gpt 6 astra ultra attempt on Math-517

Sum-Free Subsets of a Cartesian Grid

Definitions and mathematical statement

Shared notation.

Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

Write $[N]=\{1,\ldots,N\}$ and use coordinatewise addition in ${\mathbb{Z}}^d$. A set $A\subseteq[N]^d$ is sum-free if there are no $x,y,z\in A$ with $x+y=z$, including the case $x=y$. The extremal function is $s_d(N)=\max\{|A|:A\subseteq[N]^d,\ A\cap(A+A)=\varnothing\}$. The dimension $d$ is fixed while $N$ varies. [S1] [S2]

Statement recorded in the supplied source.

Fix an integer $d$. What is the largest sum-free subset of $[N]^d$? [S1]

Residual task reported by the archive.

The embedded review $2026-08-17$ records: Determine the maximum density for every fixed d. [S1]

Short English statement

For a fixed dimension, how large can a subset of an integer grid be when adding two chosen points never produces another chosen point?

Research attempt

Provenance and scope.

This individual attempt was generated by GPT-6 Astra Ultra on 1 October 2026. The method was to derive explicit partial results and test the first proposed extension adversarially. The original statement and sources below are retained. No claim of mathematical novelty or complete solution is made.

Formal target and literature check.

The equation is coordinatewise $x+y=z$ in the integer grid; it is not addition modulo $N$. Dimension is fixed before $N$ tends to infinity. Lepsveridze and Sun's current manuscript, inspected directly, determines the asymptotic densities in dimensions three and four. We therefore do not present those cases as open or the following slice construction as new. The purpose of this attempt is to calculate the slice mechanism exactly and identify a specific obstruction to a proposed fibrewise upper bound for arbitrary dimension.

A complete family of lower constructions.

For $u>0$ let \[ A_{u,N}=\{x\in[N]^d:uN\leq x_1+\cdots+x_d<2uN\}. \] If $x,y\in A_{u,N}$, their coordinate sums add to at least $2uN$, whereas every element of $A_{u,N}$ has smaller sum. Thus the set is sum-free, including when $x=y$. The strict upper endpoint is necessary: using closed endpoints would permit sums landing on the boundary. For fixed $d,u$, the proportion tends to the volume of the corresponding slice of $[0,1]^d$. To justify this without a probability assumption, partition the cube into mesh-$1/N$ cubes. All cubes not meeting either boundary hyperplane have exactly the correct membership. Each hyperplane meets $O_d(N^{d-1})$ mesh cubes, so the counting error is $O_d(N^{d-1})$.

Put $F_d(t)=\operatorname{vol}\{x\in[0,1]^d:\sum x_i<t\}$. Inclusion-exclusion applied to the nonnegative simplex gives \[ F_d(t)=\frac1{d!}\sum_{j=0}^d(-1)^j\binom dj(t-j)_+^d, \qquad V_d(u)=F_d(2u)-F_d(u). \] Indeed, the unconstrained simplex has volume $t_+^d/d!$; requiring a specified collection of $j$ coordinates to exceed one translates it to the simplex of radius $t-j$. This proves the formula, including its saturation at zero and one. On intervals avoiding the finitely many breakpoints, a stationary point satisfies $2F_d'(2u)=F_d'(u)$. This is an explicit finite algebraic optimisation for each fixed dimension; it proves the best density within this family, not global optimality.

The two-dimensional optimisation, including all ranges.

For $0\leq t\leq1$, $F_2(t)=t^2/2$, and for $1\leq t\leq2$, $F_2(t)=1-(2-t)^2/2$. If $0\leq u\leq1/2$, then $V_2(u)=3u^2/2\leq3/8$. If $1/2\leq u\leq1$, then \[ V_2(u)=-1+4u-\frac52u^2 =\frac35-\frac52(u-\tfrac45)^2. \] If $1\leq u\leq2$, then $V_2(u)=(2-u)^2/2\leq1/2$, and for $u\geq2$ the volume is zero. Hence the best slice has $u=4/5$ and density $3/5$. The calculation is independent of the established global two-dimensional theorem; it supplies only its lower-bound side. Exact counts for $N=5,10,25,100$ are recorded by the companion script using integer comparisons $4N\leq5(x+y)<8N$, so rounding cannot introduce floating-point errors.

Why the most immediate induction does not close.

Write the fibres of a general $A\subset[N]^d$ as $A_t\subset[N]^{d-1}$ according to the final coordinate. The actual condition is \[ (A_s+A_t)\cap A_{s+t}=\varnothing\qquad(s+t\leq N). \] It does not say that each $A_t$ is sum-free. For instance, choose all points with final coordinate exceeding $N/2$. The resulting set is sum-free because sums leave the last-coordinate range, while every nonempty fibre is the entire lower-dimensional grid and contains many additive triples. Thus an induction that bounds each fibre by $s_{d-1}(N)$ is false at its first step. The relevant upper-bound problem couples three different fibres. Even replacing fibres by their cardinalities discards the location of their sumsets, so the displayed condition cannot be reduced to an ordinary scalar inequality without an additional structural argument.

Verification and exact status.

For $d=1$, the same slice calculation yields density one half, and empty slices and endpoint choices behave correctly. For higher $d$, the explicit formula provides computable lower bounds and a finite search over stationary points, with all error terms taken at fixed dimension. The unresolved step is to show that arbitrary coupled fibres cannot beat the optimal slice in dimensions beyond the cases established in the cited paper. The false fibrewise induction is now ruled out by an explicit family, and the slice optimisation is proved. No general upper bound matching it has been obtained here.

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 98 (GREEN-006), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath.

• S2 Ben Green, 100 Open Problems (author’s maintained problem list). https://people.maths.ox.ac.uk/greenbj/papers/open-problems.pdf. The author-hosted document was inspected on 27 September 2026; the archive’s local code is not a problem-number locator in this paper.

• S3 S. Lepsveridze and Y. Sun, Size of the largest sum-free subset of [n]^3, [n]^4, and [n]^5, arXiv:2311.18289 $2023$. https://arxiv.org/abs/2311.18289. Bibliographic information from the supplied record.

• S4 S. Lepsveridze and Y. Sun, Size of the Largest Sum-Free Subset of [n]^3 and [n]^4, IMRN $2026$. https://doi.org/10.1093/imrn/rnag081. Bibliographic information from the supplied record.

• E1 S. Lepsveridze and Y. Sun, Size of the largest sum-free subset of $[n]^3$ and $[n]^4$, current manuscript. https://arxiv.org/abs/2311.18289. Inspected 1 October 2026.