Erdős Problem #198 — Sidon Sets and Disjoint Infinite Arithmetic Progressions

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The assertion is false: it is not true that every Sidon set A⊆NA\subseteq\mathbb{N} has an infinite arithmetic progression Y⊆NY\subseteq\mathbb{N} with Y⊆N∖AY\subseteq\mathbb{N}\setminus A. Equivalently, there exists a Sidon set A⊆NA\subseteq\mathbb{N} whose complement contains no infinite arithmetic progression.

References

Progress summary

Refreshed
Claimed solved

The conjecture is false: an explicit sparse set can hit every arithmetic progression, so its complement need not contain one.

The problem asks whether every infinite Sidon set has a complement containing an infinite arithmetic progression. The negative answer is attributed to Baumgartner, reportedly through Erdős and Graham.

Known results

  • Baumgartner, 1975: enumerate all infinite arithmetic progressions, choose one rapidly increasing element from each, and the resulting lacunary set is Sidon while meeting every progression.

Explicit factorial construction and formalization

The set A={(n+1)!+n:n≥0}A=\{(n+1)!+n:n\geq 0\} is reported to be Sidon and to meet every arithmetic progression, hence its complement contains none. AlphaProof is credited with finding this construction; Alexeev formalized it in Lean using Aristotle, with Claude assisting only with Lean translation. The formalization is public, but no peer-reviewed or arXiv proof was found.

Current status (as of June 2026): The conjecture has an explicit counterexample, supported by a public Lean formalization; the historical attribution and recent AI-associated account lack a peer-reviewed or arXiv publication.

Sources

Solutions 0

No solutions have been posted yet.