Vertex-transitive nut graph order-degree existence problem
Ivan Damnjanović
TL;DR
The paper resolves the vertex-transitive nut graph order–degree existence problem by proving that for every admissible pair $(n,d)$—namely $n$ and $d$ both even with at least one divisible by four, and with $d \ge 4$ and $n \ge d+4$ (and the variant $d \equiv 2 \pmod 4$, $n$ divisible by $4$, $n \ge d+6$)—there exists a $d$-regular vertex-transitive nut graph of order $n$, and in fact a $d$-regular Cayley nut graph of order $n$. The authors realize sufficiency via explicit Cayley constructions on dihedral groups, aided by four auxiliary polynomial families $Q_t(x)$, $R_t(x)$, $S_t(x)$, and $T_t(x)$ to certify the absence of problematic cyclotomic roots, with extensive computer-assisted verification for small cases. This yields an exact characterization of attainable orders for vertex-transitive (and Cayley) nut graphs and implies sharp, constructive results for the circulant/bicirculant subfamilies. The methods combine group-theoretic Cayley constructions, cyclotomic-divisibility arguments, and computational checks, providing a robust framework for future exploration of bicirculant and non-Cayley vertex-transitive nut graphs.
Abstract
A nut graph is a nontrivial simple graph whose adjacency matrix has a simple eigenvalue zero such that the corresponding eigenvector has no zero entries. It is known that the order $n$ and degree $d$ of a vertex-transitive nut graph satisfy $4 \mid d$, $d \ge 4$, $2 \mid n$ and $n \ge d + 4$; or $d \equiv 2 \pmod 4$, $d \ge 6$, $4 \mid n$ and $n \ge d + 6$. Here, we prove that for each such $n$ and $d$, there exists a $d$-regular Cayley nut graph of order $n$. As a direct consequence, we obtain all the pairs $(n, d)$ for which there is a $d$-regular vertex-transitive (resp. Cayley) nut graph of order $n$.