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  "body": "\n## Definition\n\nLet $n$ be a positive integer.\nLet $L = (L_{ij})$ and $M = (M_{ij})$ be [[Latin squares|44HTG620GB]] of order $n$, on symbol sets $S$ and $T$ with $n$ elements each.\nThe squares $L$ and $M$ are orthogonal if the $n^2$ ordered pairs $(L_{ij}, M_{ij})$, for $1 \\le i, j \\le n$, are pairwise distinct.\nA family of Latin squares of order $n$ is mutually orthogonal if each two distinct squares of the family are orthogonal.\n\n## Remarks\n\nSince $S \\times T$ has $n^2$ elements, orthogonality means that every pair in $S \\times T$ occurs exactly once.\nExample of order $3$: for $i, j \\in \\Z/3\\Z$, let $L_{ij} = i + j$ and $M_{ij} = 2i + j$.\nIf $(L_{ij}, M_{ij}) = (s, t)$, then $i = t - s$ and $j = 2s - t$, so each pair $(s, t)$ comes from exactly one cell, and $L$ and $M$ are orthogonal.\nNon-example: a Latin square of order $n \\ge 2$ is never orthogonal to itself, because only the $n$ pairs $(s, s)$ occur.\nThere 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|WXJAVQKFR6]].\nFor a prime power $q$ and the [[finite field|08RTDHX1SK]] $\\F_q$, the squares $L^{(a)}_{xy} = ax + y$ for $a \\in \\F_q \\setminus \\set{0}$ form $q - 1$ mutually orthogonal Latin squares of order $q$.\n",
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