Unique Factorization proof with GCD I have this math question that I am kind of stuck on.

Show that if $\gcd(m_1,m_2) = 1$ and $n$ is an integer with $m_1\mid n$
   and $m_2\mid n$, then $m_1m_2\mid n$, using unique factorization.

I'm not 100% how to start this. I already proved this using Bezout's identity. Thanks
 A: Write $n=p_1^{a_1}p_2^{a_2}\dots p_k^{a^k}$ as product of prime powers ($p_1,\dots,p_k$ pairwise distinct and $a_i>0$).
Then you can write
$$
m_1=p_1^{b_1}p_2^{b_2}\dots p_k^{b_k},\qquad
m_2=p_1^{c_1}p_2^{c_2}\dots p_k^{c_k}
$$
where $0\le b_i\le a_i$ and $0\le c_i\le a_i$ ($i=1,2,\dots,k$). Then
$$
m_1m_2=p_1^{b_1+c_1}p_2^{b_2+c_2}\dots p_k^{b^k+c_k}
$$
and $m_1m_2\nmid n$ implies $b_i+c_i>a_i$, for some $i$. In particular $b_i>0$ and $c_i>0$, so $p_i$ divides both $m_1$ and $m_2$.
A: You know that $m1$ and $m2$ are relatively prime, since their $gcd$ is 1. In other words, when written as a product of primes, they do not share any prime divisors.
Since $m1$ divides $n$, and $m2$ divides $n$, then you also know that $n$ written as a product of primes contains all those prime factors of $m1$, as well as those of $m2$. You can then conclude that $m1m2$ divides $n$ by using your definitions of what it means for an integer to divide another.
A: There is a simpler proof without the sophisticated unique factorisation tool, based on Gauß's lemma:

If a number $n$ divides a product $ab$, and $n$ and $a$ are coprime, then $n$ divides $b$.

Indeed, if $m_1$ divides $n$, we can write $n=m_1 n'$. Now $m_2$ divides $n=m_1n'$, and $m_2$ and $m_1$ are coprime. Thus, by Gauß's lemma, $m_2$ divides $n'$, i.e. $n'=m_2n''$ for some $n''$, and we have
$$n=(m_1m_2)n''.$$
