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I have been unable to find an elegant method of constructing self orthogonal Latin squares. However, I came across this question: construct a self orthogonal Latin square of order 5 using the fact that the set of elements on the main diagonal of a self orthogonal Latin from a transversal. This seems to imply that you can use this fact alone to construct self orthogonal Latin squares or at-least ones of order five.

I have been unable to figure out how I can do this question and would appreciate any help.

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    You should just be able to brute force this. Put a transversal along the diagonal, try filling in the first row/first column. Once you have that you will be pretty close to done (redo first row/column if it can't be completed). They are asking you to do this for $n=5$ because it's small enough to do in an ad-hoc way by hand. – xxxxxxxxx Apr 24 '20 at 00:03
  • Thank you @MorganRodgers, I've now managed to construct one via brute force. –  Apr 24 '20 at 00:19

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Use the elements of $\;\mathbb{Z}_v$ as the names of the rows and columns of your Latin square. Let $\boldsymbol{A}=(a_{ij})$ such that $a_{ij}=2i-j\in \mathbb{Z}_v$. This forms a self-orthogonal Latin square always when the $\gcd(v,6)=1$, so it works for the order $5$ case as well.

hirohe
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