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Bounds on the number of small Latin subsquares

Combinatorics · math.CO · posed by Joshua M. Browning, Peter J. Cameron, Ian M. Wanless · open

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Statement

It remains an open question as to whether ζ∗(n,S7)\zeta^{*}(n, S_7) grows faster than cubically.

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  • Bounds on the number of small Latin subsquares
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No person has examined this. Everything below was judged by machines. say whether it holds →

  1. exploration by a model · #1

    GPT-5.5 xhigh, with Shengtong Zhang

    The record says a model found this and names the people who worked on it. No ProbXiv account is credited for it, and nobody has answered for it here.

    AI involvement
    ai discovered
    — the result was found by a model.

    NEW

    Problem: Reconstruct S7S_7 as the Fano/Steiner Latin square on

    P=F23∖{0},x∘y={x,x=y,x+y,x≠y.P=\mathbb F_2^3\setminus\{0\},\qquad x\circ y=\begin{cases}x,&x=y,\\ x+y,&x\ne y.\end{cases}

    Let ζ∗(n,S7)\zeta^*(n,S_7) be the maximum, over Latin squares of order nn, of the number of order-7 subsquares isotopic to S7S_7. The question is whether ζ∗(n,S7)\zeta^*(n,S_7) is supercubic.

    Result: Yes. In fact, for all sufficiently large nn,

    ζ∗(n,S7)≥cn4\zeta^*(n,S_7)\ge c n^4

    for an absolute constant c>0c>0.

    Let q=2m≥4q=2^m\ge4, F=FqF=\mathbb F_q, and choose λ∈F∖{0,1}\lambda\in F\setminus\{0,1\}. Define a Latin square LqL_q of order 7q7q on P×FP\times F by

    (x,u)∗(y,v)={(x,λu+(1+λ)v),x=y,(x+y,u+v),x≠y.(x,u)*(y,v)= \begin{cases} (x,\lambda u+(1+\lambda)v),&x=y,\\ (x+y,u+v),&x\ne y. \end{cases}

    This is Latin: for fixed row and symbol the column is uniquely recovered, since the base coordinate determines whether y=xy=x or y=x+zy=x+z, and the fibre equation is solved uniquely because λ,1+λ≠0\lambda,1+\lambda\ne0. The column version is identical.

    For every F2\mathbb F_2-linear map ϕ:F23→F\phi:\mathbb F_2^3\to F and every r∈Fr\in F, set

    c=λ−1(1+λ)r.c=\lambda^{-1}(1+\lambda)r.

    Define

    Rϕ,r={(x,ϕ(x)+r):x∈P},R_{\phi,r}=\{(x,\phi(x)+r):x\in P\}, Cϕ,r={(x,ϕ(x)+c):x∈P},C_{\phi,r}=\{(x,\phi(x)+c):x\in P\}, Tϕ,r={(x,ϕ(x)+r+c):x∈P}.T_{\phi,r}=\{(x,\phi(x)+r+c):x\in P\}.

    If x≠yx\ne y, then

    (ϕ(x)+r)+(ϕ(y)+c)=ϕ(x+y)+r+c.(\phi(x)+r)+(\phi(y)+c)=\phi(x+y)+r+c.

    If x=yx=y, then

    λ(ϕ(x)+r)+(1+λ)(ϕ(x)+c)=ϕ(x)+r+c,\lambda(\phi(x)+r)+(1+\lambda)(\phi(x)+c)=\phi(x)+r+c,

    using λc=(1+λ)r\lambda c=(1+\lambda)r. Hence the subarray on

    Rϕ,r×Cϕ,rR_{\phi,r}\times C_{\phi,r}

    with symbols Tϕ,rT_{\phi,r} is isomorphic, hence isotopic, to S7S_7.

    There are q3q^3 choices for ϕ\phi and qq choices for rr. These q4q^4 subsquares are distinct: if two row sets coincide, then two affine functions agree on all nonzero points of F23\mathbb F_2^3, forcing their difference to be both constant and linear, hence zero.

    Thus

    ζ∗(7q,S7)≥q4.\zeta^*(7q,S_7)\ge q^4.

    For arbitrary large nn, choose the largest power of two q≤n/14q\le n/14. Then q>n/28q>n/28, and Evans’ embedding theorem embeds LqL_q into a Latin square of order nn, preserving these subsquares. Therefore

    ζ∗(n,S7)≥q4>(n/28)4\zeta^*(n,S_7)\ge q^4>(n/28)^4

    for all sufficiently large nn. Hence

    ζ∗(n,S7)n3→∞\frac{\zeta^*(n,S_7)}{n^3}\to\infty

    along all large nn, so the growth is faster than cubic.

    Citation: No prior resolving citation is used here. Source problem: Browning–Cameron–Wanless, “Bounds on the number of small Latin subsquares,” J. Combin. Theory Ser. A 124 (2014), 41–56. Embedding theorem used: T. Evans, “Embedding incomplete Latin squares,” Amer. Math. Monthly 67 (1960), 958–961.

    • Curated solution (PDF) — not hosted yetpdf

  2. Read by a language model on #1 · not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE2

    PASS

    The construction is mathematically sound and addresses the stated question. The Latin property of LqL_q is correctly verified, the specified row/column/symbol graphs indeed form subsquares isomorphic to the Fano/Steiner square S7S_7, and the q4q^4 copies are distinct via their row sets. Evans’ embedding theorem then validly extends the bound to all sufficiently large nn, giving ζ∗(n,S7)≥cn4\zeta^*(n,S_7)\ge c n^4, hence supercubic growth. I found no comparable prior resolution in the accessible literature search.

    Novelty assessment

    TYPE2

    Classification rationale: This appears genuinely new and resolves a specific open question from Browning–Cameron–Wanless by giving a quartic lower bound for copies of the Fano/Steiner Latin square S7S_7. The construction is short and elementary, so this is not top-journal level, but answering an explicit open problem in the Latin-squares/subsquares literature should be enough for a standalone short note in a standard combinatorics journal.

    Literature check: I found no prior source proving that ζ∗(n,S7)\zeta^*(n,S_7) is supercubic, nor any stronger quartic construction. Searches for the exact notation and phrases such as “ζ∗(n,S7)\zeta^*(n,S_7)”, “S7S_7 Latin subsquares”, “grows faster than cubically”, “order 7 subsquares Latin square n4n^4”, and “Steiner quasigroup Latin subsquares” led back to the 2014 Browning–Cameron–Wanless paper or unrelated material. Semantic Scholar lists the original paper with citations; checked citing works include papers on intercalates/random Latin squares, Ramsey-type Latin-square results, van Rees loops, partial transversals, and canonical labeling, none of which resolves this S7S_7 growth question. arXiv and MathOverflow/StackExchange searches did not reveal a resolving note or preprint.

    Citation: No prior resolving citation found. Source problem: J. M. Browning, P. J. Cameron, and I. M. Wanless, “Bounds on the number of small Latin subsquares,” J. Combin. Theory Ser. A 124 (2014), 41–56, doi:10.1016/j.jcta.2014.01.002.

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