ISSN 2148-838Xhttps://doi.org/10.13069/jacodesmath.1056511 J. Algebra Comb. Discrete Appl. 9(1) • 17–27 Received: 19 October 2020 Accepted: 8 October 2021 Journal of Algebra Combinatorics Discrete Structures and Applications Recent results on Choi’s orthogonal Latin squares Research Article Jon-Lark Kim∗, Dong Eun Ohk, Doo Young Park, Jae Woo Park Abstract: Choi Seok-Jeong studied Latin squares at least 60 years earlier than Euler although this was less known. He introduced a pair of orthogonal Latin squares of order 9 in his book. Interestingly, his two orthogonal non-double-diagonal Latin squares produce a magic square of order 9, whose theoretical reason was not studied. There have been a few studies on Choi’s Latin squares of order 9. The most recent one is Ko-Wei Lih’s construction of Choi’s Latin squares of order 9 based on the two 3 × 3 orthogonal Latin squares. In this paper, we give a new generalization of Choi’s orthogonal Latin squares of order 9 to orthogonal Latin squares of size n2 using the Kronecker product including Lih’s construction. We find a geometric description of Choi’s orthogonal Latin squares of order 9 using the dihedral group D8. We also give a new way to construct magic squares from two orthogonal non-double-diagonal Latin squares, which explains why Choi’s Latin squares produce a magic square of order 9. 2010 MSC: 05B15, 05B20 Keywords: Choi Seok-Jeong, Koo-Soo-Ryak, Latin squares, Magic squares 1. Introduction A Latin square of order n is an n×n array in which n distinct symbols are arranged so that each symbol occurs once in each row and column. This Latin square is one of the most interesting mathematical objects. It can be applied to a lot of branches of discrete mathematics including finite geometry, coding theory and cryptography [8], [9]. In particular, orthogonal Latin squares have been one of the main topics in Latin squares. The superimposed pair of two orthogonal Latin squares is also called a Graeco-Latin sqaure by Leonhard Euler (1707-1783) in 1776 [4]. It is known that the study of Latin squares was researched by Euler in the 18th century. However the Korean mathematician, Choi Seok-Jeong [Choi is a family name] (1646-1715) already studied Latin squares at least 60 years before Euler’s work [11]. A ∗ This author is supported by Basic Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2019R1A2C1088676). Jon-Lark Kim (Corresponding Author), Dong Eun Ohk, Doo Young Park, Jae Woo Park; Department of Mathematics, Sogang University, Seoul, 04107, South Korea (email: jlkim@sogang.ac.kr, tony_to@naver.com, dy9723@naver.com, 67670711@naver.com). 17 https://orcid.org/0000-0002-0517-9359 https://orcid.org/0000-0002-7737-5199 https://orcid.org/0000-0001-7404-0492 J. L. Kim et.al. / J. Algebra Comb. Discrete Appl. 9(1) (2022) 17–27 pair of two orthogonal Latin squares of order 9 was introduced in Koo-Soo-Ryak (or Gusuryak) written by Choi Seok-Jeong [2]. The Koo-Soo-Ryak was listed as the first literature on Latin squares in the Handbook of Combinatorial Designs [3]. Let K be the matrix form of the superimposed Latin square of order 9 from Koo-Soo-Ryak: K = (5,1) (6,3) (4,2) (8,7) (9,9) (7,8) (2,4) (3,6) (1,5) (4,3) (5,2) (6,1) (7,9) (8,8) (9,7) (1,6) (2,5) (3,4) (6,2) (4,1) (5,3) (9,8) (7,7) (8,9) (3,5) (1,4) (2,6) (2,7) (3,9) (1,8) (5,4) (6,6) (4,5) (8,1) (9,3) (7,2) (1,9) (2,8) (3,7) (4,6) (5,5) (6,4) (7,3) (8,2) (9,1) (3,8) (1,7) (2,9) (6,5) (4,4) (5,6) (9,2) (7,1) (8,3) (8,4) (9,6) (7,5) (2,1) (3,3) (1,2) (5,7) (6,9) (4,8) (7,6) (8,5) (9,4) (1,3) (2,2) (3,1) (4,9) (5,8) (6,7) (9,5) (7,4) (8,6) (3,2) (1,1) (2,3) (6,8) (4,7) (5,9) Then we can separate K