In the introduction, I mentioned that the first written reference to Latin squares concerned a problem of arranging the face cards of an ordinary pack of playing cards into a square so that no row or column contained more than one card of each suit and one card of each rank. Consider the following array, which is one of the solutions to this problem.
A  K  Q  J 
J  Q  K  A 
K  A  J  Q 
Q  J  A  K 

Two Latin squares L_{1} =  a_{ij} and L_{2} =  b_{ij} on n symbols 1,2,...,n are said to be orthogonal if every ordered pair of symbols occurs exactly once among the n^{2} pairs (a_{ij} ,b_{ij}), i = 1,2,...,n; j = 1,2,...,n.
Another example, a pair of orthogonal order 3 Latin squares and the 9 distinct ordered pairs that they form.



Theorem I.2.1.1  A given Latin square possesses an orthogonal mate if and only if it has n disjoint transversals.
If the Latin square in question is the multiplication table of a group, then by Thm I.1.2.2 the existence of a single transversal suffices to conclude that the square has an orthogonal mate. Furthermore,
Theorem I.2.1.2  The multiplication table of any group of odd order forms a Latin square which possesses an orthogonal mate.
Proof: By Thm I.1.2.3, a group of odd order has a complete mapping, so, by Thm I.1.2.1, the multiplication table of this group is a Latin square which has a transversal. Now, the observation above implies that this Latin square has an orthogonal mate.
Corollary I.2.1.3  There exist pairs of orthogonal Latin squares of every odd order.
The existence question for pairs of orthogonal Latin squares of even order is much more difficult to settle and has a long and famous history. To start with, there are only two Latin squares of order 2 and they are not orthogonal (prove this !). We have illustrated an example of a pair of orthogonal squares of order 4 in the introduction to this section. The next case, that of order 6, is the problem that originally interested Euler in the subject. Called the Problem of the 36 Officers, Euler stated it as follows in 1779, "Arrange 36 officers, 6 from each of 6 regiments, of 6 different ranks, into a 6x6 square, so that each row and each file contains one officer of each rank and one officer of each regiment." Clearly, the problem can be solved, as with the card problem in the introduction, by finding a pair of orthogonal latin squares of order 6. As brilliant a mathematician as Euler was, he was unable to find such a pair of squares and unable to prove that they did not exist. Based on his experience with the problem and some other pieces of evidence (such as Cor I.1.2.5, which he was aware of), Euler made a conjecture which included and went beyond this problem:
Euler's Conjecture: There does not exist an orthogonal mate for any Latin square whose order has the form n = 4k + 2 (oddly even integers as he put it).
120 years after Euler first stated the problem, Tarry in 1900 settled the problem of the 36 officers in the negative. His method was straightforward, he listed out all of the 812,851,200 Latin squares of order 6 and examined each pair for orthogonality and found none [ actually, by working with reduced squares he simplified the problem to checking only 9408 pairs  but of course this was all done by hand]. ( A short, noncomputer proof of Tarry's result was given in 1984 by Stinson, see the references).
It was beginning to look like the old master had pulled off a coupe, but in 1960 Bose, Shrikhande and Parker shocked the mathematical community by proving that if n > 6 Euler's conjecture is false. Their original method is long, complicated and involved looking at a number of special cases (see the references). We shall, after some preliminaries, present a short and elegant proof of the result published by the Chinese mathematician Zhu Lie.
As we shall see in the next section, the remaining case (of squares with orders divisible by 4) is easily settled in the affirmative, so the existence question for orthogonal Latin squares is completely settled.
Theorem I.2.1.4  For any order n 2 or 6, there exists a pair of orthogonal Latin squares of order n.



A question both interesting and important for applications we will see later is how big can a set of MOLS of a given order be? We already know that if the order is not 2 or 6, there are at least two MOLS of that order, but what more can be said?
