Existence of strongly symmetrical weakly pandiagonal graeco-latin squares

Abstract: A graeco-latin sauare is a pair of orthogonal latin squares. It is a design of experiment in which the experimental units are grouped in three di erent ways. In this paper, constructions of a pair of orthogonal latin sauares which are both strongly symmetrical and weakly pandiagonal are investigated. As a result, it is proved that there exists a pair of strongly symmetrical weakly pandiagonal orthogonal latin sauare of order n if and only if n > 4 and n ≡ 0, 1, 3 (mod 4) with only one possible exception for n = 12.


Introduction
A Latin square of order n, denoted by LS(n), is an n × n array such that every row and every column is a permutation of an n-set S. A transversal in a Latin square is a set of positions, one per row and one per column, among which the symbols occur precisely once each. A diagonal Latin square is a Latin square with the additional property that the main diagonal and back diagonal are both transversals. Two LS(n)s are called orthogonal, denoted by OLS(n), if each symbol in the rst square meets each symbol in the second square exactly once when they are superposed. A pair of orthogonal latin squares is also called a graeco-latin sauare because Euler ([7]) used Greek letters for one square of the pair and latin letters for the other. See also [5] for overview of the Latin squares.
In 1776, Euler presented a paper in which he constructed magic squares of orders 3, 4, and 5 from orthogonal latin squares. He posed the question for order 6, now known as Eulers 36 O cers Problem. He conjectured that no solution exists for order 6. Indeed he conjectured further that there exist orthogonal latin squares of all orders n except when n ≡ (mod ) ( [7,9]). Much later, Bose, Shrikhande and Parker [1] proved that the Euler conjecture was false for all orders n of the form k + except n = or . Much shorter disproofs of Euler's conjecture have been obtained. See, for example, [4,15].
Wallis and Zhu [11], Heinrich and Hilton [10], Brown, Cherry, Most, Most, Parker and Wallis [2] investigated the existence of orthogonal diagonal latin squares. They proved that there exists a pair of diagonal OLS(n)s if and only if n ∉ { , , }.
An LS(n) is called self-orthogonal if it is orthogonal to its transpose. Let n be a positive integer. Denote In = { , , · · · , n − }. An LS(n) over In A = (a i,j )(i, j ∈ In) is called strongly symmetrical if a i,j + a n− −i,n− −j = n − , i, j ∈ In. Danhof, Phillips and Wallis [6] investigated the existence of strongly symmetrical OLS(n)s. Du and Cao [8] investigated the existence of strongly symmetrical self-orthogonal LS(n). Cao and Li [3] proved that there is a strongly symmetrical self-orthogonal LS(n) if and only if n ≡ , , (mod ) and n ≠ . It results in the following.

Lemma 1.1. ([3]) A pair of strongly symmetrical OLS(n) exists if and only if n ≡ , , (mod ) and n ≠ .
Given positive integers a, n, we denote a (mod n) by a n for convenience. For a matrix A of order n and k ∈ In, we call the elements a i, k+i n , i ∈ In the k-th right pandiagonal and a i, k−i n , i ∈ In the k-th left pandiagonal. Clearly, -th right pandiagonal is the main diagonal and the (n − )-th left pandiagonal is the back diagonal. Let A be an LS(n) over In. A is weakly pandiagonal if the sum of the n numbers in each pandiagonal is the same. For example, the following is a pair of weakly pandiagonal OLS( ).
Xu and Lu [12] introduced weakly pandiagonal OLS(n) to construct pandiagonal magic squares. They proved that a weakly pandiagonal self-orthogonal LS(n) exists if and only if n ≡ , , (mod ) and n ≢ , (mod ).
Zhang, Li and Lei [13] investigated a pair of weakly pandiagonal OLS(n). They proved the following. We use SPOLS(n) to denote strongly symmetrical weakly pandiagonal OLS(n). Clearly, the above (D, D ) is a pair of SPOLS( ). In this paper, the existence of a pair of SPOLS(n) is investigated. The construction of SPOLS(n) is provided in Section 2 and in section 3 the following is proved.