into two Latin squares L and N. To get a visible effect, let us color in each square. Figure 1. A colored two Latin squares L and N, respectively We paint colors for each numbers, 1,2, · · · ,9. In details, 1,2,3 are colored in red, 4,5,6 are colored in green, and 7,8,9 are colored in blue. Then we observe that the Latin squares have self-repeating patterns. This simple structure of Choi’s Latin squares motivates some generalization of his idea. We generalize Choi’s Latin squares in three directions: the Kronecker product approach, the Dihedral group approach, and magic squares from Choi’s Latin squares. In this paper, we give a new generalization of Choi’s orthogonal Latin squares of order 9 to orthogonal Latin squares of size n2 using the Kronecker product including Lih’s construction [9]. There has been some attempt that the dihedral group D8 acts on the Latin squares [5]. We find a geometric description of Choi’s orthogonal Latin squares of order 9 using D8. We also give a new way to construct magic squares from two orthogonal non-double-diagonal Latin squares, which explains why Choi’s Latin squares produce a magic square of order 9. 2. A generalization of Choi’s orthogonal Latin squares Definition 2.1. ([9]) Let A = (aij) be a Latin square of order n(i,j ∈ {1,2, · · · ,n}) and B = (bst) be a Latin square of order m(s,t ∈ {1,2, · · · ,m}). Then the Kronecker product of A and B, which is an 18 J. L. Kim et.al. / J. Algebra Comb. Discrete Appl. 9(1) (2022) 17–27 mn×mn square A⊗B given by A⊗B = (a11,B) (a12,B) · · · (a1n,B) (a21,B) (a22,B) · · · (a2n,B) ... ... ... ... (an1,B) (an2,B) · · · (ann,B) where (aij,B) is the m×m square (aij,B) = (aij,b11) (aij,b12) · · · (aij,b1m) (aij,b21) (aij,b22) · · · (aij,b2m) ... ... ... ... (aij,bm1) (aij,bm2) · · · (aij,bmm) Lemma 2.2. ([9]) A⊗B is a Latin square if A and B are both Latin squares. Theorem 2.3. ([9]) If two Latin squares A1 and A2 of order n are orthogonal and two Latin squares B1 and B2 of order m are orthogonal, then A1 ⊗B1 and A2 ⊗B2 of order mn are orthogonal. Now it is natural to substitute m(aij −1) + bkl for the entry (aij,bkl) in A⊗B. Thus we define the substituted Kronecker product ⊗S of two Latin squares A and B by the following block matrix A⊗S B =  (m(a11 −1)×Nm + B) · · · (m(a1n −1)×Nm + B)... ... ... (m(an1 −1)×Nm + B) · · · (m(ann −1)×Nm + B)   where A = (aij) is a matrix of order n, B is a matrix of order m, and Nm is the m×m all-ones matrix. Let us return to Latin squares. Judging from Figure 1, we can expect that L is closely related to a Latin square of order 3. Let A3 = (aij) = 2 3 1 1 2 3 3 1 2 Then the following block matrix (3(a11 −1)×N3 + A3) (3(a12 −1)×N3 + A3) (3(a13 −1)×N3 + A3)(3(a21 −1)×N3 + A3) (3(a22 −1)×N3 + A3) (3(a23 −1)×N3 + A3) (3(a31 −1)×N3 + A3) (3(a32 −1)×N3 + A3) (3(a33 −1)×N3 + A3)   produces L. In other words, L = A3 ⊗S A3. Similarly, let B3 = 1 3 2 3 2 1 2 1 3 then N = B3⊗S B3. These two Latin squares A3 and B3 are elements of MOLS(3) which is the mutually orthogonal Latin squares of order 3. We recall that Lih [10] also found this relation. However he did not explain why L = A3 ⊗S A3 and N = B3 ⊗S B3 are orthogonal from the Kronecker product point of view. Corollary 2.4. Choi’s two Latin squares of order 9 are orthogonal. Proof. By the above notation, we can put Choi’s two Latin squares of order 9 by L = A3 ⊗S A3 and N = B3 ⊗S B3. Note that A3 and B3 are orthogonal. Therefore, by taking A1 = B1 = A3 and A2 = B2 = B3 in Theorem 2.3 we see that L = A3 ⊗S A3 and N = B3 ⊗S B3 are also orthogonal. 