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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, 9 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, 9 months ago

Wonderful! · 2 years, 9 months ago