$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1$, or $p$ Let $p$ be prime number ($p\gt2$) and $a,b\in\mathbb Z$ ,$a+b\neq0$ ,$\gcd(a,b)=1$ how to  prove that $$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1~~\text{or}~~ p$$
Thanks in advance .

We already now that $\frac{a^p+b^p}{a+b}=a^{p-1}-a^{p-2}b+\ldots+b^{p-1}$
Using polynomial division, we find that
$$a+b\,|\,a^{p-1}-a^{p-2}b+\ldots+b^{p-1}=(a^{p-2}-2a^{p-2}b+3a^{p-3}b^2-\ldots-pb^{p-2})+pb^{p-1}$$
Hence $\gcd(\frac{a^p+b^p}{a+b},a+b)=\gcd(pb^{p-1},a+b)=\gcd(p,a+b)=1\text{ or } p$
as desired.