Constructions for SPOLS(n)
Given an m × n matrix A and an r × s matrix B, the Kronecker Product A ⊗ B is an mr × ns matrix given as follows.
Denote Jm×n an m × n matrix with all elements 1s. A Jn×n is denoted by Jn. The Kronecker Product is used to get the following construction.
In [13] Construction 2.1 proved that (E, F) is a pair of weakly pandiagonal OLS(mn). We shall prove that E, F are also strongly symmetrical. Let Thus E is strongly symmetrical. Similarly, F is also strongly symmetrical. So (E, F) is a pair of SPOLS(mn).
Let m be an even integer and n be an integer. Let T = {T , T , · · · , T m− } be a partition of Imn such that |T i | = n, i ∈ Im. Denote S = mn(mn− ) . Given a rational number x, we use x to denote the largest integer a such that a ≤ x as usual.

Then (A, B) is a pair of strongly symmetrical OLS(mn) over Imn. Further, if φu,v satis es
and φ u,v satis es

then (A, B) is a pair of SPOLS(mn) over Imn.
Proof First we prove that A, B are LS(mn). Let u ∈ Im , i ∈ In. We consider the entries au,v(i, j) and Similarly, each column of A is also a permutation of Imn. Thus A is an LS(mn) over Imn. Similarly, B is also an LS(mn) over Imn. Now we shall prove that A, B are orthogonal. Let Thus A is strongly symmetrical. Similar, B is also strongly symmetrical by the condition (L2). Now we consider the pandiagonals of A, B. For any s ∈ Imn, the s-th right pandiagonal of A consists of the (k, k + s mn)-entry of A, k ∈ Imn.
The sum of each right pandiagonal of A is the same number S by (L3). Similarly, the sum of each left pandiagonal of A is also S by (L4). So A is weakly pandiagonal. By (L5) and (L6) B is also weakly pandiagonal. Thus (A, B) is a pair of SPOLS(mn) over Imn.  It is readily veri ed that φ, φ satisfy (L1)-(L6) in Construction 2.2. Let where u, v ∈ I , i, j ∈ I . Thus we get a pair of SPOLS( ) as Construction 2.2, (A, B) as follows.
We introduce the following de nition.
L is denoted by S( n) if it satis es the following 3 conditions. and Let Since L is a permutation, T is a partition of I n . By (R1), For xed v ∈ I , if t runs over In then φ ,v (t), φ ,v (t) run over Tv, and φ ,v (t), φ ,v (t) runs over { n − − x|x ∈ T −v }, which is exactly Tv. So, for any u, v ∈ I , φu,v is a bijection from In to Tv. By (2) φ u,v is also a bijection from In to Tv.
Let A, B be the matrices given by the Construction 2.2. We shall prove that φ and φ satisfy (L1)-(L6) in Construction 2.2. So, (A, B) is a pair of SPOLS( n) by Construction 2.2.
In fact, by (1), for any t ∈ In, Similarly, one can prove that φ ,v (t) + φ , −v (n − − t) = n − . Thus (L ) and (L ) in Construction 2.2 are satis ed. For any t ∈ In, by (1) and (R ) and (R ) it is readily veri ed the following.
By (1) we have
The following example is provided to explain Construction 2.4 and its proof. Clearly L is an S( ). By Construction 2.4 we have It is readily veri ed that (A, B) is a pair of SPOLS( ).
See similar de nition in [14]. Proof Let H be a ( , n )-CCRMR over In. De ne a penmutation σ of In: Let U, V be matrices of order n over In, where It is readily veri ed that (U, V) is a pair of OLS(n). We shall prove that U and V is a pair of SPOLS(n). First we shall prove that U, V are weakly pandiagonal. Clearly, the left pandiagonals of U and the right pandiagonals of V are transversals. For the left pandiagonal of U, xed w ∈ In, let w = s + t, s ∈ I n , t ∈ I . By (6), (8) and (9) we have We have { (i + s) + t + n|i ∈ In} = { i + t + |i ∈ I n }. In fact, since t + = t+ + t + and there is an integer q such that i + s + t+ = q · n + i + s + t+ n , we have Thus the equality (10) is . It also equals to n − i= h t+ ,i by the equality (8). Further, it is just n(n− ) by (6). So the sum of the n numbers of each right pandiagonal of U is the same. Similarly, the sum of the n numbers of each left pandiagonal of V is the same. Thus U, V are weakly pandiagonal. Now we prove that U, V are strongly symmetrical. For any i ∈ In, let q = i , r = i , then i = q + r, q ∈ I n , r ∈ I . So we have n − − i = ( n − − q) + − r. Thus by (7) and (8) we have which is just σ(n − − i + j + n). So, Thus U is strongly symmetrical. Similarly, one can prove that V is strongly symmetrical. Therefore, (U, V) is a pair of SPOSL(n).
An example is provided to explain the construction 2.5.

