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1 Diagonal (or Magic) Latin Squares

1 Diagonal (or Magic) Latin Squares

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206 Chapter 6



right-to-left diagonal. A latin square is called left semi-diagonal1 if the elements

on its main left-to-right diagonal are distinct. Similarly, if the elements on the

main right-to-left diagonal are distinct, the latin square is called right semidiagonal. A latin square is called diagonal if it is both left and right semi-diagonal

simultaneously and, of course, every such latin square is also a magic square according to our definition.

Theorem 6.1.1 For each positive integer n = 2 there exists at least one left

semi-diagonal latin square of order n and at least one right semi-diagonal latin

square of order n.

Proof. For any given n = 2 there exists an idempotent quasigroup of order n

(see Theorem 1.5.6) and the multiplication table of an idempotent quasigroup is

a left semi-diagonal latin square. By reversing the order of the columns of a left

semi-diagonal latin square one will obtain a right semi-diagonal latin square. ⊓



0

β



..

.



α

β+α

2β + α

..

.





β + 2α

2β + 2α

..

.



···

···

···

..

.



(n − 1)α

β + (n − 1)α

2β + (n − 1)α

..

.



(n − 1)β



(n − 1)β + α



(n − 1)β + 2α



···



(n − 1)β + (n − 1)α



Fig. 6.1.1.

Theorem 6.1.2 If n is any odd integer which is not a multiple of three there

exists at least one diagonal latin square of order n.

Proof. We first show that if α and β are positive integers such that α, β,

α + β and α − β are all relatively prime to n then the latin square exhibited in

Figure 6.1.1 (where the elements are taken modulo n) is a diagonal latin square.

To see this, we observe that the main left-to-right diagonal contains the elements

0, β + α, 2(β + α), . . . , (n − 1)(β + α)

and the other main diagonal contains the elements

(n − 1)α, β + (n − 2)α, 2β + (n − 3)α, . . . , (n − 1)β.

The latter elements are respectively equal to

(n − 1)α, (n − 1)α − (α − β), (n − 1)α − 2(α − β), . . . , (n − 1)α − (n − 1)(α − β).

Since α + β and α − β are both relatively prime to n, the elements i(α + β)

for i = 0, 1, . . . , n − 1, are all distinct and so also are the elements i(α − β).

Consequently, the displayed latin square is diagonal.

1 These definitions are given in Margossian(1931) but probably date from much earlier. See,

for example, Tarry(1904) where diagonal latin squares are mentioned.



Connections between latin squares and magic squares 207



Now let us choose α = 2 and β = 1. Then α − β = 1 and α + β = 3, so

it is immediately obvious that if n > 1 is odd and not a multiple of three, the

integers α, β, α − β and α + β are all relatively prime to n. The existence of a

diagonal latin square of order n follows.







The above construction fails when n is even, but our next two results will

deal with specific subcases of this situation.



Theorem 6.1.3 If n = 2k for an integer k ≥ 0 then there exists at least one

diagonal latin square of order n if and only if k = 1.

Proof. Existence for k = 0 and non-existence for k = 1 are trivial to see and

the cases k = 2 and k = 3 are settled by Figure 6.1.2, which shows that diagonal

latin squares of orders 4 and 8 exist.



1

3

4

2



2

4

3

1



3

1

2

4



4

2

1

3



1

2

8

6

4

3

5

7



2

3

6

7

8

4

1

5



3

5

4

8

7

6

2

1



4

6

1

5

2

8

7

3



5

7

2

3

6

1

4

8



6

8

5

2

1

7

3

4



7

4

3

1

5

2

8

6



8

1

7

4

3

5

6

2



Fig. 6.1.2.

For k > 3 we proceed by an induction argument. Let k = r, so that n = 2r ,

and take as induction hypothesis that there exists at least one diagonal latin

square of every order n = 2k for which 1 < k < r.

