Mutually orthogonal Latin squares

Remarks

Since $S \times T$ has $n^2$ elements, orthogonality means that every pair in $S \times T$ occurs exactly once. Example of order $3$: for $i, j \in \Z/3\Z$, let $L_{ij} = i + j$ and $M_{ij} = 2i + j$. If $(L_{ij}, M_{ij}) = (s, t)$, then $i = t - s$ and $j = 2s - t$, so each pair $(s, t)$ comes from exactly one cell, and $L$ and $M$ are orthogonal. Non-example: a Latin square of order $n \ge 2$ is never orthogonal to itself, because only the $n$ pairs $(s, s)$ occur. There are no orthogonal Latin squares of order $2$, and there is no orthogonal pair of order $6$; this answers Euler's thirty-six officers problem. For a prime power $q$ and the finite field $\F_q$, the squares $L^{(a)}_{xy} = ax + y$ for $a \in \F_q \setminus \set{0}$ form $q - 1$ mutually orthogonal Latin squares of order $q$.