III.3 END OF THE EULER CONJECTURE

The construction is just a slight modification of the prolongation method of I.1.3. We start with a special Latin square of prime order p and an arbitrary square of order q and will produce a square of order p + q.

Let L = |a_ij| be the Latin square whose elements are from GF(p) where a_ij = i + j for i,j = 0,1,2,...,p-1, and 0,-1. This is just one of the squares used in the field construction of III.1. Note that in this square it is the first column which is in natural order. Also, it will be convenient to number the rows and columns starting with 0 rather than 1. In squares of this type there are sets of disjoint transversals which are easily found, they are the broken left to right diagonals. A broken left to right diagonal starts in a cell of the 0-th column and continues diagonally downward to the right until it hits the last row where it breaks and starts up again in the top row of the next column. For example, the colored terms of this Latin square form a broken diagonal.

0	2	4	1	3
1	3	0	2	4
2	4	1	3	0
3	0	2	4	1
4	1	3	0	2

The broken diagonals are completely specified by the starting cell whose entry, for squares defined the way L is, is also the row number. Thus we may speak of the k-th broken diagonal, meaning the transversal whose cells are given by a_k,0 = k, a_k+1,1 = k + (1+), a_k+2,2 = k + 2(1+), ... , a_k+p-1,p-1 = k + (p-1)(1+) where the suffices and entries are taken modulo p. [Prove that these broken diagonals are in fact transversals of L.] We now choose an ordered selection of q of these broken diagonals. This is done by specifying an ordered set K (i.e., a vector) consisting of q starting cells, K = (k₁ ,k₂ ,...,k_q ). By arbitrarily permuting the components of K, we form another vector K' = (k₁',k₂',..., k_q'). The prolonged square which we are constructing will have q new rows and columns and q new elements which will be the values p,p+1,...,p+q-1. We fill the new rows and columns in the following way: For each of the transversals specified by K we prolong them by putting their elements in the new row corresponding to the position of this transversal in K and in the new column corresponding to the position of this transversal in K'. The cells of the transversal are then filled with a new symbol corresponding to the position of that transversal in K. When this is completed, there will remain a blank q×q subsquare, the intersection of the new rows and columns (usually placed in the bottom right corner) and this is to be filled by the given Latin square of order q whose elements have been renamed to the new elements of our square. The result is clearly a Latin square of order p + q. For example, given the following squares and choices for K and K', with p = 7 and q = 3;

0 2 4 6 1 3 5 1 3 5 0 2 4 6 2 4 6 1 3 5 0 3 5 0 2 4 6 1 4 6 1 3 5 0 2 5 0 2 4 6 1 3 6 1 3 5 0 2 4

1 2 3 3 1 2 2 3 1

K = (5,3,6) K'= (3,6,5)

we construct the following square of order 10;

0	9	7	6	8	3	5	1	2	4
1	3	9	7	2	8	6	4	5	0
2	4	6	9	7	5	8	0	1	3
8	5	0	2	9	7	1	3	4	6
4	8	1	3	5	9	7	6	0	2
7	0	8	4	6	1	9	2	3	5
9	7	3	8	0	2	4	5	6	1
5	1	4	0	3	6	2	7	8	9
3	6	2	5	1	4	0	9	7	8
6	2	5	1	4	0	3	8	9	7

The value of this construction is of course not in constructing single squares, but rather in the fact that it may be used to construct orthogonal mates.

Let L_a and L_b be latin squares of prime order p whose i,j - th entries are i + aj and i + bj respectively, where a,b are different from 0,-1 and ab. Such squares are, by Construction III.1, orthogonal mates. Let q be a positive integer for which there exists a pair of orthogonal mates of order q and such that 2q < p. Also, let K = (k₁ ,k₂ ,...,k_q) be the ordered set of broken diagonal selections and K' = (k₁',k₂',...,k_q') the permuted set used for L_a and M = (m₁ ,m₂ ,...,m_q ) and M' = (m₁',m₂',...,m_q') the corresponding sets for L_b. Carrying out the construction on both squares will give the prolongations L_a* and L_b* each of order p + q.

