Queen Domination of Even Square Boards

The queen’s graph Q m × n has the squares of the m × n chessboard as its vertices; we identify the m × n chessboard with a rectangle of width m and height n in the Cartesian plane, having sides parallel to the coordinate axes and placed so that square centers have integer coordinates. Two squares are adjacent if they are in the same row, column, or diagonal of the board. A set D of squares of Q m × n is a dominating set for Q m × n if every square of Q m × n is either in D or adjacent to a square in D . The minimum size of a dominating set of Q m × n is the domination number , denoted by γ ( Q m × n ). We give a new proof of the bound γ ( Q m × n ) (cid:62) min (cid:8) m, n, (cid:6) m + n − 2 4 (cid:7)(cid:9) , with im-plications for queen domination problems, and then consider square boards. Let n be an even integer and assume Q n × n has a dominating set D of size n/ 2 (which implies γ ( Q n × n ) = n/ 2). For p ∈ { 0 , 1 } , let D p = { ( x, y ) ∈ D : x + y ≡ p (mod 2) } . Say that D is monochromatic if D = D p for some p ; otherwise bichromatic . We show that if D is bichromatic then || D 0 |−| D 1 || (cid:54) 2 and conjecture that if n > 4 then D is monochromatic. Assume further that D is monochromatic. If n ≡ 0 (mod 4) then n ∈ { 4 , 12 } . If n ≡ 2 (mod 4) then odd integers k = n/ 2 , e, d with 1 (cid:54) d, e (cid:54) k satisfy the equation d 2 + ( k − 1) e 2 = k ( k 2 + 2) / 3. We analyze six inﬁnite sequences of solutions of this equation arising from Fermat-Pell equations, give monochromatic dominating sets of Q n × n of size n/ 2 for n = 2 , 4 , 6 , 10 , 12 , 18 , 30 (new), and 130, and show there are no others with n < 238.


Introduction
For positive integers m, n, let Q m×n denote the graph whose vertices are the squares of the m × n chessboard; two squares are adjacent if a chess queen can move from one to the other. That is, the two squares share a row, column, or diagonal of the board.
We will identify the m × n chessboard with a rectangle of width m and height n in the Cartesian plane, having sides parallel to the coordinate axes and placed so that square centers have integer coordinates. (We defer the specific choice of board placement.) We refer to board squares by the coordinates of their centers; the square (x,y) is in column x and row y. Columns and rows will be referred to collectively as orthogonals. The difference diagonal (respectively sum diagonal ) through square (x,y) is the set of all board squares with centers on the line of slope +1 (respectively −1) through the point (x,y). The value of y − x is the same for each square (x,y) on a difference diagonal, and we will refer to the diagonal by this value. Similarly, the value of y + x is the same for each square on a sum diagonal, and we associate this value to the diagonal. Orthogonals and diagonals are collectively referred to as lines of the board.
Let S = {(x i , y i ) : 1 i h} be a set of squares of Q m×n . For each i, let d i = y i − x i be the index of the difference diagonal and s i = y i + x i the index of the sum diagonal occupied by square (x i , y i ). From the definitions of these indices, we have two constraints relating the orthogonal and diagonal indices: a linear constraint and a quadratic constraint The square (x,y) is even if x + y is even, odd if x + y is odd. We write Z 2 for the set {0, 1}, frequently used in discussing parity. The parity opposite to p ∈ Z 2 is denoted p. For any set S of squares of Q m×n and p ∈ Z 2 , let S p be the set of squares of parity p in S.
A set D of squares of Q m×n is a dominating set for Q m×n if every square of Q m×n is either in D or adjacent to a square in D. The set D is independent if no two squares of D share a line.
Let γ(Q m×n ) denote the minimum size of a dominating set for Q m×n . Let i(Q m×n ) denote the minimum size of an independent dominating set of Q m×n .
Let D be a dominating set of Q m×n of size γ = γ(Q m×n ). As D may contain more than one square in an orthogonal, we need to consider multisets of square coordinates.
Given a dominating set D of Q m×n , say that a square of Q m×n is needy (meaning it needs diagonal cover by D) if it is not in any orthogonal occupied by D.
