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Could you show me an example of an order 7 pandiagonal latin square? A pandiagonal latin square is one where no broken diagonal contains repeated symbols. I have found examples for smaller order but am unable to create one for $7\times 7$.

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    Danielle, the basic recipe: $L(i,j)=i+aj\pmod p$ works again. Has this not been covered in the book you are studying? Quite$a$few of your recent questions could be solved by looking for suitable conditions for the extra parameter $a$, and these exercises in the book have the air that you are expected to apply that rather than trying to brute force your way to a suitable latin square by educated guessing.2012-07-20

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I'm not quite sure that I understood what you want, but would $L(i,j)=i+2j$ fit the bill? The diagonals have slopes $\pm1$, so they satisfy the equation $j=\pm i+k$ for some constant $k$. But $L(i,\pm i +k)=(1\pm 2)i+2k$, and as $1\pm 2\neq0$, different values for $i$ lead to different values on this diagonal.

Well, if I gave the recipe, might as well.

$ \pmatrix{ 0&2&4&6&1&3&5\cr 1&3&5&0&2&4&6\cr 2&4&6&1&3&5&0\cr 3&5&0&2&4&6&1\cr 4&6&1&3&5&0&2\cr 5&0&2&4&6&1&3\cr 6&1&3&5&0&2&4\cr } $

Notice that if you move parallel to the main diagonal one step down and one step to the right, the entry always "increases" by $3$ (modulo $7$), because $1+2=3$. But if you move one step down and one step to the left, the entries "increase" by $-1$, because $1-2=-1$. Wrap around when you reach the boundary of the square (according to the idea of broken diagonals). This is is because $i$ is the row index, so as $i$ increases, we go down, and increasing/decreasing $j$ makes us move to the right/left. [Remark: Initially I had that explained the wrong way here, sorry about that.]