Show that: $a·c \equiv b·c\ (\text{mod }m)$ with $a, b, c$ and $m$ integers with $m \ge 2$ does not imply $a \equiv b\ (\text{mod }m)$

Question

Show that $a·c\equiv b·c\ (\text{mod }m)$ with $a, b, c$ and $m$ integers with $m \ge 2$ does not imply $a \equiv b\ (\text{mod }m)$

I've seen many similar examples, but can't seem to find a step by step explanation or solution to any of them (like an actual proof).

Answer 1

$2\cdot2\equiv 2\cdot5(mod\ 2)$ but $2\not\equiv 5(mod\ 2)$

Answer 2

Here is a counter example. $a=b+1$ , $c=m$

Answer 3

Here is a counterexample: $4*4=8*4(mod16)$ but $4\neq8(mod16)$. Here $4$ is a zero-divisor of the ring $\mathbb{Z}/16\mathbb{Z}$. i.e, an element $a \neq 0$ such that there is an no zero element $b$ such that $ab=0$.