The following theorem appeared in [5]; a new and more insightful proof, relying on equation (4), is given here. Theorem 1. [5] For all positive integers m, n, γ(Q m×n ) min m, n, m + n − 2 4 .
Proof. Without loss of generality, we may assume m n. It suffices to show that if γ(Q m×n ) m − 1 then γ(Q m×n ) (m + n − 2)/4. So we assume that γ(Q m×n ) m − 1.
Let D be a dominating set of Q m×n of size γ = γ(Q m×n ). First, suppose γ(Q m×n ) = m − 1. There is then a column not containing a square of D; as each square of D can cover at most three squares of that column, we have n < 3m − 2, which implies m − 1 > (m + n − 2)/4. Thus we may take γ(Q m×n ) m − 2. Then D fails to occupy at least two columns and two rows of Q m×n . Let a be the index of the leftmost empty column, b the index of the rightmost empty column, c the index of the lowest empty row, d the index of the highest empty row. The board has a rectangular sub-board with corner squares (a, c), (a, d), (b, c), and (b, d). The box border is the set E of edge squares of this sub-board.
Let R = R(D) be the number of squares of D that have at most one diagonal that meets E. Such a square can diagonally cover at most two squares of E; other squares of D can diagonally cover at most four squares of E. Thus D can diagonally cover at most 2R + 4(γ − R) = 4γ − 2R squares of E. The diagonal waste W diag (D) of D is the difference between 4γ − 2R and the number of needy squares of E.
The orthogonal waste of D is W orth (D) = |L(col)| + |L(row)|, the difference between the maximum number of orthogonals that a square set of size γ could occupy and the number of orthogonals D occupies.
We will show that if |D| = γ(Q m×n ) min{m − 2, n − 2} then By the definition of W diag (D), 4γ − 2R − W diag (D) is the number of needy squares of E. This number can also be counted as follows. There are m − (γ − |L(col)|) columns not occupied by D, necessarily between columns a and b inclusive. These columns meet the rows of E in 2(m − γ + |L(col)|) needy squares. Similarly the columns of E have 2(n − γ + |L(row)|) needy squares. As the corners of E are counted twice so far, which simplifies to (4). As all terms on the left side of (4) are nonnegative, (4) implies (3).
For most minimum dominating sets D of Q m×n , the term R(D) is the least important in (4), but in Figure 1 we show an example on Q 9×9 with R(D) = 4 = G (9,9).
From this point, we only consider square boards, and adopt the simpler notation Q n in place of Q n×n . Figure 1: Up to isomorphism there are three minimum dominating sets of Q 4 , shown on the left. On the right is a minimum dominating set D of Q 9 from [5] with R(D) = 4 = G (9,9); the box border is outlined.
The task of determining values of γ(Q n ) for all positive integers n appears as Problem C18 in the collection [14] of unsolved problems in number theory. Finding these values has interested mathematicians for over 150 years; some of the early references are [2,11,19,20,21].
V. Raghavan and S. M. Venketesan [18] and P. H. Spencer (see [10] or [22]) independently found very similar proofs for the bound in Theorem 2 below, which also follows immediately from Theorem 1. The last part of Theorem 2 was proved in [23] and [12].
(It is conjectured that for all other odd positive integers n, γ(Q n ) ∈ {(n + 1)/2, (n + 3)/2}; this has been confirmed [6,7,8,9,13,15,17] up to n = 119. The same workers have jointly confirmed for n ≡ 1 (mod 4) that γ(Q n ) = (n + 1)/2 for n 129.) Thus the case G(n, n) = 0 is understood. As G(n, n) is even, the next smallest value possible for G(n, n) is 2, which occurs for even n such that γ(Q n ) = n/2. The rest of this paper begins the process of finding those n for which this occurs. Definition 3. Say that a set of squares of Q n is monochromatic if all its squares have the same parity; otherwise the set is bichromatic.
Up to isomorphism, Q 4 has three dominating sets of size 2, shown in Figure 1. The top set is monochromatic and the other two are bichromatic. Computer search [3,5] has shown that for each even n, 4 < n 24, if γ(Q n ) = n/2 then every minimum dominating set of Q n is monochromatic. This and other indications lead to the following conjecture.

Conjecture 4.
For each even integer n > 4, if γ(Q n ) = n/2 then every minimum dominating set of Q n is monochromatic.
