TITLE: Is there a simple proof that there is no five mutually orthogonal Latin squares of order 6?
QUESTION [7 upvotes]: 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 $6$?

REPLY [4 votes]: 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.