Since 2r = 4 · 2r−2 , it follows from the induction hypothesis that 2r can be

written as the product n1 n2 of two integers n1 , n2 such that diagonal latin squares

exist both of order n1 and of order n2 . Let us suppose now that A is a diagonal

latin square of order n1 and that B0 , B1 , . . . , Bn1 −1 are isomorphic diagonal latin

squares of order n2 such that Bi is defined on the set {in2 , in2 +1, . . . , in2 +n2 −1}

for i = 0, 1, . . . , n1 − 1. If the elements 0, 1, . . . , n1 − 1 of A are replaced by

B0 , B1 , . . . , Bn1 −1 respectively, it is easy to see that we shall obtain a diagonal

latin square of order n1 n2 = 2r . The proof can now be completed by induction

on the integer k.





Theorem 6.1.4 If n is any integral multiple of 4, there exists at least one diagonal latin square of order n.

Proof. We show first that if n is even and n = n1 n2 , where n1 , n2 are integers

such that there exists at least one left semi-diagonal latin square of order n1 ,

and also at least one diagonal latin square of order n2 , and if, further, n2 is even,

then a diagonal latin square of order n can be constructed. The construction is

similar to that used in the proof of Theorem 6.1.3 and is as follows.



208 Chapter 6



Let A be a diagonal latin square of order n2 . Let the elements 0, 1, . . . , n2 − 1

of A, excluding those which appear in the main right-to-left diagonal, be replaced

respectively by B0 , B1 , . . . , Bn2 −1 where Bi , for i = 0, 1, 2, . . . , n2 − 1, is a left

semi-diagonal latin square of order n1 defined on the set {in1 , in1 + 1, . . . , in1 +

n1 − 1}. Further, let those elements of A which are contained in the main rightto-left diagonal be replaced successively by B0∗ , B1∗ , . . . , Bn∗ 2 −1 where, for each i,

Bi∗ is defined on the same set as Bi and has the same order, but is right instead of

left semi-diagonal. Then, it is easy to see that, provided n2 is even, the resulting

square of order n1 n2 is a diagonal latin square.

We now take the special case when n2 = 4, corresponding to the case when

n = 4n1 is an integral multiple of 4. By Theorem 6.1.3, there exists a diagonal

latin square of order 4 and, by Theorem 6.1.1, there exist both left and right semidiagonal latin squares of order n1 so the conditions required for our construction

hold. This proves the theorem.





It is easy to see that the constructions of Theorem 6.1.2, Theorem 6.1.3 and

Theorem 6.1.4 do not provide the only ways of producing diagonal latin squares

since none of the above methods is applicable if n = 6, 9 or 10 and yet diagonal

squares of these orders certainly exist, as is shown by Figure 6.1.3. However, it

is easy to prove that no diagonal latin square of order 3 can exist.

Theorem 6.1.5 Diagonal latin squares of order 3 do not exist.

Proof. Suppose that ||aij || is a diagonal latin square of order 3. By the diagonal property, the elements a11 , a22 , a33 are all different and the same is true

for the elements a13 , a22 , a31 . This necessarily implies that either a11 = a13 or

a11 = a31 . However, neither equality is compatible with ||aij || being a latin

square.





Some examples of diagonal latin squares of orders 5 and 7 as well as 6, 9,

and 10, are given in Figure 6.1.3. Examples of orders 4 and 8 have already been

given in Figure 6.1.2.

In a lecture at the University of Surrey, one of the authors of the present book

(first edition) conjectured that diagonal latin squares exist for all orders n > 3.

The truth of this conjecture has since been shown independently by a number

of different people. First to give a proof was Hilton(1973), who made use of socalled cross latin squares: that is, latin squares such that all the elements of the

main left-to-right diagonal are equal and all the elements of the main right-toleft diagonal are equal. (The definition has to be modified slightly for squares of

odd order.) Shortly afterwards, Lindner(1974) found a somewhat simpler proof

which made use of prolongations.2 Later, a proof more elementary than either

of those just mentioned was obtained by Gergely(1974a). We shall now explain

Gergely’s proof, which depends essentially on the following result.

2 Both Hilton and Lindner used the term diagonal latin square to mean a left semi-diagonal

latin square and the term doubly diagonal latin square to mean a diagonal latin square.