The questions posed concern themselves only with the number of squares in a set of MOLS, not with what the squares actually are, so we may give ourselves license to alter the squares to suit our purposes as long as we don't change the number or mutual orthogonality of the squares. We know that we can change one square into an "equivalent" (i.e.,isotopic) square by permuting the rows, and/or columns and by renaming the elements. What effect do these transformations have on orthogonality? Consider a pair of orthogonal squares. It is not hard to see that we can easily destroy orthogonality by permuting rows and/or columns, unless we carry out the permutations simultaneously on both squares. Doing row and/or column permutations simultaneously on the two squares is the same thing as doing these permutations on the superimposed square, which will move the ordered pairs around but clearly will not change any of them, so the resulting squares remain orthogonal. Now consider the renaming operation, and suppose we rename the elements of the second square only. For the original pair, each element of the first square was matched in an ordered pair with all of elements of the second square. After the renaming, all these ordered pairs are still present, they just appear in different places, so the new squares are still orthogonal. These observations lead us to the following definition. Two sets of MOLS with the same number of squares are said to be equivalent sets of MOLS if one can be obtained from the other by any combination of simultaneously permuting the rows of all the squares, simultaneously permuting the columns of all the squares and renaming the elements of any square. (Homework: Show that this is in fact an equivalence relation)
Lemma I.2.2.1  Any set of MOLS is equivalent to a set where each square has the first row in natural order and one of the squares (usually the first) is reduced (i.e., it also has its first column in natural order).
Proof: Given a set of MOLS, we can convert it to an equivalent set by renaming the elements in any or all squares. If we do this to each square, we can make the first rows be anything we like, in particular, we can put them all in natural order. Now take any square and simultaneously permute the rows of all the squares so that the first column of this square is in natural order (this will not affect the first row since it is in natural order and so starts with the smallest element). The result is an equivalent set with the required properties.
A set of MOLS in the form described by this lemma is said to be in standard form, and the lemma merely says that any set of MOLS is equivalent to a set of MOLS in standard form.
(Homework: Put the set of MOLS given in the example above in standard form. There is an easy way and a hard way to do this, do it both ways.)
We may now easily answer one of the above questions.
Theorem I.2.2.2  No more than n1 MOLS of order n can exist.
Proof: Any set of MOLS of order n is equivalent to a set in standard form, which of course has the same number of squares in it. Consider the entries in first column and second row of all of the squares in standard form. No two squares can have the same entry in this cell. Suppose two squares had an r, say, in this cell, then in the superimposed square the ordered pair (r,r) would appear in this cell and also in the rth cell of the first row because both squares have the same first row, and so, the two squares can not be orthogonal contradiction. Now, we can not have a 1 in this cell, since it appears in the first column of the first row. Thus, there are only n1 possible entries for this cell and so there can be at most n1 squares.
A set of MOLS of order n containing n1 squares is called a complete set. We now have an existence question, for which orders do complete sets of MOLS exist? We know by examples that complete sets exist for orders 2 (only one square is needed), 3 and 4 and also that no complete set exists for order 6. We will construct (in the chapter following the next) complete sets of MOLS for any order which is a prime or power of a prime. However, it is an open research question of long standing whether or not a complete set of MOLS exists for any composite order. Instant fame will go to any mathematician who can settle this question.
Leaving aside the question of the existence of complete sets, it is certainly clear that for any order there is a maximum number of MOLS. Let N(n) be the maximum number of MOLS of order n. The only exact values that are known for this function are N(6) = 1 and N(p*) = p*1 ,where p* is a prime or power of a prime. We also know that N(n) 2 if n2,6. Much work has gone into finding other values of this function, mostly computer searches, but the task is immense and will not be completed by the methods currently being used. Failing to find exact values, we can at least try to get good lower bounds for this function (the upper bound is of course n1). One good attempt at this was carried out by MacNeish in the 1920's, based on the following construction.
Theorem I.2.2.3  (MacNeish [1922]) Suppose that there exist r MOLS of order n and r MOLS of order m, then there exist r MOLS of order mn.
Proof: This proof is a notational nightmare, but the idea behind it is actually very simple.
Let A^{(1)}, A^{(2)}, ... , A^{(r)} be the set of MOLS of order m and B^{(1)}, B^{(2)}, ... , B^{(r)} be the set of MOLS of order n. For e = 1,2,...,r, let (a_{ij}^{(e)}, B^{(e)}) represent the n x n matrix whose h,k entry is the ordered pair (a_{ij}^{(e)},b_{hk}^{(e)}). Let C^{(e)} be the mn x mn matrix that can be represented schematically by:
(a_{11}^{(e)} , B^{(e)} )  (a_{12}^{(e)}, B^{(e)} )  .....  (a_{1m}^{(e)}, B^{(e)} ) 
(a_{21}^{(e)}, B^{(e)} )  (a_{22}^{(e)}, B^{(e)} )  .....  (a_{2m}^{(e)}, B^{(e)} ) 
.....  .....  ..........  ..... 
