Under what conditions can we obtain $a \equiv 1 \pmod{mn}$ from $a \equiv b \pmod{m}$ and $b \equiv 1 \pmod{mn}$?

Question

If $a \equiv b \pmod{m}$ and $b \equiv 1 \pmod{mn}$, are there any conditions under which we can conclude that $a \equiv 1 \pmod{mn}$?

Here $m$ and $n$ are any integers; $a$ and $b$ are both coprime to $mn$.

Answer 1

We know that we can find integers $k,\ell$ such that

$$a = b + km$$ $$b = 1+ \ell mn$$

Hence,

$$a = 1+\ell mn+km = 1 + (\ell n+k)m$$

So if we want $a \equiv 1 \pmod{mn}$ then we need $\ell n+k \equiv k \equiv 0 \pmod{n}$. Hence

$$a \equiv b \pmod{mn}$$