A Salem-Spencer-Type Construction for Large Subsets of Integer Grids with No Isosceles Right Triangles

Gyula K\'arolyi; Jozsef Solymosi

A Salem-Spencer-Type Construction for Large Subsets of Integer Grids with No Isosceles Right Triangles

T0 review · 0 major / 6 minor · reviewed 2026-08-01 · deepseek-v4-flash

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Pith's one-line read The paper proves that every n×n integer grid contains an isosceles-right-triangle-free subset of size at least n^{1.317865}/log(n)^{140}, the first lower bound exceeding linear.

desk verdict First superlinear lower bound for IRT-free grid subsets, with a genuinely new peeling technique and a checkable finite certificate. read the letter →

arxiv 2607.22828 v1 pith:7JX5W5L7 submitted 2026-07-24 math.CO math.NT

Gyula Károlyi , Jozsef Solymosi This is my paper · ORCID

classification math.COmath.NT MSC 05D1011B30

keywords isoscelesrighttriangle-freeintegergridGaussianintegerspeelingordercarry-freedigitsetlowerboundRoth-typeproblemcomputer-assistedcertificate

verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

The paper establishes a new lower bound for F(n), the largest size of a subset of the n×n integer grid containing no nondegenerate isosceles right triangle: F(n) is at least n^{1.317865}/log(n)^{140}, up to constants. This is the first known superlinear lower bound for this quantity. The construction adapts the classic digit-based method for 3-term arithmetic progressions to two dimensions, working over Gaussian integers and using a finite alphabet of 281 digits. A key novelty is a 'peeling order' on the digits, which allows the alphabet itself to contain forbidden triangles as long as they can be removed one vertex at a time. The result rests on a computer-verified finite certificate.

What carries the argument

The central mechanism is the IRT-peeling order: an ordering p_1,...,p_q of a digit alphabet such that for each t, the only solution to a+ic=(1+i)p_t with a,c in the remaining suffix is a=c=p_t. This is paired with the IRT-carry-free condition, which says that a+ic-(1+i)b is divisible by β only when it equals zero. Together they allow the digitwise proof to go through — the carry-free condition kills any carry at each digit, and the peeling order kills any triangle whose right-angle vertex is the next digit to be peeled. A composition class (words with nearly equal digit counts) supplies the large family of encoded grid points.

What would settle it

Run the provided verifier on the 281-point list: if it reports any failed assertion — for example, a digit c in a remaining suffix for which a=(1+i)p_t−ic is also in the suffix with (a,c) not equal to (p_t,p_t), or a nonzero carry δ divisible by β — the construction collapses. A direct search over the encoded set A_m for a nondegenerate solution of a+ic=(1+i)b would also disprove the claim.

Watch

Extended reading notes

Core claim

Over the Gaussian integers, an isosceles right triangle with right angle at b is encoded by the linear equation a+ic=(1+i)b. The paper chooses the base β=51+51i and a 281-point digit set inside a carry-free diamond region. It then provides an ordering of these digits — the peeling order — such that for every suffix of the ordering, no nondegenerate triangle has its right-angle vertex at the first digit of that suffix. Encoding words with nearly uniform digit frequencies into Gaussian integers produces a set with no forbidden triangles and size roughly n^{log 281 / log|β|}/(log n)^{140}, giving the stated exponent. The geometric carry-free argument and the combinatorial peeling argument toget

Load-bearing premise

The whole construction depends on the 281-point list in Appendix A really having the claimed peeling property and being IRT-carry-free, as checked by the Python script in Appendix B; any bug in that script or mis-transcription of the list would invalidate the theorem.

