How to prove $a^{k\times \phi(n)+1} \equiv a\ (mod\ n$)?

Question

Sorry for my poor question, but I cannot prove this even if it looks so easy..
I know $a^{\phi(n)} \equiv 1\ (mod\ n)$, but how can I compute $a^{k\times \phi(n)}\ (mod\ n)$?

You know the crux part. Now just use laws of exponents. (Assuming that $a$ is coprime to $n$.) — Randall, Oct 04 '18 at 02:29
Take the $,\large k$'th power of $\ \large a^\phi\equiv 1\ $ (valid by the Congruence Power Rule) $\ $ — Bill Dubuque, Oct 04 '18 at 02:45

score 2 · Accepted Answer · answered Oct 04 '18 at 02:31

2

Hint:

$$a^{k\phi(n)} = \left(a^{\phi(n)}\right)^k\equiv ...$$

answered Oct 04 '18 at 02:31

Carl Schildkraut

32,931

Thanks for the law of exponents – baeharam Oct 04 '18 at 02:32
@baeharam No problem. – Carl Schildkraut Oct 04 '18 at 02:37
when he gets it.... – Saketh Malyala Oct 04 '18 at 02:42

How to prove $a^{k\times \phi(n)+1} \equiv a\ (mod\ n$)?

1 Answers1