Suppose that $\displaystyle a,b\in\mathbb{N}^+,\ \gcd(a,b) = 1$ and $\displaystyle p$ is an odd prime.
Show that $\displaystyle \gcd\left(a+b,\frac{a^p+b^p}{a+b}\right) \in\{1,p\}$
Note that $\displaystyle a^p + b^p = (a + b) \cdot ( a^{p-1} - a^{p-2} b + a^{p-3} b^2 ... + b^{p-1})$
(because, more generally $\displaystyle x^n - y^n = (x-y)\cdot (x^{n-1}y^0 + x^{n-2}y^1 + ... + x^1 y^{n-2} + x^0y^{n-1})$ now substitute: $\displaystyle x = a, y = -b, n = p$ )
So then $\displaystyle \frac{a^p + b^p}{a+b} = a^{p-1} - a^{p-2} b + a^{p-3} b^2 ... + b^{p-1}$ , but now $\displaystyle a^{p-1} - a^{p-2} b + a^{p-3} b^2 ... + b^{p-1} \equiv a^{p-1} - a^{p-2}\cdot (-a) + ... + (-a)^{p-1} (\bmod a + b)$ since $\displaystyle a\equiv - b(\bmod.a+b)$
That is $\displaystyle \frac{a^p + b^p}{a+b} \equiv p\cdot a^{p-1} (\bmod. a + b)$ $\displaystyle \dots$