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\begin{center}
\vskip 1cm{\LARGE\bf Self-Complementary  Magic Squares\\
\vskip .1in
of Singly Even Orders}
\vskip 1cm
\large
Gek L. Chia\\
Department of  Mathematical and  Actuarial Sciences \\
Universiti Tunku Abdul Rahman,  Sungai Long Campus,\\
43000 Cheras,  Selangor \\
Malaysia\\
\href{mailto: chiagl@utar.edu.my}{\tt chiagl@utar.edu.my} \\
\ \\
Wai Keong Kok\\
Department of Mathematical and Data Science\\
Tunku Abdul Rahman University of Management and Technology\\
53300 Kuala Lumpur \\
Malaysia\\
\href{mailto:kokwk@tarc.edu.my}{\tt kokwk@tarc.edu.my}
\end{center}

\vskip .2 in

\begin{abstract}
A {\em self-complementary magic square  of order $n$\/} is a magic square $M$ whose entries consist of the first $n^2$ natural numbers such that when each entry $i$ of $M$ is replaced by $n^2+1-i$, the resulting square  is equivalent to $M$ (under rotation or reflection).  The purpose of this paper is twofold.  For any given singly even natural number $n$, (i) we extend the method of Strachey  to produce many magic squares (instead of just one)  of order $n$, and (ii) by making a variation of this extended method, we construct many self-complementary magic squares of order $n$.
\end{abstract}


\section{Introduction}
A {\em magic square} $M$ is an $n\times n$  array of integers (which may overlap) so that the sum of the entries in each row, column and the diagonal is a constant. The constant sum is called the {\em magic sum\/}. In the case  that the entries of $M$ are the consecutive integers from $1, 2, \ldots, n^2$, then $M$ is said to be {\em of order $n$}. It is easy to see that  the magic sum of an $n$-order magic square is $\frac{n(n^2 + 1)}{2}$. A {\em doubly even order magic square\/} is a magic square matrix of order $n$, where $n$ is divisible by $4$.  A {\em singly even order magic square\/} is a magic square matrix of order $n$, where $n$ is even but not divisible by $4$.




Suppose $M = (a_{i,j})$ is a magic square of order $n$. Then $M$ is said to be {\em ro-symmetrical\/} if
$a_{i,j} + a_{n+1-i, n+1-j} = n^2 +1$ for all $1 \leq i, j\leq n$. It is known that the magic square constructed using the De la Loub\`{e}re method is ro-symmetrical and that ro-symmetrical magic square of singly even order does not exist (see \cite[p.~203]{baco:refer}).


On  the other hand, $M$ is said to be {\em ref-symmetrical\/} if $a_{i,j} + a_{n+1-i, j} = n^2 +1$ for all $1 \leq i, j\leq n$.
Alternatively, these two notions maybe expressed in the following way.


 Let $J_k$ denote the $k \times k$ matrix where all entries are $1$ and  let $\sigma(M)$ (respectively  $\rho(M)$) denote the $180$-degree clockwise rotation  on $M$ (respectively  reflection on the central horizontal of $M$). Then $M$ is ro-symmetrical (respectively ref-symmetrical)  if $M + \sigma(M) = (n^2+1) J_n$ (respectively $M + \rho(M) = (n^2+1) J_n$).
 Note that equivalently, in this definition, we could replace  $\rho(M)$ by $\pi(M)$  which denotes  the  reflection on the central vertical of $M$, in which case we have   $a_{i,j} + a_{i, n+1-j} = n^2 +1$ for all $1 \leq i, j\leq n$ instead. % However, in this paper rotation will refer to $180$-degree clockwise rotation and that reflection will refer to $180$-degree reflection on the central  horizontal of $M$.




The methods of construction for  magic squares of singly even order appear to be relatively scarce when compared to magic squares of other orders. The earliest record for singly even order case dates back to a letter of Strachey to Coxeter in $1918$.  %Detail of this construction is described in \cite[pp.~196--199]{baco:refer}.
The magic square of odd order  constructed by the De la Loub\`{e}re method is used in the Strachey method. For ease of reference, the De la Loub\`{e}re and the Strachey methods of construction are given in Section \ref{ls}. The magic squares which are constructed by these two methods are called the
{\em De la Loub\`{e}re square\/} and the {\em  Strachey square\/} respectively.


