ProbXiv
sign in
Problem archiveProblem record

Improved Bound for Colorings Without Symmetrically Colored k-APs

Combinatorics · posed by Deng, Tidor and Zhao · partial

1 comment

Statement

Deng, Tidor and Zhao asked whether [N][N] admits a coloring with No(1)N^{o(1)} colors and no symmetrically coloured 4-term arithmetic progression, giving an O(Nlog⁡223)O(N^{\log_{22}3}) coloring. The paper gives an Ok(N4/k2)O_k(N^{4/k^2}) coloring of [N][N] avoiding symmetrically coloured kk-APs for every even k≥4k\ge4, improving the exponent.

Record

Source
Added

Comments

No person has examined this. Nothing here has been checked at all. say whether it holds →

  1. construction · #1

    Ruizhe Shi and Yiqi Dong, using GPT-5

    That credit came with the record as it was imported. No ProbXiv account is credited for this work, and nobody has answered for it here.

    AI involvement
    ai co developed
    — a person and a model developed the result together.

    The acknowledgements state the key construction idea was suggested during an interaction with GPT-5.

    Improves the exponent rather than answering the asked question: whether an No(1)N^{o(1)} colouring exists remains open.

Sign in with an institutional address to take part in the discussion. Reading every thread stays open to everyone.

Sign in

Solve with an agent

Open the statement in a chat, with the problem and the ground rules already written into the prompt.

open in Claude →open in ChatGPT →

This opens a third-party site. Nothing is posted back to ProbXiv and nothing you write there is recorded here — what a model gives you is an attempt, which a person still has to check.