Assume $(a,b)$=1, and let $p$ be an odd prime, prove that $(a+b,\frac{a^p+b^p}{a+b})=1 \text{ or } p$.
I thought of letting $p=2k+1,k\in\mathbb{Z}$,
then use the identity
$a^{2k+1}+b^{2k+1}=(a+b)(a^{2k}-a^{2k-1}b+a^{2k-2}b^2-\ldots+b^{2k})$
So dividing throughout,
$\frac{a^p+b^p}{a+b}=a^{2k}-a^{2k-1}b+a^{2k-2}b^2-\ldots+b^{2k}$
However, after that, I am kind of stuck..
Sincere thanks for any help!