If for all $n\in\Bbb{N}, a^n-n$ divides $b^n-n$ then $a=b$. 
Exercise: Let $a,b\in\Bbb{N}$, show that if for all $n\in\Bbb{N}, \quad a^n-n$ divides $b^n-n$, then $a=b$.

I don't have lot of knowledge on this subject, I am aware about some elementary result but here I am stuck. Any hint to start this problem will be great. 
 A: Let $p$ be a prime number and let $n=k(p-1)+1$. Then by Fermat, $a^n -n \equiv a - n \equiv a + k - 1 \mod p$ and similarly $b^n - n \equiv b+k-1 \mod p$. It follows that if $p \mid a+k-1$, then $p \mid b+k-1$. In other words, for any $i>0$ any prime divisor of $a+i$ is also a divisor of $b+i$. If $a+i$ is prime, this implies that $a+i=b+i$ $($and hence $a=b)$ or $b+i \geq 2(a+i) \iff b \geq 2a + i$. Now take $i$ large to conclude that the first must apply.
A: Partial Answer
Let $p$ be a prime dividing $a$, then $p|a^p-p$ which implies that $p|b$. 
Now let $p$ be a prime which doesn't divide $a$. Then, by Fermat Little Theorem we have
$$p|a^{p-1}-1 \Rightarrow p| a^{p-1}+p-1$$
This implies that 
$$p| b^{p-1}+p-1$$
and hence $p \nmid b$.
This shows that $a,b$ have the same prime divisors.
What I would try next is to pick one of the primes $p$ and write $a=p^km$ with $p \nmid m$.
Now $p| m^{p-1}+p-1$ therefore $p^{k+1}|p^km^{p-1}+p^{k+1}-p^k$. If you can somehow get from here something like $p^{k+1}|a^{p^k}-p^{k}$ or $p^{k+1}|a^{mp^k}-mp^{k}$ you can complete the proof by the same argument as before....
A: The statement immediately implies that $a \leq  b$
Choose a prime $p>b+1$. It means that $p$ is co-prime with $a$ and $b$ and the number $(p-a+1)(p-1)+1$ is natural.
Now let $n=(p-a+1)(p-1)+1$.
By Fermat's Little Theorem:
$a^n=(a^{p-1})^{p-a+1} \cdot a \equiv a$ (mod p)
Similarly $b^n \equiv b$ (mod p).
Furthermore, $n=(p-a+1)(p-1)+1=p^2-ap+p-p+a-1+1 \equiv a$ (mod p).
Therefore we get $a^n-n \equiv a-a \equiv 0$ (mod p).
But then $b^n-n \equiv 0$ (mod p). On the other hand, $b^n-n \equiv b-a$ (mod p).
Thus $p \mid b-a$, which implies that either $p \leq |b-a|$ or $b-a=0$. The former doesn't hold because $p>b>b-a\geq 0$, thus $a-b=0$ and $a=b$.
Note that for the 'plus' version of this problem when $a^n+n \mid b^n+n$ for every natural number, it suffices to choose $n=(a+1)(p-1)+1$.