Theorem III.3.1 - Let H be a vector consisting of the 2q distinct elements of K and M arranged in some order. Also, let H⁺ be the vector consisting of the first q elements of H and H^– the vector consisting of the last q elements of H. Then, if there exists such a vector H so that,

Proof: When the prolongated squares are superimposed, the ordered pairs formed by the new rows are:

1. h_t + (1+a)s = k_t + (1+a)j and h_t + (1+b)s = m_t + (1+b)j, as well as,
2. h_t' + (1+a)s = - k_t' + (1+a)j and h_t' + (1+b)s = -bm_t' + (1+b)j.

To continue we will need some number theory results.

Lemma III.3.2 - If p is a prime, p7, p1 mod 3, then there exists a value of such that ³1 mod p and 1 or 2^-1.

If p is a prime, p > 7, p2 mod 3, then there exists a value of such that ³10 mod p.

Proof: By Fermat's Theorem, we have a^p-1 1 mod p for all a, and primes p. In the case when p = 3k + 1, a^3k1 mod p. Consequently, = a^k is a solution of the equation ³ = 1. A second solution is = a^2k. If a 1 then these two solutions are the roots of the equation ² + + 1 = 0. Now = 1/2 is a root of this equation only if 7/4 = 0, which can happen under our assumptions only if p = 7. For this prime however, = 2 is a root of the equation, proving the first stated result.

In the case when p = 3k - 1, let be a generator of GF(p) and let b = ^t be any non-zero element. If b has a cube root, say, ^x in the field, then ^{3x t} mod p, or since the multiplicative group is cyclic of order p - 1, 3xt mod (p-1). For each t this last congruence has a unique solution for x, since if 3yt mod (p-1), then 3x 3y mod (p-1) which implies that x = y since 3 does not divide p - 1. We have shown that every non-zero element has a unique cube root. Consider the elements 1³,2³,...,(p-1)³ which must all be distinct. If p 11, then one of these must be 10, i.e., there exists a such that ³ = 10.

Theorem III.3.3 - For any prime p 1 mod 3, p 7, there exists a pair of orthogonal mates of order p + 3 with mutually orthogonal latin subsquares in their lower right-hand corners.

Proof: By lemma III.3.2, there exists GF(p) such that ³ = 1 and 1 or 1/2. Let a = ^-1(- 1) and b = (- 1)^-1 = a^-1. Also, let K = K' = (1,,²) and M = -K, M' = -^-1K. We shall show that with H⁺ = ²K and H^– = M, the conditions of Thm. III.3.1 are satisfied. Note first that K and M are disjoint as sets so the elements of H are all distinct. Now consider the vector equation (1+b)K - (1+a)M = (b-a )H⁺.

(1+ a^-1)(1,,²) - (1+a)(-,- ²,-1) = (1++ a^-1+ a,+ ²+ a^-1 + a², ²+1+ a^-12+ a)
= (²,1,) = H⁺,

= + ² + a^-1 + a² = -1 + a^-1 - a - a = -1 + ( - 1)^-12 - a - a
= -1 + ( - 1)^-12 - + 1 - a = ( - 1)^-1[² - ( - 1)] - a
= ( - 1)^-1 - a = b - a.

Theorem III.3.4 - For any prime p 2 mod 3, p > 7, there exists a pair of orthogonal mates of order p + 3 with mutually orthogonal latin subsquares in their lower right-hand corners.

Proof: By lemma III.3.2, there exists GF(p) such that ³ = 10. Let a = -3, b = -1/3(1 + 2), K = K' = (0,1,), M = (c,d,e) and M' = (c,e,d), where

d = -½(1 + b)
c = (1 + b)(3 + b)/4b
= (1 + b)/2b
e = 1 + (b² - 1)/4b.

3b³ + 3b² + b + 9 = (3b+3)b² + b + 9 = (2-2)(1+4+4 ²)/9 - 1/3 - 2/3 + 9
= (2 + 6 - 8³)/9 - 1/3 - 2/3 + 9 = 0.

2c - (b+3) = 0
(1 + b) + 2d = 0
(1 + b) + 2e - (b + 3) = 0
-2bc - (b + 3)d = 0
3(1 + b) - 2be - (b + 3)c = (-3b³-3b²-b-9)/4b = 0
and 3(1 + b) - 2bd - (b + 3)e = (3b³ + 3b² + b + 9)/4b = 0.