 Recently, in \cite{chialee:refer}, a method of producing a magic square of order $2(2m+1)$ with the reflectional symmetrical  property was presented. Earlier, in \cite{cc:refer}, magic squares of singly even orders were constructed which involves  a special kind of composition (called the Yang-Hui composition). While the Strachey method produces only one magic square for a given singly even order, we show (in Section \ref{stra}) that the method can be extended to produce (i) many magic squares of a given singly even order,  and (ii) we can use any magic square of odd order in the construction (instead of restricting to the De la Loub\`{e}re  magic square).



Two magic squares $M_1$ and $M_2$ are said to be {\em equivalent\/} if $\phi(M_1)=M_2 $ for some $\phi \in D_4$ where $D_4$ denotes the dihedral group of order 8. Suppose  $M$ is a magic square of order $n$.  The {\em complement\/} of $M$ is the magic square (of order $n$) obtained by replacing each entry $i$ of $M$ by $n^2+1-i$. We say that $M$ is {\em self-complementary\/} if $M$ is equivalent to its complement.  Self-complementary magic squares of order $n$ have been characterized in \cite{chialee:refer} where it was shown that (i) if   $n$ is odd, then $M$ is self-complementary if and only if $M$ is ro-symmetrical,  whereas (ii) if $n$ is even, then $M$ is self-complementary if and only if either $M$ is ref-symmetrical or else $M$ is  ro-symmetrical.    Note that if $M$ is ro-symmetrical (respectively ref-symmetrical), then  $\sigma(M)= \overline{M} $ (respectively $\rho(M)= \overline{M}$), where $\overline{M}$  denotes  the complement of $M$.



Subsequently, in \cite{chia:refer}, a construction for self-complementary magic squares of doubly even order was given; it converts a ro-symmetrical magic square (constructed using a well-known construction called the Generalized Doubly Even Method,  \cite[pp.~199--200]{baco:refer}   and  \cite[p.~527, Section 34.21]{km:refer}) into a ref-symmetrical magic square.
In quite the same spirit as in  \cite{chia:refer}, in Section \ref{4m+2}, we provide  a new method of constructing self-complementary magic squares of singly even order.  This time we make use of a ro-symmetrical magic square  (which may be constructed using the De la Loub\`{e}re method) on a variation of the extended Strachey method  to produce a ref-symmetrical square of singly even order.

\section{De la Loub\`{e}re and Strachey squares} \label{ls}

Let $n$ be an odd integer. A magic square of order $n$ can be constructed by the {\em De la Loub\`{e}re method\/} in which the sequence of integers $1, 2, \ldots, n^2$ are filled successively  into an $n\times n$ square according to the following rule.

\begin{description}
  \item[L(i)] Put $1$  in the middle cell of the top row of square; fill the next integer  diagonally in the north-east direction.
  \item[L(ii)] When the top row is reached, the next integer is written in the bottom row as if it came immediately above the top row (for example, the integers $2$ and $9$ in the magic square below for the case $n=5$).
 \item[L(iii)] When the last column is reached, the next integer is written in the first column as if it immediately succeeded the last column (for example, the integers $4$ and $10$ in the magic square below).
 \item[L(iv)] When the next cell to be filled is already occupied or when the top-right cell is reached, the move drops
     directly below it  and continue to mount again (for example, the integers $6, 16$ and $21$ in the magic square below).
\end{description}
$$
 \begin{bmatrix}
  17 & 24 & 1  & 8  & 15\\
  23 &  5 & 7  & 14 & 16\\
   4 &  6 & 13 & 20 & 22\\
  10 & 12 & 19 & 21 &  3\\
  11 & 18 & 25 &  2 &  9
 \end{bmatrix} $$

Now consider the case $n=2(2m+1)$. A magic square of order $n$ can be constructed by the {\em Strachey method}. In this case, we begin by partitioning an $n \times n$ square into four $(2m+1) \times (2m+1)$ subsquares $A, B, C$ and $D$ as shown below. Then apply the following rule.


