How to show $h\equiv 1\bmod{(p-1)(q-1)(r-1)}\implies a^h\equiv a \bmod{pqr}$?

Asked Jun 21 '23 at 15:14

Active Jun 21 '23 at 15:30

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Let $p, q, r$ be distinct primes,
let $n = pqr$, and
let $m = (p − 1)(q − 1)(r − 1).$

Show that for any $a\in \mathbb Z$ and $h\in \mathbb N$, we have
$(h\equiv 1\mod m)$ implies $(a^h\equiv a \mod n).$

I know I need to approach this with Fermat's Little Theorem, but am not sure what to do.

edited Jun 21 '23 at 15:30

Anne Bauval

34,650

asked Jun 21 '23 at 15:14

Robin

2

Quick beginner guide for asking a well-received question + please avoid "no clue" questions, and format your formulas (and fix your typoS in the title). – Anne Bauval Jun 21 '23 at 15:22
Thanks for sharing that! Sorry pretty new to the platform, plus I'm also new to this topic, so appreciate that! – Robin Jun 21 '23 at 15:26
1

Are you familiar with Chinese Remainder Theorem? If so, can you state how you'd apply it to this problem? (Just how you've get started is fine, not expecting that you'd fully solve it) – Calvin Lin Jun 21 '23 at 15:29
would we need to show that $a^h$ satisfies the congruences
$a^ h ≡a(modp)$, $a ^ h≡a(modq)$, $a^ h ≡a(modr)$,
– Robin Jun 21 '23 at 15:39
@RyanJoshi Yup! Do you see why that is true? – Calvin Lin Jun 21 '23 at 15:49
Yep thanks! (since n = pqr right?) thus if we prove each of those it gives us what we want – Robin Jun 21 '23 at 15:57
$a^{m(p-1)}\equiv1^m\pmod p$. – Piquito Jun 21 '23 at 16:43
Special case $,e=1, $ in the Theorem in the linked dupe. – Bill Dubuque Jun 22 '23 at 05:19

How to show $h\equiv 1\bmod{(p-1)(q-1)(r-1)}\implies a^h\equiv a \bmod{pqr}$?

0 Answers0