Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 16 down vote accepted

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

share|cite|improve this answer

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.

share|cite|improve this answer
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. – Tobia Tesan Jul 19 '12 at 18:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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