Divisibility by $(a-b)^{2}$

Question

Let $a$ and $b$ be coprime positive natural numbers, and $n$ a positive natural number . We know that $(a-b)|(a^n-b^n)$ is true. Also, if $(a-b)|n$, $(a-b)^{2}|(a^n-b^n)$ is true. My question is

Is there a positive integer $n$, such that $(a-b)^{2}|(a^n-b^n)$ is true but $(a-b)|n$ is NOT true?

Answer 1

Let $a-b=d,(a,b)=1\implies(b,d)=1$

$a^n-b^n=(b+d)^n-b^n\equiv\binom n1b^{n-1}d\pmod{d^2}$

As $(b,d)=1,$ here we need $d|n$