Editorial extensions

If this is right

If correct, this gives F(n) ≥ n^{1.317865-o(1)}, the first lower bound with exponent greater than 1, shrinking the gap to the known upper bound n^2/(log n)^{1+c}.
The method is general: any Gaussian base β with an IRT-carry-free, peelable alphabet of size q yields F(n) ≫ n^{log q/log|β|}/(log n)^{(q-1)/2}, so improving the digit search in other bases could improve the exponent.
The peeling-order idea applies beyond isosceles right triangles to other rotation-invariant or matrix-coefficient equations, suggesting new digit-encoding constructions in higher dimensions or with other forbidden shapes.
The finite certificate is independently checkable by a standalone program, so the mathematical reduction is separated from the computational discovery and can be verified directly from the paper's source.
Even the smallest nontrivial example — a 3-digit alphabet in base 2+2i — already yields an exponent above 1 (about 1.0566), showing the phenomenon is not unique to the large 281-digit construction.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

The exponent 1.317865 is almost certainly not the end of the story; larger carry-free regions, asymmetric shapes, or block digits could plausibly raise it, while the large gap to the upper bound suggests room for further constructions.
If the peeling technique could be combined with a density increment or Fourier-based upper-bound argument, one might approach exponents closer to 2, though the paper itself does not attempt this.
The credibility of the theorem hinges on the correctness of the verifier and the transcribed 281-point list; a human-readable or formally verified proof of the finite certificate would make the result fully independent of computational trust.
The paper notes that 281 is not proved to be maximal in the diamond region; an exact optimization of the alphabet size in that region could either produce a better exponent or reveal a flaw in the current certificate.

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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Desk Editor's Note

read the letter

The big news: this paper finally breaks the linear barrier for F(n), giving F(n) = n^{1.3179 - o(1)}. That alone would make it worth reading, but the construction is also more than a routine digit trick. The Salem–Spencer idea is adapted to Gaussian integers, with a carry-free diamond region that makes the digitwise argument work, and the genuinely new twist is the peeling order: digits can contain isosceles right triangles as long as they can be eliminated one by one. That feels like the right conceptual step, and it is likely to be reused for other forbidden configurations.

The proof itself is in good shape. The geometric Lemma 3.3 cleanly establishes the carry-free condition for the diamond region. The induction in Theorem 2.6 propagates the peeling order through digit positions via composition classes, and the counting argument is standard. The finite 281-point certificate is checked by a short Python verifier included in the appendix, with a SHA-256 digest to prevent transcription errors. I ran through the verifier logic mentally and the peeling check directly tests the required suffix property; the carry check is in fact redundant given Lemma 3.3, but harmless. The reliance on an external program is the weakest link, but for a finite certificate of this size it is acceptable—the code is simple enough that a skeptical reader can verify it by hand or with a different implementation.

Soft spots worth noting, none fatal. The verifier itself is not formally proved, so someone could object that a bug in the code could sink the theorem; in practice this is a minor concern. The exponent 140 in the log factor is ugly and almost certainly improvable; the authors do not claim otherwise. The gap between this lower bound and the n^2/(log n)^{1+c} upper bound remains enormous, but that is a statement about the problem's difficulty, not a flaw in the paper. I also appreciate that Remark 4.3 explicitly disclaims that the search proves optimality of the alphabet.

The reader's ACCEPT verdict and the stress-test note both hold up on reading the actual manuscript. The citation pattern looks fine: the relevant prior work is cited, and the AlphaEvolve tool is presented honestly as a discovery mechanism, with the mathematical certificate independently checked.

Who should read this: anyone working on Roth-type problems in higher dimensions, additive combinatorics with matrix coefficients, or computer-assisted combinatorial constructions. It deserves a serious referee—the result is new, the method is interesting, and the verification is transparent. I would send it to a good combinatorics journal and expect it to survive review with at most minor revisions.

Referee Report

0 major / 6 minor

Summary. The paper gives a lower bound for F(n), the maximum size of a subset of the n×n grid containing no nondegenerate isosceles right triangle. The construction works over the Gaussian integers, where an IRT with right angle at b is encoded by a + ic = (1+i)b. A digit alphabet P is called IRT-carry-free if this equation has no carry modulo β, and is equipped with an IRT-peeling order if points can be removed one by one so that each removed point is not the right-angle vertex of any IRT in the remaining suffix. Theorem 2.6 shows that from such an alphabet of size q in base β one obtains F(n) ≫ n^{log q / log|β|} / (log n)^{(q-1)/2}. A small example gives exponent about 1.0566. Section 3 proves a geometric lemma supplying carry-free diamond regions for bases β=(2k+1)(1+i). The main construction uses β=51+51i and a 281-point alphabet found by an AlphaEvolve-assisted search; the peeling and carry-free properties of the list are verified by a standalone Python program in Appendix B. This yields Theorem 4.2: F(n) ≫ n^α/(log n)^{140} with α = log 281 / log|51+51i| ≈ 1.317865485534, so F(n) ≥ n^{α-o(1)}.

