If $\gcd(m, n) = d$, prove that the system $$ \begin{cases} x\equiv a \pmod m \\ x\equiv b \pmod n \\ \end{cases} $$
has a soution iff $a \equiv b \pmod d$.
I'll go $\leftarrow$ just to see if I am going in the right direction:
$mu + nv = 1$ implies
$mu \equiv 0 \pmod m$,
$nv \equiv 1 \pmod m $,
$nv \equiv 0 \pmod n$,
$mu \equiv 1 \pmod n$
$b - a = dk$ and $mu + nv = d \rightarrow muk + nvk = b - a$. Then,
$t = muk + nvk = k \pmod m$
$t = muk + nvk = k \pmod n$
Does this $t$ make sense?