If true, this would give a short proof of the main theorem of [23]: that a dominating set of Q 4k+3 with size 2k + 1 is monochromatic.
The best we can do here is to prove that a bichromatic dominating set of Q n with size n/2 must have nearly equal numbers of squares of each color.
Before doing this, we note it is known for each of the values n = 3, 11 (those for which γ(Q n ) = (n − 1)/2) that there is one isomorphism class of minimum dominating sets, and the sets are monochromatic. So Conjecture 4 could be stated: if G(n, n) 2 then except for n = 4, every minimum dominating set of Q n is monochromatic. However, for those n with G(n, n) 4 (that is, γ(Q n ) (n + 1)/2), bichromatic dominating sets are common. In particular, γ(Q n ) = (n + 1)/2 for n = 5, 7, 9, 13; as shown in [5], for n = 5, 7 of the 37 isomorphism classes of minimum dominating sets are monochromatic; for n = 7, it is 4 of 13 classes; for n = 9, it is 6 of 21, and for n = 13, it is 14 of 41.
2 Bichromatic dominating sets of size k for Q 2k Definition 5. Distinct orthogonals of the same type are parallel. The distance between parallel orthogonals is the absolute value of the difference of their indices.
Our next definition requires a lemma. Lemma 6. Let n be an even integer, n 4, and D a bichromatic dominating set of size n/2 for Q n . For t ∈ {row, column} there exist parallel empty t-orthogonals at odd distance.
Proof. As D is bichromatic, for at least one t ∈ {row, column} there are occupied torthogonals of opposite parity and thus empty t-orthogonals of opposite parity. Without loss of generality we may assume this is true for columns. Place the n × n board with lower left corner having center (1, 1). If the desired conclusion does not hold, all the empty rows have indices of the same parity; by flipping the board across its horizontal midline if necessary, we may assume rows 1, 3, . . . , n − 1 are occupied. Then any empty column with index of parity p contains k = n/2 needy squares of parity p. At most two of these squares can be covered by each square of D, so D contains at least k/2 squares of each parity. As |D| = k, we see that k is even and D contains exactly k/2 squares of each parity. This implies that for every square q of parity p in D and every empty column with index of parity p, each diagonal of q covers a needy square in that column.
As k is even, n is a multiple of four. We have already seen that each dominating set of size 2 for Q 4 is either monochromatic or satisfies the conclusion of this lemma, and γ(Q 8 ) = 5, so we may assume that n 12. Thus there are at least six unoccupied columns, implying there are at least three unoccupied columns with indices c 1 < c 2 < c 3 of the same parity. Then s = (c 2 , n) is the top needy square in column c 2 . If s is covered by D along its difference diagonal, that diagonal passes above all needy squares of column c 3 , which is not possible by the preceding paragraph. Similarly, if s is covered along its sum diagonal, that diagonal passes above all needy squares of column c 1 and this is not possible. Thus the conclusion holds.
For the rest of this section, we assume that n is even, n > 4, and D is a bichromatic dominating set of Q n of size k = n/2. Definition 7. Since n > 4 and |D| = n/2, for each t ∈ {row, column} there are at least three empty t-orthogonals, so there are empty t-orthogonals at even distance. Thus we can make the following definition: among all even distances between parallel empty t-orthogonals, let m(t, 0) be the maximum.
As D is bichromatic and n > 4, there exist parallel empty orthogonals at odd distance. By Lemma 6, for each t ∈ {row, column} we can define m(t, 1) to be the maximum such odd distance. As m(t, 0) and m(t, 1) have different parity they are not equal; given D, we For t ∈ {row, column}, For t ∈ {row, column} and any positive integer d, let P(t, d) be the partition of E(t) into arithmetic progressions with step d and maximal length. Let o(t, d) be the number of progressions in P(t, d) having odd length. Let f (t, d) be the number of pairs of empty t-orthogonals at distance d.
Theorem 8. Let n be an even integer, n > 4, and let D be a bichromatic dominating set of size n/2 for Q n . For t ∈ {row, column}, let d be a positive integer such that f (t, d) > 0. Then any two empty t-orthogonals at distance d contain a set S(t, d) of |E(t)| + o(t, d) needy squares, with no three of these squares covered by any one square of D.
