Prove that $\gcd(ab,m)\mid\gcd(a,m)\gcd(b,m)$

Question

Prove that if $a,b,m\in\mathbb N\setminus\{0\}$, then $$\gcd(ab,m)\mid\gcd(a,m)\cdot\gcd(b,m)$$

Answer 1

I am frustrated because the OP has run off with an inelegant idea using prime decompositions. So I thought I would at least try to rectify this by giving the following cryptic hint.

Hint: $\gcd(a, m)\cdot\gcd(b, m)=abX+mY$ for some $X,Y\in\mathbb{Z}$.

Answer 2

If you already know ring theory: Show $(A+M)(B+M) \subseteq AB+M$ for ideals $A,B,M$ of a ring. Apply this to $A=(a),B=(b),C=(c)$ in $\mathbb{Z}$.

Answer 3

Assume $d = \gcd(ab,m)$ and then we get $ab = c d $ and $ m = \overline{m} d $ with $\gcd(c,\overline{m}) = 1 $.

Furthermore let $d_{a} = \gcd(a,m) $ and $d_{b} = \gcd(b,m)$,So you want to prove $d | d_{a}d_{b}$,Please Try it again.