A pair of MOLS of order 15?

Question

Is there any place where I can find a pair of MOLS(mutually orthogonal latin squares) of order 15? I can't seem to find a place where it's spelled out explicitly.

Answer 1

For any odd $n$, a diagonally cyclic Latin square construction works. Here's GAP code that implements it:

n:=15;

# Diagonally cyclic Latin square
L:=List([1..n],i->List([1..n],j->0));
for j in [1..n] do L[1][j]:=((2*(j-1)) mod n)+1; od;
for i in [2..n] do for j in [1..n] do L[i][j]:=(L[i-1][((j-2) mod n)+1] mod n)+1; od; od;
L;

# Fixed diagonal Latin square
M:=List([1..n],i->List([1..n],j->((j-i) mod n)+1));
M;

and this is what the output looks like:

gap> L;
[ [ 1, 3, 5, 7, 9, 11, 13, 15, 2, 4, 6, 8, 10, 12, 14 ], 
  [ 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13 ], 
  [ 14, 1, 3, 5, 7, 9, 11, 13, 15, 2, 4, 6, 8, 10, 12 ], 
  [ 13, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11 ], 
  [ 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 2, 4, 6, 8, 10 ], 
  [ 11, 13, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9 ], 
  [ 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 2, 4, 6, 8 ], 
  [ 9, 11, 13, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7 ], 
  [ 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 2, 4, 6 ], 
  [ 7, 9, 11, 13, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5 ], 
  [ 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 2, 4 ], 
  [ 5, 7, 9, 11, 13, 15, 2, 4, 6, 8, 10, 12, 14, 1, 3 ], 
  [ 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 2 ], 
  [ 3, 5, 7, 9, 11, 13, 15, 2, 4, 6, 8, 10, 12, 14, 1 ], 
  [ 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15 ] ]

and

gap> M;
[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 ], 
  [ 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 ], 
  [ 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 ], 
  [ 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ], 
  [ 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ], 
  [ 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ], 
  [ 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9 ], 
  [ 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8 ], 
  [ 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7 ], 
  [ 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6 ], 
  [ 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5 ], 
  [ 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4 ], 
  [ 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3 ], 
  [ 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2 ], 
  [ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1 ] ]

We can readily see that they're orthogonal (the forwards broken diagonals of the first square are $1,2,\ldots,n$ cyclically permuted, whereas the forwards broken diagonals of the second are $x,x,\ldots,x$, for some $x$). The only thing we should check is that the first square is always a Latin square, which comes from its first row being an orthomorphism.

Answer 2

The Wikipedia page on Greco-Latin squares shows squares of order 3 and 5. Gerry Myerson gave an answer to your question /mutually-orthogonal-latin-squares-of-order-mn-from-order-m-and-order-n which shows how to make one. What don't you understand?