If gcd$(a,b)=$1 and $p$ is a odd prime then show that gcd$\left(a+b,\frac{a^p+b^p}{a+b}\right)=1$

Question

I only know the expansion of $a^p+b^p=(a+b)(a^p-1 -a^p-2×b^1......b^p-1)$ but I don't know how to proceed furthur.thankhs in advance.

Answer 1

The question is not always true; if $p \mid a+b$ the $\gcd$ may not be $1$, e.g. when $(a,b,p) = (1,2,3)$ we get $\frac{a^p+b^p}{a+b} = 3 = a+b$. So we further assume that $\gcd(a+b, p) = 1$, and show that $\displaystyle \gcd \left( a+b, \frac{a^p+b^p}{a+b} \right) = 1$. Suppose that some prime $q$ divides both $a+b$ and $\frac{a^p+b^p}{a+b} = a^{p-1} - a^{p-2}b + a^{p-3}b^2 \ldots + b^{p-1}$. Since $q \mid a+b$, we have $a \equiv -b \pmod q$. So we therefore obtain $\frac{a^p+b^p}{a+b} \equiv (-b)^{p-1} - (-b)^{p-2}b + \ldots + b^{p-1} \equiv pb^{p-1} \pmod q$. But since $q \mid a+b$, and $a$ and $b$ are coprime, we have $q \not \mid b \implies q \not \mid pb^{p-1}$, and so $\frac{a^p+b^p}{a+b} \equiv pb^{p-1} \not \equiv 0 \pmod q$, yielding a contradiction.