Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Claim. The answer to the question of Problem 960 is no: the threshold is neither nor . Theorem 2.1 of Alexeev, Putterman, Sawhney, Sellke and Valiant gives, for every , and , a set of points in the plane with no on a line and at least
ordinary lines, in which no points have all their connecting lines ordinary. The graph whose vertices are the points and whose edges are the ordinary lines is bipartite (Proposition 2.5(2) of the paper, resting on Proposition 2.4 for the base set), so it has no triangle, which is the case and hence every . Writing with , the base set of points is taken in a cyclic torsion subgroup of order of the real points of the elliptic curve , with the residue class of modulo removed, so that collinearity becomes arithmetic in the group; when does not divide , the set adds back chosen points of the removed class, and Proposition 2.5 checks that no four points of are collinear and that the ordinary-line graph stays bipartite. Together with the upper bound from Turán's theorem, which the site's commentary records, the threshold is of order for every fixed and ; its exact value is not determined. The remaining parameters are degenerate, as Section 2.1 of the paper states: for no valid set exists, for every line determined by the set is ordinary and the graph is complete, and for with the Sylvester-Gallai theorem supplies an ordinary line and so the pair. The problem's discussion notes that the lines spanned by the points must be ordinary with respect to the whole set, which is how the theorem reads the question.
Claimant. The result is Theorem 2.1 of Boris Alexeev, Moe Putterman, Mehtaab Sawhney, Mark Sellke and Gregory Valiant, Short proofs in combinatorics, probability and number theory II, arXiv:2604.06609, posted on 2026-04-08. The paper states that each of its proofs is due to an internal model at OpenAI; the site credits the bound to that model through the paper. Erdős posed the question in his 1984 research-problem note (card, p. 102), where he conjectured the bound and suggested a linear one.
Acceptance. Thomas Bloom, the site's curator, marks the problem disproved
and credits this bound on the problem's page at erdosproblems.com, last edited
2026-04-09, whose label and credit showed on 2026-10-06; that credit is the
reviewed evidence.
The paper is a preprint, with no refereed version found on 2026-10-06, and no
Lean proof is recorded, so the claim is neither refereed nor formalized.
What remains. The order of is settled at for and , between and ; the constant is open. The site points to Problem 209 as related.