Example 2. Let n = , and
Clearly, H is a ( , )-CCRMR. By Construction 2.5, de ne a permutation of σ over I : That is σ = .
Let (U, V) be a pair of OLS( ) given by (9), where It is readily veri ed that (U, V) is a pair of SPOLS( ).

Proof of Theorem 1.3
In this section we consider the exitence of a pair of SPOLS(n). There doesn't exist a pair of SPOLS(n) if n ≡ (mod ) by Lemma 1.1. To prove Theorem 1.3 it remains to check the following three cases: (1) n ≡ , (mod ); (2) n ≡ (mod ); (3) n ≡ (mod ).
(U, V) is a pair of OLS(n) over In. Now we prove that U, V are strongly symmetrical and weakly pandiagonal. It is enough to prove U is strongly symmetrical and weakly pandiagonal. Similarly we have V.
For any i, j ∈ In, we have We consider the pandiagonals of U. For any k ∈ In, since gcd( , n) = , we have It shows that each right pandiagonal of U is a transversal. Similarly, each left pandiagonal of U is also a transversal. Thus U is weakly pandiagonal.
There is a ( , n)-CCRMR for odd n ≥ .
. For s ≥ , supppose that A(s, w) is a ( , s + w)-CCRMR. Then the row sum of A(s, w) is The sum of each pair of symmetrical elements is s + w − . Let R(s, w)).
It is readily veri ed that the elements of A(s + , w) run over I n+ . By calculation, for any i ∈ I , the sum of the elements of the i-th row of A(s + , w) is n( n− ) + n + ( n) + = ( (s+ )+w)( (s+ )+ w− ) , which is independent on i. The sum of each pair of symmetrical elements of A(s + , w) is s + w + = (s + ) + w − , which is also independent on i. Thus A(s + , w) is a ( , (s + ) + w)-CCRMR. There is a ( , n)-CCRMR for any odd n ≥ by induction.

Case 3. n ≡ (mod )
Let n = k, k ≥ . We should deal with the case k ≡ (mod ) and the case k ≡ , , (mod ).

Lemma 3.5.
There is a pair of SPOLS( k) for any positive integer k ≡ (mod ).
Proof The (D, D ) given in Section 1 is a pair of SPOLS( Proof There is an S( k) for k = , , as follows.
Suppose that there is an S( k) for any k ≥ , k ≡ , , (mod ), where the j-position of L i is l i (j), i ∈ I , j ∈ I k . Let Clearly, the elements of L run over I (k+ ) , and it satisfy the conditions (R1)-(R3). Thus L is an S( (k + )). The proof is completed by induction.
Example 3. By using the S( ) given in Lemma 3.6 we can get an S( ) as follows: L = .
Proof Let k ≡ , , (mod ), k > . There is a pair of strongly symmetrical OLS(k) by Lemma 1.1, and there is an S( k) by Lemma 3.6. Thus there is a pair of SPOLS( k) by Construction 2.4.  Let e = a + a + a + a , Then e + f = e + g = , hence f = g. Since A and B are strongly symmetrical, we have b + c = c + b = b + c = c + b = . Thus f + g = . So, f = g = . On the other hand, since A is weakly pandiagonal, b + c + b + c = b + c + b + c = . So c + c = c + c , c + c + c + c = (c + c ) = , it follows that c + c = . Previously we have proved that c + b = , so c = b , a contradiction. Thus there is no SPOLS( ).

Proof of Theorem1.3
The proof follows by Combining Theorem 3.4, Theorem 3.8 and Lemma 3.9.