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Kovacs 2024 note erdos s mysterious remark
Zoltán Kovács, A note on Erdős's mysterious remark. arXiv:2412.05190v2 (2024).
Kovács re-proves, by polynomial elimination in the computer algebra system Giac, that a 6-point set in the plane in which every triple spans an isosceles triangle must be the vertices of a regular pentagon together with its center, and that no 7-point set exists. The method encodes the isosceles condition for each of the 20 triples as a degree-6 polynomial, enforces non-degeneracy with Rabinowitsch's trick, and computes an elimination ideal; the full 6-point system is computationally difficult, with no successful run known, so the author eliminates for the 5-point subproblem, obtaining a 33-point solution set (the possible positions of the third point) whose configurations are 5-point subsets of a regular pentagon with its center and squares with their centers, then rules out a sixth point for the square configuration by a second elimination that returns the unit ideal. This settles in the plane Erdős's Problem E 735, whose generalization to R^n is problem #503 (the largest size of a set in R^n all of whose triples are isosceles), giving an algebraic alternative to L. M. Kelly's 1947 geometric argument. For problem #91 (non-similar minimizers of the number of distinct distances), the paper checks the surviving 5-point configurations and confirms Erdős's remark that for n = 5 the regular pentagon is the unique minimizer, since the other candidates realize three distinct distances rather than two. The n >= 3 space case of #503 is not resolved here.
Source: https://arxiv.org/abs/2412.05190. The arXiv record (https://arxiv.org/abs/2412.05190, read 2026-10-02) names the Creative Commons Attribution 4.0 license.
Results to transcribe.
- Six-point isosceles classification: The only S ⊂ R^2 with |S| = 6 all of whose triples form isosceles triangles is a regular pentagon together with its center, proved by elimination ideals in Giac.
- Seven points impossible: No 7-point planar set has all triples isosceles: any two of its 6-point subsets are pentagon-plus-center and must coincide, forcing |S| = 6.
- Five-point elimination: Eliminating for 5 points yields a 33-point solution variety (the possible positions of the third point) whose configurations are 5-point subsets of a regular pentagon with its center and squares with their centers.
- Erdős's remark on #91 verified: Among 5-point sets only the regular pentagon achieves 2 distinct distances; the alternative configurations give 3, confirming uniqueness for n = 5.