I would like to prove the claim that pandiagonal latin squares, which are of form
0 a 2a 3a ... (n-1)a b b+a b+2a b+3a ... b+ (n-1)a . . . . . . . . . (n-1)b (n-1)b +a (n-1)b +2a (n-1)b +3a ... (n-1)(b+a)
for some $a,b\in (0,1...n-1)$ cannot exist when the order is divisible by 3.
I think we should be able to show this if we can show that the pandiagonal latin square of order $n$ can only exist iff it is possible to break the cells of a $n \times n$ grid into $n$ disjoint superdiagonals. Then we would show that an $n\times n$ array cannot have a superdiagonal if $n$ is a multiple of $2$ or $3$. I am, however, not able to coherently figure out either part of this proof. Could someone perhaps show me the steps of both parts?
A superdiagonal on an $n \times n$ grid is a collection of $n$ cells within the grid that contains exactly one representative from each row and column as well as exactly one representative from each broken left diagonal and broken right diagonal.
EDIT: Jyrki, could you please explain how the claim follows from #1?