19 J. L. Kim et.al. / J. Algebra Comb. Discrete Appl. 9(1) (2022) 17–27 Hence it appears that Choi might know how to get the orthogonal Latin squares of order 9 by expanding orthogonal Latin squares of order 3. It is natural to generalize Choi’s approach to obtain orthogonal Latin squares by copying a smaller Latin square several times. If A is a Latin square of order n, we call A⊗S A Choi type Latin square of order n2. Since there exists a pair of orthogonal Latin squares of order n ≥ 3 and n 6= 6, the following is immediate. Corollary 2.5. There exists a pair of Choi’s type Latin squares of order n2 which are orthogonal whenever n ≥ 3 and n 6= 6. We remark that Lih’s construction [10] gives only the case when n = 3. Corollary 2.5 extends this result to any n ≥ 3 and n 6= 6. 3. Latin squares acted by the dihedral group D8 We have noticed that L is symmetric to N with respect to the 5th column of L. In other word, if we let L = (lij), then N = (li(n+1−j)). So we define some operation. Definition 3.1. Let A = (aij) be an n×n matrix (or square or array). Define the n×n matrix s2(A) by s2(A) = (ai(n+1−j)). We can consider more symmetries. The dihedral group of degree n denoted by D2n is a well-known group of order 2n consisting of symmetries on a regular n-polygon consisting rotations and reflections. In this case, we concentrate on a square, so the dihedral group of order 8, denoted by D8, is needed. In D8, there are eight elements, s0,s1,s2,s3,r1,r2,r3,r4. Note that si for i = 0,1,2,3 denotes a reflection. More precisely, s0 is a horizontal reflection, s1 is a main diagonal reflection, s2 is a vertical reflection, and s3 is an antidiagonal reflection. Note that r0 denotes the rigid motion and ri’s (i = 1,2,3) denote counterclockwise rotations by 90, 180, 270 degrees respectively so that r2 = r21 and r3 = r 3 1. We can define a set D8(A) = {A,r1(A),r2(A),r3(A),s0(A),s1(A),s2(A),s3(A)} for a given Latin square A. Definition 3.2. Let Ln be the set of all Latin squares of order n. Then σ ∈ D8 is a function with σ : Ln → Ln defined by r0(A) = (aij), r1(A) = (aj(n+1−i)), r2(A) = (a(n+1−i)(n+1−j)), r3(A) = (a(n+1−j)i), s0(A) = (a(n+1−i)j), s1(A) = (aji) , s2(A) = (ai(n+1−j)), s3(A) = (a(n+1−j)(n+1−i)) where A ∈ Ln and A = (aij). Then we can regard an element in D8 as a function acting on Ln. In fact, the dihedral group D8 acts on Ln (or Ln is a D8-set) as follows. Lemma 3.3. Ln is a D8-set. Proof. Let σ ∈ D8 and A ∈ Ln. Since A = (aij) is a Latin square of order n, {a1j,a2j, · · · ,anj} = {ai1,ai2, · · · ,ain} = {1,2, · · · ,n} for all i,j = 1,2, · · · ,n. Thus by definition, σ(A) is a Latin square. If σ = r0, then r0(A) = A for any A ∈ Ln. Suppose that σ1,σ2 ∈ D8. Let σ3 = σ1 ◦σ2 ∈ D8. It is straightforward to check that σ3(A) = σ1(σ2(A)) by Definition 3.2. 20 J. L. Kim et.al. / J. Algebra Comb. Discrete Appl. 9(1) (2022) 17–27 In the Choi’s Latin squares, N = s2(L) (or L = s2(N)). Since L and N are orthogonal, we can say that L and s2(L) are orthogonal. Then we can have some questions. Is L orthogonal to σ(L) for another σ in D8? And how many mutually orthogonal Latin squares are in the set D8(L)? Moreover, for any Latin square A, what is the maximum number of mutually orthogonal Latin squares in the set D8(A)? Lemma 3.4. Suppose A and B are Latin squares of order n and take an arbitrary σ ∈ D8. Then A is orthogonal to B if and only if σ(A) is orthogonal to σ(B). By Lemma 3.4, we have a criteria when two Latin squares in the set D8(A) are orthogonal. If two Latin squares A and B are orthogonal, we denote it by A⊥B: r0(A)⊥s0(A) ⇐⇒ r1(A)⊥s1(A) ⇐⇒ r2(A)⊥s2(A) ⇐⇒ r3(A)⊥s3(A) r0(A)⊥s1(A) ⇐⇒ r1(A)⊥s2(A) ⇐⇒ r2(A)⊥s3(A) ⇐⇒ r3(A)⊥s0(A) r0(A)⊥s2(A) ⇐⇒ r1(A)⊥s3(A) ⇐⇒ r2(A)⊥s0(A) ⇐⇒ r3(A)⊥s1(A) r0(A)⊥s3(A) ⇐⇒ r1(A)⊥s0(A) ⇐⇒ r2(A)⊥s1(A) ⇐⇒ r3(A)⊥s2(A) r0(A)⊥r1(A) ⇐⇒ r1(A)⊥r2(A) ⇐⇒ r2(A)⊥r3(A) ⇐⇒ r3(A)⊥r0(A) ⇐⇒ s0(A)⊥s1(A) ⇐⇒ s1(A)⊥s2(A) ⇐⇒ s2(A)⊥s3(A) ⇐⇒ s3(A)⊥s0(A) r0(A)⊥r2(A) ⇐⇒ r1(A)⊥r3(A) ⇐⇒ s0(A)⊥s2(A) ⇐⇒ s1(A)⊥s3(A) Thus for finding mutually orthogonal Latin squares in D8(A), we should look at the orthogonality of A = r0(A) and σ(A) for σ ∈ D8. Lemma 3.5. For any A ∈ Ln, A is not orthogonal to r2(A). Proof. Let A = (aij) and r2(A) = (bij). Suppose that A and r2(A) are orthogonal. Then we have {(aij,bij) |i,j = 1,2, · · · ,n} = {(x,y) |x,y = 1,2, · · · ,n} . Therefore there exist some integers sk, tk such that (asktk,bsktk) = (k,k) for each nonnegative integer k = 1,2, . . .n. Let n + 1 − sk = s′k and n + 1 − tk = t ′ k. Since bsktk = as′kt′k and bs′kt′k = asktk, so (asktk,bsktk) = (bsktk,asktk) = (as′kt′k,bs′kt′k). It means that the two ordered pairs (asktk,bsktk) and (as′ k t′ k ,bs′ k t′ k ) are the same in the set {(aij,bij)}. Since A and r2(A) are orthogonal, we have (sk, tk) = (s′k, t ′ k). That is, sk = s ′ k and tk = t ′ k. This implies that n = 2sk − 1 = 2tk − 1, that is, sk = tk for any k. It contradicts. Lemma 3.6. Let A ∈ Ln and n be even. Then A is not orthogonal to either s0(A) or s2(A). Proof. Suppose that A is orthogonal to s0(A). Let s0(A) = (a(n+1−i)j) = bij. By the similar argument of proof of Lemma 3.5, there exist integer u and v such that auv = buv = k for some k. So auv = buv = a(n+1−u)v. Since A is a Latin square, the entries in the v-th column are all distinct. Thus auv = a(n+1−u)v implies u = n+1−u and so u = (n+1)/2. However, n is even so that u is not an integer. It contradicts. Hence A is not orthogonal to s0(A). We can show that A is not orthogonal to s2(A) in a similar manner. Theorem 3.7. Let A ∈ Ln and n be odd. Then the maximum number of mutually orthogonal Latin squares of order n in the set D8(A) is less than or equal to 4. And if we assume that n is even, then the maximum number of mutually orthogonal Latin squares in the set D8(A) is 2. Proof. Let M be the set of mutually orthogonal Latin squares, which has the maximum number of mutually orthogonal Latin squares in the set D8(A). By Lemma 3.5, we can get r0(A) 6⊥ r2(A), r1(A) 6⊥ r3(A), s0(A) 6⊥ s2(A) and s1(A) 6⊥ s3(A). If we take three or more elements of M from the set 21 J. L. Kim et.al. / J. Algebra Comb. Discrete Appl. 9(1) (2022) 17–27 {r0(A),r1(A),r2(A),r3(A)}, then there should appear a pair of non-orthogonal Latin squares. Similarly, we cannot take three or more elements from the set {s0(A),s1(A),s2(A),s3(A)}. It means that the set M can be M = {ri1(A),ri2(A),sj1(A),sj2(A)}. Therefore we have that the maximum number of mutually orthogonal Latin squares in the set D8(A) is less than or equal to four. Suppose n is even and M = {ri1(A),ri2(A),sj1(A),sj2(A)}. It is possible that M does not contain r0(A), however, we can get the set of 4 mutually orthogonal Latin squares containing r0(A) by the group action. So without loss of generality, assume that ri1 = r0. By Lemma 3.6, sj1,sj2 should be 1 and 3. However s1(A) 6⊥ s3(A) by Lemma 3.5, so |M | 6= 4. Now suppose that M = {r0(A),ri2(A),sj1(A)}. Note that i2 = 1,3 and j1 = 1,3. However, Lemma 3.6 also implies that r1(A) 6⊥ s1(A), r1(A) 6⊥ s3(A), r3(A) 6⊥ s1(A) and r3(A) 6⊥ s3(A). Thus |M | 6= 3. Therefore |M | = 2 if n is even. Corollary 3.8. Let L be one of Choi’s Latin squares of order 9. Then the maximum number of mutually orthogonal Latin squares in D8(L) is two. Proof. By Theorem 3.5 there are at most 4 mutually orthogonal Latin squares in D8(L). Without loss of generality, we may assume that L is one of them. We first show that there are