The question is as per the subject line. Does the fundamental theorem of arithmetic imply (prove?) that if I multiply two primes, then those two primes are the only factors of the product? I.e. 17 x 23 = 391 and the only factors of 391 are 17 and 23. I have searched, but can't find an answer to this precise question.
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closed as off-topic by figlesquidge, DrLecter, AFS, Thomas, e-sushi♦ Apr 25 '14 at 16:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
Yes, that is true (unless you count the trivial factors of 391, namely 1 and 391). The fundamental theorem of arithmetic states that any postive integer is a product of a unique multiset of primes, and that no other multiset will multiply together to form that integer.
And so, since $391$'s multiset of primes is $(17, 23)$ (because both $17$ and $23$ are prime, and $17 \times 23 = 391$), that implies that no other prime can possibly divide $391$. If there were such a prime $p$, then we'd have an alternative multiset that consisted of $p$ and a multiset that was the factorization of $391/p$; that would be a different multiset (because, by assumption, $p$ is neither $17$ nor $23$) with the same product, and we know that can't happen.