Cancellability of Congruences

Theorem

$c a \equiv c b \pmod n \iff a \equiv b \pmod {\dfrac n d}$

where $d = \gcd \left\{{c, n}\right\}$.

Corollary

Let $c \perp n$.

Then:

$c \perp n \iff c a \equiv c b \pmod n \implies a \equiv b \pmod n$

Proof

• Let $c a \equiv c b \pmod n$.

Then we have that $c a - c b = k n$ for some $k \in \Z$ by definition of congruence.

Now $d = \gcd \left\{{c, n}\right\}$, so from Divide by GCD for Coprime Integers we have:

$\exists r, s \in Z: r \perp s: c = dr, n = ds$

So we substitute for $c$ and $n$ in $c a - c b = k n$:

$d r a - d r b = k d s$

which leads us to:

$r \left({a - b}\right) = k s$

So $s \backslash r \left({a - b}\right)$ and as $r \perp s$, from Euclid's Lemma $s \backslash \left({a - b}\right)$.

So $a \equiv b \pmod s$ where $s = \dfrac n d$.

• Now suppose $a \equiv b \pmod {\dfrac n d}$ where $d = \gcd \left\{{c, n}\right\}$.

Then:

$\exists k \in \Z: a - b = k \dfrac n d$

Hence:

$c a - c b = \dfrac {k c} d n$

As $d = \gcd \left\{{c, n}\right\}$ we have $d \backslash c$ and so $\dfrac c d \in \Z$.

So:

$c a \equiv c b \pmod n$

$\blacksquare$

Proof of Corollary

Follows directly from the fact that $c \perp n$ means $\gcd \left\{{c, n}\right\} = 1$.

$\blacksquare$