# Prove that one of $2^n +1$ and $2^n-1$ must be composite for $n$ is a natural number greater than $2$ [duplicate]

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For natural numbers $n>2$ prove that at least one of $2^n-1$ and $2^n+1$ must be composite. How would one conduct this proof? I had the idea of trying to prove by contradiction since for this to be true these would have to be twin primes however this did not seem to go anywhere. Thanks

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## marked as duplicate by lab bhattacharjee, Old John, Sujaan Kunalan, Calvin Lin, Dominic MichaelisNov 15 '13 at 6:27

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## 1 Answer

Hint: Consider the numbers $$2^{n} - 1, 2^n, 2^{n} + 1$$ These are three consecutive numbers, so what does the pigeonhole principle say about division by $3$?

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I fail to see how the pigeonhole principle is relevant here though maybe i am missing something...? edit: nvm –  user108878 Nov 14 '13 at 23:01
@user108878 There are three numbers, and three possibilities for a remainder when dividing by $3$. –  user61527 Nov 14 '13 at 23:02