Proof. Let h, h + d be the indices of two empty t-orthogonals. Let S be the set of 2|E(t)| squares at which empty t-orthogonals meet t-orthogonals h, h + d. All squares of S are needy; we wish to reduce S to a subset S(t, d) no three of whose squares are covered by any one square of D. It suffices to arrange that no diagonal contains two squares of S(t, d). To do this, for any square of side length d whose corner squares are in t-orthogonals h, h + d and also in empty t-orthogonals, we remove two corner squares adjacent along a t-orthogonal, as follows.
For each arithmetic progression A = {i, i + d, . . . , i + (l − 1)d} of length l in P(t, d), we remove the squares in t-orthogonals h, h+d that are in t-orthogonals i+jd for odd j. Thus we remove l−1 squares from progressions of odd length l and l squares from progressions of even length l.
Corollary 9. Let D be a bichromatic dominating set of size n/2 for Q n . Let t ∈ {row, column} and let d be an odd integer such that f (t, d) > 0. Then for each q ∈ Z 2 , Proof. Since d is odd, each empty type t orthogonal that meets the set S(t, d) of Theorem 8 does so at one even and one odd square. Thus S(t, d) contains equal numbers of squares of each parity. As a square of parity q in D covers at most two squares in S(t, d), each of parity q, 2|D q | |S(t, d)|/2. By Theorem 8, |S(t, d)| = |E(t)| + o(t, d); the conclusion then follows.
Lemma 10. Let n be an even integer, n > 4, and let D be a bichromatic dominating set of size n/2 for Q n . Then |E(t)| = (n/2) + |L(t)|.
By the definition of m(t, v t ) there is at least one pair of indices of empty t-orthogonals at distance m(t, v t ). If there were two such pairs, the lowest and highest of the indices involved would come from a pair of empty t-orthogonals at distance greater than m(t, v t ), a contradiction.
Thus there is exactly one such pair, implying that the maximal arithmetic progressions If there were three pairs of indices of empty t-orthogonals at distance m(t, v t ), two pairs would have lower index of the same parity, and the lowest and highest indices involved in these pairs would come from a pair of empty t-orthogonals at distance of parity v t and greater than m(t, v t ), a contradiction.
If either there is one such pair, or there are two coming from a progression of length three, then o(t, m(t, v t )) = |E(t)| − 2. Otherwise there are two pairs coming from two (maximal length) progressions of length two, so o(t, m(t, v t )) = |E(t)| − 4.
If H(t) ∈ {−1, 0} and k is even, The first possibility and the preceding ones are included in (A), so assume the second occurs. By the definition of H, |L(t)| = 0 and h = 2. The latter implies by Lemma 10 that m(t, 1) = m(t, v t ). That is, v t = 0, which means m(t, 0) > m(t, 1).
We may then reverse the roles of t and t to conclude that |L(t)| = 0, so W orth (D) = 0, and m(t, 0) > m(t, 1).
3 Monochromatic dominating sets of size k for Q 2k Definition 12. For each positive integer n, let γ m (Q n ) be the minimum size of a monochromatic dominating set of Q n . For n ∈ {4, 5, 6}, let i m (Q n ) be the minimum size of an independent monochromatic dominating set of Q n (such sets do not exist for n ∈ {4, 5, 6}).
Proposition 13. Let n be a positive integer and D a dominating set of Q n of size n/2 . Then D is monochromatic if and only if it is possible to place the board so that the squares of D include exactly one from every even-indexed orthogonal.
Proof. The converse direction is clear, so assume that D is a monochromatic dominating set of Q n of size n/2 . Choose a square in D and place the board so the center of that square is the origin. Then every square of D is even, so no odd square of the board is diagonally covered. If there is an empty orthogonal with even index, we may assume without loss of generality that it is a column. Then every odd square of that column is covered along its row, which has odd index. So each of the n/2 or more odd-indexed rows is occupied, as well as row 0, but this cannot be done by n/2 squares. Thus each of the even-indexed rows and columns is occupied by D, and since |D| = n/2 there must be n/2 of each, each occupied once.
Definition 14. Let n be an even positive integer such that Q n has a monochromatic dominating set D of size k = n/2. By Proposition 13, we may place the board in the Cartesian plane with upper right corner square (k, k) and with the orthogonals occupied by D exactly those with the same parity as k. We assume this standard placement of board and dominating set in what follows.
