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Given a pair of mutually orthogonal latin squares (MOLS) of order $m$ and a pair of MOLS of order $n$, how would we construct a pair of MOLS of order $mn$?

[EDIT: MOLS means Mutually Orthogonal Latin Squares]

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@Gerry: Yes, I added the parenthetical "MOLS" right after the first use to clarify that (original post contained only "MOLS", no explanation) – Arturo Magidin Jul 13 '12 at 3:35

A proof is given here, if you click on MOLS Produce More MOLS (and maybe scroll up one slide).

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Let $M_1$ and $M_2$ represent the pair of MOLS of order $m$. Let $N_1$ and $N_2$ represent the pair of MOLS of order $n$.Take the cross product of $M_1$ and $N_1$ to produce one of the MOLS of order $mn$ and then take the crossproduct of $M_2$ and $N_2$ to produce the $2$-nd MOL of order $mn$. These two will then give you a pair of MOLS of order $mn$.

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