1 Introduction

For the notions of latin squares and orthogonal squares see [9] and [8]. Latin squares of order 6 are studied in [11] and [4]. A MOL(6), i.e., a pair of orthogonal latin squares of order 6, is a linear space on 24 points with 4 blocks of length 6 and 36 blocks of length 4. Each 4-block is a quadruple \(\{(r,c,d,t), r,c,d,t \in \{1,2,3,4,5,6\}\}\) consisting of a row r, a column c, a digit d and a transversal t. Two 4-blocks are called disjoint, if they differ in rows, columns, digits and transversals. We prove that the maximal number of pairwise disjoint 4-blocks in a MOL(6) is 3 (Theorem 1). This simplifies the proofs in [3, 5, 7, 12] for the non-existence of two orthogonal latin squares of order 6 (the problem of Euler [10] and Tarry [13]).

2 Disjoint Quadruples

In order to show that the maximal number of pairwise disjoint 4-blocks in a MOL(6) is 3, we start—by contradiction—with a quadruple of pairwise disjoint 4-blocks. This means that the MOL(6) has up to isomorphism a substructure:

$$\begin{aligned} \begin{array} {|cccc|c|c|} \hline 11&{} &{} &{} &{} &{} \\ &{} 22&{} &{} &{} &{} \\ &{} &{} 33&{} &{} &{} \\ &{} &{} &{} 44&{} &{} \\ \hline &{} &{} &{} &{} &{} \\ \hline &{} &{} &{} &{} &{} \\ \hline \end{array} \end{aligned}$$
(1)

Therefore the MOL(6) can be generated by the following point tactical decomposition scheme:

$$\begin{aligned} \begin{array}{c|ccc|} \rightarrow &{} 4&{} 4&{} 32\\ \hline 16&{} 1&{} 1&{} 5\\ 8&{} 1&{} 0&{} 6\\ \hline \end{array} \end{aligned}$$
(2)

Proposition 1

The point tactical decomposition scheme (2) refines uniquely to the following tactical decomposition scheme:

$$\begin{aligned} \begin{array}{c|cccc|} \downarrow &{} 4&{} 4&{} 8&{} 24\\ \hline 16&{} 4&{} 4&{} 4&{} 2\\ 8&{} 2&{} 0&{} 0&{} 2\\ \hline \end{array} \cong \begin{array}{c|cccc|} \rightarrow &{} 4&{} 4&{} 8&{} 24\\ \hline 16&{} 1&{} 1&{} 2&{} 3\\ 8&{} 1&{} 0&{} 0&{} 6\\ \hline \end{array} \end{aligned}$$
(3)

Proof

Refining the decomposition (2) (computer program TDO-method, [1, 2]) we get one (point and block)-tactical decomposition (3). \(\square \)

Remark

The refining step (done by the computer program automatically) can also be carried out by hand: one has to establish and to solve some system of linear equations, see the “Appendix”.

Proposition 2

The substructure (1) extends (up to isomorphism) uniquely to the following partial MOL(6):

$$\begin{aligned} \begin{array} {|cccc|c|c|} \hline 11&{} 43&{} 24&{} &{} &{} \\ 34&{} 22&{} &{} 13&{} &{} \\ 42&{} &{} 33&{} 21&{} &{} \\ &{} 31&{} 12&{} 44&{} &{} \\ \hline &{} &{} &{} &{} &{} \\ \hline &{} &{} &{} &{} &{} \\ \hline \end{array} \end{aligned}$$
(4)

Proof

We extend the subscheme of (3):

$$\begin{aligned} \begin{array}{c|ccc|c|c|} \rightarrow &{} 4&{} 4&{} 8\\ \hline 16&{} 1&{} 1&{} 2\\ \hline \end{array} \,\,\,\rightarrow \,\,\, \begin{array} {c|cccc|c|} \rightarrow &{} 4&{} 4&{} 8&{} (4)\\ \hline 16&{} 1&{} 1&{} 2&{} 1\\ \hline \end{array}. \end{aligned}$$

This extended tactical decomposition generates exactly one geometry:

$$\begin{aligned} \begin{array} {|cccc|c|c|} \hline 11&{} 43&{} 24&{} (32)&{} &{} \\ 34&{} 22&{} (41)&{} 13&{} &{} \\ 42&{} (14)&{} 33&{} 21&{} &{} \\ (23)&{} 31&{} 12&{} 44&{} &{} \\ \hline &{} &{} &{} &{} &{} \\ \hline &{} &{} &{} &{} &{} \\ \hline \end{array} \end{aligned}$$

This is the MOL(4), (pair of orthogonal latin squares of order 4). On the main diagonal there is the quadruple we started with, and on the secondary diagonal there is the quadruple (in brackets) we added. We now delete this added quadruple and get the proposition.