Connections between latin squares and magic squares 209



1

4

2

5

3



1

6

7

5

9

8

3

4

2



2

3

9

7

8

1

4

5

6



2

5

3

1

4



3

2

8

6

5

9

7

1

4



3

1

4

2

5



4

8

5

9

7

6

1

2

3



4

2

5

3

1



5

7

6

1

4

3

2

9

8



1

6

2

3

5

4



5

3

1

4

2



6

5

4

8

1

2

9

3

7



7

9

3

4

2

5

6

8

1



8

1

2

3

6

4

5

7

9



2

5

4

1

3

6



3

4

6

5

1

2



4

3

1

2

6

5

1

4

10

9

2

3

5

7

8

6



9

4

1

2

3

7

8

6

5



5

2

3

6

4

1

2

10

8

4

9

5

6

1

7

3



1

7

6

5

4

3

2



6

1

5

4

2

3

3

8

2

10

6

1

7

5

4

9



4

7

6

8

1

10

3

9

2

5



2

3

1

7

6

4

5



3

4

2

1

5

7

6



5

6

7

1

3

9

2

4

10

8



6

9

5

3

8

7

4

10

1

2



4

5

7

6

2

1

3

7

5

1

2

4

6

9

8

3

10



5

6

3

4

7

2

1



6

1

4

2

3

5

7



7

2

5

3

1

6

4



8

3

4

7

5

2

10

6

9

1



9

1

3

6

10

4

8

2

5

7



10

2

9

5

7

8

1

3

6

4



Fig. 6.1.3.

Theorem 6.1.6 For every n ≥ 3, there exist left semi-diagonal latin squares of

order n which possess at least one transversal disjoint from the main left-to-right

diagonal.

Proof. Gergely’s original proof of Theorem 6.1.6 made use of prolongations of

cyclic group tables, but we can deduce the result very simply as follows. It follows

from the results of Bose, Shrikhande and Parker(1960) described in Chapter 5

that, for all orders n except 2 and 6, there exist latin squares of order n which have

n disjoint transversals. By rearranging the rows, we may easily arrange that one

of the transversals lies along the main left-to-right diagonal. For the case n = 6,

the truth of the theorem is shown by the example in Figure 6.1.4 which actually

has three transversals disjoint from each other and from the main left-to-right

diagonal. [Four disjoint transversals is the maximum number possible for a latin

square of order 6. See Wanless(2007).]





1

4

5d

6b

2c

3



2

6

3c

5

1d

4b



3b

5c

2

4d

6

1



4c

1

6

3

5b

2d



Fig. 6.1.4.



5

3d

1b

2

4

6c



6d

2b

4

1c

3

5



210 Chapter 6



We are now able to prove our main result.

Theorem 6.1.7 For n ≥ 4, there exists at least one diagonal latin square of

order n.

Proof. The cases n = 4 and n = 5 are settled by the examples given in

Figure 6.1.2 and Figure 6.1.3 respectively, so we may assume that n ≥ 6.

We suppose first that n is even. We first construct a left semi-diagonal latin

square D1 of order k = n/2 on the symbols 0, 1, 2, . . . , k − 1 having the property

described in Theorem 6.1.6. We construct a second latin square D2 of order k

by reversing the order of the columns of D1 and then adding k to each of its

elements.

With the aid of the squares D1 and D2 , we construct a latin square L of

order n of the form

L = D1 D2

D4 D3

where the subsquares D3 and D4 are yet to be defined.

Let τ1 denote the off-diagonal transversal of D1 whose existence is guaranteed

by Theorem 6.1.6, and let τ2 denote the corresponding transversal of D2 . We

choose the elements of the main left-to-right diagonal of D3 to be the elements

of the transversal τ2 which occur in the corresponding columns of D2 . Similarly,

we choose the elements of the main right-to-left diagonal of D4 to be the elements

of the transversal τ1 which occur in the corresponding columns of D1 . Now, for

(i) (i)

(i)

i ∈ {1, 3} let d11 d22 · · · dkk denote the elements of the main left-to-right diagonal

of Di .

