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Number Theory #3

Prove that for all even positive integers $$n$$, $$n^{2}-1|2^{n!}-1$$.

Note by Victor Loh
2 years, 7 months ago

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$$n^2-1=(n-1)(n+1)$$, 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. · 2 years, 7 months ago

Wonderful! · 2 years, 7 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 · 1 year, 5 months ago

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