Let bf x^m=x_1,…,x_m and bf y^m=y_1,…,y_m. Let pi(bf x^m)=∏_j=1^m x_j, let F_n(bf x^m)=b_n B_n(f_1,f_2,…), where b_n is independent of bf x^m and f_k is defined by the complete Bell-polynomial generating function. Let bf s_i^m be the ith row of the matrix bf S(bf x^m,bf y^m). Define V_n(bf x^m)=fracF_n(bf x^m)pi(bf x^m),qquad U_n(bf x^m,bf y^m)=V_n(bf x^m)-∑_i=1^m y_i^m-n-1V_n(bf s_i^m). Generalized Bell-polynomial relation. The relation U_n(bf x^m,bf y^m)=0, holds for 0≤ n≤ m-1. This is a proposed extension of analogous relations for particular families of symmetric polynomials; the source provides no resolution status beyond presenting it as a conjecture.

Generalized Bell-polynomial relation for symmetric functions

Let ${\bf x}^m=\{x_1,\ldots,x_m\}$ and ${\bf y}^m=\{y_1,\ldots,y_m\}$ . Let $\pi({\bf x}^m)=\prod_{j=1}^m x_j$ , let $F_n({\bf x}^m)=b_n\mathbb B_n(f_1,f_2,\ldots)$ , where $b_n$ is independent of ${\bf x}^m$ and $f_k$ is defined by the complete Bell-polynomial generating function. Let ${\bf s}_i^m$ be the $i$ th row of the matrix ${\bf S}({\bf x}^m,{\bf y}^m)$ . Define

V_n({\bf x}^m)=\frac{F_n({\bf x}^m)}{\pi({\bf x}^m)},\qquad U_n({\bf x}^m,{\bf y}^m)=V_n({\bf x}^m)-\sum_{i=1}^m y_i^{m-n-1}V_n({\bf s}_i^m).

Generalized Bell-polynomial relation. The relation

U_n({\bf x}^m,{\bf y}^m)=0,

holds for $0\leq n\leq m-1$ . This is a proposed extension of analogous relations for particular families of symmetric polynomials; the source provides no resolution status beyond presenting it as a conjecture.

Progress summary

Solved

A 2025 preprint and a separate posted argument claim a complete proof, but neither has been independently verified.

The conjecture asserts that the stated Bell-polynomial expression vanishes for $0\leq n\leq m-1$ . Boris Y. Rubinstein presented it as part of a broader claim for arbitrary homogeneous symmetric polynomials in an October 2025 preprint.

October 2025 claimed proof

Rubinstein’s manuscript states the relation as valid and claims the broader homogeneous-symmetric-polynomial result. The retrieved record contains no independent verification, referee report, counterexample, withdrawal, or correction.

Posted attempt

A complete proof is claimed for every symmetric homogeneous polynomial of degree $n<m$ , using cancellation of residues at the possible poles and homogeneity to force the remaining negative-degree polynomial to vanish. This argument has not been independently verified.

Current status (as of August 2026): The relation for $0\leq n\leq m-1$ is claimed proved by the 2025 preprint and a posted argument, but remains unverified.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Boris Y. Rubinstein, “A New Class of Relations for Homogeneous Symmetric Polynomials”, arXiv:2510.25749 (2025).

bell-polynomials symmetric-polynomials

1 week ago

The identity holds, more generally, for every symmetric homogeneous polynomial of degree $n<m$ .

Let $S$ be symmetric and homogeneous of degree $n$ in $m$ variables. Put

D_{ij}=y_ix_j-y_jx_i, \qquad (s_i)_j= \begin{cases} x_i,&j=i,\\ D_{ij},&j\ne i, \end{cases}

and define

U= \frac{S(x)}{\prod_{r=1}^m x_r} - \sum_{i=1}^m \frac{y_i^{m-n-1}S(s_i)} {x_i\prod_{j\ne i}D_{ij}}.

Work over $K=\mathbb Q(y_1,\ldots,y_m)$ . As a rational function of $x$ , the only possible poles are the simple linear factors $x_i$ and $D_{ij}$ .

On $x_i=0$ , homogeneity gives

S(s_i)=y_i^nS(x_1,\ldots,x_{i-1},0,x_{i+1},\ldots,x_m),

\prod_{j\ne i}D_{ij} =y_i^{m-1}\prod_{j\ne i}x_j.

Therefore the residue of the $i$ -th summand equals the residue of $S(x)/\prod_r x_r$ , and their opposite signs cancel.

On $D_{ij}=0$ , set $c=y_j/y_i$ . Then

x_j=cx_i, \qquad D_{jr}=cD_{ir}\quad(r\ne i,j), \qquad s_j=c\,\tau_{ij}(s_i),

where $\tau_{ij}$ interchanges coordinates $i,j$ . Symmetry and homogeneity imply

S(s_j)=c^nS(s_i).

Since $D_{ji}=-D_{ij}$ , the ratio of the $j$ -th residue to the $i$ -th residue is

-c^{(m-n-1)+n-(m-1)}=-1.

Hence these residues cancel as well.

All possible poles are removable, so

U\in K[x_1,\ldots,x_m].

Every term is homogeneous of degree $n-m$ in $x$ . If $0\le n<m$ , that degree is negative, forcing

U=0.

Taking $S=F_n$ , the symmetric homogeneous Bell polynomial from the conjecture, proves the desired identity for every stated $m,n$ .

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Shivam Patel · 2 days ago