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I know that $6\times 6$ MOLS cannot be constructed, but if I am not mistaken we can draw up two MOLS that are $6 \times 6$ with $34$ distinct pairs of symbols. However, I am not able to find this construction of MOLS. Could someone show me what it would be like?

[EDIT: MOLS = Mutually Orthogonal Latin Squares]

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Please, if you are using an acronym like MOLS, add at least once the full terminology. Also, your title was pleonasmic: "MOLS Latin squares" would be "mutually orthogonal latin squares latin squares". That's like going to the ATM machine. –  Arturo Magidin Jul 13 '12 at 3:52
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And entering your PIN number. –  Gerry Myerson Jul 13 '12 at 3:58
    
Along the same lines, it's also peculiar to say "two mutually orthogonal Latin squares", rather than "two orthogonal Latin squares". –  Douglas S. Stones Jul 14 '12 at 5:13

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It was found by Joseph Douglas Horton, Sub-latin squares and incomplete orthogonal arrays, Journal of Combinatorial Theory, Series A, 16 (1974) 23-33. Unfortunately, I don't have access to this journal. Interlibrary loan could probably help you.

Edit (Douglas S. Stones): The JDH paper gave a pair of orthogonal partial Latin squares, which I edited to give this pair:

5 6 3 4 1 2
2 1 6 5 3 4
6 5 1 2 4 3
4 3 5 6 2 1
1 4 2 3 5 .
3 2 4 1 . 6

1 2 5 6 3 4
6 5 1 2 4 3
4 3 6 5 1 2
5 6 4 3 2 1
2 4 3 1 6 .
3 1 2 4 . 5
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Thanks, @Douglas. –  Gerry Myerson Jul 14 '12 at 6:57

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