only two mutually orthogonal Latin squares among L,r1(L),r2(L),r3(L). By Lemma 3.3, L is not orthogonal to r2(L). This also implies that r1(L) is not orthogonal to r3(L). On the other hand, we have checked by enumerating all ordered pairs that L is orthogonal to both r1(L) and r3(L). Therefore we have only two cases {L,r1(L)} and {L,r3(L)} among rotations. We can easily check that L is orthogonal to both s0(L) and s2(L) while L is neither orthogonal to s1(L) nor to s3(L) because the two diagonal reflections do not change the value of 5 in the main diagonal. However s0(L) cannot be orthogonal to s2(L) because they reduce to L and r2(L) which are not orthogonal by Lemma 3.3. Therefore we have the following four possibilities. 1. {L,r1(L),s0(L)} 2. {L,r1(L),s2(L)} 3. {L,r3(L),s0(L)} 4. {L,r3(L),s2(L)} r1(L) = 132798465 321987654 213879546 798465132 987654321 879546213 465132798 654321987 546213879 s0(L) = 978312645 789123456 897231564 312645978 123456789 231564897 645978312 456789123 564897231 s2(L) = 132798465 321987654 213879546 798465132 987654321 879546213 465132798 654321987 546213879 r3(L) = 978312645 789123456 897231564 312645978 123456789 231564897 645978312 456789123 564897231 We have checked that r1(L) is not orthogonal to s0(L) because (4,6) is repeated and r1(L) is not orthogonal to s2(L) because (4,4) is repeated. Similarly, r3(L) is not orthogonal to s0(L) because (6,6) is repeated and r3(L) is not orthogonal to s2(L) because (6,4) is repeated. These are visualized by pairing the bold face numbers in r1(L),s0(L),s2(L),r3(L). Therefore, we have {L,r1(L)}, {L,s0(L)}, {L,s2(L)}, or {L,r3(L)} as a maximal mutually orthog- onal Latin square subset of D8(L). Hence the maximum number of mutually orthogonal Latin squares in D8(L) is two. 22 J. L. Kim et.al. / J. Algebra Comb. Discrete Appl. 9(1) (2022) 17–27 If a Latin square A of order n is orthogonal to σ(A) for some σ ∈ D8(A), we call such A a dihedral Latin square. We recall that a Latin square A is self-orthogonal if it is orthogonal to its transpose [12]. Since the transpose of A can be represented as s1(A) (s1 is a main diagonal reflection), the concept of a dihedral Latin square includes the concept of a self-orthogonal Latin square. For example, Choi’s two Latin squares L,N of order 9 are dihedral since N = s2(L). Let us take another example as follows. A = 1 2 3 4 3 4 1 2 4 3 2 1 2 1 4 3 r1(A) = 4 2 1 3 3 1 2 4 2 4 3 1 1 3 4 2 s1(A) = 1 3 4 2 2 4 3 1 3 1 2 4 4 2 1 3 Then A and r1(A) are a pair of orthogonal Latin squares. So A is a dihedral Latin square. Similarly, A and s1(A) are orthogonal. So A is self-orthogonal too. However r1(A) is not orthogonal to s1(A) since (1,4) is repeated. By the previous theorem, the maximum number of mutually orthogonal Latin squares in the set D8(A) is 2. Consider Choi’s type Latin squares A⊕S A, r1(A) ⊕S r1(A), and s1(A) ⊕S s1(A). Then A⊕S A is orthogonal to both r1(A)⊕S r1(A) and s1(A)⊕S s1(A). 4. Magic squares from Latin squares Definition 4.1. A magic square of order n is an n×n array (or matrix) of the n2 consecutive integers with the sums of each row, each column, each main diagonal, and each antidiagonal are the same. For example, 4 9 2 3 5 7 8 1 6 is a magic square of order 3 since the sums of each row, column, main diagonal and antidiagonal are the same. Similarly, for order n Latin square, we assume the symbols are {1,2, · · · ,n2}. Then the question is what the relation between Latin squares and magic squares is. We need the following definition. Definition 4.2. Let A be a Latin square of order n. Then, A is called a double-diagonal Latin square [6], [7] if the n entries in main diagonal