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.


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


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.

  • 4
    $\begingroup$ Don't worry, happens to everyone :) $\endgroup$ – Zev Chonoles Jul 19 '12 at 18:22
  • 2
    $\begingroup$ @ZevChonoles Thanks for not embellishing your answer with a much deserved Nelson Muntz "ha-ha", anyway. $\endgroup$ – Tobia Tesan Jul 19 '12 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.