Can a number have a prime factor that isn't a part of the number's prime factorization? Is it possible to have a number, $x$, divisible by some prime, such that that prime does not appear in the unique prime factorization of $x$?
 A: If we are working in a unique factorisation domain, then this is impossible: if $p$ divides $x$, then $x = p y$ for some $y$, and so $p$ (or an associate) must appear in the prime factorisation of $x$, which is unique by hypothesis.
In fact, if we go by the strict definition of prime, then if $p$ is prime and divides $x$, it (or an associate) must appear in any factorisation of $x$ into irreducibles, regardless of whether the factorisation is unique.
However, there are rings which are not unique factorisation domains. For example, in the ring $\mathbb{Z}[\sqrt{-5}]$, $(1 - \sqrt{-5})(1 + \sqrt{-5}) = 6 = 2 \cdot 3$, but $2$, $3$, and $1 \pm \sqrt{-5}$ are all irreducible and non-associate. (Of course, this also shows that none of them are prime.)
A: let $p$ divide $x$, where $x = p_1p_2\cdots p_n$ so we have $\frac {x}{p} = \frac {p_1p_2\cdots p_n}{p}=k , k \in \mathbb N$ but if $p \neq p_1 , p \neq p_2 \cdots p \neq p_n  $ that means one of the primes $p_1p_2\cdots p_n$ was divisble by p ! contradicting the fact that it was a prime. So no it is not possible.
A: If $p|x$, then write the prime factorization of $\frac{x}{p} = p_1p_2....p_n$.  But then $x=pp_1p_2...p_n$ is a prime factorization of $x$, and unique factorization shows it is the only prime factorization of $x$.
A: This cannot happen in the natural numbers.  You said your self that factorizations are unique.  If prime $p$ divides $x$, but $p$ doesn't appear in the unique factorization of $x$, then write $x = kp$. Then factor $k$.  
Now you have a second factorization of $x$ that does include $p$ violating your uniqueness assumption.
