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Let $a$ and $b$ be positive integers such that $b^n + n$ is a multiple of $a^n + n$ for all natural numbers $n$. Prove that $a = b$.

Here is my start:
$a^n + n | b^n + n$,
$a^n + n | b^n - a^n$
$a^n + n|(b - a)(b^{n-1} + ab^{n-2}... + a^{n-2} b + a^{n-1})$
Now we attempt to prove $b - a = 0$ by proving $a^n +n$ does not divide $(b^{n-1} + ab^{n-2}... + a^{n-2} b + a^{n-1})$. If we prove this successfully then we have $a^n + n | b-a$ for all $n$ and clearly $a=b$. I do not know if this method is valid, though I have had a hard time proving $a^n +n$ does not divide $(b^{n-1} + ab^{n-2}... + a^{n-2} b + a^{n-1})$ which leads me to suspect I am missing a more clever approach.

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