If \(a ≡ b \bmod m\), then there is an integer \(k\) such that:
$$ mk = b - a $$
If \(n|m\), then there is an integer \(q\) such that:
$$ qn = m $$
substituting \(m\):
$$\begin{aligned} (qn)k &= b - a \\ n(qk) &= b - a \implies a ≡ b \bmod n \end{aligned}$$