It is well known that there is a projective plane of order $n$ if and only if there exist a set of $n-1$ mutually orthogonal Latin squares. The first nontrivial case is $n=6$, which fails because of Bruck-Ryser theorem. The history of the problem mentions Thomas Clausen and Gaston Tarry who proved that there are no two mutually orthogonal Latin squares of order $6$. Their proof consists of a lot separate cases and Stinson gives some short proof of that in 1984, but it also contains some cases discussion. Is there a simple proof (unlike long cases discussion) that there is no five (instead of two) mutually orthogonal Latin squares of order $6$?
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$\begingroup$ There is not even a pair of MOLS of order 6. If Euler couldn't find one and Stinson saw fit to publish a (fairly self contained) 3 page proof, it seems unlikely that there is an obvious shorter proof. You might want to look up "thirty-six officer problem". Some information is at math.stackexchange.com/questions/356793/… $\endgroup$– Aaron MeyerowitzCommented Aug 8, 2014 at 23:36
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$\begingroup$ @AaronMeyerowitz I hoped that there is some simple proof because of increasing number of MOLS from 2 to 5. $\endgroup$– ArimakatCommented Aug 9, 2014 at 0:55
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2$\begingroup$ Sossinsky gives this an exercise in his book, so apparently by Russian standards there is a "simple proof". Unfortunately, there is no "solution manual" :) --- ium.mccme.ru/postscript/f11/sossinskii-GeoBook-part1.pdf --- exercise 14.9 $\endgroup$– Carlo BeenakkerCommented Aug 9, 2014 at 14:59
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$\begingroup$ I agree with OP and Peter Mueller, but I suspect there is no proof that would make me happy. $\endgroup$– Peter DukesCommented Nov 7, 2014 at 11:12
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1 Answer
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I agree with the OP that Stinson's paper, while short, doesn't give a clean and conceptual proof of the non-existence of a pair of 2 MOLS of order $6$.
If the OP is happy with another proof of the non-existence of planes of order $6$, there is an alternative to Bruck-Ryser suggested by Assmus: One can show that there is no plane of order $n$ for $n\equiv6\pmod{8}$. (That's a special case of Bruck-Ryser.) The proof is contained in the second edition of Lineare Algebra by Huppert/Willems. They, however, rely on Gleason's Theorem about the weight enumerator of binary doubly-even selfdual codes. A slightly different treatment avoiding this theorem can be found in this script on coding theory.
answered Aug 9, 2014 at 21:28