are all distinct and the n entries in antidiagonal are also all distinct. A construction of orthogonal double-diagonal Latin squares has been actively studied [8], [1], [12]. Theorem 4.3. ([9]) Suppose a pair of orthogonal double-diagonal Latin squares of order n exist. Then a magic square of order n can be constructed from them. Definition 4.4. Suppose A = (aij) and B = (bij) are orthogonal Latin squares of order n. Then define an n×n square A +S B by A +S B = (n(aij −1) + bij). This A +S B is not necessarily a magic square since its sums of two main diagonals is not the same as its sums of columns or rows. Theorem 4.3 states that if the two Latin squares A and B are orthogonal and double-diagonal, then A +S B is a magic square. 23 J. L. Kim et.al. / J. Algebra Comb. Discrete Appl. 9(1) (2022) 17–27 And the another noticeable point is that the pair of Choi’s orthogonal Latin squares is not double- diagonal. However, Choi’s squares also can produce a magic square even though they are not double- diagonal. Theorem 4.5. If there is a pair of orthogonal Latin squares A and B of order n such that the sum of main diagonal of each of A and B is n(n + 1)/2 and the sum of antidiagonal of each of A and B is n(n + 1)/2, then A +S B is a magic square of order n. Proof. Suppose A = (aij) and B = (bij),(i,j ∈{1,2, · · · ,n}) are orthogonal Latin squares such that n∑ i=1 aii = n∑ i=1 bii = n(n + 1) 2 and n∑ i=1 ai(n+1−i) = n∑ i=1 bi(n+1−i) = n(n + 1) 2 . Now define M = (mij) by M = (mij) = (n(aij − 1) + bij). We want to show that M is a magic square. Since 1 ≤ aij, bij ≤ n for all (i,j), 1 ≤ n(aij − 1) + bij ≤ n2. We first show that each mij is distinct. Suppose n(auv −1) + buv = n(ast −1) + bst. Then n(auv −ast) = bst −buv. So n | (bst −buv). However, 1 ≤ bij ≤ n for all (i,j), so 1 − n ≤ bst − buv ≤ n − 1. Thus n | (bst − buv) implies bst − buv = 0 and so auv = ast. Since A and B are orthogonal Latin squares, (u,v) = (s,t). Hence if (u,v) 6= (s,t) then n(auv −ast) 6= bst − buv so all mij are distinct. Now calculate the sums. n∑ i=1 {n(aij −1) + bij} = n n∑ i=1 aij −n2 + n∑ i=1 bij = n(n2 + 1) 2 , n∑ j=1 {n(aij −1) + bij} = n n∑ j=1 aij −n2 + n∑ j=1 bij = n(n2 + 1) 2 , n∑ i=1 {n(aii −1) + bii} = n n∑ i=1 aii −n2 + n∑ i=1 bii = n(n2 + 1) 2 , and similarly, n∑ i=1 {n(ai(n+1−i) −1) + bi(n+1−i)} = n(n2 + 1) 2 . Thus the sums are the same. Hence M is a magic square. We have an existence theorem satisfying Theorem 4.5. Theorem 4.6. For any odd number n ≥ 3, there exists a pair of orthogonal Latin squares each of whose sum of main diagonal (and antidiagonal respectively) is n(n + 1)/2. 24 J. L. Kim et.al. / J. Algebra Comb. Discrete Appl. 9(1) (2022) 17–27 Proof. Suppose n = 2k − 1 where k ≥ 2. Let An = (aij) be a matrix where each descending diagonal from left to right is constant like following matrix: An = k n k −1 n−1 ... k + 2 2 k + 1 1 1 k ... ... ... ... k + 2 2 k + 1 k + 1 ... k ... ... n−1 ... k + 2 2 2 ... ... ... n k −1 n−1 ... k + 2 ... ... ... 1 k n ... ... ... k −2 ... 2 k + 1 1 ... ... ... n−1 n−1 k −2 ... 2 ... ... k ... k −1 k −1 n−1 k −2 ... ... ... ... k n n k −1 n−1 k −2 ... 