Significance. If the finite certificate is accepted, this is the first superlinear lower bound for F(n), improving the trivial linear lower bound and complementing the upper bound F(n) ≪ n^2/(log n)^{1+c}. The main mathematical novelty is the peeling order, which allows the digit alphabet itself to contain forbidden triangles, and the use of composition classes so that peeling propagates through every digit position. The separation of a clean geometric carry-free argument from a finite computer-verified search is methodologically sound. A reproducible Python verifier and a SHA-256 digest are included, which is a real strength: the mathematical derivation does not depend on the search algorithm once the certificate is accepted.

minor comments (6)

[§2, proof of Theorem 2.6] The size estimate |W_m| ≫ q^m / m^{(q-1)/2} is introduced immediately after an averaging argument over all compositions, but the average gives only q^m / m^{q-1}. The better bound follows by choosing a composition with nearly equal coordinates and applying Stirling's formula to that multinomial coefficient. Please make this choice explicit so the logical step is unambiguous.
[§3, first sentence] The text refers to Theorem 2.4, but the referenced result is Theorem 2.6. Please correct the cross-reference.
[§4, before Theorem 4.2] The sentence 'Theorem 2.6 applied to this construction yields or main result' contains a typo: 'or' should be 'our'.
[Theorem 4.2] The statement says 'For every integer n>1', but the proof establishes the bound only for sufficiently large n, with the usual Vinogradov convention. Either state 'for all sufficiently large n' or explain how the finitely many small n are absorbed by adjusting the implied constant.
[Appendix B, verifier] In the version of the manuscript I received, the verbatim Python listing appears to have lost whitespace in import statements and after keywords such as 'return', which would make the code syntactically invalid as printed. If this is not a rendering artefact, please ensure the published source contains valid Python. Also confirm that the machine-readable \PointData macro is actually embedded in the TeX source, since the verifier reads directly from it.
[Remark 3.2] The assertion that the displayed function attains its maximum for positive integers at k=5 is stated without proof or numerical evidence. A one-line verification or a reference to a short computation would be helpful.

Circularity Check

0 steps flagged · score 0.0 of 10

No significant circularity: the proof is self-contained, with the finite certificate independently verified.

full rationale

The derivation chain is not circular. The central theorem (Thm 4.2) follows from the general peeling construction (Thm 2.6) applied to the specific alphabet P of size 281 in the carry-free diamond C for β=51+51i. Lemma 3.3 independently proves the IRT-carry-free condition for the whole region C; Proposition 4.1 delegates only the finite peeling check to the Python verifier in Appendix B, which directly tests exactly the required suffix property and carry condition without reference to F(n). The exponent α = log 281 / log|51+51i| is a consequence of q and N(β), not a fitted parameter; AlphaEvolve is explicitly used only as a discovery mechanism and the certificate is independently checked. The paper does not invoke any uniqueness result, and Remark 4.3 expressly disclaims optimality, so there is no imported self-citation chain forcing the answer. The one self-citation ([11]) is contextual and not load-bearing. The reliance on a short, transparent verifier for a finite certificate is acceptable computer-assisted proof practice, not circularity.

Assumptions & free parameters 3 free parameters · 3 assumptions · 0 invented entities

The central result depends on a geometric lemma (proved), an explicit finite certificate (verified by a script), and standard analytic estimates. The construction parameters (β, q, ordering) are chosen to maximize the exponent and are part of the construction, not fitted to data.

free parameters (3)

Base β = 51+51i
Chosen to make the exponent log q/log|β| large; near-diagonal Gaussian base with modulus ≈72.12.
Alphabet size q = 281
The 281 points in Appendix A found by AlphaEvolve; gives exponent ≈1.3179.
Peeling order = the specific order in Appendix A
The peeling order is essential; not every set of 281 points is peelable.

assumptions (3)

standard math Standard properties of Gaussian integers and residue systems modulo β
Used throughout, e.g., Observation 2.1 and Lemma 3.3.
standard math Stirling's approximation and asymptotic of multinomial coefficients
Used in Theorem 2.6 to bound the size of A_m.
ad hoc to paper The Python verifier in Appendix B correctly checks the peeling and carry-free conditions
The finite certificate is accepted on the basis of this program; no formal proof of the program's correctness is given.

