My original question is: Find all the number $x$ such that, for all $n \in \mathbb{N}, n\ge 5$, we have $ 2^n-n^2 | x^n -n^x$.
The first thing to notice is that, if $n$ is even, then $2^n-n^2$ is even and thus, $x$ must be even. So I denote $x=2k$.
Further transformations yields $2^n-n^2 | k^n -(2^{k-1})^n , \forall n \in \mathbb {N}, n \ge 5, n$ odd.
The problem will be completed if I have the following statement:
There are infinitely many primes of the form $2^n-n^2$
Any help for either of the problem is appreciated!
OK, this should work for most of it. Since it's meant to be a challenge I have left one part unexplained.
I claim that x must be a power of 2. To see this, let $p$ be odd and suppose that $p\mid x$. Let $x=pk$ for some $k\in \mathbb{N}$.
[Here's the unexplained bit.] Choose $n$ so that $p\mid(2^n-n^2)$. Notice that $p\nmid n$ in this case. But $p\mid (x^n-n^x)=(p^n\cdot k^n-n^x)$. Thus $p\mid n$, a contradiction.
Hence no odd prime can divide $x$. (Notice that if $p=2$ then all this breaks down.) Note that $2$ and $4$ work. I know $8$ doesn't, but I haven't got a good proof for the rest of the powers of $2$.