- What is the lowest positive, what the highest possible value for the determinant of a standard-magic-square-matrix of order $n$?
- Are there singular standard-magic-square-matrices of any order greater than $3$?
First of all, the determinant of a standard-magic-square-matrix must be a multiple of $\frac{n^2(n^2+1)}{2}$ for odd $n$. This follows easily by the following process:
Add all the columns to the last column. Then every entry in the last column is $\frac{n(n^2+1)}{2}$,the constant of the standard-magic-square.
Now extract this constant and add all the rows to the last one. Then every entry in the last row is again the constant, beside the last entry, which is $n$.
Since $n$ is for odd n a divisor of the constant, it can be extracted as well. For even $n$, only $\frac{n}{2}$ can be extracted, so the determinant is only a multiple of $\frac{n^2(n^2+1)}{4}$. This gives lower bounds for the absolute value of the determinant of regular magic-square-matrices.
For size $3$, the only possible determinant (ignoring the sign) is $360$.
For size $4$, my personal minimum for the absolute non-zero determinant is $2176$ and my maximum is $17408$.
For size $5$, my best results are $325$ and $6\ 547\ 775$.
For sizes $4$ and $5$, I also found matrices with determinant $0$, but for $n = 6$ I found none. OEIS claims that the magic square of order $6$ produced by Matlab has determinant $0$ (By the way, the sequence seems to contain a typo because in the list $-360$ appears for $n=2$ instead of $n=3$).
My pascal program generating random magic squares did not find a magic square with order $6$ and determinant $0$. Since I do not have Matlab, I cannot verify the magic square produced by it.
