GCD of a and b Prime and Odd: 1 or p?

[elimqiu]

Show that if $a,b\in\mathbb{N}^+,\ \gcd(a,b) = 1$ and $p$ is an odd prime,
then $\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)\in \{1,p\}$

 

[elimqiu]

Suppose $1\le d \mid \gcd(a+b,\frac{a^p+b^p}{a+b})$, then we have the following

$1\le d \mid a+b\implies b \equiv -a\ (\text{mod }d)\implies \sum_{k=0}^{p-1}(-1)^k a^{p-1-k}b^k \equiv pa^{p-1}(\text{mod }d)$

$\frac{a^p+b^p}{a+b} \equiv pa^{p-1}(\text{mod }d)$. Now since $\gcd(d,a)=1$, this means that $d \mid p$