\begin{description}
  \item[S(i)] Let $A$ be the De la Loub\`{e}re square of order $2m+1$ and let $B = A + (2m+1)^2 J_{2m+1}$, $C= A + 2(2m+1)^2 J_{2m+1}$ and $D= A + 3(2m+1)^2 J_{2m+1}$.
  \item[S(ii)] Divide $A$ as shown below where $A_1, A_3$ are $m \times m$ subsquares which are at the upper left and the bottom left corners of $A$ respectively, $A_4$ is the first cell of the middle row of $A$ while $A_2$ is the $1 \times m$ rectangle in the middle row, next to $A_4$. Divide $D$ in a similar manner and obtain $D_1, D_2, D_3$ and $D_4$.
 \item[S(iii)] Interchange $A_i$ with $D_i$ for $i=1,2,3$.
 \item[S(iv)] Interchange the last $m-1$ columns of $C$ with those of $B$.
\end{description}
\[
\begin{tabular}{|ccc|ccc|}
  \hline
  & & & & & \\
  & $A$  & & & $C$ &\\
  & & & & & \\
  \hline
  & & & & & \\
  & $D$ & & & $B$ &\\
  & & & & & \\
  \hline
\end{tabular}
\; \; \; \; \; \; \; \; \; \; \; \; \; \; \;
A= \begin{tabular}{|llll|ccccccc|}
  \hline
  & & & & & & && & & \\
  & & & & & & && & & \\
  &$A_1$& & & & & && & & \\
  & & & & & & && & & \\
  & & & & & & && & & \\
  \hline
  $A_4$\vline  &&\multicolumn{2}{l}{$A_2$} & & \vline & && & & \\
  \hline
  & & & & & & && & & \\
  & & & & & & && & & \\
  &$A_3$& & & & & && & & \\
  & & & & & & && & & \\
  & & & & & & && & & \\
  \hline
\end{tabular}
\]

For $n=10$, the Strachey method yields the following magic square of order $10$.
$$
 \begin{bmatrix}
  92 & 99 & 1  & 8  & 15 & 67 & 74 & 51 & 58 & 40 \\
  98 & 80 & 7  & 14 & 16 & 73 & 55 & 57 & 64 & 41 \\
  4  & 81 & 88 & 20 & 22 & 54 & 56 & 63 & 70 & 47 \\
  85 & 87 & 19 & 21 & 3  & 60 & 62 & 69 & 71 & 28 \\
  86 & 93 & 25 & 2  & 9  & 61 & 68 & 75 & 52 & 34 \\
  17 & 24 & 76 & 83 & 90 & 42 & 49 & 26 & 33 & 65 \\
  23 & 5  & 82 & 89 & 91 & 48 & 30 & 32 & 39 & 66 \\
  79 & 6  & 13 & 95 & 97 & 29 & 31 & 38 & 45 & 72 \\
  10 & 12 & 94 & 96 & 78 & 35 & 37 & 44 & 46 & 53 \\
  11 & 18 & 100& 77 & 84 & 36 & 43 & 50 & 27 & 59
 \end{bmatrix}.
$$









\section{Beyond Strachey}\label{stra}

If every entry of an $n$-th order magic square is reduced by $1$, the resulting square is called a  {\em reduced magic square}.


\begin{definition}  \label{b}
Suppose $B_1$ and $B_2$ are reduced magic squares of order $2m+1$. Let $B$ denote the $2(2m+1) \times 2(2m+1)$    magic square defined by
$$
 B =
 \begin{bmatrix}
  B_1 & B_2 \\
  B_1 & B_2
 \end{bmatrix}.
$$
\end{definition}


