Factorization of a particular polynomial in $F_p[x]$

Question

For an odd prime $p$, define the polynomial $f_p(x) = \sum\limits_{k=1}^{p-1} k^{(p-1)/2}x^k$ in $\mathbb{F}_p[x]$. Show that $(x-1)^{(p-1)/2}$ divides $f_p(x)$, but that $(x-1)^{(p+1)/2}$ does not. Can we say anything interesting about the factorization of the quotient that remains after dividing $f_p(x)$ by $(x-1)^{(p-1)/2}$? I generated a lot of data but couldn't find a discernible pattern.

Answer 1

I can prove the first claim. My proof is based on the following fact (well known and easy to prove):

For all primes $p$ and all $j=1,2,\ldots,p-2$ we have the power sum identities (in $\Bbb{F}_p$): $$ \sum_{k=1}^{p-1}k^j=0, $$ but $$ \sum_{k=1}^{p-1}k^{p-1}=-1. $$ The latter being an immediate consequence of Little Fermat.

We can utilize this as follows. Let $L$ denote the differential operator that maps a polynomial $g(x)$ to $Lg(x):=x g'(x)$. By an obvious induction we have $$L^t f_p(x)=\sum_{k=1}^{p-1}k^{t+\frac{p-1}2}x^k$$ for all natural numbers $t$.

The listed power sum results imply that $$ L^tf_p(1)=\begin{cases} 0,&\ \text{when $t=0,1,\ldots,(p-3)/2$, and}\\ -1,&\ \text{when $t=(p-1)/2$.}\end{cases} $$ It is easy to see that $L_tf_p(x)$ is a sum containing the term $x^tf_p^{(t)}(x)$ as well as lower order derivatives multiplied by monomials. Therefore this implies the following:

Corollary. The order $t$ derivative $f^{(t)}_p(x)$ of the Fekete-polynomial vanishes at $x=1$ when $t=1,2,\ldots,(p-3)/2$, but is non-zero, when $t=(p-1)/2$.

The claim follows from this by the familiar result relating the multiplicity of a zero of a polynomial to the vanishing of the prescribed order of derivatives. Some care needs to be applied when using this result, familiar from characteristic zero, $\Bbb{R}[x]$ in particular. It works in characteristic $p$ as long as the multiplicity of the zero is strictly less than $p$, which is the case here.