Each odd square lies in exactly one orthogonal whose index has the same parity p as n/2, thus is covered exactly once by D. Each square (x, y) with x ≡ y ≡ p (mod 2) is covered twice orthogonally by D. It remains to see how D covers those squares (x, y) with x ≡ y ≡ p (mod 2). The set of these squares induces a copy U of Q n/2 that is symmetric across row 0 and across column 0; this symmetry is useful later.
Proof. Suppose that γ m (Q n ) = n/2. The first assertion is clear. If n ∈ {4, 12} then Theorem 2 implies γ(Q n−1 ) n/2. By Proposition 13 there exist an edge row and an edge column of Q n that do not contain any squares of D. Thus D dominates a copy of Q n−1 , so γ(Q n−1 ) = n/2. If n ∈ {2, 10} then γ(Q n+1 ) (n/2) + 1 by Theorem 2. Adding an edge row and edge column to Q n and adding the new corner square to D then shows γ(Q n+1 ) = (n/2) + 1.
Definitions. Let n be an even positive integer and let D be a monochromatic dominating set of Q n with size n/2 with standard orientation.
Among the integers congruent to 0 modulo 4, let d 0 be the least that is the absolute value of the index of an unoccupied difference diagonal and let s 0 be the least that is the absolute value of the index of an unoccupied sum diagonal.
Among the integers congruent to 2 modulo 4, let d 2 be the least that is the absolute value of the index of an unoccupied difference diagonal and let s 2 be the least that is the absolute value of the index of an unoccupied sum diagonal.
Let n > 2 be an even integer and let D be a monochromatic dominating set of Q n with size k = n/2 that has standard orientation.
As row 1 − k and column 1 − k are not occupied by D, the square (1 − k, 1 − k) can only be covered by a square in difference diagonal 0, which thus contains a square of D.
Assume for purposes of contradiction that sum diagonal 0 does not contain a square of D. For j = 1, . . . , (k+1)/2 , we examine how D covers the squares ±(2j−1−k, k+1−2j) of sum diagonal 0. These squares are in empty orthogonals, so must be covered along their difference diagonals.
With j = 1 we get the squares ±(1 − k, k − 1) which are in difference diagonals ±(2k − 2). Each of these diagonals has only one square in orthogonals with the parity of k, so those squares, which are (k, 2 − k) and (2 − k, k), are in D. If k > 2 we go to j = 2, and see that D must occupy difference diagonals ±(2k − 6). However, the only squares of those diagonals in orthogonals of the parity of k and not already occupied by D are (k − 2, 4 − k) and (4 − k, k − 2), which thus are in D.
Continuing, we find that for successive values of j, the difference diagonals ±(2k + 2 − 4j) that must be occupied each have only one square that lies in orthogonals of the parity of k that are not already occupied by D, namely (2j −k, k +2−2j) and (k +2−2j, 2j −k). So these squares are in D.
This finally implies that the members of D are exactly the squares in sum diagonal 2 whose orthogonals have the parity of k. Then if k is even, all difference diagonals of squares of D have indices congruent to 2 modulo 4, so D does not contain a square of difference diagonal 0, contradicting a previous conclusion. If k is odd, all difference diagonals of squares of D have indices congruent to 0 modulo 4. The needy square (−2, 0) (on the board since k 3) is thus not covered by D, a contradiction. Therefore D contains a square of sum diagonal 0.
(B) This is easily verified. (C) As there are n/2 squares in D, this follows from (B).
For any even n, the definitions of d 0 and d 2 imply that the left sides of the inequalities (7) are congruent to 2 modulo 4. But for n ≡ 0 (mod 4), the right sides of those inequalities are congruent to 0 modulo 4. Thus we have d 0 + d 2 n + 2 and s 0 + s 2 n + 2 if n ≡ 0 (mod 4).
