When does $a^k \equiv b^k \pmod m \implies a \equiv b \pmod m$?

Question

I know that $\forall{k}\in\mathbb{N}: a \equiv b \pmod m \implies a^k \equiv b^k \pmod m$.

But the converse is not always true.

A few simple counterexamples:

• $2^2 \equiv 1^2 \pmod 3$
• $2^3 \equiv 1^3 \pmod 7$

I am wondering if there are any theorems where the converse actually holds with some particular conditions being imposed on the variables.

I tried googling but hadn't found anything useful on it yet.

Answer 1

If $a^n=b^n, \pmod{p}$ implies $a=b \pmod p$ then for this to be always true, then $\gcd(n,p-1)=1$. This means, for example, $2^5=x^5 \pmod 7$ implies that $x-2$ is a multiple of 7.

Answer 2

A sufficient condition is that the kernel of the groups-homomorpgism $$\phi _k (x)=x^k $$ is trivial. If it is not trivial you will always get counter examples.