 \begin{example} Let $B_1$ be the reduced magic square of the  De la Loub\`{e}re magic square of order $5$, and let $B_2$ be the square obtained from $B_1$  by taking a $90$-degree clockwise rotation. Then
$$
 B =
 \begin{bmatrix}
  16 & 23 & 0  & 7  & 14 & 10 & 9  & 3  & 22 & 16 \\
  22 & 4  & 6  & 13 & 15 & 17 & 11 & 5  & 4  & 23 \\
  3  & 5  & 12 & 19 & 21 & 24 & 18 & 12 & 6  & 0 \\
  9  & 11 & 18 & 20 & 2  & 1  & 20 & 19 & 13 & 7 \\
  10 & 17 & 24 & 1  & 8  & 8  & 2  & 21 & 15 & 14 \\
  16 & 23 & 0  & 7  & 14 & 10 & 9  & 3  & 22 & 16 \\
  22 & 4  & 6  & 13 & 15 & 17 & 11 & 5  & 4  & 23 \\
  3  & 5  & 12 & 19 & 21 & 24 & 18 & 12 & 6  & 0 \\
  9  & 11 & 18 & 20 & 2  & 1  & 20 & 19 & 13 & 7 \\
  10 & 17 & 24 & 1  & 8  & 8  & 2  & 21 & 15 & 14
 \end{bmatrix}.
$$
\end{example}

\begin{definition}  \label{a}
 Let $A$ denote the $2(2m+1) \times 2(2m+1)$     square defined by
$$
 A =
 \begin{bmatrix}
  A_0 & A_2 \\
  A_3 & A_1
 \end{bmatrix}
 $$
 where

\begin{description}
  \item[(i)   ] $A_0$ is a $(2m+1) \times (2m+1)$ matrix whose entries consist of only $0$ and $3$ such that each row has only $(m+1)$ $0$'s while each diagonal has only $(m+1)$ $3$'s,
  \item[(ii) ] $A_1$ is a $(2m+1) \times (2m+1)$ matrix whose entries consist of only $1$ and $2$ such that each row and each diagonal has only $(m+2)$ $1$'s,  and
  \item[(iii)] $A_2 = 3J_{2m+1} - A_1$  and $A_3 = 3J_{2m+1} -A_0$.
\end{description}
  \end{definition}

It is easy to construct the matrices $A_0$ and $A_1$.  Figure \ref{a00} depicts $3$ examples of the matrix  $A_0$  with $m=2$ while Figure \ref{a11} depicts $3$ examples of the matrix  $A_1$. More general constructions of $A_0$ and $A_1$ will be discussed in Section \ref{remark}.



\begin{figure}[h]
$$
 \begin{bmatrix}
  3 & 3 & 0 & 0 & 0\\
  3 & 3 & 0 & 0 & 0\\
  0 & 3 & 3 & 0 & 0\\
  3 & 3 & 0 & 0 & 0\\
  3 & 3 & 0 & 0 & 0
 \end{bmatrix}     \qquad \qquad
 \begin{bmatrix}
  0 & 3 & 0 & 0 & 3\\
  0 & 3 & 0 & 0 & 3\\
  0 & 0 & 3 & 3 & 0\\
  0 & 3 & 0 & 0 & 3\\
  0 & 3 & 0 & 0 & 3
 \end{bmatrix}        \qquad \qquad
 \begin{bmatrix}
  0 & 0 & 0 & 3 & 3\\
  0 & 0 & 0 & 3 & 3\\
  0 & 0 & 3 & 3 & 0\\
  0 & 0 & 0 & 3 & 3\\
  0 & 0 & 0 & 3 & 3
 \end{bmatrix}
 $$

\qquad \qquad \qquad \qquad (i) \qquad \qquad \qquad \qquad \qquad (ii)  \qquad  \qquad \qquad \qquad \ \ \quad (iii)

  \caption{ $A_0$  with $m=2.$} \label{a00}
\end{figure}


\begin{figure}[h]
$$
 \begin{bmatrix}
  1 & 1 & 1 & 1 & 2\\
  1 & 1 & 1 & 1 & 2\\
  1 & 1 & 1 & 1 & 2\\
  1 & 1 & 1 & 1 & 2\\
  1 & 1 & 1 & 1 & 2
  \end{bmatrix} \qquad \qquad
 \begin{bmatrix}
  1 & 1 & 1 & 2 & 1\\
  1 & 1 & 1 & 2 & 1\\
  2 & 1 & 1 & 1 & 1\\
  1 & 1 & 1 & 2 & 1\\
  1 & 1 & 1 & 2 & 1
 \end{bmatrix} \qquad \qquad
 \begin{bmatrix}
  2 & 1 & 1 & 1 & 1\\
  2 & 1 & 1 & 1 & 1\\
  2 & 1 & 1 & 1 & 1\\
  2 & 1 & 1 & 1 & 1\\
  2 & 1 & 1 & 1 & 1
 \end{bmatrix}
 $$

