tell that what you meant to ask is whether every number of the form $3k + 1$, with $k \geq 0$ and $k \in \mathbb{Z}$ can be uniquely factored into numbers of that form. The answer is no.

The first few numbers are $$1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, \ldots$$

Just looking at these few numbers, one may be led astray and think that there is indeed unique factorization among them. Mwahahaha!

But here's one way to find a bunch of counterexamples: think of a prime number $p$ of the form $3k - 1$. Then $p^2 \equiv 1 \pmod 3$, and so is $2p$. Then $4p^2 = (2p)^2$, readily leading to the examples $100 = 4 \times 25 = 10^2$, $484 = 4 \times 121 = 22^2$, $1156 = 4 \times 289 = 34^2$, etc.

Not too different from $4k + 1$.