Suppose you create an *n* × *n* Latin square filled with the first *n* letters of the Roman alphabet. This means that each letter appears exactly once in each row and in each column.

You could repeat the same exercise only using the Greek alphabet.

Is it possible to find two *n* × *n* Latin squares, one filled with Roman letters and the other filled with Greek letters, so that when you combine corresponding entries, each combination of Roman and Greek letters appears exactly once? If so, the combination is called a Greco-Latin square.

Whether Greco-Latin squares of size *n* exist depends on *n*. But before we explore that, let’s give an example.

Here are two Latin squares, one filled with the first four Roman letters and the other filled with the first four Greek letters.

A B C D α β γ δ
D C B A γ δ α β
B A D C δ γ β α
C D A B β α δ γ

If we concatenate the two matrices, we get

Aα Bβ Cγ Dδ
Dγ Cδ Bα Aβ
Bδ Aγ Dβ Cα
Cβ Dα Aδ Bγ

and each of the two-letter entries is unique. So we’ve shown that a Greco-Latin square exists for *n* = 4.

## Mutually Orthogonal Latin Squares (MOLS)

The more modern name for Greco-Latin squares is “mutually orthogonal Latin squares,” abbreviated MOLS. We say two Latin squares *M* and *N* are mutually orthogonal if the list of pairs (*M*_{ij}, *N*_{ij},) contains no duplicates.

## Euler’s work

Euler showed how to construct Greco-Latin squares when *n* is odd and when *n* is a multiple of 4. He conjectured that no Greco-Latin squares exist when *n* = 4*k* + 2.

His conjecture was incorrect. For example, there is a Greco-Latin square of size 10. And in fact, Greco-Latin squares exist for all *n* except *n* = 2 and *n* = 6.

## Magic squares

Suppose you have two orthogonal Latin squares *M* and *N* of size *n*, and suppose that the diagonal elements of both squares contain no duplicates.

Use the numbers 0 through *n*-1 rather than Greek or Latin letters to fill the squares and interpret the Latin squares as matrices. Then the matrix

*nM* + *N*

is a magic square. This is equivalent to taking the corresponding Greco-Latin square and interpreting the entries as base *n* numbers.

For example, using the Latin squares above, replace *A* and α with 0, B and β with 1, C and γ with 2, and *D* and γ with 3. Then the Greco-Roman square

Aα Bβ Cγ Dδ
Dγ Cδ Bα Aβ
Bδ Aγ Dβ Cα
Cβ Dα Aδ Bγ

becomes

00 11 22 33
32 23 10 01
13 02 31 20
21 30 03 12

with the entries being base 4 numbers. The rows and columns clearly have the same sum because they all have the same set of numbers in the 1s place and in the 4s place. Written in base 10, the magic square above is

0 5 10 15
14 11 4 1
7 2 13 8
9 12 3 6

It’s conventional for magic squares to be filled with consecutive numbers starting with 1, so you could add 1 to every entry above if you’d like.

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