We define a permutation πL of the symbols of D1 by

(1)



πL =



d11

(3)



d11 − k



···

···



(1)



djj

(3)



djj − k



(1)



···

···



dkk

(3)



dkk − k



Similarly, we define a permutation πR of the symbols of D2 by

(2)



πR =



d1k

(4)



d1k + k



···

···



(2)



dj,k+1−j

(4)



dj,k+1−j + k



···

···



(2)



dk1

(4)



dk1 + k



where in this case the permutation is expressed in terms of the elements of the

main right-to-left diagonals of D2 and D4 .

Each element of D3 not on the main left-to-right diagonal is defined to be

the transform of the corresponding element of D1 by the permutation πL . Each

element of D4 not on the main right-to-left diagonal is defined to be the transform

of the corresponding element of D2 by the permutation πR . Finally, to make the

square L into a latin square, we increase each of the elements of the transversal τ1

of D1 by k and at the same time reduce each of the elements of the transversal τ2

of D2 by k. Then L is of order n and is the diagonal latin square whose existence

was to be shown. We note further that the elements in the cells of the transversal



Connections between latin squares and magic squares 211



τ1 in D1 together with the elements in the corresponding positions in D3 form

an off-diagonal transversal of L.

Using this fact, it is easy to construct from L a diagonal latin square L∗ of

order n + 1, where n + 1 is odd. (We illustrate the procedure below.) We can

thereby construct squares of the odd orders required to complete the proof. ⊓



To illustrate the construction of L∗ from L, we shall give it in detail for the

special case n = 10.

0 1 2

1 2 3

D1 = 2 3 4

3 4 0

4 0 1



3 4

4 0

0 1

1 2

2 3



9

5

D2 = 6

7

8



8

9

5

6

7



7

8

9

5

6



6

7

8

9

5



5

6

7

8

9



πL =



0

7−5



2

5−5



4

8−5



1

6−5



3

9−5



=



0

2



1

1



2

0



3

4



4

3



πR =



5

2+5



7

0+5



9

3+5



6

1+5



8

4+5



=



5

7



6

6



7

5



8

9



9

8



4

0

1

α

3

7

2

6

5

9

8



6

8

5

7

9

α

1

4

2

0

3



7

3

9

5

6

4

0

α

8

2

1



1

7

8

9

5

2

4

3

α

6

0



0

1

2

3

9

L=

8

7

6

5

4



6

2

3

4

0

9

8

7

1

5



2

8

4

0

1

5

9

3

7

6



3

4

5

1

2

6

0

9

8

7



4

0

1

7

3

2

6

5

9

8



9

5

6

2

8

7

1

0

4

3



8

9

0

6

7

1

5

4

3

2



7

3

9

5

6

0

4

8

2

1



1

7

8

9

5

4

3

2

6

0



5

6

7

8

4

3

2

1

0

9



0

1

2

3

α

L∗ = 9

8

7

6

5

4



α

2

3

4

0

6

9

8

7

1

5



2

α

4

0

1

8

5

9

3

7

6



3

4

α

1

2

5

6

0

9

8

7



9

5

6

2

8

3

7

1

0

4

α



8

9

0

6

7

1

α

5

4

3

2



5

6

7

8

4

0

3

2

1

α

9



W.Taylor(1972) discussed the analogue of a diagonal latin square in higher

dimensions. He also remarked that he and Faber had devised yet another proof

of Theorem 6.1.7.

As we remarked at the beginning of this section, all diagonal latin squares

are magic squares. However, the more usual type of magic square contains n2

different integers, and usually these are required to be consecutive integers. In

the next section, we explain how to construct the latter type of magic square

with the aid of pairs of orthogonal latin squares.



212 Chapter 6



6.2



Construction of magic squares with the aid of orthogonal latin

squares.



