Parity obstruction in the minimum-determinant problem for Latin squares
partialconfidence 70%
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Precise statement
A Mathematics Stack Exchange question posted on 3 August 2014 asks when the standard divisibility lower bound for determinants of Latin square matrices is attained. For an $n\times n$ Latin square $L$ with entries $1,\ldots,n$, let
$$
b_n=\begin{cases}
n^2(n+1)/2,&n\text{ odd},\\
n^2(n+1)/4,&n\text{ even}
\end{cases}.
$$
For which positive integers $n$ does there exist such an $L$ with $|\det L|=b_n$? The question conjectures that $n=4,6$ are the only orders for which this minimum cannot be attained.
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fidelity, correctness, priority, or novelty.
What AI did
GPT-5.4
Under the author's direction, OpenAI's ChatGPT, using the GPT-5.4 model, generated the central mathematical development of this work, including the ordinary-to-centered determinant reduction, the exact binary rank and adjugate criteria governing the additional factor of two, and the all-order construction producing an odd determinant quotient for every $n\equiv2\pmod4$, $n\ge6$. It also assisted with the development of the exact verification code and the manuscript. The author selected the research direction, checked the mathematical derivations and certified outputs, established the public claim boundaries, commissioned adversarial reviews, and takes responsibility for the final content.
The paper and public repository contain complete proofs, exact certified datasets, and a deterministic verifier that currently passes all 12 public artifacts. The release also received an artifact-oriented adversarial audit. These checks establish internal consistency and reproducibility, not independent expert endorsement of the headline theorem; no domain expert has yet endorsed it. The appropriate VibeMathed verification label is therefore Unreviewed.
Determinant Divisibility of Centered Latin Squares
For even $n$, let $q(L)=\det(L)/b_n$. The work proves that $q(L)$ is even exactly when the stronger centered divisibility $n^2\mid\det(E_{\mathrm{std}})$ holds. For $n\equiv2\pmod4$, this is equivalent to $\operatorname{rank}_{\mathbb F_2}(A\bmod2)<n-1$; for $n\equiv0\pmod4$, it is equivalent to $\operatorname{adj}(A\bmod2)\mathbf1=0$. An explicit family gives odd $q(L)$ for every $n\equiv2\pmod4$, $n\ge6$. This removes a universal extra-factor-two obstruction, but it does not prove $|q(L)|=1$. Exact minimum attainment and the separate singularity question remain open.
Known method families
argument (source-reported)
Source-reported tools: argument.
Independent: unknown · difference confidence: 0
What remains uncertain
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VibeMath has not independently audited the mathematical statement, proof, or novelty claim.