[Math] Can a number have different prime factorizations

Question

Given a number $n$, can $n$ be decomposed into different prime factors? For example:

$c\times d = n$, where $c$ and $d$ are both prime factors.

Could there also be a $e\times f = n$ where $e$ and $f$ are also prime and do not equal $c$/$d$?

Answer 1

No:

We first need a Lemma: if $p$ is prime and $p$ divides $ab$ then $p$ divides $a$ or $p$ divides $b$.

Proof: Assume that $a$ and $n$ are relatively prime and that $n$ divides $ab$. By Bezout's identity, there exist $r$ and $s$ such that $ra + sn = 1$ which upon multiplying by $b$ yields $rab + snb = b$. $n$ divides $ab$ (by hypothesis) which itself divides the first term on the left. Further, $n$ divides the second term on the left. Therefore, $n$ divides $b$. To prove the lemma, take $p = n$ and note that either $p$ divides $a$ (in which case we're done) or $p$ is coprime to $a$ (as it is prime) and therefore by the above discussion $p$ divides $b$.

Now to answer your question:

Assume by way of contradiction that $n = p_1...p_n = q_1...q_n$ with the $p_i$ and $q_i$ here not necessarily distinct and further assume without loss of generality that $p_1<p_2<...<p_n$, $q_1<q_2<...<q_n$ and that $p_1<q_1$. $p_1$ clearly divides $n$. Thus, by our lemma above, $p_1$ divides $q_i$ for some $i$. But $p_1<q_i$ are both prime (therefore they are coprime) so that $p_1 = 1$, a contradiction.