Completeness of the rational-point list on the obstruction curve

Let F(s,a)F(s,a) be the obstruction polynomial, let F^(S,A,Z)\widehat{F}(S,A,Z) be its homogenization, and let

C:F^(S,A,Z)=0PQ2\overline{C}:\quad \widehat{F}(S,A,Z)=0\subset\mathbb{P}^{2}_{\mathbb{Q}}

be the projective closure. The computation lists the 88 rational points

{(1:2:1), (0:1:0), (0:0:1), (0:6:1), (1:0:0), (1:2:1), (1:0:1), (1:1:0)}.\{(1:2:1),\ (0:1:0),\ (0:0:1),\ (0:6:1),\ (1:0:0),\ (-1:2:1),\ (1:0:1),\ (-1:1:0)\}.

Completeness of the rational-point list. These are all rational points on C\overline{C}, so C(Q)\overline{C}(\mathbb{Q}) consists of exactly these 88 points. If true, this would turn the height-bounded computational evidence into a complete exclusion of affine rational points with s>0s>0 and s1s\neq 1 on the obstruction curve. The source presents the assertion in a conjecture environment, and no proof of completeness is supplied.

Sources & referencesView supporting material

Primary source

Valery Asiryan and Randall L. Rathbun, “Computational Evidence Against Quadratic-Cubic Factorization for the Second Cuboid Quintic”, arXiv:2601.07899 (2026).

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