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Claim. The answer to Problem 659 is yes. Benjamin Grayzel, Solution to a Problem of Erdős Concerning Distances and Points, arXiv:2601.09102, first posted as a comment with a working note on the site's discussion thread on 13 January 2026 and on arXiv on 14 January 2026 (v2 on 16 January 2026), proves in its Theorem 1 that for every integer $n\ge2$ there is a set $P$ of $n$ points in $\mathbb R^2$ such that every four-point subset of $P$ determines at least three distinct distances and the number of distinct distances of $P$ is $O(n/\sqrt{\log n})$. The set is any $n$-point subset of the $m\times m$ box $P_m=\{(i,\sqrt2\,j):0\le i,j\le m-1\}$ in the lattice $\mathbb Z\times\sqrt2\,\mathbb Z$ with $m=\lceil\sqrt n\rceil$. Squared distances in the lattice are values of the form $u^2+2v^2$, so Bernays' asymptotic for the integers up to $x$ represented by a primitive positive definite binary quadratic form of nonsquare discriminant, here $-8$, bounds the distances of $P_m$ by $O(m^2/\sqrt{\log m})$ (Corollary 4). The local condition (Theorem 5) follows from Perucca's classification of the six similarity types of four-point sets with two distances: five contain a square or an equilateral triangle, which the lattice excludes because a perpendicular or a sixty-degree rotation of a nonzero lattice vector leaves the lattice (Lemmas 6 and 7), and the sixth, four vertices of a regular pentagon, has its diagonal and side in the golden ratio $\varphi$, so its two squared distances are in the irrational ratio $\varphi^2=(3+\sqrt5)/2$, which lattice squared distances, all integers, cannot realize (Lemma 8). The paper's acknowledgment attributes the core idea, in particular the pentagon exclusion, to Gemini 3.0 and the drafting to GPT-5.2 and Gemini 3.0, with the author taking responsibility for every claim; Grayzel's first comment on the thread (13 January 2026) calls the approach "fully ideated" by that system. The lattice construction with its distance count is older: the paper credits Moree and Osburn and a 2014 blog post by Sheffer, and the site credits Moree and Osburn and, independently, Lund and Sheffer, who also noted the absence of squares and equilateral triangles; the new step is the pentagon exclusion that completes the local condition. The source card is grayzel_2026_solution_problem_erdos_concerning_distances_points, whose Theorem 1 page holds the corpus's own-words proof chain with Bernays' asymptotic and Perucca's classification as stated external premises.

Acceptance. The site's curator, Thomas Bloom, confirmed on the discussion thread on 13 January 2026 that the problem is solved and that the thread records its history, and the site labels the problem proved, with a Lean marker; its commentary credits the earlier constructions and Grayzel's comment on the thread (made using Gemini) with the pentagon exclusion that completes the solution (problem page last edited 16 January 2026); that confirmation is the reviewed evidence. The note is an arXiv preprint with no journal publication recorded, so no refereed evidence is listed. The corpus's own reconstruction of the complete chain on the source card is author-recorded, with an independent review reported but not retained, and awards no standing here.

Formalization. Boris Alexeev's repository plby/lean-proofs holds, at the pinned commit linked above, a Lean 4 file whose header declares it a formalization of a solution to Problem 659 found by Grayzel using Gemini, auto-formalized by Aristotle (Harmonic), under Lean 4.24.0 and the Mathlib revision the header names. Its final theorem, erdos_659, is stated in the formal-conjectures form: a sequence of finite subsets of $\mathbb R^2$ with $n$ points, every four of which determine at least three distances, and $O(n/\sqrt{\log n})$ distinct distances. It is discharged from the intermediate main_theorem, which takes Perucca's classification and Bernays' theorem as hypotheses; the classification is proved in the file (PeruccaClassificationStatement_proof, which the header says Aristotle proved by itself), while Bernays' theorem is declared as an axiom, since Mathlib has no proof of it, and #print axioms erdos_659 lists bernays, propext, Classical.choice and Quot.sound. Alexeev announced it on the thread on 14 January 2026, and Terence Tao accepted on the thread a formalization that takes an uncontroversial published theorem as an axiom. The corpus has not built the file or audited its statement, and a proof from an axiom is not a kernel check of the whole statement, so the file is a link here and not formalized evidence. Feng and coauthors' report on the Aletheia research agent gives an independent proof of the same answer on a different lattice, generated before this note was written; it has its own page, Feng and coauthors.