**Let p and q be distinct primes. If x \equiv a \ (\text{mod } p) and x \equiv a \ (\text{mod } q), then x \equiv a \ (\text{mod } pq).**

**Proof.**Since x \equiv a \ (\text{mod } p) and x \equiv a \ (\text{mod } q), we have that

\begin{align*} p &\mid (x - a) \\ q &\mid (x - a). \end{align*}

\begin{align*} x - a = \sum_{i = 1}^l r_i^{m_i} \end{align*}

\begin{align*} pq \mid (x - a). \end{align*}