Number Theory #3

Prove that for all even positive integers nn, n2−1∣2n!−1n^{2}-1|2^{n!}-1.

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Note by Victor Loh
6 years, 11 months ago

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Comments

n2−1=(n−1)(n+1)n^2-1=(n-1)(n+1), since nn is even, (n−1),(n+1)(n-1),(n+1) are relatively prime.

Hence n2−1∣2n!−1  ⟺  n−1∣2n!−1,n+1∣2n!−1n^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 22 and n±1n\pm 1 are relatively prime, by Euler's theorem 2ϕ(n±1)≡1(modn±1)2^{\phi (n\pm 1)}\equiv 1\pmod {n\pm 1}. Since it's clear that ϕ(n±1)≤n\phi (n\pm 1)\le n(equality when n+1n+1 is a prime, therefore ϕ(n±1)∣n!⇒n±1∣2ϕ(n±1)−1∣2n!−1\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 - 6 years, 11 months ago

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Wonderful!

Victor Loh - 6 years, 11 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 - 5 years, 10 months ago
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