If $a$ and $b$ are positive integers such that $a^n+n\mid b^n + n$ for all positive integers $n$, prove that $a=b$. I ran into this problem in a math camp, but I can't seem to solve it via elementary techniques. 
If $a$ and $b$ are positive integers such that $a^n+n\mid b^n + n$ for all positive integers $n$, prove that $a=b$.

Assume the contrary and consider a prime $p$ that does not divide $b-a$, By the Chinese Remainder Theorem we can find a postive integer $n$ such
$$\begin{cases}n\equiv 1\pmod{p-1}\\
n\equiv -a\pmod p
\end{cases}$$
Then by Fermat's little theorem
$$a^n+n\equiv a+n\equiv a-a\equiv 0\pmod p$$
and 
$$b^n+n\equiv b+n\equiv b-a\pmod p$$
It follows that $p$ divides $a^n+n$,But does not divide $b^n+n$.Contradiction.