Proposition 3

The partial MOL(6) (4) extends uniquely (up to isomorphism) to

$$\begin{aligned} \begin{array} {|cccc|c|c|} \hline 11&{} 43&{} 24&{} 55&{} &{} \\ 34&{} 22&{} 56&{} 13&{} &{} \\ 42&{} 65&{} 33&{} 21&{} &{} \\ 66&{} 31&{} 12&{} 44&{} &{} \\ \hline &{} &{} &{} &{} &{} \\ \hline &{} &{} &{} &{} &{} \\ \hline \end{array} \end{aligned}$$
(5)

Proof

In order to complete the upper half of the decomposition (3), we have to add the 24 vertical pairs which are not yet joined. These come from the quadruple of elements of the MOL(4) which we have deleted. For instance the element (1,4,3,2) gives the 6 pairs (1,4,.,.),(1,.,3,.), (1,.,.,2), (.,4,3,.),(.,4,.,2),(.,.,3,2). Similarly for the elements (2,3,4,1), (3,2,1,4) and (4,1,2,3). These are the 24 4-blocks of the MOL(6) and we get the unique incidence structure of the first 16 variables:

$$\begin{aligned} \begin{array}{c|cccc|cccc|cccccccc|cccccccccccccccccccccccc|} \rightarrow &{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1.1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1\\ \hline r1&{} 1&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .\\ r2&{} 1&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .\\ r3&{} 1&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .\\ r4&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} 1&{} 1&{} .&{} .&{} .\\ \hline c1&{} .&{} 1&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} 1&{} 1&{} .\\ c2&{} .&{} 1&{} .&{} .&{} .&{} 1&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .\\ c3&{} .&{} 1&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .\\ c4&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} 1&{} .&{} 1&{} .&{} .&{} 1&{} .&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .\\ \hline d1&{} .&{} .&{} 1&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} 1&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .\\ d2&{} .&{} .&{} 1&{} .&{} .&{} 1&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} 1&{} .&{} 1\\ d3&{} .&{} .&{} 1&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} 1&{} .&{} .&{} 1&{} .&{} 1&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .\\ d4&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} 1&{} 1&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} 1&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .\\ \hline t1&{} .&{} .&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .\\ t2&{} .&{} .&{} .&{} 1&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} 1&{} .&{} .&{} 1&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} .&{} .\\ t3&{} .&{} .&{} .&{} 1&{} .&{} .&{} 1&{} .&{} 1&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} 1\\ t4&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} 1&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} 1&{} .&{} 1&{} 1&{} .&{} .&{} .&{} .&{} .&{} .\\ \hline r5&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} \\ r6&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} \\ \hline c5&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} \\ c6&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} \\ \hline d5&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} \\ d6&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} \\ \hline t5&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} \\ t6&{} .&{} .&{} .&{} 1&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} .&{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} &{} \\ \hline \end{array} \end{aligned}$$

In order to extend this structure to the lower part (8 variables) of (3), we have to replace the points in the 24 4-blocks by numbers 5 or 6. In particular the four empty positions in (4) have to be filled by the four elements (5, 5), (5, 6), (6, 5), (6, 6). For the following proposition each permutation of these elements can be taken. \(\square \)

Remark

The automorphism group of MOL(4) has a subgroup which acts as symmetry group \(S_4\) on the secondary diagonal. Therefore we can reach the permutation as given in the proposition.

Proposition 4

The partial MOL(6) (5) cannot be extended to a MOL(6).

Proof

Let the position (1, 4) be occupied by, say, (5, 5). Then in the first row the two digits 3 and 6 are missing and also the two transversals 2 and 6. Since 66 is already used, we have only the combination 36 and 62. The same happens in the fourth column and there are the pairs 36 and 62 missing and must be inserted. But then the pair \((d,t)=(3,6)\) (and also (6, 2)) occurs twice, a contradiction.

(6)

\(\square \)

So we have proved:

Theorem 1

The maximal number of pairwise disjoint 4-blocks in a MOL(6) is 3.

3 Latin Squares of Shape C

In [3] there were derived 3 different shapes A), B) and C) for latin squares of order 6. It was proved that squares of shape A) and B) have no or at most 4 transversals, therefore they cannot have an orthogonal square. So only squares of shape C have to be considered for the problem of Euler and Tarry. Let \(d\in \{1,2,3\}\) and \(D\in \{4,5,6\}\) then we give in the following figure the shape C) and a special example:

(7)

There are the following conditions for the transversals in Case C), see [3, Lemma p. 425]:

Proposition 5

(a) The three elements with low digits in the \((3 \times 3)\)-subsquare left down have three different transversals. (b) In each of the two \((2 \times 3)\)-rectangles in the first two rows there are six different transversals.

Now we can show

Theorem 2

For each latin square of shape C there does not exist an orthogonal square.

