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 $m$
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 $m$
$2^{2p}=4^p=3m+1\equiv 1 \pmod m$ 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 $p\mid m-1$.