Let $p$ be an odd prime and let $m = 2p$. Prove that $a^{m−1} ≡ a \pmod{m}$ for all natural numbers $a$.

Question

Let $p$ be an odd prime and let $m = 2p$. Prove that $$a^{m−1} ≡ a \pmod{m}$$ for all natural numbers $a$.

My attempts

If $m$ is relatively prime to $a$, then $a^{\phi(m)}\equiv1\pmod{m}$ so $$a^{p-1}\equiv1\pmod{m}$$ $$a^{p}\equiv a\pmod{m}$$ Together implies $a^{2p-1}\equiv a\pmod{m}$ that $a^{m-1}\equiv a\pmod{m}$

If $m$ is not relatively prime to $a$, then exists some prime divisor $d$ such that $d$ divides $a$ and $d$ divides $m$, since $m=2p$ we have $d=2$ or $d=p$
Suppose $d=2$ that $\dots$

Now i'm stucked here, any ideas about how to prove this? Thanks for your help.

By the dupe $,a^p\equiv a,$ so scaling that by $,a^{p-1},$ yields $,a^{2p-1}\equiv a^p\equiv a\ \ $ — Bill Dubuque, Feb 24 '20 at 00:06
Or by this Theorem: $\ 2p\mid a(a^{\large (p-1)n}-1)$ for all $n\ge 0.\ $ OP is case $,n=2\ \ \ $ — Bill Dubuque, Feb 24 '20 at 00:15

acat3 · Accepted Answer · 2020-02-24T01:17:43.760

2

How about this

$a^{m-1}-a=(a^{p}-a)(a^{p-1}+1)$

According to Fermat's little theorem, $p|a^{p}-a$. In addition, $a^{p}-a$ is divisible by $2$.

Since $p$ is an odd prime, $2p|a^{m-1}-a$ aka $m|a^{m-1}-a$.

edited Feb 24 '20 at 01:17

answered Feb 23 '20 at 23:58

acat3

11,897

Welcome to Mathematics Stack Exchange. Nice answer +1. Isn’t $a^p-a$ always divisible by $2$? – J. W. Tanner Feb 24 '20 at 01:05
Well yes haha, thanks. – acat3 Feb 24 '20 at 01:17

Let $p$ be an odd prime and let $m = 2p$. Prove that $a^{m−1} ≡ a \pmod{m}$ for all natural numbers $a$.

1 Answers1