\qquad \qquad \qquad \qquad (i) \qquad \qquad \qquad \qquad \qquad (ii)  \qquad  \qquad \qquad \qquad \ \ \quad (iii)

  \caption{ $A_1$  with $m=2.$} \label{a11}
\end{figure}


\begin{example}
Suppose  $A_0$ is the matrix in Figure \ref{a00} (iii) and $A_1$ is the matrix in Figure \ref{a11} (iii). Then
  $$
 A =
 \begin{bmatrix}
  0 & 0 & 0  & 3  & 3 & 1 & 2  & 2  & 2 & 2 \\
  0 & 0 & 0  & 3  & 3 & 1 & 2  & 2  & 2 & 2 \\
  0 & 0 & 3  & 3  & 0 & 1 & 2  & 2  & 2 & 2 \\
  0 & 0 & 0  & 3  & 3 & 1 & 2  & 2  & 2 & 2 \\
  0 & 0 & 0  & 3  & 3 & 1 & 2  & 2  & 2 & 2 \\
  3 & 3 & 3  & 0  & 0 & 2 & 1  & 1  & 1 & 1 \\
  3 & 3 & 3  & 0  & 0 & 2 & 1  & 1  & 1 & 1 \\
  3 & 3 & 0  & 0  & 3 & 2 & 1  & 1  & 1 & 1 \\
  3 & 3 & 3  & 0  & 0 & 2 & 1  & 1  & 1 & 1 \\
  3 & 3 & 3  & 0  & 0 & 2 & 1  & 1  & 1 & 1
 \end{bmatrix}.
$$
\end{example}


With  the matrices $A$ and $B$ as given in the above definitions, we  obtain a reduced magic square of order $2(2m+1)$ as follows.


\begin{definition}\label{ab} Suppose $A$ and $B$ are as defined in the Definitions \ref{a} and \ref{b} respectively.
Let $[ A; B]$ denote the $2(2m+1) \times 2(2m+1)$ matrix obtained by superimposing $A$ and $B$ so that the entries of $A$ are the radix digits while those of $B$ are the unit digits.
\end{definition}


\begin{example}
Suppose $A$ and  $B$ are the $10 \times 10$  matrices in Examples $2$ and $1$ respectively. Then
 $$
 [A; B] =
 \begin{bmatrix}
  {\bf 0} 16 &{\bf 0}23  & {\bf 0}\ 0 &{\bf 3}\ 7 &{\bf 3}14  &{\bf 1} 10 &{\bf 2}\ 9 &{\bf 2}\ 3 &{\bf 2}22  &{\bf 2}16 \\
  {\bf 0} 22 &{\bf 0}\ 4 & {\bf 0}\ 6 &{\bf 3}13  &{\bf 3}15  &{\bf 1} 17 &{\bf 2} 11 &{\bf 2}\ 5 &{\bf 2}\ 4 &{\bf 2}23 \\
  {\bf 0}\ 3 &{\bf 0}\ 5 & {\bf 3}12  &{\bf 3}19  &{\bf 0}21  &{\bf 1} 24 &{\bf 2} 18 &{\bf 2} 12 &{\bf 2}\ 6 &{\bf 2}\ 0 \\
  {\bf 0}\ 9 &{\bf 0}11  & {\bf 0}18  &{\bf 3}20  &{\bf 3}\ 2 &{\bf 1}\ 1 &{\bf 2} 20 &{\bf 2} 19 &{\bf 2}13  &{\bf 2}\ 7 \\
  {\bf 0} 10 &{\bf 0}17  & {\bf 0}24  &{\bf 3}\ 1 &{\bf 3}\ 8 &{\bf 1}\ 8 &{\bf 2}\ 2 &{\bf 2} 21 &{\bf 2}15  &{\bf 2}14 \\
  {\bf 3}16  &{\bf 3}23  & {\bf 3}\ 0 &{\bf 0}\ 7 &{\bf 0}14  &{\bf 2}10  &{\bf 1}\ 9 &{\bf 1}\ 3 &{\bf 1}22  &{\bf 1}16 \\
  {\bf 3}22  &{\bf 3}\ 4 & {\bf 3}\ 6 &{\bf 0}13  &{\bf 0}15  &{\bf 2}17  &{\bf 1}11  &{\bf 1}\ 5 &{\bf 1}\ 4 &{\bf 1}23 \\
  {\bf 3}\ 3 &{\bf 3}\ 5 & {\bf 0}12  &{\bf 0}19  &{\bf 3}21  &{\bf 2}24  &{\bf 1}18  &{\bf 1}12  &{\bf 1}\ 6 &{\bf 1}\ 0 \\
  {\bf 3}\ 9 &{\bf 3}11  & {\bf 3}18  &{\bf 0}20  &{\bf 0}\ 2 &{\bf 2}\ 1 &{\bf 1}20  &{\bf 1}19  &{\bf 1}13  &{\bf 1}\ 7 \\
  {\bf 3}10  &{\bf 3}17  & {\bf 3}24  &{\bf 0}\ 1 &{\bf 0}\ 8 &{\bf 2}\ 8 &{\bf 1}\ 2 &{\bf 1}21  &{\bf 1}15  &{\bf 1}14
 \end{bmatrix}.
 $$
\end{example}

