If x = a (mod p) and x = a (mod q), then x = a (mod pq)

The Mathematician — Tue, 15 Nov 2022 13:00:30 +0000

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*}

By the fact that each integer has a unique prime factorization, we see that

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

where

r_i

are distinct primes and

p = r_j

and

q = r_k

for some

i \in \{1,2,\ldots,l\}

. Therefore, we get

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

which implies that

x \equiv a \ (\text{mod } pq)

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then x = a (mod pq) Archives - Epsilonify

If x = a (mod p) and x = a (mod q), then x = a (mod pq)