## January 2016 riddle

UPDATE (1 February): As there have not been any correct solutions so far, this riddle will run another month, in parallel to a new February riddle. However, to make the riddle easier next month, solvers can also answer about a largest set of mutually orthogonal Latin squares of order p for a prime p. Separate solver lists will be kept for each of the two cases.

This month's problem is a well-known one, so no Internet searching, please.

I thank Ian Wanless for pointing it out to me.

Let us begin with some definitions:

A Latin Square is a matrix of size n×n whose entries are all in {1,...,n}, such that no entry (a.k.a. "symbol") repeats more than once in any row or in any column.

Here is an example of a Latin square of order 3 (which is to say: n=3):

 1 2 3 3 1 2 2 3 1

Two Latin squares, Q and R, are said to be Orthogonal if the function mapping the row/column pair (i,j) to the symbol pair (Qij, Rij) is a bijection.

An example of an order 3 Latin square that is orthogonal to the first one given is:

 3 2 1 2 1 3 1 3 2

A set of Latin squares is called Mutually Orthogonal if every pair in the set is a pair of orthogonal Latin squares.

Now, to the question.

Let k be some natural. Find, with proof, a largest set of mutually orthogonal Latin squares of order 2k. Your proof must show that the set you found is mutually orthogonal and also that no larger set exists with this property.

### List of solvers:

#### Prime order:

Jim Boyce (1 February 07:42)
Harald Bögeholz (8 February 09:13)

#### Power of two order:

Jim Boyce (1 February 07:42)
Harald Bögeholz (10 February 07:27)

Elegant and original solutions can be submitted to the puzzlemaster at riddlesbrand.scso.com. Names of solvers will be posted on this page. Notify if you don't want your name to be mentioned.

The solution will be published at the end of the month.

Enjoy!