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

Question

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

Answer 1

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 $3$?