Erdős Problem #42 — Sidon sets difference-disjoint from a maximum Sidon set

About 31 years old · traced to

Let 1≤a1<a2<⋯<ah≤n1 \leq a_1 < a_2 < \cdots < a_h \leq n be a maximum Sidon sequence. Can one find a Sidon sequence b1<b2<⋯<br≤nb_1 < b_2 < \cdots < b_r \leq n for every rr and n>n0(r)n > n_0(r) so that the differences aj−aia_j - a_i, bv−bub_v - b_u are all distinct, i.e., so that

aj−ai≠bv−bufor all i<j and u<v?a_j - a_i \neq b_v - b_u \qquad \text{for all } i < j \text{ and } u < v?
References

Additional references

P. Erdős, Some of my favourite problems in number theory, combinatorics, and geometry, Resenhas IME-USP 2 (1995), 165-186.

Progress summary

Refreshed
Open

A complete solution has been claimed, but no public mathematical artifact independently verifies it; smaller cases and a sparse-set version are established.

The problem asks whether every Sidon set AA in {1,…,N}\{1,\ldots,N\} eventually admits an MM-element Sidon set BB with no nonzero difference in common. The cases M=1M=1 and M=2M=2 are settled, and a full proof has been claimed but remains under examination.

Known results

  • M=1M=1 and M=2M=2: settled; the source gives no mathematician or year.
  • M=3M=3: proved by Sedov using ChatGPT and Codex.
  • Sparse regime: for fixed MM and Sidon AA with ∣A∣=o(N)|A|=o(\sqrt N), the assertion holds for sufficiently large NN.
  • A claimed quantitative extension gives ∣B∣≫(log⁡log⁡Nlog⁡log⁡log⁡N)1/2|B|\gg\left(\frac{\log\log N}{\log\log\log N}\right)^{1/2}; its proof is not verified.

Claimed full solution

Sandhu prompted GPT 5.5 Pro to produce a proof for all MM. A separate Fourier--compactness write-up states the same theorem, but provides neither independent verification nor publication details; the Lean formulation still displays sorry.

Current status (as of February 2026): M=1M=1, M=2M=2, M=3M=3, and a sparse-set subcase are recorded, while the claimed all-MM solution remains unverified.

Sources

Solutions 0

No solutions have been posted yet.