If $a \mid c$ and $b \mid c$, but $\gcd(a,b) = 1$, then $ab \mid c$. If $a | c$ and $b | c$ and $a$ and $b$ are relatively prime prove that
  $ab|c$.


What I did was since $(a,b)=1$ then we can find integers $m,n$ such that $ma + nb=1$. Now since $a|c$ then $a = mc$. Similary, since $b|c$ then $b=nc$. Then I got that: 

$$ma+nb=1$$
$$c(m+n)=1$$
$$m +n = 1/c$$
$$abm +abn=\frac{ab}{c}.$$

Is this correct?
The key ideas are present in your proof.
We know that $a \mid c$ and $b \mid c$. As $a \mid c$, we have that $c = ak$ for some integer $k$. In this form, we see that $b \mid c = ak$, but $\gcd(a,b) = 1$. By the theorem below, we have that $b \mid k$, so that $k = b \ell$. Now $c = ak = ab\ell$, and now it's clear that $ab \mid c$.

Theorem
If $a \mid bc$ and $\gcd(a,b) = 1$, then $a \mid c$.

This can be proved from your application of Bezout's Theorem in your attempted solution. And in fact this is the content of this question and its answers.
If you want to avoid prime factorization, try saying $c = kab + r$ with $0 \leq r &lt; ab$. Then, because $a | c$ and $b|c$, we have that $a | r$ and $b | r$. Since $(a,b) = 1$, what can you say about $r$?