$\forall a,b \in \mathbb{Z}, p\in \mathbb{P}$, let $$F_p(a,b) = \frac{(a+b)^p-a^p-b^p}{p}$$


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

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?

  • 2
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jun 22 '18 at 21:17
  • 1
    $\begingroup$ (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. $\endgroup$ – Jyrki Lahtonen Jun 22 '18 at 21:22
  • 1
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jun 23 '18 at 18:09
  • 1
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jun 23 '18 at 18:11
  • 1
    $\begingroup$ I think I have a proof for $F_3$ dividing $F_p$ for all primes $p>3$. Shall I write it as an answer? $\endgroup$ – 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 $(x^2+x+1)\nmid G_p(x)$.

As for proving the residual factors are irreducible, that appears to be a hard problem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.