If (a ≡ b mod m) And (a ≡ b mod n) Then (a ≡ b mod [m, n])

If \((a ≡ b \bmod m)\) and \((a ≡ b \bmod n)\), then we can use this lemma:

$$\begin{gather} m|(b-a), \quad n|(b-a) \\ \therefore [m, n]|(b-a) \end{gather}$$