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Let $m = \frac{4^p - 1}{3}$

Where $p$ is a prime number exceeding $3$. how to prove that $2^{m-1}$ has reminder $1$ when divided by $m$

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    [Wolfram](http://tinyurl.com/crv5qz7) gave up after the 6th prime...2012-07-03

1 Answers 1

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$2^{2p}=4^p=3m+1\equiv 1 \pmod m$ so the result follows if $2p\mid m-1$.

Since $m$ is odd $2\mid m-1$, and by Fermat's Little Theorem $p\mid 4^p-4=3(m-1)$. Since $p>3$ is prime we must have $p\mid m-1$.

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    @Zander, thanks for your lucid explanation.2012-07-06