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Cite this review

Pith. "Pith review of A Salem-Spencer-Type Construction for Large Subsets of Integer Grids with No Isosceles Right Triangles." pith.science (2026). https://pith.science/paper/7JX5W5L7

@misc{pith2026260722828,
  author       = {Pith},
  title        = {Pith review of: A Salem-Spencer-Type Construction for Large Subsets of Integer Grids with No Isosceles Right Triangles},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/7JX5W5L7}},
  note         = {Machine review of arXiv:2607.22828}
}

abstract

Let $F(n)$ be the largest size of a subset of $\{0,1,\ldots,n-1\}^2$ containing no nondegenerate isosceles right triangle. We give a modified Salem--Spencer-type construction over the Gaussian integers showing that $F(n)=\Omega(n^{1.3})$. The best known upper bound is $F(n)\ll n^2/(\log n)^{1+c}$ for some absolute constant $c>0$, so there is still a large gap between the bounds.

Figures

Figures reproduced from arXiv: 2607.22828 by the authors.

**Figure 1.** A square-free forbidden configuration Theorem 2.6. Let β ∈ Z[i] be a nontrivial element, and let P be an IRT-carry-free alphabet in base β with an IRT-peeling order p1, . . . , pq. For every integer m ≥ 1, there is a set Am ⊂ Z[i] contained in a circle of radius ≪ β m, which does not contain any nondegenerate isosceles right triangle and satisfies |Am| ≫ q m m(q−1)/2 . 3 [PITH_FULL_IMAGE:figures/full_fig_p003_1.png] view at source ↗

**Figure 2.** The 281 Gaussian digits in the carry-free diamond C. Proposition 4.1. The ordered list in Appendix A consists of 281 distinct points of C and is an IRT-peeling order. Its canonical SHA-256 digest is 9ebeedde004c4d30a1da52f5f8fae4f148fee3d524f540c7e35369f192e83724. Proof. This is a finite deterministic check. For each t, form Rt = {pt , . . . , p281}. For each c ∈ Rt , Equation (2.3) determines the only possible endp… view at source ↗

Discussion (0). Continue with ORCID to comment.

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Works this paper leans on

11 extracted references · 3 linked inside Pith

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I. D. Shkredov and J. Solymosi,Tilted corners in integer grids, in: Number Theory and Combinatorics: A Collection in Honor of the Mathematics of Ronald Graham, 2022, 329–338. A The ordered 281-point certificate The tth item below is the Gaussian digitpt =x +iy represented by t...

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Pith tools

Reviewed August 1, 2026 · model on record in the stance chip above.

[1] [1]

F. A. Behrend,On sets of integers which contain no three terms in arithmetical progression, Proceedings of the National Academy of Sciences of the United States of America32(1946), 331–332

1946

[2] [2]

T. F. Bloom and O. Sisask,The Kelley–Meka bounds for sets free of three-term arithmetic progressions, Essential Number Theory2(2023), 15–44

2023

[3] [3]

T. F. Bloom and O. Sisask,An improvement to the Kelley–Meka bounds on three-term arithmetic progressions, arXiv:2309.02353

arXiv

[4] [4]

Erdős and P

P. Erdős and P. Turán,On some sequences of integers, Journal of the London Mathematical Society11(1936), 261–264

1936

[5] [5]

Kelley and R

Z. Kelley and R. Meka,Strong bounds for 3-progressions, in: 2023 IEEE 64th Annual Symposium on Foundations of Computer Science (FOCS), IEEE Computer Society, 2023, 933–973

2023

[6] [6]

Novikov, Ngân V˜ u, M

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Pilatte,New bound for Roth’s theorem with generalized coefficients, Discrete Analysis 2022, Paper No

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