We first consider the case n ≡ 0 (mod 4). Then k = n 2 ≡ 0 (mod 2), so we need to see how the odd-odd squares are covered. By the definitions of d 0 and s 2 , at least one of the difference diagonals with indices ±d 0 and at least one of the sum diagonals with indices ±s 2 are unoccupied by D. If difference diagonal d 0 and sum diagonal s 2 are unoccupied, then the odd-odd square ( s 2 −d 0 2 , s 2 +d 0 2 ) at which these diagonals meet must be off the board, implying s 2 +d 0 2 k. However the two sides of this inequality have different parity, implying that s 2 +d 0 2 k + 1, so d 0 + s 2 n + 2. By the symmetry of U , the other possibilities of unoccupied diagonals imply the same result. An argument with d 2 and s 0 works similarly, so we have d 0 + s 2 n + 2 and d 2 + s 0 n + 2 for n ≡ 0 (mod 4).
Theorem 17. Among the positive integers n, n ≡ 0 (mod 4), only for n = 4, 12 does Q n have a monochromatic dominating set of size n/2. (8), (9) we have d 2 s 2 . The other pairs of inequalities from (8), (9) then imply d 0 = s 0 and d 2 = s 2 , and d 0 + d 2 = n + 2 and s 0 + s 2 = n + 2. From Lemma 16(C), e d = n 2 − ( d 0 2 − 1) − ( d 2 2 − 1), which here reduces to 1. Similarly, this is the value of e s . By the linear constraint (1), the sum over D of all difference diagonal indices is zero. Also the sum of the required difference diagonal indices is zero, so the single excess difference diagonal index is zero, repeating a required difference diagonal index. The sum of all sum diagonal indices is 2k and the sum of the required sum diagonal indices is zero, so the single excess sum diagonal index is 2k. The only board square on sum diagonal 2k is (k, k), so this square is in D. It also is in difference diagonal 0.

Proof. From the first inequalities in
However, as D also contains another square in difference diagonal 0, the set D = D \ {(k, k)} dominates the copy of Q 2k−1 obtained by removing the rightmost column and top row of Q 2k . By Theorem 2, this implies that 2k − 1 is either 3 or 11, so n = 2k is either 4 or 12.
Conversely From this point we will assume n ≡ 2 (mod 4), with k = n 2 ≡ 1 (mod 2). We ask how even-even squares are covered here. By the definitions of d 0 and s 0 , at least one of the difference diagonals with indices ±d 0 and at least one of the sum diagonals with indices ±s 0 are unoccupied by D. If difference diagonal d 0 and sum diagonal s 0 are unoccupied, then the even-even square ( s 0 −d 0 2 , d 0 +s 0 2 ) at which these diagonals meet must be off the board, implying d 0 +s 0 2 k. The two sides of this inequality have different parity, so d 0 +s 0 2 k + 1, implying d 0 + s 0 n + 2. By symmetry, the other possibilities of unoccupied diagonals imply the same result. An argument with d 2 and s 2 works similarly, so we have d 0 + s 0 n + 2 and d 2 + s 2 n + 2 for n ≡ 2 (mod 4).
Theorem 18. Let n > 2 be an integer, n ≡ 2 (mod 4), and let D be a monochromatic dominating set of Q n of size n/2 with standard orientation. Then D has no excess difference diagonals; D has two excess sum diagonals, say with indices a 0 and a 2 , where a i ≡ i (mod 4) and a 0 + a 2 = n. Here d 0 , d 2 , s 0 , s 2 are related by the equations s 0 + s 2 = n, d 0 = s 2 + 2, d 2 = s 0 + 2. Let k = n/2, d = |a 0 − a 2 |/2, and e = |s 0 − s 2 |/2. Then Proof. Since n ≡ 2 (mod 4), Lemma 16(C) implies e d and e s are even. Using (10)  2. As the sum of the required sum diagonal indices is zero, and the sum of all sum diagonal indices equals n by (1), there exist excess sum diagonal indices and their sum is n. So we have two nonnegative even integers e d and e s whose sum does not exceed 2, and e s is not zero: we may conclude that e d = 0 and e s = 2, and the two excess sum diagonal indices a 0 , a 2 have sum n. As all indices of occupied orthogonals are odd, all indices of occupied diagonals are even, so the indices a 0 , a 2 of the excess sum diagonals are even. From a 0 + a 2 = n ≡ 2 (mod 4), we see that just one of them is congruent to 0 modulo 4. We will assume henceforth that a 0 ≡ 0 (mod 4) and then a 2 ≡ 2 (mod 4). As n is the greatest possible index for a sum diagonal, necessarily a 0 , a 2 n, which implies a 0 , a 2 0.
