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Under which conditions does $a^n \equiv 1\mod(b) \Rightarrow\ a^{n^m} \equiv 1\mod(b) $? What about viceversa?

What is the strongest result(s) that can be proved regarding this kind of thing?

I'm kind of getting stuck with following a passage in an example, and anyway I have a general confusion in my head about this which I haven't been able to dispel with googling, searching and grepping, hence asking as my last resort.

Thanks a lot.

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2 Answers

For any $m\geq 1$, if $a^n\equiv 1\bmod b$, then $$a^{(n^m)}=a^{n\cdot (n^{m-1})}=(a^n)^{n^{m-1}}\equiv (1)^{n^{m-1}}\equiv 1\bmod b.$$

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I am a silly person, because $a\equiv b\pmod{n}, c\equiv d\pmod{n} \Rightarrow ac\equiv bd\pmod{n}$ means that by substituting 1 for b and d and $a^n$ for a, c it can be shown that the first congruence holds.

It's hot and my brain is toast. Sorry.

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Don't worry, happens to everyone :) –  Zev Chonoles Jul 19 '12 at 18:22
@ZevChonoles Thanks for not embellishing your answer with a much deserved Nelson Muntz "ha-ha", anyway. –  tobiatesan Jul 19 '12 at 18:26
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