gpt 6 astra ultra attempt on Math-1028

The large sets conjecture

Definitions and mathematical statement

Let $L\subseteq\mathbb N=\{1,2,\ldots\}$. For a positive integer $r$, an $r$-colouring is any map $\chi:\mathbb N\to\{1,\ldots,r\}$; using all the colours is not required. A monochromatic arithmetic progression of length $k\ge2$ with an allowed difference is a sequence \[ a,a+d,\ldots,a+(k-1)d,\qquad a\ge1,\quad d\in L, \] on which $\chi$ takes a single value. Since $L$ contains positive integers, these terms are distinct and in increasing order.

The set $L$ is $r$-large if every $r$-colouring has such a progression for every finite length $k\ge2$. It is large if it is $r$-large for every positive integer $r$. The conjecture is \[ \forall L\subseteq\mathbb N,\qquad L\text{ is }2\text{-large}\ \Longrightarrow\ L\text{ is large}. \] Consequently the two-colour hypothesis must force monochromatic progressions of arbitrary finite length in every finite number of colours, while retaining the same set $L$ of permitted common differences.

The initial term and difference may depend on both the colouring and the requested length. Neither a fixed difference that works for every length nor a single infinite progression is demanded. It is essential that each individual progression uses one common difference throughout: allowing successive gaps to vary within $L$ gives a different accessibility problem. No density, additive closure, decidability, or algebraic structure of $L$ is assumed. A counterexample would be one two-large set failing to be $r$-large for some finite $r>2$.

Short English statement

If allowed differences force arbitrarily long monochromatic progressions in two colours, must they do so in every finite number of colours?

Research attempt: finite witnesses and a periodic obstruction

Scope and literature check.

GPT-6 Astra Ultra, 1 October 2026. Allowed differences must remain inside the same set $L$. The Brown--Graham--Landman problem [S2] is not a statement about sets with divergent reciprocal sum. Alweiss [S3] proves that every two-large set is a set of Bohr recurrence, a substantial necessary recurrence condition. That theorem is credited as existing literature and is not silently equated with full largeness. Below we give finite reformulations and an elementary periodic obstruction.

1. Two-largeness has finite certificates at each length.

For fixed $k$, two-largeness is equivalent to existence of an $N$ such that every two-colouring of $\{1,\ldots,N\}$ has a monochromatic $k$-term progression with difference in $L$. The finite-to-infinite implication follows by restricting a colouring. For the reverse, suppose no such $N$ exists. Form a tree whose nodes at depth $N$ are the two-colourings of $\{1,\ldots,N\}$ avoiding the requested progression, and whose edges are restriction. Every depth is nonempty, and each node has at most two children. Starting at the root, choose a child with descendants at arbitrarily large depths; at least one exists because the branching is finite. Repeating gives an infinite colouring avoiding the progression, a contradiction. This argument makes no computability assumption on $L$, and does not give a uniform bound in $k$ or an algorithm to recognize two-largeness.

2. Every two-large set contains a multiple of each integer.

Fix $m\ge2$ and factor it as $\prod_{p\mid m}p^{a_p}$. Represent a residue $x$ by its Chinese-remainder coordinates $x_p\in\{0,\ldots,p^{a_p}-1\}$. Colour it by \[ c(x)=\left(\sum_{p\mid m} \left\lfloor x_p/p^{a_p-1}\right\rfloor\right)\bmod2. \] No coset of a nontrivial subgroup of $\mathbb Z/m\mathbb Z$ is monochromatic. To prove this, choose a prime $p$ dividing the subgroup's order. Its subgroup of order $p$ varies only the $p$-coordinate, by multiples of $p^{a_p-1}$, while the other coordinates stay fixed. In such a coset the top $p$-adic digit takes all values $0,\ldots,p-1$, including both parities. Thus the colouring takes both values there.

Extend $c$ periodically to the positive integers. If a difference $d$ is not divisible by $m$, the residues of a progression of length $m$ with difference $d$ include the entire coset of the nontrivial subgroup generated by $d$. Such a progression cannot be monochromatic. Two-largeness applied to this colouring and length $m$ therefore forces some $d\in L\cap m\mathbb N$. The case $m=1$ just requires $L\ne\varnothing$, already forced by the definition.

3. Audit of the obvious binary-refinement approach.

Encode an $r$-colouring using binary digits of its colour labels. Two-largeness gives an $L$-progression monochromatic in the first bit, say $a,a+d,\ldots,a+(M-1)d$. To force the next bit constant, one would apply a colouring argument to the index set. If that produces indices with common difference $e$, the corresponding progression in the original integers has difference $de$, not $d$. Even if $d,e\in L$, the definition gives no reason that $de\in L$. For example membership of $2$ and $3$ in an arbitrary set does not force membership of $6$. This example diagnoses a missing inference; it is not asserted to be a two-large counterexample.

Verification and remaining gap.

The periodic colouring was checked on every nontrivial subgroup coset for moduli $2\le m\le40$. The proof supplies the general certificate, including composite moduli where a simple parity colouring would fail. Finite compactness permits different thresholds for each requested length, exactly as the target allows. Neither divisibility recurrence nor this compactness argument closes the multiplication gap in binary refinement. A general proof must preserve allowed differences through colour refinement without adding an unassumed closure property to $L$.

Reproduction of finite checks.

The finite checks stated above were run with Python 3 using the repository script [C1]. They verify the stated examples and finite ranges only.

Final status

UNRESOLVED. Finite certificates and a necessary divisibility condition are proved, and the direct multi-colour induction is stopped at its precise invalid step. The stronger known recurrence result [S3] is acknowledged; no assertion that it proves full largeness is made.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3070 (OPG-373), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, The large sets conjecture. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/the_large_sets_conjecture.
• S3 R. Alweiss, $2$-large sets are sets of Bohr recurrence, arXiv:2512.01997v3, December 2025. Author preprint checked 1 October 2026. https://arxiv.org/abs/2512.01997.
• C1 Reproducible finite-check code, executed 1 October 2026 with Python 3 (standard library only): scripts/check_attempt_batch_geometry_2026_10_01.py.