A Latin square with trace-sum $11$ is : $$\begin{bmatrix} 1&4&2&3\\ 2&3&4&1\\ 4&1&3&2\\ 3&2&1&4\end{bmatrix}$$ Is trace sum $7/9/5/13/15$ possible in some other arrangement?
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By swapping numbers in the given arrangement, we can trivially derive trace sums of $7$ (main diagonal reads $2113$), $9$ ($3114$) and $13$ ($2443$).
But trace sums of $5$ and $15$ are not possible, for there would have to be three $1$s or three $4$s on the main diagonal respectively, forcing the fourth number to be the same as the other three by the Latin square property and thereby not yielding the desired trace sum.
Parcly Taxel
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Can you answer a similar question : Is a 5 cross 5 latin square matrix possible whose trace-sum is 6 or 7 or 9 or 22 or 23 or 24 ? – Shalini Tomar Apr 04 '20 at 05:00
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@ShaliniTomar Please ask that in another question and upvote and accept my answer. – Parcly Taxel Apr 04 '20 at 05:01
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I did that ....:) – Shalini Tomar Apr 04 '20 at 05:18
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@ Parcly Taxel Here is the new question : https://math.stackexchange.com/questions/3608980/is-a-5%c3%975-latin-square-possible-whose-trace-sum-is-6-or-7-or-9-or-22-or-23-or-24 – Shalini Tomar Apr 04 '20 at 05:29