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 $6$?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ 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 MeyerowitzAug 8, 2014 at 23:36
-
$\begingroup$ @AaronMeyerowitz I hoped that there is some simple proof because of increasing number of MOLS from 2 to 5. $\endgroup$– ArimakatAug 9, 2014 at 0:55
-
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 BeenakkerAug 9, 2014 at 14:59
-
$\begingroup$ I agree with OP and Peter Mueller, but I suspect there is no proof that would make me happy. $\endgroup$– Peter DukesNov 7, 2014 at 11:12
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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