Does $ak \equiv bk \textrm{ mod }m \implies a \equiv b \textrm{ mod } m$?

Question

I'm wondering the following: I know that $a \equiv b \textrm{ mod } m \implies ac \equiv bc \textrm{ mod } m$ for $a,b,c,m \in \mathbb{Z}$ and $m \neq 0$. I'm wondering that if we know that $\gcd(c,m) = 1$ and that $ac \equiv bc \textrm{ mod } m$, is $a \equiv b \textrm{ mod } m$? If it is true, how would we prove it?

Answer 1

You know that $(a-b)c$ is divisible by $m$, and you want to know if $a-b$ is divisible by $m$.

If $m$ is prime and does not divide $c$ this is certainly true. It is false in general, since the factors of $m$ could be shared between $a-b$ and $c$.

If $gcd(m,c)=1$ the same conclusion is true, just because none of the prime factors of $m$ are present in $c$ so they must be present in $a-b$.