Cancellability of Congruences

Theorem
Let $a, b, c, n \in \Z$ be integers.

Then:


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

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

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 Integers Divided by GCD are Coprime we have:
 * $\exists r, s \in Z: r \perp s: c = d r, n = d s$

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

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

Now suppose $a \equiv b \pmod {n / d}$ 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 $d \mathrel \backslash c$ and so $\dfrac c d \in \Z$.

So:
 * $c a \equiv c b \pmod n$