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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.

user729424
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Bob Markus
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1 Answers1

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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, $$n^2-1\mid2^{n!}-1$$

Rushabh Mehta
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