Congruence by Factors of Modulo

Theorem

Let $a, b \in \Z$.

Let $r$ and $s$ be coprime integers.

Then:

$a \equiv b \pmod {r s}$ if and only if $a \equiv b \pmod r$ and $a \equiv b \pmod s$

where $a \equiv b \pmod r$ denotes that $a$ is congruent modulo $r$ to $b$.

Proof

Necessary Condition

This is proved in Congruence by Divisor of Modulus.

Note that for this result it is not required that $r \perp s$.

$\Box$

Sufficient Condition

Now suppose that $a \equiv b \pmod r$ and $a \equiv b \pmod s$.

We have by definition of congruence:

$a \equiv b \pmod r \implies \exists k \in \Z: a - b = k r$

From $a \equiv b \pmod s$ we also have that:

$k r \equiv 0 \pmod s$

As $r \perp s$, we have from Common Factor Cancelling in Congruence:

$k \equiv 0 \pmod s$

So:

$\exists q \in \Z: a - b = q s r$

Hence by definition of congruence:

$a \equiv b \pmod {r s}$

$\blacksquare$

Examples

$n \equiv 7 \pmod {12}$ implies $n \equiv 3 \pmod 4$

Let $n \equiv 7 \pmod {12}$.

Then:

$n \equiv 3 \pmod 4$

Warning

Let $r$ not be coprime to $s$.

Then it is not necessarily the case that:

$a \equiv b \pmod {r s}$ if and only if $a \equiv b \pmod r$ and $a \equiv b \pmod s$

where $a \equiv b \pmod r$ denotes that $a$ is congruent modulo $r$ to $b$.