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Try this: $n^2-1|2^n+1$

$2^{n}+1$ is odd.So,$n^{2}-1$ is odd.So,$n$ is even.Let $n=2k$.Then $2^{n}+1=(2^{k})^{2}+1$.So,every divisor of $2^{n}+1$ is of the form $4m+1$.
So,$n^{2}-1 \equiv 1 (mod 4)$.
Or,$n^{2} \equiv 2 (mod 4)$,which is impossible.
So there is no solution.

tanmoy wrote:$2^{n}+1$ is odd.So,$n^{2}-1$ is odd.So,$n$ is even.Let $n=2k$.Then $2^{n}+1=(2^{k})^{2}+1$.So,every divisor of $2^{n}+1$ is of the form $4m+1$.
So,$n^{2}-1 \equiv 1 (mod 4)$.
Or,$n^{2} \equiv 2 (mod 4)$,which is impossible.
So there is no solution.

You have to prove it.

Okay.Suppose,$p$ is an odd divisor of $2^{n}+1$.So, $p$ is either of the form $4m+1$ or $4m+3$,where $m$ any integer.Let $p$ is of the form $4m+3$.Then $2^{n} \equiv -1 (mod p)$.Or $2^{p-1} \equiv (2^{n})^{2m+1} \equiv -1 (mod p)$,which contradicts Fermat's little theorem.
So,$p$ is of the form $4m+1$.
Now,if $n^{2}-1$ is prime,we are done! But if $n^{2}-1$ is composite,then it has two or more odd prime divisors which have the form $4m+1$.So,there product is also of the form।But $n^{2}-1$ is of the form $4m+3$।So,contradiction.Is it correct now?

now it should be correct