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Problem 3 of Dubickas (2006): Is $\sqrt{3} \in \mathcal{Z}$?

Answers Problem 3 and generalizes it: the classification $\sqrt{m} \in \mathcal{S} \iff m = 2$ covers every square root, and a further theorem replaces parity by divisibility by any $p \ge 2$. Note the scope of the machine-checking, which is narrower than the paper: the author states that the case $m = 3$ is what is verified in Lean, and the repository flags the thickness computation of section 4.1 and all of section 8 as not formalized.

Exact FrontierDelta

Prior state unknownproved

Scope and record

Occurred: Aug 21, 2026

Delta type: SOURCE CLAIM

Assumptions: VibeMathed verification: lean-checked. Publication: preprint. AI contribution: ai-discovered. Imported under CC BY 4.0.

Canonical aliases: Problem 3 of Dubickas (2006): Is $\sqrt{3} \in \mathcal{Z}$? · Dubickas Problem 3

Confidence: Not scored

Registry verification: lean checked · preprint · candidate

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Attribution

VibeMathed
registry · event recorded by

Ralf Stephan
human · human collaborator

Fable 5
model · ai model contributor · Anthropic

Opus 5
model · ai model contributor · Anthropic

Artifacts and verifiers

Compute record

No linked compute attempts recorded.

Lineage and corrections

This event attributed to Ralf Stephan

This event attributed to Opus 5

This event attributed to Fable 5

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Problem 3 of Dubickas (2006): Is $\sqrt{3} \in \mathcal{Z}$? — Mathematical Frontier Network