For our first construction, we shall need to consider only pairs of orthogonal

latin squares which are isotopic to the square which represents the multiplication

table of the cyclic group Cn . The method which we shall describe is effective for

all odd integers n. Our procedure is substantially equivalent to that of De la

Hire3 but the technique of our proof involves the use of latin squares.

0



1



2 ···



1



2



..

.



..

.



..

.



..



n−1



0



1



···



3 ···



1

2 (n

1

(n

2



− 3)

− 1)



1

2 (n

1

(n

2



..

.



.

1

(n

2



− 5)



1

(n

2



− 1) · · ·

+ 1)



···



..

.



..



.



− 3) · · ·



n−2



n−1



..

.



..

.



n−3



n−2



n−1



0



Fig. 6.2.1.

Let L∗ denote the latin square exhibited in Figure 6.2.1, which is the unbordered Cayley table of the cyclic group Cn , for odd n, when represented as

an additive group. By interchanging the elements n − 1 and 12 (n − 1) in each

of the rows of L∗ , we can transform it into a latin square L whose row, column

and diagonal sums are each equal to 12 n(n − 1) and which can be characterized as follows. The main left-to-right diagonal forms a transversal of L (so L

is left semi-diagonal) and each broken diagonal parallel to the main left-to-right

diagonal also forms a transversal. The main right-to-left diagonal contains the

element 12 (n − 1) duplicated n times and each broken diagonal parallel to the

main right-to-left diagonal has all its elements equal.

A latin square L′ orthogonal to L can be obtained from L by reversing the

order of its columns. The orthogonality follows from the fact that all the elements

of each broken diagonal parallel to the main left-to-right diagonal of L′ are the

same, whereas in L they are all different. Also, the elements of each broken

diagonal parallel to the main right-to-left diagonal of L′ form a transversal,

whereas in L they are all the same.

We illustrate the above construction in Figure 6.2.2 by exhibiting the squares

L∗ , L and L′ for the case n = 5.

We can express our result in the form of a theorem as follows:

Theorem 6.2.1 Let Cn denote the cyclic group of an odd order n, with elements

represented by the integers 0, 1, . . . , n − 1 under addition modulo n and let its

3 The work of the French mathematician, astronomer, physicist, naturalist and painter

Philippe de la Hire (1640–1718) is frequently quoted, without giving an explicit reference,

in the literature on magic squares.



Connections between latin squares and magic squares 213



0

1

L∗ = 2

3

4



1

2

3

4

0



2

3

4

0

1



3

4

0

1

2



4

0

1

2

3



0

1

L= 4

3

2



1

4

3

2

0



4

3

2

0

1



3

2

0

1

4



2

0

1

4

3



2

0

L′ = 1

4

3



3

2

0

1

4



4

3

2

0

1



1

4

3

2

0



0

1

4

3

2



Fig. 6.2.2.

isotope under the isotopism ρ = (α, β, γ) be denoted by ρ(Cn ). Let ρ1 = (ǫ, ǫ, γ)

and ρ2 = (ǫ, β, γ) where ǫ is the identity permutation,

β=



0



1



n−1



n−2



···



···



1

(n

2

1

(n

2



1

(n

2

1

(n

2



− 3)

+ 1)



− 1)

− 1)



···

···



n−1

0



and γ simply transposes 12 (n−1) and n−1. Then the latin squares which represent

the unbordered multiplication tables of ρ1 (Cn ) and ρ2 (Cn ) are orthogonal and

have the structure described above.

We can use this result to build magic squares, as follows:

Theorem 6.2.2 Let L = ||aij || and L′ = ||bij || be two orthogonal latin squares

of odd order n formed in the manner described in Theorem 6.2.1 and having

as their elements the integers 0, 1, . . . , n − 1. Let a square matrix M = ||cij ||

be constructed from L and L′ by putting cij = naij + bij . Then the sum of the

elements of each row, column, and diagonal of M is equal to 12 n(n2 − 1) and the

elements of M are the consecutive integers 0, 1, . . . , n2 − 1.

