I just saw this question, and I recalled a similar problem.
Problem. Find all integers $a,b>1$ such that $a^n-n^a$ divides $b^n-n^b$ for all sufficiently large integers $n$.
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Partial results:
Clearly $a=b$ works. As $a>b$ implies $a^n-n^a>b^n-n^b$ for $n$ large enough, we are left with the case $a<b$.
Let $p$ be a prime and let $v_p(x)$ denote the exponent of $p$ in $x$; so $v_p(xy)=v_p(x)+v_p(y)$, $v_p(x^y)=yv_p(x)$ and $v_p(x+y)\ge\min\{v_p(x),v_p(y)\}$ (with equality if $v_p(x)\ne v_p(y)$).
Assume $p\mid a$. Let $n$ be a large multiple of $p$. Then $p\mid a^n-n^a$, hence also $p\mid b^n-n^p$ and ultimately $p\mid b$.
Assume $p\nmid a$. Then $a^{p-1}\equiv 1\pmod p$. Now pick large $n$ that is $\equiv 0\pmod{p-1}$ and $\equiv 1\pmod p$. With this we get $p\mid a^n-n^a$, hence $p\mid b^n-n^b$, so that $b^n\equiv 1\pmod p$. Especially, $p\nmid b$.
So we have $$ p\mid a\iff p\mid b$$
Hagen von Eitzen
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