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Given $(a,m) = 1$, I want to show that $a ^e \equiv a^f \pmod m \implies e \equiv f \pmod {\phi(m)}$. I already know how to show the converse, and so far I have the following:

Suppose wlog that $e \leq f$. Then, $a^{f-e} \equiv 1 \pmod m$, so by Euler-Fermat we have $a^{f-e} \equiv a^{\phi(m)} \pmod m$. Now how do I finish the proof?

b_pcakes
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    It's not true. $4^2\equiv 4^4\pmod 5$, but $2\not\equiv 4\pmod {4}$. – Thomas Andrews Sep 29 '16 at 19:12
  • The other direction is false in general. Consider the case $a=-1$. We have $a^f\equiv a^e\pmod m$ if and only if $f\equiv e\pmod 2$. For any $m>2$. – Jyrki Lahtonen Sep 29 '16 at 19:13
  • I see, is there a stronger hypothesis that would make it true? – b_pcakes Sep 29 '16 at 19:13
  • For some $m$ there is no $a$ such that conclusion would be true. There do exist such $a$ when $m$ or $m/2$ is a power of an odd prime (or $m\mid 4$). What holds in general is that $a^e\equiv a^f\implies e\equiv f\pmod \ell$, where $\ell$ is the smallest positive integer such that $a^\ell\equiv1\pmod m$. And we know that always $\ell\mid \phi(m)$. – Jyrki Lahtonen Sep 29 '16 at 19:16
  • @JyrkiLahtonen Actually, I want to use this in a bigger problem that i've posted here: http://math.stackexchange.com/questions/1946983/prove-that-an-1-equiv-1-pmod-n-implies-n-is-prime

    Can you take a look at that?

     – b_pcakes Sep 29 '16 at 19:32

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