The question is in which case(s) `f(x) = x^n mod m` is a bijection for `0 <= x < m`. My intuition is to guess that `gcd(n,m) = 1` is at least a sufficient cond

The question is in which case(s) `f(x) = x^n mod m` is a bijection for `0 <= x < m`. My intuition is to guess that `gcd(n,m) = 1` is at least a sufficient condition (correct math term? In german it's "hinreichende Bedingung"). Maybe it's the sole necessary condition. But I did only think about this for 5 to 10s and did not spend too much time on looking for a counter example, let alone a proof strategy (my number theory is a bit rusty -- ah, but on second thought, if the theory is correct, then I bet the Chinese remainder theorem is involved).