(a_{m1}^{(e)}, B^{(e)} )  (a_{m2}^{(e)}, B^{(e)} )  .....  (a_{mm}^{(e)} , B^{(e)} ) 
We will show that C^{(1)},C^{(2)},...,C^{(r)} is a set of MOLS of order mn.
To see that C^{(e)} is a Latin square, note first that in a given row, two entries in different columns are given by (a_{ij}^{(e)} ,b_{uv}^{(e)}) and (a_{ik}^{(e)},b_{uw}^{(e)}), and so are distinct since A^{(e)} and B^{(e)} are Latin squares. In a given column two entries in different rows are distinct by the same reasoning.
To see that C^{(e)} and C^{(f)} are orthogonal, suppose that
MacNeish used this construction to prove:
Theorem I.2.2.4  Suppose that n = p_{1}^{a} p_{2}^{b} p_{3}^{c} ...p_{s}^{t} is the prime power decomposition of n, n > 1, and r is the smallest of the quantities
Proof: For each prime power p* in the decomposition we know that there are p*  1 MOLS of that order. Thus there are r MOLS for each p*, since r is the smallest of these values. Repeated use of Thm I.2.2.3 yields r MOLS of order n.
We may now tidy up a result of the last section, where we left open the existence of orthogonal pairs for orders divisible by 4. This is taken care of by:
Corollary I.2.2.5  If n is not of the form 4k + 2, then N(n) 2.
Proof: For n of this type, either 2 is not a divisor or its power is greater than 1. In either case, the smallest possible value of p*  1 is 2.
MacNeish believed that his theorem actually gave the upper bound for N(n) as well (this is true for prime powers). This conjecture was put to rest in 1959 when E.T.Parker showed that N(21) 4 by constructing a set of 4 MOLS of order 21.
Recent work on the lower bounds of N(n) has shown that
We would like to look at some results which deal with the structure of individual squares.
Theorem I.2.3.1  If a Latin square L of order 4k + 2 contains a Latin subsquare of order 2k + 1, then L has no orthogonal mate.
Proof: It is easily seen that if a Latin square has an orthogonal mate, then any isotope of it also has a mate, so we can, without loss of generality, assume that the subsquare occupies the first 2k+1 rows and columns for if not then row and column permutations will put it there. The square L is thus partitioned into 4 (2k+1) × (2k+1) submatrices which we will label as:
A  B 
C  D 
Now consider a transversal of L. Say that there are h cells of which appear in A. Since 2k + 1 cells of must appear in the first 2k + 1 rows of L, and h of these are in A, the remaining 2k + 1  h must appear in B. A similar argument for the first 2k + 1 columns shows that there must be 2k + 1  h cells of in C. In these cells of B and C, all the elements of Q must appear exactly once. Since there are 2k + 1 elements of Q, we have
This theorem was proved by H.B. Mann in 1944, but it seems probable that the result was known to Euler and may have influenced him to make his conjecture. The result can be generalized and in order to do so we need the following definition.
A Latin square L of order n = mq is said to be of qstep type if it can be represented by a matrix C of the form shown below, where for each fixed choice of k, the A_{ij}^{(k)} are Latin subsquares of L all of which contain the same q elements (but need not be identical).


Below are examples of a 2step type and a 3step type Latin square.
















Theorem I.2.3.2  [Maillet, 1894] A Latin square L of order n = mq and of qstep type has no transversals if m is even and q is odd.
Proof: Since L is of qstep type, we may assume it has the form of matrix C and that its elements are the integers 1,2,...,n. By renaming if necessary, we can arrange that for a fixed k, the elements that appear in the subsquares A_{ij}^{(k)} are those of the form kq + b where 1 b q. This is possible since every integer 1,2,...,n can be uniquely represented in the form aq + b with 0 a m1 and 1 b q. Also note that for any subsquare A_{ij}^{(k)} we can see that k i + j  2 (mod m).
