I would like some advice if I have approached this problem correctly please:
let $a,b,m,n \in \mathbb{Z}$ and $m,n > 0$. Prove that if $a\equiv b \pmod n$ and $m|n$, then $a\equiv b\pmod m$
Proof:
Assume $a\equiv b \pmod n$ and $m|n$, then by definition:
$$n=lm \tag{$l\in \mathbb{Z}$}$$ and $$ a-b=kn\tag{$k\in \mathbb{Z}$}$$
$$\therefore a-b=m(lk)$$ Without loss of generality let $$p = lk\tag{$p\in \mathbb{Z}$}$$ $$\therefore a-b=mp$$ Therefore $m|(a-b)$ and so $a\equiv b\pmod m$. Hence Proof.