If $n \in \mathbb Z^+$ , $a,b$ are integers such that $d=g.c.d.(a-b,n)$ , then $d^2|a^n-b^n$ ?

Question

If $n \in \mathbb Z^+$ , $a,b$ are integers such that $d=g.c.d.(a-b,n)$ , then is it true that $d^2|a^n-b^n$ ?

Answer 1

Note that $$ a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots + ab^{n-2}+b^{n-1})$$ so we see a factor $a-b$ that is a multiple of $d$ imediately. Remains to show that the other factor is a multiple of $d$. From $a\equiv b\pmod d$ we see $a^{n-k}b^{k+1}\equiv a^{n-1}\pmod d$. Thus the second factor is $\equiv n\cdot a^{n-1}\equiv 0\pmod d$.