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

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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\}
 
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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
 

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