2 k + 1 1 k In particular, if n = 3 and k = 2, we get Latin square A3 in Section 2. Since Latin square B3 in Section 2 is obtained by reflecting A3 along the 2nd column of A3, it is natural to reflect An along the kth column of An as follows. The sum of main diagonal and the sum of antidiagonal of An are n(n+1)/2 since ∑n i=1 aii = n×k = n(n + 1)/2 and ∑n i=1 ai(n+1−i) = ∑n i=1 i = n(n + 1)/2. Recall that s2(A) is the Latin square obtained by reflecting along the middle vertical line of An. Then s2(An) has the same sum of the main diagonal (and antidiagonal respectively) of An since the trace of An, tr(An) is the sum of antidiagonal (and main diagonal respectively) of s2(An). Now it remains to show that A and s2(An) are orthogonal. There is a one-to-one correspondence between pandiagonals of An and line equations; let y = x+α be a line with α ∈ Zn. Then each constant pandiagonal corresponds to each equation of line. For example, y = x corresponds to the diagonal constant k in An since k = aij ⇔ i = j in Zn. (i,e. (i,j) is a root of y = x in Zn). Similarly, x = y − 2 corresponds to the constant k−1, · · · , x = y−(n−1) corresponds to the constant 1. And x = y+(n−1) corresponds to the constant n, x = y+(n−3) corresponds to the constant n−1, · · · , x = y+2 corresponds to the constant k + 1. Then we can do this to s1(A); similarly, x = −y corresponds to the constant k, · · · , x = −y− (n−1) corresponds to the constant 1. And x = −y + (n−1) corresponds to the constant n, · · · , x = −y + 2 corresponds to the constant k + 1. Any two lines x = y + α and x = −y + β have exactly one unique root. It means that an entry (aij,ai(n+1−j)) appears only once. By the above theorem, we get a magic square constructed from a pair of orthogonal Latin squares which are not double-diagonal. Although there are many other ways to construct magic squares, our method is the way Choi obtained magic squares from two orthogonal non-double-diagonal Latin squares. However, we can ask a question ”What does happen if n is even?” It is well known that a pair of orthogonal Latin square does not exist when n = 2 and n = 6, and so it is more difficult to get an even order magic square consisting of a pair of Latin squares. So we construct magic squares of some even order cases in a different way. Lemma 4.7. Suppose that a Latin square A1 of order n has main diagonal and antidiagonal sums n(n + 1)/2 respectively and that a Latin square B1 of order m has main diagonal and antidiagonal sums m(m+1)/2 respectively. Then A1⊗B1 is a Latin square of order mn with main diagonal and antidiagonal sums nm(nm + 1)/2 respectively. Proof. The fact that A1 ⊗S B1 is a Latin square of order mn follows from Theorem 2.2. It remains to show that the two sums give nm(nm + 1)/2. 25 J. L. Kim et.al. / J. Algebra Comb. Discrete Appl. 9(1) (2022) 17–27 First we consider the sum of main diagonal of A1⊗S B1. By definition of A1⊗S B1, its main diagonal sum is equal to m{m n∑ i=1 aii −mn}+ ( m∑ i=1 bii)n = m 2 { n(n + 1) 2 −n } + mn(m + 1) 2 = mn(mn + 1) 2 . Similarly its antidiagonal sum is equal to m{m n∑ i=1 ai(n+1−i) −mn}+ ( m∑ i=1 bi(n+1−i))n = m 2 { n(n + 1) 2 −n } + mn(m + 1) 2 = mn(mn + 1) 2 . This completes the proof. Theorem 4.8. For every n with n ≡ 2 (mod 4), there exists a pair of orthogonal Latin squares each of whose sum of main diagonal (and antidiagonal, respectively) is n(n + 1)/2. Proof. Define four Latin squares by A1 = 1 4 3 2 4 1 2 3 2 3 4 1 3 2 1 4 A2 = 3 1 2 4 4 2 1 3 2 4 3 1 1 3 4 2 B1 = 1 5 8 4 2 6 7 3 3 8 5 2 4 7 6 1 8 3 2 5 7 4 1 6 6 2 3 7 5 1 4 8 2 6 7 3 1 5 8 4 4 7 6 1 3 8 5 2 7 4 1 6 8 3 2 5 5 1 4 8 6 2 3 7 B2 = 1 4 5 8 2 6 3 7 4 1 8 5 6 2 7 3 3 8 1 6 5 7 2 4 8 3 6 1 7 5 4 2 7 5 4 2 8 3 6 1 5 7 2 4 3 8 1 6 6 2 7 3 4 1 8 5 2 6 3 7 1 4 5 8 Then A1 and A2 are orthogonal. B1 and B2 are also orthogonal. So we can construct two orthogonal Latin squares of order 4k (where k is an integer) using the following way. If we want to construct of orthogonal Latin squares of order 4t with t odd, then we can make two Latin squares A1 ⊗S C1 and A2 ⊗S C2 where C1 and C2 are orthogonal Latin squares of order t and the sums of diagonal and antidiagonal