An $n\times n$ matrix is defined to be a "latin square" if each row and column is a permutation of the first $n$ natural numbers. Two squares of same order are orthogonal if the $n^2$ pairs $(x_{ij},y_{ij})$ are "all distinct".
Proof is required that the maximal number of mutually orthogonal pairs of latin squares is $n-1$.
I have put "all distinct" in quotes, because if interpreted one way this would be easier than the book projects it to be (my intuition is usually wrong in these cases) If we fix an element $x_{ij}$ then there are $n-1$ choices for $y_{ij}$ so there can be atmost $n-1$ mutually orthogonal latin squares. I suspect this is incorrect. If so, any insight would be appreciated. This is incorrect.
For $n=1$ no pairs possible. For $n=2$ the possible latin squares are
$\begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix}\qquad \mbox{and}\qquad \begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}$
By my assumed definition the above squares would be orthogonal but clearly I am asked to look at all $n^2$ pairs so my reasoning is false. There are no mutually orthogonal pairs for $n=2$ either. I would appreciate any help/pointers/hints.