On $\gcd(a-b, (a^n-b^n)/(a-b))$

Question

Let $a,b$ be two coprime integers. Show that the gcd of the numbers $a-b, (a^n-b^n)/(a-b)$ divides $n$ for all $n\in\mathbb{N}$.

Answer 1

$$\frac{a^n-b^n}{a-b}\equiv a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\equiv na^{n-1}\equiv nb^{n-1}\pmod{a-b}$$

Now apply Euclidean algorithm:

$$\gcd\left(\frac{a^n-b^n}{a-b},a-b\right)=\gcd\left(na^{n-1},a-b\right)=\gcd\left(nb^{n-1},a-b\right)$$

$$=\gcd\left(\gcd\left(na^{n-1},nb^{n-1}\right),a-b\right)=\gcd\left(n\gcd(a,b)^{n-1},a-b\right)$$

If $\gcd(a,b)=1$, then this is a divisor of $n$ by the definition of $\gcd$.

Your claim is not necessarily true if $a,b$ aren't coprime: e.g., let $(a,b,n)=(6,2,2)$.

I used some $\gcd$ properties that aren't difficult to prove.