Prove or disprove if ma ≡ mb (mod n), then a ≡ b (mod n) for all positive integers a, b, m, n.

Question

How can I prove/disprove if ma ≡ mb (mod n), then a ≡ b (mod n) for all positive integers a, b, m, n. I know the fact that

if (m,n) = d, then ma ≡ mb (mod n) if and only if a≡b (mod n/d)

I have no idea what should I do?

$x\to mx,$ is injective ($1$-to-$1)!!\iff!! m,$ is invertible $\bmod n!!\iff!! \gcd(m,n) = 1,,$ see my answer in linked dupe — Bill Dubuque, Dec 12 '19 at 20:09

score 1 · Accepted Answer · answered Dec 12 '19 at 08:13

This is not true in the general case. A counter example:

Modulo $6$ we have

$$3\cdot 1\equiv 3\cdot 5\pmod 6$$

The case where it works is when $m$ is not a zero divisor modulo $n$. It is the case when $(m,n)=1$ that is to say $m$ and $n$ are relatively prime.

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