(1+b)(0,1,) - (1+a)(c,d,e) - (b-a)(,0,1) = 0 and
- a(1+b)(0,1,) + b(1+a)(c,e,d) - (b-a )(d,c,e) = 0

In order to apply Thm. III.3.1 we still need to show that the 6 elements 0,1,,c,d and e are all distinct. Clearly, (1+b) - (1+a) = (b-a) and -a(1+b) + b(1+a) = (b-a). If any two of x,y,z are equal and they satisfy (1+b)x - (1+a)y = (b-a)z then all three must be equal. Similarly, if any two of x',y',z' are equal and they satisfy -a(1+b)x' + b(1+a)y' = (b-a)z' then again all three must be equal. Consequently, by considering the above vector equations, none of the elements in the following triples can be equal: (0,c,), (1,d,0), (,e,1), (0,c,d), (1,e,c) and (,d,e). From this it follows that the only possible elements that could be equal are 0 and e. Suppose e = 0, then b² + 4b - 1 = 0. This equation together with 3b³ + 3b² + b + 9 = 0 implies that b = 0 [prove it]. That would mean that = -1/2 and since ³ = 10 we must have, -1/8 = 10 or 80-1 mod p. But, if 810 mod p, p would have to divide 81 which is impossible for p > 7.

All the ingredients have now been assembled for:

Theorem III.3.5 - For every integer v = 4t + 2 > 6, there exists a pair of orthogonal latin squares of order v.

Proof: There will be two cases to consider, depending on whether or not 3 divides v.

Case I: v is not a multiple of 3. In this case, the odd number v - 3 = 4t - 1 is not divisible by 3 but must be divisible by an odd prime of the form 4s - 1. [ v-3 is only divisible by odd numbers, and so the divisors can only be of the forms 4s + 1 or 4s - 1. Not all of the divisors can be of the form 4s + 1, since the product of two numbers of that form is of that form, i.e., (4s+1)(4r+1) = 16rs + 4r + 4s + 1 = 4(4rs+r+s) + 1 = 4t + 1.] Let this odd prime be p. Then p3, since 3 does not divide v - 3, so p 7. Then, v - 3 = pm and so v = pm + 3.

If m > 1, and since m is odd and not divisible by 3, there exist at least 3 MOLS of order m by MacNeish's Theorem (I.2.2.4). Thus by Sade's singular direct product construction (Thm III.2.3.3) there exists a pair of orthogonal latin squares of order v = pm + 3.

If m = 1, v = p + 3, and so, by either Thm III.3.3 or Thm III.3.4 there exists a pair of orthogonal latin squares of order p + 3.

Case II: v is a multiple of 3. If v has odd prime factors other than 3, we may write v = u3^h , where u is not divisible by 3. Now u must have the form 2(2s + 1) = 4s + 2 since v has only one 2 factor and the product of odd numbers is odd. Also, u > 6, so by Case I, there exists a pair of orthogonal latin squares of order u. Now by MacNeish's Thm I.2.2.3, there exists a pair of orthogonal mates of order v, since there exist pairs of order 3^h.

If v = 2(3^h) > 6, we have v = 18(3^h-2). We can obtain a pair of orthogonal mates of order 18 by taking p = 13 and q = 5 in Theorem III.3.1 and defining K = (1,2,3,4,5), K' = (4,1,2,3,5), M = -K', M' = -K where H⁺ = (3,-2,5,-1,-4) and H^– = (-3,2,-5,1,4). [Carry out this construction]. Hence, again by MacNeish's Theorem we can construct a pair of orthogonal mates of order v.

REFERENCES

The field construction of Bose is found in

Bose, R.C., "On the application of the properties of Galois fields to the construction of hyper-Graeco-Latin squares", Sankhya 3(1938), pp. 323-338.

The generalized version is in,

Mann,H.B., "The construction of orthogonal Latin squares", Annals of Mathematical Statistics, 13(1942), pp.418-423.

Sade's singular direct product construction is located in,

Sade,A., "Produit direct-singulier de quasigroupes orthogonaux et anti-abéliens" , Ann. Soc. Sci. Bruxelles, Sér. I, 74(1960), pp. 91-99.

- If your French is not so hot, this article is reviewed in, Mathematical Reviews 25(1963), #4017.

The proof presented here, disproving Euler's conjecture is from,

Lie, Z., "A Short Disproof of Euler's Conjecture Concerning Orthogonal Latin Squares", Ars Combinatoria 14(1982), pp. 47-55.

- A small warning, this article has a number of typographical errors and must be read with care.