If an integer is divisible by a prime then that number must appear in it's unique factorization.
If we don't already know (prime) factorizations are unique, how do we know that, if an integer is divisible by a prime, that prime must appear in every possible factorization of that integer?
"Prime factorizations are unique" is equivalent, in a straightforward way, to "If p is a prime factor of x, then p appears in every prime factorization of x". It's easy to deduce either of these from the other. But neither of these statements is obviously true.
They turn out to be true for the integers, but seeing why they are true requires an extra, not at all obvious insight (the reasoning invoked above via the Euclidean algorithm).
no, whether or not it is obvious is under contention. I hope no one doubts what I said.
I also never said it's obvious, I said the statement
> if it's divisible by a number, the number must appear in its factorization
isn't true, and wouldn't be claimed.
PepeGomez claimed "If it's divisible by a number, the number must appear in its factorization and vice versa." in apparent support of the argument that uniqueness of prime factorizations is therefore obvious.
In response to this, I wrote my initial comment. When, in it, I asked "Who says 'if it's divisible by a number, the number must appear in its factorization'? Why is that true?", that was in response to PepeGomez, a rhetorical way of engaging with the claim they made.
I further noted in my comment that this claim about divisibility and factorizations wasn't quite correct for arbitrary numbers, but was true for prime numbers, but is nonetheless non-obvious for prime numbers. It appears you agree with me on all of this, so... great.