<s>[INST] Show that if $n$ divides $a^n-b^n$ then $n$ divides $\frac{a^n-b^n}{a-b}$ 
Let $a,b,n \in \mathbb Z^+$. Show that if $n$ divides $a^n-b^n$ then $n$ divides $\frac{a^n-b^n}{a-b}$.

This is from Apostol’s Introduction to Analytic Number Theory, Chapter $5$, exercise $13$.
It is trivial when $\gcd(n,a-b) = 1.$
It is also easy when $n$ divides $a-b$ since $\frac{a^n-b^n}{a-b}=\sum_{k=0}^{n-1} a^kb^{n-1-k} \equiv na^{n-1}\pmod n \equiv 0\pmod n$.
But after that, I am stuck. Any hint or help is appreciated, thank you.
 [/INST] Let $d=\gcd(n,a-b)$. The explicit expression
$$\frac{a^n-b^n}{a-b}=a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1},$$
shows that, because $a\equiv b\pmod{d}$, we also have
$$a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1}\equiv na^{n-1}\equiv0\pmod{d}.$$
That is to say $d$ divides $\tfrac{a^n-b^n}{a-b}$.
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