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If $A_1$ and $A_2$ are orthogonal latin squares of order $n$ and $B_1$ and $B_2$ are orthogonal latin squares of order $m$ then the latin squares $A_1 \times B_1$ and $A_2 \times B_2$ are orthogonal.

It seems that this statement is related to quasi-groups but I don't know how...

multiplication of latin squares are described here in the "tensor product of linear maps" section.

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    How do you multiply two latin squares? – 2017-01-05
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    @TonyK i added a link up there – 2017-01-05
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    But a latin square is not a linear map. – 2017-01-05
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    By the way, once we have sorted this out, I expect that you will want $A_1 \times B_1$ and $A_2 \times B_2$, not $A_1 \times A_2$ and $B_1 \times B_2$. – 2017-01-05
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    @TonyK nope it's not a linear map but the multiplication is like that , look at this photo https://pictub.club/image/scllSW – 2017-01-05
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    @TonyK it's 6.6.2 proposition in Peter J. Cameron's Combinatorics: Topics, Techniques, Algorithms – 2017-01-05

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