Erdős Problem #960 — Let r,k≥2r,k\geq 2 be fixed.

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Let r,k≥2r,k\geq 2 be fixed. Let A⊂R2A\subset \mathbb{R}^2 be a set of nn points with no kk points on a line. Determine the threshold fr,k(n)f_{r,k}(n) such that if there are at least fr,k(n)f_{r,k}(n) many ordinary lines (lines containing exactly two points) then there is a set A′⊆AA'\subseteq A of rr points such that all (r2)\binom{r}{2} many lines determined by A′A' are ordinary. Is it true that fr,k(n)=o(n2)f_{r,k}(n)=o(n^2), or perhaps even ≪n\ll n?

References

Progress summary

Refreshed
Claimed solved

A 2026 manuscript claims the conjecture is false in every nontrivial case, but its proof has not yet been independently verified.

The problem asks whether sufficiently few ordinary lines in a planar point set with no kk collinear points force an rr-point subset whose all connecting lines are ordinary. The proposed negative answer says even quadratically many ordinary lines need not force such a subset.

Known results

  • Turán's theorem gives the general upper bound fr,k(n)≤(1−1r−1)n22+1f_{r,k}(n)\le \left(1-\frac{1}{r-1}\right)\frac{n^2}{2}+1.

April 2026 claimed disproof

A manuscript claims that for r≥3r\ge 3, k≥4k\ge 4, and n≥72n\ge 72, an elliptic-curve construction has no four collinear points, at least n212−103n\frac{n^2}{12}-\frac{10}{3}n ordinary lines, and a bipartite ordinary-line graph, hence no required KrK_r. It attributes the proof to an internal OpenAI model; the claim remains unverified.

Current status (as of August 2026): A manuscript claims a complete negative resolution for all r≥3r\ge 3 and k≥4k\ge 4, but independent verification is not recorded, so the problem is not settled.

Sources

Solutions 0

No solutions have been posted yet.