Let us suppose that the theorem is false and that is a transversal of L. Then will contain just q cells belonging to the set of subsquares in the first column of C, since they represent q columns of L. Let the entries in these cells be given by a_{11}q + b_{11} , a_{21}q + b_{21} , ... , a_{q1}q + b_{q1} and let the subscripts of the squares they belong to be c_{11}1, c_{21}1, ..., c_{q1}1 respectively, note that the integers c_{ij} are not necessarily distinct.
Now, will also contain just q cells belonging to the subsquares of the jth column of C. As before, we will represent the entries of these cells by a_{1j}q + b_{1j} , a_{2j}q + b_{2j} , ... , a_{qj}q + b_{qj} and the corresponding subscripts of the squares by c_{1j} j, c_{2j} j,..., c_{qj} j. Now since a_{ij}q + b_{ij} is an element of the subsquare A_{cij}^{(aij)} we must have a_{ij} c_{ij} + j 2 (mod m) as noted above.
The entries in the n cells of are equal, in some order, to the integers 1,2,...,n. When these integers are expressed in the form a_{ij}q + b_{ij} , there will be exactly q of them with the same a_{ij}, each with a different b_{ij}. So, if we sum the a_{ij} 's of the entries in the cells of ,we get:
We have already shown that a_{ij} = q(m)(m1)/2, so this sum can be congruent to 0 mod m only if 2 divides q(m1). Thus, if q is odd and m is even, no transversal can exist.
Corollary I.2.3.3  [Euler,1779] A cyclic Latin square of even order has no transversals, and hence no orthogonal mate.
Proof: A cyclic Latin square can be thought of as a 1step type Latin square. With q = 1 and m even, the above theorem applies.
(Homework: Prove that Thm I.2.3.2 implies Thm I.2.3.1.)
Other books with good sections on orthogonal Latin squares are
M. Hall, Combinatorial Theory, Blaisdell, 1967 , Chapter 13.
H. Ryser, Combinatorial Mathematics, Carus Mathematical Monographs #14, Mathematical Association of America, 1963, Chapter 7.
C. Liu, Introduction to Combinatorial Mathematics, McGraw  Hill, 1968.
The original work of enumeration the Latin squares of order six, by Tarry is in French.
G. Tarry, "Le problème des 36 officiers", C.R. Assoc. France Av. Sci., 29(1900) part 2, pp. 170  203.
A theoretical approach for order 6 is contained in the paper,
D.R. Stinson, "A Short Proof of the Nonexistence of a Pair of Orthogonal Latin Squares of Order Six", Journal of Combinatorial Theory, Series A, 36(1984), pp. 373376.
The original proof of the falsity of Euler's Conjecture is given in a series of papers.
Bose,R.C. & Shrikhande,S.S., "On the Falsity of Euler's Conjecture About the Nonexistence of Two Orthogonal Latin Squares of Order 4t + 2", Proceedings of the National Academy of Science, 45(1959), pp. 734737.
Bose,R.C.,& Shrikhande,S.S.," On the Construction of Sets of Mutually Orthogonal Latin Squares and Falsity of a Conjecture of Euler", Transactions of the American Mathematical Society, 95(1960), pp.191209.
Bose,R.C.,Shrikhande,S.S. & Parker, E.T., "Further Results on the Construction of Mutually Orthogonal Latin Squares and the Falsity of Euler's Conjecture", Canadian Journal of Mathematics, 12(1960), pp. 189203.
A short proof of the result can be found in,
Z. Lie, "A Short Disproof of Euler's Conjecture Concerning Orthogonal Latin Squares", Ars Combinatoria, 14(1982), pp. 4755.
MacNeish's theorem and conjecture can be found in,
H.F. MacNeish, "Euler Squares", Annals of Mathematics, 23(1922) pp. 221227.
Parker's disproof of MacNeish's conjecture is contained in,
E.T. Parker, "Construction of some sets of mutually orthogonal latin squares", Proceedings of the American Mathematical Society, 10(1959), pp. 946949.
Maillet's construction and theorem for qstep type Latin Squares is in
E.Maillet, "Sur les carrés latins d'Euler", C.R. Assoc. France Av. Sci. 23(1894), part 2, pp. 244252.
The specialization by Mann is a bit more accessible.
H.B.Mann, "On Orthogonal Latin Squares", Bulletin of the American Mathematical Society, 50(1944), pp. 249  257.