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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. The answer to the question of Problem 960 is no: the threshold fr,k(n)f_{r,k}(n) is neither o(n2)o(n^2) nor ≪n\ll n. Theorem 2.1 of Alexeev, Putterman, Sawhney, Sellke and Valiant gives, for every r≥3r\geq3, k≥4k\geq4 and n≥72n\geq72, a set of nn points in the plane with no kk on a line and at least

n212−103 n\frac{n^2}{12}-\frac{10}{3}\,n

ordinary lines, in which no rr points have all their (r2)\binom r2 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 r=3r=3 and hence every r≥3r\geq3. Writing n=6m+sn=6m+s with 0≤s≤50\leq s\leq5, the base set A0A_0 of 6m6m points is taken in a cyclic torsion subgroup of order 7m7m of the real points of the elliptic curve y2=x3−x+1y^2=x^3-x+1, with the residue class of 00 modulo 77 removed, so that collinearity becomes arithmetic in the group; when 66 does not divide nn, the set A=A0∪TsA=A_0\cup T_s adds back ss chosen points of the removed class, and Proposition 2.5 checks that no four points of AA are collinear and that the ordinary-line graph stays bipartite. Together with the upper bound fr,k(n)≤(1−1r−1)n22+1f_{r,k}(n)\leq(1-\tfrac1{r-1})\tfrac{n^2}2+1 from Turán's theorem, which the site's commentary records, the threshold is of order n2n^2 for every fixed r≥3r\geq3 and k≥4k\geq4; its exact value is not determined. The remaining parameters are degenerate, as Section 2.1 of the paper states: for k=2k=2 no valid set exists, for k=3k=3 every line determined by the set is ordinary and the graph is complete, and for r=2r=2 with k≥4k\geq4 the Sylvester-Gallai theorem supplies an ordinary line and so the pair. The problem's discussion notes that the lines spanned by the rr 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 o(n2)o(n^2) 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 fr,k(n)f_{r,k}(n) is settled at n2n^2 for r≥3r\geq3 and k≥4k\geq4, between n2/12−O(n)n^2/12-O(n) and (1−1r−1)n22+1(1-\tfrac1{r-1})\tfrac{n^2}2+1; the constant is open. The site points to Problem 209 as related.