Proof

Start with the three different transversals of the three low digits left down. Take a fourth transversal. This transversal occurs in the \((2 \times 3)\)-rectangle right above. Since in this rectangle there are only high digits, we get 4 pairwise disjoint blocks in contradiction to Theorem 1. \(\square \)

4 Stinson’s Proof for the Non-existence of a MOL(6)

In [12] Stinson gave a proof for the non-existence of a MOL(6) using coding theory. He derived one situation where a MOL(6) might occur and then excluded this by geometric reasoning. This single situation can be described by the following tactical decomposition scheme:

Columns 5 and 10 of the scheme lead to the geometry at the right side, and here we have indicated four pairwise disjoint 4-blocks, a contradiction to Theorem 1.

5 The Two Tactical Decompositions in [5]

In [5] we reduced the problem of the existence of a MOL(6) to the two tactical decompositions TDO I and TDO II and then proved that these cannot generate a geometry. By TDO we mean a tactical decomposition by ordering, a notion which was introduced in [6].

$$\begin{aligned} TDO I:&\begin{array}{cccccccccccccccc} \rightarrow &{} 1&{} 1&{} 1&{} 1&{} 6&{} 3^{*}&{} 3&{} 3&{} 3&{} 3&{} 3&{} 3&{} 3^{*}&{} 3&{} 3\\ \hline 3|&{} 1&{} 0&{} 0&{} 0&{} 2&{} 1&{} 1&{} 1&{} 1&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0\\ 3|&{} 0&{} 1&{} 0&{} 0&{} 2&{} 1&{} 0&{} 0&{} 0&{} 1&{} 1&{} 1&{} 0&{} 0&{} 0\\ 3|&{} 0&{} 0&{} 1&{} 0&{} 2&{} 0&{} 1&{} 0&{} 0&{} 1&{} 0&{} 0&{} 1&{} 1&{} 0\\ 3|&{} 0&{} 0&{} 0&{} 1&{} 2&{} 0&{} 0&{} 1&{} 0&{} 0&{} 1&{} 0&{} 1&{} 0&{} 1\\ \hline 3|&{} 1&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 1&{} 1&{} 1&{} 1&{} 1&{} 1\\ 3|&{} 0&{} 1&{} 0&{} 0&{} 0&{} 0&{} 1&{} 1&{} 1&{} 0&{} 0&{} 0&{} 1&{} 1&{} 1\\ 3|&{} 0&{} 0&{} 1&{} 0&{} 0&{} 1&{} 0&{} 1&{} 1&{} 0&{} 1&{} 1&{} 0&{} 0&{} 1\\ 3|&{} 0&{} 0&{} 0&{} 1&{} 0&{} 1&{} 1&{} 0&{} 1&{} 1&{} 0&{} 1&{} 0&{} 1&{} 0\\ \end{array}\nonumber \\\end{aligned}$$
(8)
$$\begin{aligned} TDO II:&\begin{array}{cccccccccccccccc} \rightarrow &{} 1&{} 1&{} 1&{} 1&{} 6&{} 3^{*}&{} 3&{} 3&{} 3&{} 3&{} 3&{} 3&{} 3^{*}&{} 3&{} 3\\ \hline 3|&{} 1&{} 0&{} 0&{} 0&{} 2&{} 1&{} 1&{} 1&{} 1&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0\\ 3|&{} 0&{} 1&{} 0&{} 0&{} 2&{} 1&{} 0&{} 0&{} 0&{} 1&{} 1&{} 1&{} 0&{} 0&{} 0\\ 3|&{} 0&{} 0&{} 1&{} 0&{} 2&{} 0&{} 1&{} 0&{} 0&{} 1&{} 0&{} 0&{} 1&{} 1&{} 0\\ 3|&{} 0&{} 0&{} 0&{} 1&{} 2&{} 0&{} 0&{} 1&{} 0&{} 0&{} 1&{} 0&{} 1&{} 0&{} 1\\ \hline 1|&{} 1&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 3&{} 0&{} 0&{} 0&{} 0&{} 3\\ 1|&{} 1&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 3&{} 0&{} 0&{} 3&{} 0\\ 1|&{} 1&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 0&{} 3&{} 3&{} 0&{} 0\\ 3|&{} 0&{} 1&{} 0&{} 0&{} 0&{} 0&{} 1&{} 1&{} 1&{} 0&{} 0&{} 0&{} 1&{} 1&{} 1\\ 3|&{} 0&{} 0&{} 1&{} 0&{} 0&{} 1&{} 0&{} 1&{} 1&{} 0&{} 1&{} 1&{} 0&{} 0&{} 1\\ 3|&{} 0&{} 0&{} 0&{} 1&{} 0&{} 1&{} 1&{} 0&{} 1&{} 1&{} 0&{} 1&{} 0&{} 1&{} 0\\ \end{array}\nonumber \\ \end{aligned}$$
(9)

In TDO I columns 6 and 13 define 6 pairwise disjoint 4-blocks. And in TDO II the three disjoint 4-blocks of column 6 together with one 4-block from column 13 define 4 pairwise disjoint 4-blocks, a contradiction to Theorem 1.