I know this has been hinted at a previous page but I can't seem to find a complete answer.
we know that $\gcd(a,m) = ax_1+mx_2$ from the euclidean algorithm. In a similar way, we know that $\gcd(b,m)=bx_2+mx_3$ and $\gcd(ab,m)=abx_5+mx_6$, and so
I don't understand how we can say that it divides without a remainder.
this is not homework. I'm doing this for sports.