$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1$, or $p$

Question

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 .

Answer 1

$\begin{align}&{\bf Hint}\ \ \rm\displaystyle {\left(a\!-\!b,\frac{a^p\!-b^p}{a-b}\right)} = \overbrace{(a\!-\!b,\,a^{p-1}\!+\cdots+b^{p-1}) = (a\!-\!b,\color{#C00}{pb^{p-1}})}^{\large \bmod a-b:\ \ a\ \equiv\ b} = (a\!-\!b,p)\\[.2em] &\rm\ \ \ by\ \ \ (a,b) = 1\:\Rightarrow\: (a\!-\!b,b) = (a,b)=1,\,\ therefore\ \,\rm (a\!-\!b,b^n) = 1\ \ by\ Euclid's\ Lemma. \end{align}$

Remark $\ $ It's a special case $\rm\ f(x) = x^p,\,\ x = a\ \ (so\ \ \color{#0A0}{f'(b)} = \color{#C00}{pb^{p-1}})\:$ of

Theorem $\rm\displaystyle\quad \frac{f(x)-f(b)}{x-b} \equiv\ \color{#0A0}{f\:'(b)}\pmod {x\!-\!b}\quad$ for $\rm\ f(x)\in \mathbb Z[x]$

Proof $\ $ By Taylor, $\rm\,\ f(x)\ =\ f(b) +\: f\:'(b)\ (x\!-\!b) \,+\, (x\!-\!b)^2\: g(x)\ \ $ for some $\rm\ g(x) \in \mathbb Z[x]$

Answer 2

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.