Proposition 19. The only positive even integers n for which i m (Q n ) = n/2 are n = 2, 10.
Proof. For n ≡ 0 (mod 4), Theorem 17 says that only for n = 4, 12 does Q n have a monochromatic dominating set of size n/2, and the proof of Theorem 17 shows that each such set has two squares on difference diagonal 0, so is dependent. Suppose then that n ≡ 2 (mod 4) and D is a monochromatic independent dominating set of Q n with standard orientation. If n = 2 then D = {(1, 1)}, so we will assume n 6. By independence the excess sum diagonal indices a 0 , a 2 are distinct from the required sum diagonal indices, which implies a 0 s 0 and a 2 s 2 . Since a 0 + a 2 = n = s 0 + s 2 , we have a 0 = s 0 and a 2 = s 2 .
Let l = (n/2)+1. Make a copy of Q n+1 by adjoining a row and a column, each indexed l, to Q n . We show that D dominates this Q n+1 .
Change the coordinates by subtracting one from each row and column index, thus moving the origin of the coordinate system to the square formerly labeled (1, 1). It is then not difficult to verify that for each of the four types of lines, the sets of indices of occupied lines are symmetric across zero. This implies that the set of undominated squares is symmetric under rotation by a half-turn about the origin. But we know that the left and bottom edges are covered, so also the top and right-hand edges are.
Since a 0 , a 2 are even and not congruent modulo 4, d = |a 0 − a 2 |/2 is odd. Similarly, e = |s 0 − s 2 |/2 is odd. We may regard each triple (k, e, d) satisfying (11), with k, e, d each an odd integer and 1 d, e k, as a candidate for solutions of γ m (Q 2k ) = k. From (11) and 1 d k it follows that This implies that for any odd integer k 3 there is at most one value of e giving a candidate (k, e, d), and if there is one, it is (k 2 − 2k)/3 .  The argument leading to (11) is not valid for k = 1, but we can take the solution (k, e, d) = (1, 1, 1) of (11) as corresponding to the dominating set {(1, 1)} of Q 2 .
The fifteen candidates with least values of k are shown in Table 1.
Below we identify six infinite sequences of candidates. Each sequence is derived from solutions of an equation of Fermat-Pell type. The following lemma collects the facts we need about such equations; proofs may be found in [1].
Lemma 20. Let C be a fixed integer and consider the Fermat-Pell equation (A) Pair (x, y) is a solution of (12) if and only if H(x, y) = (2x + 3y, x + 2y) is; (B) There are only finitely many positive integral solutions of (12) (the minimal solutions) that cannot be obtained by (A) from smaller positive integral solutions.
Thus the positive integral solutions of (12) are the members of a finite number of infinite sequences derived from the minimal solutions by (A).
The equation X 2 − 3Y 2 = −11 has two minimal positive solutions (x, y) = (1, 2) and (4,3), and again iteration with H gives solutions whose x-values alternate in parity, so we obtain the desired solutions with even x by applying the iteration of the case d = k. Here we want  H 2i+1 (1, 2) and H 2i (4, 3) for i 0.
By (14) and (15), For each candidate (k, e, d) with k 3, we will construct an upper bound F (k) for the left side of (16). For some k we get F (k) < 0, implying γ m (Q 2k ) > k.
We begin by constructing a sum F 1 (k) involving the absolute values of the indices of the required diagonals. Note that since all occupied orthogonals have odd indices, for each occupied square one of its diagonals has index congruent to 0 modulo 4 and the other has index congruent to 2 modulo 4. So in what follows we seek to pair off absolute sum diagonal indices congruent to 0 mod 4 (respectively 2 mod 4) with absolute difference diagonal indices congruent to 2 mod 4 (respectively 0 mod 4) to achieve an upper bound for (x,y)∈D max{|y − x|, |y + x|}.
We examine four cases. As they are similar, we only describe the first in detail.
Case s 2 ≡ −2 (mod 8). Then there is a positive integer h with s 2 = 8h − 2 and d 0 = 8h. The number of indices of occupied difference diagonals that are congruent to 0 modulo 4 is 4h − 1, and necessarily this is also the number of indices of occupied sum diagonals that are congruent to 2 modulo 4; this includes the extra sum diagonal index a 2 . In this and the next case, we treat a 2 as the least absolute sum diagonal index for the moment.
