TITLE: What are the factors of this quotient given by Fermat's Little Theorem?
QUESTION [10 upvotes]: $\forall a,b \in \mathbb{Z}, p\in \mathbb{P}$, let 
$$F_p(a,b) = \frac{(a+b)^p-a^p-b^p}{p}$$
Note:  

$F_3 = ab(a+b)$  
$F_5 = ab(a+b)(a^2+ab+b^2)$  
$F_7 = ab(a+b)(a^2+ab+b^2)^2$

According to data from Matlab for $p < 31$, I have the following conjectures:

$\forall p>3, F_3|F_p$  
$\forall p>5, F_5|F_p$  
$\forall p>7, F_7|F_p$ iff $p\equiv 1\pmod{6}$  
$\forall p>7, F_p$ will be an irreducible polynomial times $ F_5\text{ or }F_7$

What are the possible factors of $F_p$? What techniques can I use to attack this problem?

REPLY [6 votes]: The $ab$ factor is obvious.
Forget the factor of $p$, and dehomogenise:
$$G_p(x)=(x+1)^p-x^p-1.$$
If $p$ is odd $G_p(-1)=0-(-1)^p-1=0$: $(x+1)\mid G_p(x)$
and so $(a+b)\mid F_p(a,b)$.
Let $\omega$ be a primitive cube root of unity. If $p\equiv1\pmod 6$
then
$$G_p(
\omega)=(1+\omega)^p-x^p-1=(-\omega^2)^p-\omega^p-1=-\omega^2-\omega-1=0.$$
The same is true when $p\equiv5\pmod 6$ and both $(x-\omega)$, $(x-\omega^2)\mid G_p(x)$. So $(x^2+x+1)=(x-\omega)(x-\omega^2)\mid G_p(x)$.
Again, let $p\equiv1\pmod6$. Then
$$G_p'(\omega)=(p-1)(\omega+1)^{p-1}-(p-1)=(p-1)(1-1)=0.$$
Thus $(x-\omega)^2\mid G_p(x)$ and we get $(x^2+x+1)\mid G_p(x)$.
But if $p\equiv5\pmod6$,
$$G_p'(\omega)=(p-1)(\omega+1)^{p-1}-(p-1)=(p-1)((-\omega^2)^{p-1}-1)
\ne0.$$
Then $(x^2+x+1)\nmid G_p(x)$.
As for proving the residual factors are irreducible, that appears
to be a hard problem.