1
$\begingroup$

I am trying to prove that for every even positive integer $n$, $n^2 − 1$ divides $2^{n!} − 1$.

My attempt: I am thinking of using Euler's Theorem and totient function to get $2^{n!} \equiv 1$ (mod $n^2 - 1$). We would have to show $\text{gcd}(2^{n!} - 1, n^2 − 1) = 1$ however and I'm not sure how to proceed with this.

$\endgroup$
2
$\begingroup$

Note that $2^{n!}\equiv1\bmod{n+1}$ and $2^{n!}\equiv1\bmod{n-1}$

This is because $\phi(n+1),\phi(n-1)<n+1$, so $\phi(n+1),\phi(n-1)\mid n!$, and you know that $2^{\phi(n-1)}\equiv1\bmod{n-1}$ by Euler's Theorem (and similarly for $n+1$).

Since $\operatorname{gcd}(n+1,n-1)=1$ since $n$ is even, by CRT, $2^{n!}\equiv1\bmod{n^2-1}$. So, $$n^2-1\mid2^{n!}-1$$

$\endgroup$
1
  • $\begingroup$ @RobArthan I removed all group theory, so it should be easier to understand. $\endgroup$ – Don Thousand Feb 24 '20 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.