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Conjecture · version 1

Normal five-edge-colorings for bridgeless cubic graphs

Statement

Every finite loopless bridgeless cubic graph has a proper edge-coloring with at most five colors in which the union of the palettes at the endpoints of every edge has either three or five colors.

Assumptions

  • A poor edge has three colors across its endpoint stars; a rich edge has five. Four is forbidden.

Evidence

Source reports

Connections

In this problem · publication v7 · Selected context

Petersen coloring conjecture →

Equivalent formulation · Disproved

Equivalent statements

Jaeger’s theorem gives equivalent existence conditions for each loopless cubic graph.

The positive universal assertions are both refuted by the reported witness; equivalence remains a valid theorem.

Details & sources
Every bridgeless cubic graph admits a Petersen coloring →

Disproved · version 1

Every finite loopless bridgeless cubic graph admits an edge map to the Petersen graph that maps each incident edge-star bijectively onto a target vertex’s edge-star.

  • Parallel edges are allowed; the simple connected witness already lies in this class.

Scope: Parallel edges are allowed; the simple connected witness already lies in this class.

Normal five-edge-colorings for bridgeless cubic graphs →

Disproved · version 1

Every finite loopless bridgeless cubic graph has a proper edge-coloring with at most five colors in which the union of the palettes at the endpoints of every edge has either three or five colors.

  • A poor edge has three colors across its endpoint stars; a rich edge has five. Four is forbidden.

Scope: A poor edge has three colors across its endpoint stars; a rich edge has five. Four is forbidden.

Petersen and normal-five colorings are equivalent →

Relationship statement · Proved · version 1

For each finite loopless cubic multigraph, the existence of a Petersen coloring is equivalent to the existence of a normal five-edge-coloring.

  • The paper gives an explicit bijection using P = KG(5,2). Bridgelessness is not needed for this equivalence.

Scope: The paper gives an explicit bijection using P = KG(5,2). Bridgelessness is not needed for this equivalence.

Relationship not independently reviewed.

Connection in Petersen coloring conjecture →

Relationship not independently reviewed. This concerns its use in this problem.

Sources

A 112-Vertex Counterexample to the Petersen Coloring Conjecture →

Bryce Putman · Preprint · Publication date Aug 8, 2026

Source-reported results and external checks are separate from independent OpenMath assessment.

Revision & attribution

Source revision: v1. Observed Oct 14, 2026.

Putman reports extensive OpenAI-model assistance in discovery, search, verification and preparation; he reviewed the final claims and artifacts and accepts responsibility.

History

Statement versions

Statement versions and problem publication versions are separate. A new publication can update context or sources while retaining the same statement.

Assessment history · 1

Assessments retained in these publication records. Earlier statement versions retain their own source histories.

Record details
Claim ID
claim-petersen-normal-five
Statement version
1
SHA-256
75f863c9f83abba4b1443d06f873c626a0ae1d1f5a26e85af7b94bf3d831bd84

Publication records