Here, the boldface integers are the  radix digits. By taking the entries to be integers represented in base $5^2$,  we have the following reduced magic square of order $10$.

$$
  \begin{bmatrix}
   16 & 23  &  0  & 82  & 89  & 35 & 59 & 53 & 72  & 66 \\
   22 &  4  &  6  & 88  & 90  & 42 & 61 & 55 & 54  & 73 \\
    3 &  5  & 87  & 94  & 21  & 49 & 68 & 62 & 56  & 50 \\
    9 & 11  & 18  & 95  & 77  & 26 & 70 & 69 & 63  & 57 \\
   10 & 17  & 24  & 76  & 83  & 33 & 52 & 71 & 65  & 64 \\
   91 & 98  & 75  &  7  & 14  & 60 & 34 & 28 & 47  & 41 \\
   97 & 79  & 81  & 13  & 15  & 67 & 36 & 30 & 29  & 48 \\
   78 & 80  & 12  & 19  & 96  & 74 & 43 & 37 & 31  & 25 \\
   84 & 86  & 93  & 20  &  2  & 51 & 45 & 44 & 38  & 32 \\
   85 & 92  & 99  &  1  &  8  & 58 & 27 & 46 & 40  & 39
 \end{bmatrix}.
$$



\begin{theorem} \label{strachey}
Suppose $A$ and $B$ are as defined in the Definitions \ref{a} and \ref{b} respectively.  Then
$[A; B]$ yields a reduced magic square of order $2(2m+1)$ when its entries are considered as integers represented in base $(2m+1)^2$.
\begin{proof}
Clearly $B$ is a  magic square with magic sum $4m(m+1)(2m+1)$.

Clearly the column sum of $A$ is $3(2m+1)$.

 Note that the row sum of $A_0$ and likewise of  $A_1$ is $3m$. This implies that the row sum of $A_3$ and likewise of $A_2$ is $3(m+1)$. Hence the row sum of $A$ is  $3(2m+1)$.

Clearly the diagonal sum of $A_0$ is $3(m+1)$ and that of $A_1$ is $3m$. This implies that the diagonal sum of $A_3$ is $3m$ (since the diagonal has only $m$ $3$'s), and that the diagonal sum of $A_2$ is $3(m+1)$ (since the diagonal has $(m-1)$ $1$'s and $(m+2)$ $2$'s). Hence the diagonal sum of $A$ is $3(2m+1)$.


Since $A$ and $B$ are both magic square, it follows that $[A; B]$ is also a magic square.  Moreover, since the entries of $A$ and $B$ are the radix and unit digits of $[A; B]$ respectively, the magic sum of $[A; B]$ is the magic sum of $A$ multiply with $(2m+1)^2$ plus the magic sum of $B$ yielding $(2m+1)(4(2m+1)^2 -1)$.

