Let $a$ and $m$ be a positive integers with $a < m$. Suppose that $p$ and $q$ are prime divisors of $m$. Suppose that $a$ is divisible by $p$ but not $q$.
Is there necessarily an integer $k>1$ such that $a^k \equiv a \pmod{m}$?
Or is it that the best we can do is say there are $n>0$ and $k>1$ such that $a^{n+k} \equiv a^n \pmod{m}$
What can be said about $n$ and $k$?
EDIT: Corrected to have $k>1$ rather than $k>0$.
EDIT: The following paper answers my questions about $n$ and $k$ very nicely.
A. E. Livingston and M. L. Livingston, The congruence $a^{r+s} \equiv a^r \pmod{m}$, Amer. Math. Monthly $\textbf{85}$ (1978), no.2, 97-100.
It is one of the references in the paper Math Gems cited. Arturo seems to say essentially the same thing in his answer.