Proof. In order to show that the row, column, and diagonal sums are each

equal to 12 n(n2 −1), we derive the following equalities directly from the structure

of the latin squares L and L′ as described in Theorem 6.2.1. For each fixed i,

cij =



(naij + bij ) = n



j



j



aij +

j



bij = (n + 1)

j



1

2 n(n − 1)



= 12 n(n2 − 1),



and, for each fixed j,

cij = n

i



aij +

i



bij = (n + 1)



1

2 n(n



i



− 1) = 12 n(n2 − 1).



Similar calculations applied to the main diagonals show that



i



cii = 12 n(n2 − 1) =



ci,n+1−i .

i



To complete the proof, we have to show that no two elements of M are equal

and that 0 ≤ cij ≤ n2 − 1 for all values of i, j. Suppose first that two distinct



214 Chapter 6



elements cij and ckl of M were equal. This would imply that n(aij − akl ) + (bij −

bkl ) = 0. Since the elements of L′ are all less than n, the latter equation could

only hold if aij = akl and then, since L is a latin square, we would necessarily

have i = k and j = l. It would follow that aij and akl were in the same broken

diagonal parallel to the main right-to-left diagonal. However, in L′ the elements

of each such diagonal are all different, so bij = bkl . It follows that cij = ckl .

Finally, since 0 ≤ aij ≤ n − 1 and 0 ≤ bij ≤ n − 1, the largest value that cij can

take is n(n − 1) + n − 1 = n2 − 1, so 0 ≤ cij ≤ n2 − 1 as required.





We demonstrate the above procedure in Figure 6.2.3, where we give the final

step of the construction for the example exhibited in Figure 6.2.2.

0

5

nL = 20

15

10



5 20 15 10

20 15 10 0

15 10 0 5

10 0 5 20

0 5 20 15



2

5

M = nL + L′ = 21

19

13



2

0

L′ = 1

4

3



3

2

0

1

4



4

3

2

0

1



1

4

3

2

0



0

1

4

3

2



8 24 16 10

22 18 14 1

15 12 3 9

11 0

7 23

4 6 20 17



Fig. 6.2.3.

In the following theorem, we give another construction which uses orthogonal

latin squares.

Theorem 6.2.3 If n is any integer for which an orthogonal pair of diagonal

latin squares of order n exists, then an n × n magic square whose entries are the

consecutive integers 0 to n2 − 1 can be constructed.

Proof. We first write the two squares in juxtaposed form, as for example in

Figure 6.2.6. Since the sums of the elements of each row, column, and main diagonal are all equal for each of the two squares, the same is true in the juxtaposed

form. Moreover, this is true regardless of the number base selected. If we take

the number base as the integer n we get a magic square M whose entries are

the integers 0 to n2 − 1 as required. Thus, to get the matrix M exhibited in

Figure 6.2.6 the number base 4 has been selected.

We may express the construction in another way by saying that M is related

to L1 and L2 by the matrix equation M = nL1 + L2 . (Compare the proof of

Theorem 6.2.2.)





The construction of magic squares by means of orthogonal pairs of diagonal

latin squares or by means of orthogonal pairs of latin squares of the type given by



Connections between latin squares and magic squares 215



Theorem 6.2.1 have been known and used for more than two centuries. See, for example, Euler(1779), Maillet(1894a) and Maillet(1894c) or (1896) [Maillet(1894c)

and Maillet(1896) are two publications of the same paper], Barbette(1896),

Tarry(1904,1905), Kraichik(1942,1953) and Ball(1939). These constructions are

very useful but they are not the only methods available. For example, neither

of these constructions can be used to obtain a magic square of order 6 and yet

magic squares of this order certainly exist, as is demonstrated by Figure 6.2.4.

35

3

31

8

30

4



1

32

9

28

5

36



6 26 19 24

7 21 23 25

2 22 27 20

33 17 10 15

24 12 14 16

29 13 18 11



Fig. 6.2.4.

Because of their importance in connection with the construction of magic

squares, many authors have tried to answer the question: “For which orders n

distinct from 2, 3 and 6 do there exist orthogonal pairs of diagonal latin squares?”

