Grayzel 2026 solution problem erdos concerning distances points

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corollary_4: Uses Bernays' represented-integer asymptotic to construct n planar points with only order n over the square root of log n distinct distances.

lemma_6: Excludes nondegenerate squares because a perpendicular side vector cannot have the required integer and square-root-of-two coordinates.

lemma_7: Excludes equilateral triangles by showing that a sixty-degree rotation of a nonzero lattice vector cannot return to the lattice.

lemma_8: Rules out the four-vertex regular-pentagon configuration because its two squared distances have an irrational ratio.

theorem_1: Constructs n planar points with few total distances while every four points determine at least three distances.

theorem_5: Combines Perucca's classification with three lattice exclusions to prove the local four-point condition in every finite box.

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Benjamin Grayzel, Solution to a Problem of Erdős Concerning Distances and Points. arXiv preprint (2026). arXiv:2601.09102.

Grayzel answers a 1997 question of Erdos affirmatively: Theorem 1 gives, for every integer n >= 2, a planar set P of n points in which every 4-point subset determines at least 3 distinct pairwise distances while the total number of distinct distances is O(n / sqrt(log n)). The construction is an m-by-m box P_m in the anisotropic lattice L = Z times sqrt(2) Z; squared distances are values of the binary quadratic form u^2 + 2v^2, so Bernays's asymptotic for integers represented by a primitive, positive definite integral form of non-square discriminant (Theorem 3) bounds the distance count (Corollary 4), following earlier work of Moree-Osburn on lattices with few distances, with a pointer to a related discussion by Sheffer. Theorem 5 verifies the local constraint by using Perucca's classification of the six similarity types of 4-point two-distance sets and ruling out each in L: no nondegenerate square (Lemma 6), no equilateral triangle (Lemma 7), and no regular-pentagon trapezoid (Lemma 8). Against the review notes, the paper is confirmed unrelated to problems 660, 1088 and 100: it settles the planar 4-point/3-distance question (indexed as problem 659 on erdosproblems.com), not the R^3 convex-polyhedron distance question, not f_d(n), and not the separate distance problems those entries concern.

Version read. The copy read for this card is arXiv:2601.09102v2. The official arXiv history identifies v2 from 16 January 2026 as the latest version. The arXiv record names arXiv's non-exclusive distribution license (arXiv:2601.09102), every other right reserved.

Source: https://arxiv.org/abs/2601.09102.

Bears on. #659: Theorem 1 constructs, for every integer n >= 2, an n-point planar set in which every 4-point subset determines at least 3 distinct distances and the total number of distinct distances is O(n / sqrt(log n)), the configuration the problem asks for; Corollary 4 gives the distance bound and Theorem 5, through Lemmas 6 to 8, the four-point condition.

Compiled results.

• Theorem 1 gives the full solving construction by combining the global and local branches. Its living verification record is the current review record for the complete chain.
• Corollary 4 proves the $O(n/\sqrt{\log n})$ global bound for an $n$-point subset of $P_{\lceil\sqrt n\rceil}$, stating Bernays' asymptotic as an external premise.
• Theorem 5 proves the local four-point condition, stating Perucca's six-type classification as an external premise.
• Lemma 6, Lemma 7, and Lemma 8 exclude, respectively, squares, equilateral triangles, and the regular-pentagon trapezoid from the lattice.

The complete proof chain is covered by the living verification record on Theorem 1. Bernays and Perucca remain precise external interfaces whose proofs are not compiled here. The reported Lean formalization assumes Bernays' theorem as an axiom; no local build is claimed.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.