We will consider both conditions at once. Since \(2\) and \(n\pm 1\) are relatively prime, by Euler's theorem \(2^{\phi (n\pm 1)}\equiv 1\pmod {n\pm 1}\). Since it's clear that \(\phi (n\pm 1)\le n\)(equality when \(n+1\) is a prime, therefore \(\phi (n\pm 1)\mid n!\Rightarrow n\pm 1\mid 2^{\phi (n\pm 1)}-1\mid 2^{n!}-1\) which is what we want to prove so we are done.
–
Xuming Liang
·
2 years, 9 months ago

Euler's phi func Φ(n)<=n-1 and n|2^Φ(n) - 1 and it is easy to prove that Φ(n-1)|n! and Φ(n+1)|n!
first, to find gcd(n-1,n+1),,,
let gcd=d => d|n+1 and d|n-1 => d|2, which is false as d|(an odd integer)
now, n-1|2^Φ(n-1) - 1 (which is true by euler) => n-1|2^n!-1
also, n+1|2^Φ(n+1) - 1 => n+1|2^n!-1
combining we get the required result
–
Sayak Chakrabarti
·
1 year, 8 months ago

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TopNewest\(n^2-1=(n-1)(n+1)\), since \(n\) is even, \((n-1),(n+1)\) are relatively prime.

Hence \(n^2-1\mid 2^{n!}-1\iff n-1\mid 2^{n!}-1,n+1\mid 2^{n!}-1\).

We will consider both conditions at once. Since \(2\) and \(n\pm 1\) are relatively prime, by Euler's theorem \(2^{\phi (n\pm 1)}\equiv 1\pmod {n\pm 1}\). Since it's clear that \(\phi (n\pm 1)\le n\)(equality when \(n+1\) is a prime, therefore \(\phi (n\pm 1)\mid n!\Rightarrow n\pm 1\mid 2^{\phi (n\pm 1)}-1\mid 2^{n!}-1\) which is what we want to prove so we are done. – Xuming Liang · 2 years, 9 months ago

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– Victor Loh · 2 years, 9 months ago

Wonderful!Log in to reply

Euler's phi func Φ(n)<=n-1 and n|2^Φ(n) - 1 and it is easy to prove that Φ(n-1)|n! and Φ(n+1)|n! first, to find gcd(n-1,n+1),,, let gcd=d => d|n+1 and d|n-1 => d|2, which is false as d|(an odd integer) now, n-1|2^Φ(n-1) - 1 (which is true by euler) => n-1|2^n!-1 also, n+1|2^Φ(n+1) - 1 => n+1|2^n!-1 combining we get the required result – Sayak Chakrabarti · 1 year, 8 months ago

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