Parity obstruction in the minimum-determinant problem for 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.

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.