It remains to show that the entries in $[A; B]$ consists of distinct ordered pairs of entries $(x, y)$ where  $x \in A$ and $y \in B$.

 Since the entries in $A_0 \cup A_3$ consist of only integers from $\{0, 3\}$ and that the entries of $A_1 \cup A_2$ consist of only integers from  $\{1, 2\}$, $(x_1, y_1) \neq (x_2, y_2)$ for any integers $x_1 \in A_0 \cup A_3$, $x_2 \in A_1 \cup A_2$ where $y_1 \in B_1$ and $y_2 \in B_2$.

 Moreover $(x_1, y) \neq (x_2, y)$ for any integers $x_1 \in A_0$ and $x_2 \in A_3$ where $y  \in B_1$ because $x_1 + x_2 = 3$ by Definition \ref{a}(iii).

 $(x_1, y) \neq (x_2, y)$ for any integers $x_1 \in A_1$ and $x_2 \in A_2$ where $y  \in B_2$ because $x_1 + x_2 = 3$ by Definition \ref{a}(iii).

  This completes the proof.	
\end{proof}
\end{theorem}

\section{Ref-symmetric magic squares}  \label{4m+2}

Recall that, if $N$ is a matrix, then $\pi(N)$ (respectively $\rho(N)$) denote the matrix obtained by taking a central vertical (respectively horizontal) reflection on $N$.



\begin{definition}  \label{br}
Let  $M$ be a  reduced ro-symmetric magic squares of order $2m+1$. Write
$$  M = \begin{bmatrix}  M_1 \\  M_2 \\ M_3 \end{bmatrix}          $$
where  $M_1$ and $M_3$ are $m \times (2m+1)$ matrices and $M_2$ is a $1 \times (2m+1)$ matrix. Let $B_1$  be obtained from $M$ by replacing $M_3$ by $\pi(M_3)$. Also, let $B_1^*$ be obtained from $B_1$ by replacing $M_2$ by $\pi(M_2)$. Now, let $B$  denote the  $2(2m+1) \times 2(2m+1)$ matrix  defined by
$$
 B =
 \begin{bmatrix}
  B_1   & \pi(B_1)    \\
  B_1^* & \pi(B_1^*)
 \end{bmatrix}.
$$
\end{definition}


\begin{example}
Let $M$ be the reduced magic square of the  De la Loub\`{e}re magic square of order $5$.   Then
$$
 B =
 \begin{bmatrix}
  16 & 23 & 0  & 7  & 14     & 14 & 7  & 0  & 23 & 16 \\
  22 & 4  & 6  & 13 & 15     & 15 & 13 & 6  & 4  & 22 \\
  3  & 5  & 12 & 19 & 21     & 21 & 19 & 12 & 5  & 3 \\
  2  & 20 & 18 & 11 & 9      & 9  & 11 & 18 & 20 & 2 \\
  8  & 1  & 24 & 17 & 10     & 10 & 17 & 24 & 1  & 8 \\

  16 & 23 & 0  & 7  & 14     & 14 & 7  & 0  & 23 & 16 \\
  22 & 4  & 6  & 13 & 15     & 15 & 13 & 6  & 4  & 22 \\
  21 & 19 & 12 & 5  & 3      & 3  & 5  & 12 & 19 & 21 \\
  2  & 20 & 18 & 11 & 9      & 9  & 11 & 18 & 20 & 2  \\
  8  & 1  & 24 & 17 & 10     & 10 & 17 & 24 & 1  & 8
 \end{bmatrix}.
$$
\end{example}

 It is routine to verify that  $B$ is a magic square.  Moreover, it has  the   ref-symmetrical property (horizontally).


\begin{definition}  \label{asc}
Let $A^*$ denote the $2(2m+1) \times 2(2m+1)$     square defined by
$$
 A^* =
 \begin{bmatrix}
  A_0 & A_2 \\
  A_3 & A_1
 \end{bmatrix}
 $$
 where $A_0, A_1, A_2, A_3$ as  defined in Definition \ref{a} with the following additional conditions.

\begin{description}
  \item[(i)  ] $\rho(A_0) = A_0$ and $\rho(A_1) = A_1$, and
  \item[(ii)] the $(m+1)$-th row of $A_0$ (and  of $A_1$) is symmetrical about the  central cell.
\end{description}