That such pairs exist for the case n = 4 has been known at least since

1723, as has been pointed out by Ball(1939), page 190. The fact that a solution

exists whenever n is odd and not a multiple of three (see Theorem 6.2.4) has

also been known at least since the nineteenth century. Using both these results,

Tarry(1905) proved at the beginning of the twentieth century that orthogonal

pairs of diagonal latin squares exist for every order n which is a multiple of 4.

Moreover, he proved in Tarry(1904) that if n = 8m, where m is not a multiple of

two or three, then the method of Theorem 6.2.3 can be used to obtain a magic

square with the additional property that the sums of the squares of the elements

in each row, column and main diagonal are equal. A magic square with the latter

property is called bimagic. We discuss such squares further in the next section.

The complete resolution of the existence question was not achieved until 1993.

We shall now explain the main steps.

Theorem 6.2.4 There exist orthogonal pairs of diagonal latin squares of every

odd order which is not a multiple of 3.

Proof. It follows directly from Theorem 6.1.2 that, if α = 2 and β = 1, the

latin square exhibited in Figure 6.1.1 is a diagonal latin square provided that n

is odd and not divisible by 3. Its transpose is also a diagonal latin square and

we shall show that the two squares so obtained are orthogonal, again under the

condition that n is not divisible by 3.

Let us denote the square obtained from Figure 6.1.1 when α = 2 and β = 1

by L1 and its transpose by L2 (the squares L1 and L2 for the special case when

n = 5 are exhibited in Figure 6.2.5). We shall show first that the contents of the



216 Chapter 6



cells of L1 which correspond to the cells containing the element 0 in L2 are all

different.

0

1

L1 = 2

3

4



2

3

4

0

1



4

0

1

2

3



1

2

3

4

0



3

4

0

1

2



0

2

L2 = 4

1

3



1

3

0

2

4



2

4

1

3

0



3

0

2

4

1



4

1

3

0

2



Fig. 6.2.5.

Rows and columns will be indexed with the set {0, 1, . . . , n − 1} and all

calculations will be performed modulo n. In the r-th row of L2 , the element 0

appears in the (n − 2r)-th column. In the cell of the r-th row and (n − 2r)-th

column of L1 , the element r + 2(n − 2r) appears. To see this, observe that the

entry in the cell of the r-th row and 0-th column of the square L1 is r and that for

each step taken to the right along this row, the cell entry is increased by 2. Since

r + 2(n − 2r) ≡ −3r mod n and since −3 is relatively prime to n, the elements

−3r are all different modulo n as r varies through the set {0, 1, 2, . . . , n − 1}.

This proves the result.

By a similar argument, we could show that the contents of the cells of L1

which correspond to the cells containing the element i in L2 are all different and

that this is true for each choice of i in the range 0 ≤ i ≤ n − 1. It follows that

L1 and L2 are orthogonal diagonal latin squares.





It is an obvious consequence of Theorem 6.1.5 that orthogonal pairs of diagonal latin squares of order 3 do not exist. However, our next theorem, which

comes from Ball(1939), gives a further set of values of n for which orthogonal

pairs of diagonal latin squares exist.

Theorem 6.2.5 Orthogonal pairs of diagonal latin squares of order n can be

constructed whenever n is odd or a multiple of 4 except, possibly, when n is a

multiple of 3 but not of 9.

Proof. We first point out that there exist orthogonal pairs of diagonal latin

squares of orders 4, 8, 9 and 27. Examples of such pairs corresponding to the

first three of these values of n are exhibited in Figure 6.2.6, Figure 6.2.7 and

Figure 6.2.8 respectively and a construction which gives a pair of order 27 is

described by Ball(1939), page 192. (In Figure 6.2.6 the diagonal squares L1 and

L2 are first shown separately and then in juxtaposition. In the remaining cases,

only the juxtaposed form is shown.)

Then, making use of Theorem 6.2.4 and of the construction described in the

proof of Theorem 6.1.3, it is easy to deduce the truth of the present theorem

which comes from Ball(1939), page 192.







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