are t(t + 1)/2 (By Theorem 4.6, we can get such pair of Latin squares). Then A1 ⊗S C1 and A2 ⊗S C2 are orthogonal by Theorem 2.2 and each sum of their diagonal and antidiagonal is 4t(4t + 1)/2 by Lemma 4.4. Or if we want to construct orthogonal Latin squares of order 2p with p ≥ 3, we recursively use the substituted Kronecker products of A1,B1,A2, and B2. So we can construct orthogonal Latin Squares of an even order n which is not of the form of 2r (r is odd) each of whose sum of diagonal (and antidiagonal, respectively) is n(n + 1)/2. Corollary 4.9. For any integer n with n 6= 2r where r is odd, there exists a pair of non-double-diagonal orthogonal Latin Squares of order n such that the pair of Latin squares can produce a magic square of order n. Proof. By Theorems 4.2, 4.3, and 4.5, we can construct a magic square of order n where n 6= 2r (r is odd). Therefore, Choi’s orthogonal Latin squares of various orders give a new way to construct magic squares based on non-double-diagonal orthogonal Latin squares. 26 J. L. Kim et.al. / J. Algebra Comb. Discrete Appl. 9(1) (2022) 17–27 References [1] J. W. Brown, F. Cherry, L. Most, M. Most, E. T. Parker, W. D. Wallis, Completion of the spectrum of orthogonal diagonal Latin squares, Graphs, Matrices and Desings, Dekker (1993) 43–49. [2] S. J. Choi, Gusuryak, Seoul National University Kyujanggak Institute for Korean Studies. [3] C. J. Colbourn, J. H. Dinitz, Handbook of combinatorial designs, CRC Press, Second Edition (2007). [4] L. Euler, De Quadratis Magicis, Commentationes Arithmeticae Collectae 2 (1849) 593-602 and Opera Omnia 7 (1911) 441–457. [5] M. A. Francel , D. J. John, The dihedral group as the array stabilizer of an augmented set of mutually orthogonal Latin squares, Ars Combin. 97 (2010) 235–252. [6] A. J. W. Hilton, Some simple constructions for double diagonal Latin squares, Sankhya: The Indian Journal of Statistics 36(3) (1974) 215–229. [7] A. J. W. Hilton, S. H. Scott, A further construction of double diagonal orthogonal Latin squares, Discrete Mathematics 7 (1974) 111–127. [8] A. D. Keedwell, J. DÃľnes, Latin squares and their applications, Academic Press, Second Edition (2015). [9] C. F. Laywine, G. L. Mullen, Discrete mathematics using Latin squares, John Wiley & Sons, New York (1998). [10] K. W. Lih, A remarkable Euler square before Euler, Mathematics Magazine 83(3) (2010) 163–167. [11] H. Y. Song, Choi’s orthogonal Latin squares is at least 67 years earlier than Euler’s, Global KMS Conference, Jeju, Korea (2008). [12] Y. Zhang, K. Chen, N. Cao, H. Zhang, Strongly symmetric self-orthogonal diagonal Latin squares and Yang Hui type magic squares, Discrete Mathematics 328 (2014) 79–87. 27 https://doi.org/10.1201/9780203719916-4 https://doi.org/10.1201/9780203719916-4 https://doi.org/10.1201/9781420010541 https://scholarlycommons.pacific.edu/euler-works/795/ https://scholarlycommons.pacific.edu/euler-works/795/ https://mathscinet.ams.org/mathscinet-getitem?mr=2721801 https://mathscinet.ams.org/mathscinet-getitem?mr=2721801 https://www.jstor.org/stable/25051907 https://www.jstor.org/stable/25051907 https://doi.org/10.1016/S0012-365X(74)80023-3 https://doi.org/10.1016/S0012-365X(74)80023-3 https://doi.org/10.1016/C2014-0-03412-0 https://doi.org/10.1016/C2014-0-03412-0 https://books.google.ru/books?id=VwqN86g68sIC https://books.google.ru/books?id=VwqN86g68sIC https://doi.org/10.4169/002557010x494805 https://doi.org/10.1016/j.disc.2014.04.002 https://doi.org/10.1016/j.disc.2014.04.002 Introduction A generalization of Choi's orthogonal Latin squares Latin squares acted by the dihedral group D8 Magic squares from Latin squares References