What are the factors of this quotient given by Fermat's Little Theorem?

$\forall a,b \in \mathbb{Z}, p\in \mathbb{P}$, 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?

Solution 1:

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.