# 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?

• The polynomial $G_p(x)$ studied by Lord Shark satisfies the identities $G_p(-x-1)=G_p(x)$ and $x^pG_p(1/x)=G_p(x)$. Therefore if $\alpha\neq0$ is one of its zeros, we get the familiar sextet of zeros: $\alpha,-\alpha-1,1/\alpha, -1/(\alpha+1), -\alpha/(\alpha+1)$ and $-(\alpha+1)/\alpha$. If there are repetitions among those, then we are in a case already covered. – Jyrki Lahtonen Jun 22 '18 at 21:17
• (cont'd) My limited testing suggests that each such sextet of zeros contains a pair of complex conjugates with real part $\alpha=-1/2$ (with $\overline{\alpha}=-1-\alpha$). I only tested $p=11,13,17$ though. – Jyrki Lahtonen Jun 22 '18 at 21:22
• Anyway, the elements of such a sextet are zeros of the polynomial $$P_\Delta(x)=x^6+3x^5+\Delta x^4+(2\Delta-5)x^3+\Delta x^2+3x+1,$$ where $$\Delta=-(1+3\alpha-5\alpha^3+3\alpha^5+\alpha^6)\alpha^{-2}(1+\alpha)^{-2}$$ is an invariant of that group of six fractional linear transformations. When $p=11$ we have $\Delta=7$, with $p=13$ we have $\Delta=8$. When $p=17$ there are two sextets with corresponding $\Delta=(17\pm\sqrt{21})/2$, when $p=19$ we get $\Delta=(19\pm\sqrt{37})/2$. Those had to be irrational for otherwise the degree twelve polynomial would factor. – Jyrki Lahtonen Jun 23 '18 at 18:09
• From that point on it becomes more complicated. Doesn't look like that would lead anywhere :-( A further possibility is that this polynomial shows up in the definition of Witt vector arithmetic. – Jyrki Lahtonen Jun 23 '18 at 18:11
• I think I have a proof for $F_3$ dividing $F_p$ for all primes $p>3$. Shall I write it as an answer? – Haran Jul 4 '18 at 15:11

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 $(x^2+x+1)\nmid G_p(x)$.