TITLE: Does the product of two primes only have those primes, itself, and 1 as factors?
QUESTION [5 upvotes]: Happened to review RSA key generation today, and it has led me to this question. I've tried it with a few but obviously that doesn't make it mathematically proven.

REPLY [3 votes]: If you're only looking at positive integers, then yes, of course, the product of two positive primes has only $1$, the two primes and itself as divisors.
Call $\sigma_0(n)$ the function that tells you how many positive divisors a positive number $n$ has. Then, if $p$ and $q$ are primes, then $\sigma_0(pq) = 4$ and those four divisors are $1, p, q, pq$. Example: $65 = 5 \times 13$. The only positive divisors are $1, 5, 13, 65$.
If you broaden your view to the negative integers, you can just take the positive divisors and multiply each by $-1$ to get the negative divisors: $-65, -13, -5, -1$.
And if you further broaden your view to a domain in which those numbers are not primes, things get more interesting: $(2 - i)(2 + i)(2 - 3i)(2 + 3i) = 65.$