Suppose $a|m$, $b|m$ and $\gcd(a,b) = 1$. Prove, without appealing to the fundamental theorem of arithmetic, that $ab|m$.
I know that $\gcd(a,b)=1$ means they are relatively prime. I also know that $a|m$ means $m=as$ and that $b|m$ means $m=bt$ and lastly that $\gcd(a,b)=1$ means $1=ax+by$. I just have no idea what to do next. Any help would be appreciated. Thanks!
If $\gcd(a,b)=1$, then there exists integers $s$ and $t$ such that $as + bt = 1$. If $a|m$ and $b|m$ then there exists integers $S$ and $T$ such that $m = Sa = Tb.
\begin{align} as + bt &= 1\\ mas + mbt &= m\\ Tbas + Sabt &= m\\ ab(Ts + St) &= m\\ \end{align}
It follows that $ab | m$