# Does there exist an $a$ such that $a^n+1$ is divisible by $n^3$ for infinitely many $n$?

It is well known that there are infinitely many positive integers $n$ such that $2^n+1$ is divisible by $n$.

Also it is well known that there exist infinitely many positive integers $n$ such that $4^n+1$ is divisible by $n^2$.

But I still cannot find any positive integer $a$ for which there exist infinitely many positive integers $n$ such that $a^n+1$ (or $a^n-1$) is divisible by $n^3$.

How can I find such and $a$ or prove that it doesn't exist?

• I got as far as existence of $a$ such that $a^n - 1$ is divisible by $n^3$ for infinitely many $n$ is weaker than existence of $a$ such that $a^n + 1$ is divisible by $n^3$ for infinitely many $n$. (Use $a^n + 1$ divides $a^{2n} - 1$, e.g. taking $a^2$ instead of $a$.) Also note the trivial case: when $a=1$, $a^n - 1$ is divisible by $n^3$ for every $n$. – hardmath Apr 13 '14 at 15:28
• please improve the title by making it more descriptive of the question – jimjim Oct 18 '14 at 18:16

When $a$ is even, it is not easy, but not difficult either, to prove that there exist an infinite number of $n$ such that $n \mid a^n + 1$, and all these $n$ are given by powers of $a+1$. Can this or its proof help you?