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

Question

I have been stuck with this one for a while now.

$a,b,m$ are natural numbers not including $0$. $\{1,2,3,4,\ldots\}$.

I have managed to prove that $ \gcd(ab,m) < ( \gcd(a,m) \cdot \gcd(b,m) )$ but I keep failing on proving that $$\gcd(ab,m) \mid ( \gcd(a,m) \cdot \gcd(b,m) )$$

Thanks.

Answer 1

1) if $(a,m)=1$ then $(ab,m)=(b,m)$.

2) if $(a,m)=k>1$, then $(ab,m)=k((a/k)b,m/k)=k(b,m/k)=(a,m)(b,m/k) \mid (a,m)(b,m)$.

Answer 2

$(a, m) \cdot (b, m) = x_1 x_2 a b + x_1 y_2 a m + y_1 x_2 b m + y_1 y_2 m^2$ for some integers $x_1, x_2, y_1, y_2$. We have

$$ x_1 x_2 a b + x_1 y_2 a m + y_1 x_2 b m + y_1 y_2 m^2 = x_1 x_2 \cdot a b + (x_1 y_2 a + y_1 x_2 b + y_1 y_2 m) \cdot m, $$ and the result follows.