No IMO 2025/2026 Math Found; Only IMO 1988 P6 Vieta Jumping — E8 Intelligence Research

FINDING: No substantive mathematical content — search returned video titles, a robotics competition, and no actual IMO 2025/2026 problem statements or solutions. | MATH: Only identifiable constant: IMO 1988 P6 (the "legendary" problem) — statement: for positive integers a,b, if ab+1 divides a²+b², then (a²+b²)/(ab+1) is a perfect square. This is a Vieta-jumping / quadratic-residue problem, not a geometric-harmony problem. | CONNECTION: None supported. Vieta jumping exploits the discriminant structure of a quadratic in one variable — a lattice/root-system adjacency in disguise (integer points on a hyperbola), but no 0.618/1.618/base-60/crystallographic constant appears in the evidence. | DEPTH: 1 FINDING: The 2011 IMO P2 "windmill" problem is a combinatorial-geometry invariant problem (a line rotates about a point, sweeping points alternately across it). | MATH: Invariant: parity of point-count on each side of the rotating line is preserved under the reflection rule; the process is per Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com

Authors

Publication Details

Journal
Zenodo (CERN European Organization for Nuclear Research)
Published
2026-10-30
DOI
https://doi.org/10.5281/zenodo.23796984
Primary Topic
Mathematics and Applications
Type
preprint
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No IMO 2025/2026 Math Found; Only IMO 1988 P6 Vieta Jumping — E8 Intelligence Research

Andrew Stewart Caldin
10 links
Zenodo (CERN European Organization for Nuclear Research)
Mathematics and Applications
preprint

No IMO 2025/2026 Math Found; Only IMO 1988 P6 Vieta Jumping — E8 Intelligence Research

Andrew Stewart Caldin
preprint en

Abstract

FINDING: No substantive mathematical content — search returned video titles, a robotics competition, and no actual IMO 2025/2026 problem statements or solutions. | MATH: Only identifiable constant: IMO 1988 P6 (the "legendary" problem) — statement: for positive integers a,b, if ab+1 divides a²+b², then (a²+b²)/(ab+1) is a perfect square. This is a Vieta-jumping / quadratic-residue problem, not a geometric-harmony problem. | CONNECTION: None supported. Vieta jumping exploits the discriminant structure of a quadratic in one variable — a lattice/root-system adjacency in disguise (integer points on a hyperbola), but no 0.618/1.618/base-60/crystallographic constant appears in the evidence. | DEPTH: 1 FINDING: The 2011 IMO P2 "windmill" problem is a combinatorial-geometry invariant problem (a line rotates about a point, sweeping points alternately across it). | MATH: Invariant: parity of point-count on each side of the rotating line is preserved under the reflection rule; the process is per Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com

Zenodo (CERN European Organization for Nuclear Research)
Mathematics and Applications
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