If $(a^{n}+n ) \mid (b^{n}+n)$ for all $n$, then $ a=b$ I happened to receive this from my friend.

Let $a,b \in \mathbb{N}$, such that $a^{n}+n \: \bigl|\: b^{n}+n$ for all $n \in \mathbb{N}$. Prove that $a=b$. How do we proceed?
Hint$\;$ from my $ $ sci.math post $ $ 2006/4/4 $ $ here or here (see there for much motivation)

$$\rm\quad\begin{eqnarray}
  p-1 \!\!\! &amp;&amp;\mid\rm \color{#c00}{n-1} \\
 \rm p\!\!\! &amp;&amp;\mid\rm \color{#0a0}{a+n} &amp;&amp; \\
\end{eqnarray}\! \Rightarrow\,\ 
p\mid \overbrace{a^{\large \color{#c00}n^{\phantom{I}}}\!\!\!-\!a+\color{#0a0}{a\!+\!n}}^{\Large a^n+n}\mid \overbrace{b^{\large \color{#c00}n^{\phantom{I}}}\!\!\!-\!b+\color{blue}{b\!-\!a}+\color{#0a0}{a\!+\!n}}^{\Large b^n +n} \ \,\Rightarrow\,\ p\mid \color{blue}{b-a}\qquad\qquad\quad  $$
UPDATE: the proof seems incorrect as b=4, a=2, fail when n=4 (check the comments below).

First, there is a counter example when n=1, where a=6 and b=13 (as $ (6+1) | (13+1) $ ). However I shall continue for all n>1.

If $(a^n+n) | (b^n+n)$ then there is a natural number k $\geq 1$ such that $b^n+n=k(a^n+n)$. Clearly $a=b \mbox{ iff } k=1$. We shall proceed to prove that k is always 1 for all n>1.

Continuing from $b^n+n=k(a^n+n)$,

$b^n+n \equiv 0 \quad (\mbox{mod } k)$

If we allow n to vary for the same a and b, then the only value for k that will make that equation hold for all n>1 is k=1, which implies that a=b.