# 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 $a = b$. [duplicate]

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.