No pandiagonal latin squares with order divisible by 3? 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?

It is weakly pandiagonal Latin square of order 9
http://forumimage.ru/uploads/20171204/151240677122376874.png
Weakly pandiagonal Latin square is pandiagonal as magic square.
I found weakly pandiagonal Latin square of order 12.
But I have not yet found the weakly pandiagonal Latin square of order 15.
It is semi-pandiagonal Latin square of order 9 (pandiagonal in one direction)
http://forumimage.ru/uploads/20171217/151350437058089933.png
I don't know the semi-pandiagonal Latin square of order n > 9 yet.
See more here
https://boinc.progger.info/odlk/forum_thread.php?id=178&postid=7105
https://boinc.progger.info/odlk/forum_thread.php?id=178&postid=7107
