It is easy to show that a latin square of size n x n has a determinant, which is a multiple of $\large \frac{n^2(n+1)}{2}$, if n is odd and $\large \frac{n^2(n+1)}{4}$, if n is even. This is a lower bound for the absolute value of the determinant, if the matrix is regular. But not always can the minimum be attained : For n = 4, the minimum is $80$ instead of $20$ and for n = 6 , it is $126$ instead of $63$.
My questions :
- For which n does a singular latin square matrix exist ?
For which n does a latin square matrix with a determinant of the given limit exist ?
My conjecture for the first question is that n = 1,2,3,5 are the only values for which there is no singular latin square matrix.
My conjecture for the second quesion is that n = 4,6 are the only values for which there is no latin square matrix with a determinant fulfilling the given limits.