To maximize the terms contributed to the bound S max here, we pair off the top 2h absolute difference diagonal indices with the bottom 2h absolute sum diagonal indices (including a 2 ), and the top 2h − 1 absolute sum diagonal indices with the bottom 2h − 1 absolute difference diagonal indices. This gives The least value used comes from a sum diagonal with absolute index (s 2 /2) − 1.
To maximize the terms contributed to the bound S max here, we pair off the top 2h + 1 absolute difference diagonal indices with the bottom 2h + 1 absolute sum diagonal indices (including a 2 ), and the top 2h absolute sum diagonal indices with the bottom 2h absolute difference diagonal indices. This gives the sum (s 2 − 2)(3s 2 + 2)/8 as in the previous case. The least value used comes from a difference diagonal with absolute index (s 2 /2) − 1.
Case s 0 ≡ 0 (mod 8). There is a positive integer h with s 0 = 8h and d 2 = 8h + 2. In this and the next case, we treat a 0 as the least absolute sum diagonal index temporarily.
To maximize the terms contributed to the bound S max here, we pair off the top 2h absolute difference diagonal indices with the bottom 2h absolute sum diagonal indices (including a 0 ), and the top 2h absolute sum diagonal indices with the bottom 2h absolute difference diagonal indices. This gives The least value used comes from a sum diagonal with absolute index s 0 /2.
To maximize the terms contributed to the bound S max here, we pair off the top 2h + 2 absolute difference diagonal indices with the bottom 2h + 2 absolute sum diagonal indices (including a 0 ), and the top 2h absolute sum diagonal indices with the bottom 2h absolute difference diagonal indices. This gives the sum s 0 (3s 0 − 4)/8 as in the previous case. The least value used comes from a difference diagonal with absolute index s 0 /2.
In our construction of the sum F 1 (k) we ignored the extra sum diagonal indices a 0 , a 2 by assuming they had smaller absolute values than all those used in the sum. Now we consider a 0 , a 2 .
If a 0 > (s 2 /2) − 1 then we need to replace the lowest index used in the sum in the appropriate one of the first two cases above, which is (s 2 /2)−1, with a 0 . (In the first case, we just replace the absolute sum diagonal index (s 2 /2) − 1 with absolute sum diagonal index a 0 . In the second case, we can replace the absolute sum diagonal index which was paired with, and less than, the absolute difference diagonal index (s 2 /2)−1, with absolute sum diagonal index a 0 .) And if a 2 > s 0 /2 then we need to replace the lowest index used in the sum in the appropriate one of the latter two cases above, which is s 0 /2, with a 2 . At least one of these changes is necessary, as the inequalities a 0 (s 2 /2) − 1 and a 2 s 0 /2 would imply n = a 0 + a 2 ((s 2 /2) − 1) + (s 0 /2) = (n/2) − 1, which is not possible.
With the changes described, we can find the desired function F and know that those candidates (k, e, d) with F (k) < 0 have γ m (Q 2k ) > k.
Let d = k−2. We will need to consider two possibilities for the initial pair (k 0 , e 0 ); from each, the recursion (18) allows us to find an infinite sequence of candidates, and together these include all candidates with d = k − 2. With j = 1 in (17), Starting with (k 0 , e 0 ) = (9, 5), we may use (18) to obtain (s 0 ) i = k i + (−1) i+1 e i , and then F (k i ) = [(−1) i e i − 1]/2 for i 0. So here γ m (Q 2k i ) > k i for odd i.
In Table 2 we show the number of candidates in certain ranges, along with the number for which we can already conclude γ m (Q 2k ) = k is not possible (because either F (k) < 0, or d = e but k > 5), and the number coming from Fermat-Pell equations.

Domination numbers and sets
For each of the sets mentioned below to show that a parameter of Q n has value (n/2) + 1 or is in the set {n/2, (n/2) + 1}, there is p ∈ Z 2 such that n/2 of the squares (x, y) in the set have x ≡ y ≡ p (mod 2) and jointly occupy all orthogonals with indices of parity p, and one square has x ≡ y ≡ p (mod 2).