I am looking for a certain kind of latin square (nxn).
Rules:
No repeats in any column or row (Definition of Latin Square)
No repeats in any diagonal including others than the main diagonal.
So $\begin{matrix} 1&2\\2&1 \end{matrix}$ would not qualify nor would $\begin{matrix} 1&2&(3)&4\\4&(3)&2&1\\2&1&4&3\\3&4&1&2 \end{matrix}$
So far I have found that no solution exists for $1\lt n\lt 5$ or for $n=6$.
For $n=1$ we have the trivial solution: $\begin{matrix} 1 \end{matrix} $.
For $n=5$ we have $2$ solutions
$\begin{matrix} 1&2&3&4&5\\3&4&5&1&2\\5&1&2&3&4\\2&3&4&5&1\\4&5&1&2&3\end{matrix}$ $\begin{matrix} 1&2&3&4&5\\4&5&1&2&3\\2&3&4&5&1\\5&1&2&3&4\\3&4&5&1&2\end{matrix}$
and homomorphisms of these for a total of $240$ solutions.
For larger odd dimensions the $n=5$ set of solutions can be generalized bacause the same pattern visible in the $5x5$ works as a $7x7,11x11,13x13,17x17,19x19,...$ so we have for n$\ge$5 where $n$ is odd and $n$ is not a multiple of $3$, we always have at least $2*n!$ solutions which are the homomorphisms of squares with elements $2y+x+1$(mod n) where $y$ is the row index and $x$ is the column index of each element. So my question is are there any other squares which satisfy these constraints? Ex: any even dimensioned square, or any odd dimensioned square not following these rules.
Any ideas on which way to go next would be appreciated. Including perhaps ideas on how to write a program to solve it. I have found my current solutions simply by method of exhaustion.