Modular Exponentiation and Primes

Question

I have tried to search similar posts to this, but unfortunately I may not have enough keywords for it. I'm looking for some help so check the following:

Given a prime number $p$, let $k, m \in\mathbb N \backslash\{0\}$ such that $k \equiv m(\bmod (p-1))$. Now given $ u, v \in\mathbb Z$ such that $u \equiv v\pmod p$, check that $u^k \equiv v^m\pmod p$.

Any tips to show this or where should I start? Thank you!

Would it be something like applying the Theorem to one of the terms to get to another? — Rodri, May 02 '21 at 16:09
Just note that $u^{h(p-1) + k} = [\color{red}{u^{p-1}}]^h\cdot u^k$ and use FLT to figure out what $ [\color{red}{u^{p-1}}]^h\pmod p$ is. — fleablood, May 02 '21 at 16:46
Thank you so much! Needed a bit of time to digest this, but it makes sense! I just get confused where to start with these things. — Rodri, May 02 '21 at 19:58
Apply mod order reduction as described in the linked dupe, using Fermat's little Theorem, and the Congruence Power Rule — Bill Dubuque, May 03 '21 at 01:33

score 1 · Answer 1 · answered May 02 '21 at 16:40

This is a direct result of Fermat's Little Thereom.

If $p$ is prime and $p\not \mid u$ then $u^{p-1}\equiv 1 \pmod p$.

We can conclude: For any $k \equiv m \pmod {p-1}$ then there is an $h \in \mathbb Z$ so that $m = k + h(p-1)$, Wolog lets assume $h \ge 0$ and $m \ge k$. Then $u^{m} = u^{h(p-1) + k} = (u^{p-1})^hu^k \equiv 1^h u^k\equiv u^k\pmod p$.

And if $p$ does divide $u$ then $u\equiv 0\pmod p$ and $u^k \equiv 0 \equiv u^j$ for all $k,j\in \mathbb N$.

Thus for any $u, v: u\equiv v\pmod p$ and $k,m: m\equiv k \pmod{p-1}$ then $u^k\equiv u^m \equiv v^m \pmod p$.

Please strive not to add more dupe answers to dupes of FAQs. See recently updated site policy here. — Bill Dubuque, May 03 '21 at 01:35

Modular Exponentiation and Primes

1 Answers1

Linked