Why $x^{a\cdot b} \equiv x \pmod{n}$ where $b$ is the inverse of $a$ in $\mathbb{Z}_n$?

Question

I'm currently studying the RSA algorithm and while I understand why if $b$ is the inverse of $a$ in $\Bbb Z_n $ then $ab \equiv 1\pmod{n}$. I don't quite understand why $x^{ab} \equiv x \pmod{n}$. Can I just replace $ab$ with $1$? A proof showing why this holds would be really helpful.

When $a\in\mathbb Z_n$, the expression $x^a\pmod n$ is not well-defined... — Kenta S, Apr 07 '21 at 19:01
You misread, it should say $,ab\equiv 1\pmod{\color{#c00}{\phi(n)}},$ and proofs are in the linked dupe. Post further question in comments here or there if anything remains unclear. — Bill Dubuque, Apr 07 '21 at 19:05
@BillDubuque I didn't misread, they mistyped. My comment is legitimate as a response to the question they asked. — Kenta S, Apr 07 '21 at 19:28
@KentaS Now you have misread. My prior comment is to the OP, not you (if it was addressed to you it would begin like this comment at-you) — Bill Dubuque, Apr 07 '21 at 19:30
@BillDubuque Yes,I did actually misread the proof...It's clear now,thanks a lot! — sliiime, Apr 08 '21 at 00:31

Hagen von Eitzen · Answer 1 · 2021-04-07T19:50:14.773

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Rather, we have $$ x^{ab}\equiv x\pmod n$$ if $$ ab\equiv 1\pmod{\phi(n)}\quad\text{and}\quad \gcd(x,n)=1.$$

edited Apr 07 '21 at 19:50

answered Apr 07 '21 at 18:58

Hagen von Eitzen

374,180

You mean $\gcd(x,n)=1$, right? – Mike Apr 07 '21 at 19:07
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@Mike Yes, off by one key :) – Hagen von Eitzen Apr 07 '21 at 19:50
another connection for $a,b$ is when written in bsse $\phi(n)+1$ they have a product of digital roots that is congruent to $1\bmod\phi(n)$ – Roddy MacPhee Apr 07 '21 at 23:26
Thanks a lot for the comment.This,and the post that was linked as the original question made things clear.I had actually misread the proof..This was bothering my mind for 2 days in a row.I need to be more focused. – sliiime Apr 08 '21 at 00:30

Why $x^{a\cdot b} \equiv x \pmod{n}$ where $b$ is the inverse of $a$ in $\mathbb{Z}_n$?

1 Answers1