Divisibility proof with GCD condition

Question

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!

duplicate of http://math.stackexchange.com/questions/1416335/ — user84413, Sep 01 '15 at 00:43
Multiply by $m$ and argue that $ab$ divides each term of the right-hand side. — André Nicolas, Sep 01 '15 at 00:46

score 0 · Answer 1 · answered Oct 04 '15 at 20:08

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 $ab | m$

Divisibility proof with GCD condition

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