Does $n\mid(a^n-b^n)$ imply $n\mid(a^n-b^n)/(a-b)$? Finding my previous question quite naive, I improve my question:  

Given that $n,a,b \in  \mathbb{N}$ and $n\mid(a^n-b^n)$ , can we prove or disprove $n\mid(a^n-b^n)/(a-b)$ ? 

 A: Let $p$ be an odd prime divisor of $n$. And let $v_p(n)$ denote the highest power of $p$ in the prime decomposition of $n$. Now assume that $v_p(n) = k$
Now if $p \not \mid a-b$ then as $p^k \mid (a^n - b^n)$ we have that $p^k \mid \frac{a^n - b^n}{a-b}$
If $p \mid a-b$ and $p \not \mid a$ then we have that $p \not \mid b$. Now using the following lemma we have:
$$v_p\left(\frac{a^n - b^n}{a-b}\right) = v_p(a^n - b^n) - v_p(a-b) = v_p(a-b) + v_p(n) - v_p(a-b) = v_p(n) = k $$
$$\implies p^k \mid \frac{a^n - b^n}{a-b}$$
If $p \mid a-b$ and $p\mid a$ then we have that $p \mid b$. Let $m$ be the highest power of $p$ dividing them both. Then we have:
$$\frac{a^n - b^n}{a-b} = p^{m(n-1)}\frac{(a_1)^n - (b_1)^n}{a_1-b_1}$$
Now obviously $p^k \mid \frac{a^n - b^n}{a-b}$, as $k \le m(n-1)$
NOTE: If $p=2$ only the second part has to be altered, but that's easy again using the lemma mentioned above.
Now since for every prime divisor of $n$ we have that $p^k \mid \frac{a^n - b^n}{a-b}$ we have that $n \mid \frac{a^n - b^n}{a-b}$. Q.E.D.
