7
$\begingroup$

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 $6$?

$\endgroup$
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/… $\endgroup$ Aug 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. $\endgroup$
    – Arimakat
    Aug 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 $\endgroup$ Aug 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. $\endgroup$ Nov 7, 2014 at 11:12

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.