3
$\begingroup$

Because $97$ is a prime number of the form $3.2^n+1$, The order of $2$ modulo $97$ is either $3$, $2^k$ or $3\cdot2^k$ for some $0\le k\le n$.

Since $2^{2^k}-1=F_0F_1F_2\cdots F_{k-1}$,The order of $2$ modulo $97$ is not divisible by $3$ if and only if $97$ divides a Fermat number $F_k$ with $0\le k\le n-1$. Now order of $2$ modulo $97$ is $48$, so there is no Fermat number $F_k$ with $0\le k\le n-1$ that is divisible by $97$. but I don't how obtain this result for$k\ge n$?

$\endgroup$
2

1 Answer 1

Reset to default
2
$\begingroup$

I think there is a small thing you're overlooking. We have that $97\mid 2^e-1$ if and only if $\mbox{ord}_{97}(2)\mid e$. If $97\mid F_k$, then $97\mid2^{2^{k+1}}-1$, thus...

$\endgroup$

You must log in to answer this question.