If (a ≡ b mod m), (a ≡ b mod n) And gcd(m, n) = 1, Then (a ≡ b mod mn)

If \((a ≡ b \bmod m)\) and \((a ≡ b \bmod n)\):

$$\begin{gather} m|(b-a) \\ n|(b-a) \end{gather}$$

Since \(\gcd(m, n) = 1\), we can use this lemma here:

$$ mn|(b-a) $$

This means \((a ≡ b \bmod mn)\).

Styles

(uses cookies)