Number Theory problem mod 2p

Question

Let $p$ be an odd prime, and let $a$ be an odd integer such that $p \nmid a .$ Prove that $$ a^{p-1} \equiv 1 \quad(\bmod 2 p) $$

I thought about Fermat Little Theorem could be useful but for that we need prime in modulo. But her we have 2p so it is an even integer. Can somebody give me hint for how to proceed?

$a^{p-1} = 1 mod2$ ; $a^{p-1} = 1 mod p$ then what @ParclyTaxel — maths student, Apr 14 '20 at 11:39
@mathsstudent you have $2|a^{p-1}-1$ as well as $p|a^{p-1}-1$ and as $gcd(2,p)=1$ $2p|a^{p-1}-1$ — , Apr 14 '20 at 11:42

score 3 · Accepted Answer · answered Apr 14 '20 at 11:43

3

You have $2|a^{p-1}-1$ and $p|a^{p-1}-1$ by fermat's little theorem, and as $gcd(2,p)=1$ you have $2p|a^{p-1}-1$ as well

answered Apr 14 '20 at 11:43

score 1 · Answer 2 · answered Apr 14 '20 at 12:20

1

Since $a\in \mathbb Z_{2p}^*$ and $|\mathbb Z_{2p}^*|=\varphi(2p) = p-1$, then \begin{equation} a^{p-1} \equiv 1 \quad\bmod 2p \end{equation}

answered Apr 14 '20 at 12:20

Menezio

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Number Theory problem mod 2p

2 Answers2