Zero-of-complete-symmetric-polynomial conjecture

Let $\mathbb{F}_q$ be a finite field, let $3\leq k\leq q-2$ (or $4\leq k\leq q-3$ if $q$ is even), and let $h_e(x_1,\ldots,x_k)$ denote the complete symmetric polynomial of degree $e$ . For coefficients $a_0,\ldots,a_m\in\mathbb{F}_q$ , set

h(x_1,\ldots,x_k)=\sum_{e=0}^m a_eh_e(x_1,\ldots,x_k)

and let $N_q^*(h)$ count zeros of $h$ in $\mathbb{F}_q^k$ with pairwise distinct coordinates. Zero-of-complete-symmetric-polynomial conjecture. For a complete symmetric polynomial $h$ of positive degree $m$ , $N_q^*(h)\geq 1$ if and only if the reduction of

x^{k-1}\left(\sum_{e=0}^m a_ex^e\right)

modulo $(x^q-x)$ is not a polynomial of degree $k-1$ .

The reduction condition is necessary, while the source identifies sufficiency as the difficult direction; no general resolution is given.

The conjecture asserts that the necessary reduction condition for a zero of a complete symmetric polynomial is also sufficient. The relevant paper leaves the general case open.

Known results

The reduction condition is necessary for all stated parameters.
The conjecture is proved when $k\leq p$ , where $q=p^r$ .
It is also proved when $k\geq\left\lfloor(q+1)/2\right\rfloor$ , within the stated admissible ranges.
In particular, the first range covers every admissible $k$ when $q$ is prime.

2024 journal article

A 2024 article titled “Zeros of Complete Symmetric Polynomials over Finite Fields” is directly relevant, but the available record does not establish whether it proves the full conjecture or only another partial case. No verified general proof, counterexample, or claimed AI solution was found.

Current status (as of August 2026): The conjecture is settled in the ranges $k\leq p$ and $k\geq\left\lfloor(q+1)/2\right\rfloor$ , while the intermediate range remains open.

Zero-of-complete-symmetric-polynomial conjecture

Primary source

Progress summary

Solutions 0