Prove that if $a\mid m$, $b\mid m$ and $(a,b)=1$, then $ab\mid m.$
I'm currently studying Elementary Number Theory, and there's a simple proof for that, but I'd like to check if the proof that I did, that is not as trivial as the one from the book, is correct.
Here it is:
If $m = 0$ then the theorem is proved. If $m \neq 0$, let $qa=m$ and $rb=m$ hence $abqr=m^2$. Since $(a,b)=1$ it follows that $xa + yb = 1$, hence: $$ x\left(\frac{m}{q}\right)+y\left(\frac{m}{r}\right)=1 \rightarrow m\left(\frac{x}{q}+\frac{y}{r}\right)=1 $$ Now, multiply both sides of $abqr=m^2$ by the previous equation: $$ m\left(\frac{x}{q}+\frac{y}{r}\right)abqr=m^2\\ \left(\frac{x}{q}+\frac{y}{r}\right)abqr=m\\ ab(xr+yq)=m \rightarrow ab | m\\ $$
I'm just a beginner so I apologize for any huge mistakes... Thanks!
I see no mistakes. However, note that, when you wrote “it follows that $xa + yb = 1$”, you should have written “it follows that there are integers $x$ and $y$ such that $xa + yb = 1$”.