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Source. T. Feng, T. Trinh, G. Bingham et al., Semi-Autonomous Mathematics Discovery with Gemini: A Case Study on the Erdős Problems, arXiv:2601.22401v3 (5 February 2026); Section 2.1, the problem as posed and Remark 2.1 on p. 9, the solution on pp. 10--11. The result is unnumbered; the paper's Theorem 1 (p. 10) is the Pach--Sharir incidence bound it quotes. The artifact is identified on the source card.

Read depth. Claims checked: the problem, the assertion and the proof (pp. 9--11) were read in full on the print; the cited incidence theorem was not checked against its source. Nothing here is independently reviewed. A preprint.

Statement

For points x1,…,xn∈R2x_1,\ldots,x_n\in\mathbb R^2 let R(xi)=#{∣xj−xi∣:j≠i}R(x_i)=\#\{|x_j-x_i|:j\ne i\}, with the points ordered so that R(x1)≤⋯≤R(xn)R(x_1)\le\cdots\le R(x_n), and let αk\alpha_k be minimal such that for all large enough nn some nn-point set has R(xk)<αkn1/2R(x_k)<\alpha_k n^{1/2} (p. 9). The paper proves (p. 10) that αk=Ω(k1/4)\alpha_k=\Omega(k^{1/4}) as k→∞k\to\infty, and hence that αk→∞\alpha_k\to\infty.

Proof pointer

Fix kk and, for large nn, an nn-point set with R(xk)<αkn1/2R(x_k)<\alpha_k n^{1/2}. The circles centred at x1,…,xkx_1,\ldots,x_k through the other points number fewer than kαkn1/2k\alpha_k n^{1/2}, and each of the n−kn-k remaining points lies on kk of them. The Pach--Sharir bound (the paper's Theorem 1, quoted from Pach and Sharir 1998, Theorem 1.1), applied to circles with k=3k=3, s=2s=2 and exponents (3/5,4/5)(3/5,4/5), bounds the incidences above; dividing by nn and letting n→∞n\to\infty gives k≤C(αk4/5k4/5+1)k\le C(\alpha_k^{4/5}k^{4/5}+1) (pp. 10--11). Remark 2.1 (p. 9) states that the agent's original output used wrong exponents (2/3,2/3)(2/3,2/3) from a reference the authors could not find, and an αk+ϵ\alpha_k+\epsilon limit step; the displayed solution corrects both.

Dependencies

Pach and Sharir, the incidence bound for curves with kk degrees of freedom and multiplicity type ss (cited, not held).

Bears on

  • Problem 652: the statement answers the question as posed, with the rate k1/4k^{1/4}; the paper notes (p. 9) that the solution is an immediate reduction to the literature and that a later solution by others uses a theorem of Mathialagan instead.