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Number theory · math.CO · math.NT · posed by A. Hajnal, Bernd Voigt, Jenö Lehel, Martin Gavalec, András Recski, Fred Galvin · open

1 attempt · 1 machine check

Statement

For every 2-coloring of the positive integers there exist positive integers a,d such that the elements of the arithmetic progression a,a+d,...,a+d2a,a+d,...,a+d^{2} all are colored the same.

Context

Candidate 6 of the open problems stated in "Problems", extracted for the Scalable Mathematical Discovery run.

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Attempts

1 attempt

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
    AI involvement
    ai discovered
    — the result was found by a model.
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    GPT-5.5 xhigh

    NEW

    Problem: Let N+={1,2,… }\mathbb N^+=\{1,2,\dots\}. The literal conjecture is:

    For every 2-coloring c:N+→{0,1}c:\mathbb N^+\to\{0,1\}, there exist a,d∈N+a,d\in\mathbb N^+ such that

    c(a)=c(a+d)=⋯=c(a+d2),c(a)=c(a+d)=\cdots=c(a+d^2),

    i.e. the arithmetic progression {a+id:0≤i≤d}\{a+id:0\le i\le d\} is monochromatic.

    Result: The literal statement is true.

    Proof. Let c:N+→{0,1}c:\mathbb N^+\to\{0,1\} be any 2-coloring.

    If there exists n≥1n\ge1 with c(n)=c(n+1)c(n)=c(n+1), then take a=na=n and d=1d=1. The progression is n,n+1n,n+1, hence monochromatic.

    Otherwise c(n)≠c(n+1)c(n)\ne c(n+1) for every nn. Since there are only two colors, this forces

    c(n+2)=c(n)for all n≥1.c(n+2)=c(n)\qquad\text{for all }n\ge1.

    Thus c(1)=c(3)=c(5)c(1)=c(3)=c(5). Taking a=1a=1 and d=2d=2, the progression is

    1,3,5=1,1+2,1+22,1,3,5=1,1+2,1+2^2,

    so it is monochromatic.

    Hence every 2-coloring contains such a monochromatic progression.

    Citation: No external citation used; this resolves the literal reconstructed statement directly.

    Reviews

    1 machine check

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    • Machine check · not human verification

      machine: correct

      Recorded from GPT-5.5 xhigh (SMD judge 1) ·

      scope Full solution as submitted; SMD novelty classification TYPE1

      PASS

      The proof correctly attacks the supplied literal statement. Since d=1d=1 is allowed, any same-colored adjacent pair gives the required progression. If no adjacent pair has the same color, the coloring must alternate, so 1,3,51,3,5 are monochromatic, giving a=1,d=2a=1,d=2. This is rigorous and complete for the stated conjecture.

      Novelty assessment

      TYPE1

      Classification rationale: The accepted proof is a correct resolution of the literal statement, but it is entirely degenerate: d=1d=1 handles any equal adjacent pair, and otherwise the coloring alternates, giving 1,3,51,3,5 with d=2d=2. This is a routine observation and not publishable as a standalone combinatorics result. It also should not be viewed as progress on any intended nondegenerate version requiring larger dd, d>1d>1 in all cases, arbitrarily large dd, etc.

      Literature check: I found no accessible literature source presenting this exact trivial two-color observation as a theorem. Searches for the exact pattern a,a+d,…,a+d2a,a+d,\ldots,a+d^2, “length d+1d+1 common difference dd,” and related Ramsey/arithmetic-progression formulations mainly point back to the original problem context or to broader Ramsey results that do not materially change the assessment. The absence of a citation is unsurprising given the one-paragraph nature of the argument.

      Citation: No prior citation for the resolved literal observation located. Original problem source: A. Hajnal et al., “Problems,” 1981 Winter School on Abstract Analysis, DML-CZ handle 10338.dmlcz/701252.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

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