  \end{definition}

 Figure \ref{a01} depicts an example each for the squares $A_0$ and $A_1$.



\begin{figure}[h]
$$
 \begin{bmatrix}
  3 & 0 & 0 & 0 & 3\\
  3 & 3 & 0 & 0 & 0\\
  3 & 0 & 0 & 0 & 3\\
  3 & 3 & 0 & 0 & 0\\
  3 & 0 & 0 & 0 & 3
 \end{bmatrix} \ \qquad  \qquad \qquad
 \begin{bmatrix}
  1 & 2 & 1 & 1 & 1\\
  2 & 1 & 1 & 1 & 1\\
  1 & 1 & 2 & 1 & 1\\
  2 & 1 & 1 & 1 & 1\\
  1 & 2 & 1 & 1 & 1
 \end{bmatrix}
 $$

 \qquad \ \ \qquad \qquad    \qquad \qquad \qquad  $A_0$ \qquad \ \qquad \qquad \qquad \qquad \qquad $A_1$
  \caption{ $A_0$ and $A_1$  with $m=2.$} \label{a01}
\end{figure}


 \begin{example} Suppose  $A_0$ and $A_1$ are the matrices in Figure \ref{a01}.  Then
  $$
 A^* =
 \begin{bmatrix}
  3 & 0 & 0  & 0  & 3 & 2 & 1  & 2  & 2 & 2 \\
  3 & 3 & 0  & 0  & 0 & 1 & 2  & 2  & 2 & 2 \\
  3 & 0 & 0  & 0  & 3 & 2 & 2  & 1  & 2 & 2 \\
  3 & 3 & 0  & 0  & 0 & 1 & 2  & 2  & 2 & 2 \\
  3 & 0 & 0  & 0  & 3 & 2 & 1  & 2  & 2 & 2 \\
  0 & 3 & 3  & 3  & 0 & 1 & 2  & 1  & 1 & 1 \\
  0 & 0 & 3  & 3  & 3 & 2 & 1  & 1  & 1 & 1 \\
  0 & 3 & 3  & 3  & 0 & 1 & 1  & 2  & 1 & 1 \\
  0 & 0 & 3  & 3  & 3 & 2 & 1  & 1  & 1 & 1 \\
  0 & 3 & 3  & 3  & 0 & 1 & 2  & 1  & 1 & 1
 \end{bmatrix}.
$$
\end{example}

\begin{definition} \label{absc} Suppose $A^*$ and $B$ are as defined in the Definitions \ref{asc} and \ref{br} respectively.
Let $[ A^*; B]$ denote the $2(2m+1) \times 2(2m+1)$ matrix obtained by superimposing $A^*$ and $B$ so that the entries of $A$ are the radix digits while those of $B$ are the unit digits.
\end{definition}


\begin{example} Suppose $A^*$ and  $B$ are the $10 \times 10$  matrices in Examples $5$ and $4$ respectively. Then $[A^*; B]$ is a $10 \times 10$ matrix which yields a reduced magic square of order $10$ when its entries are considered as integers represented in base $5^2$.
Moreover it is a ref-symmetrical magic square (with respect to the central horizontal line) which is as depicted below.
 $$
   \begin{bmatrix}
   91 & 23  &  0  & 7   & 89  & 64 & 32 & 50 & 73  & 66 \\
   97 & 79  &  6  & 13  & 15  & 40 & 63 & 56 & 54  & 72 \\
   78 &  5  & 12  & 19  & 96  & 71 & 69 & 37 & 55  & 53 \\
   77 & 95  & 18  & 11  &  9  & 34 & 61 & 68 & 70  & 52 \\
   83 &  1  & 24  & 17  & 85  & 60 & 42 & 74 & 51  & 58 \\
   16 & 98  & 75  & 82  & 14  & 39 & 57 & 25 & 48  & 41 \\
   22 &  4  & 81  & 88  & 90  & 65 & 38 & 31 & 29  & 47 \\
   21 & 94  & 87  & 80  &  3  & 28 & 30 & 62 & 44  & 46 \\
    2 & 20  & 93  & 86  & 84  & 59 & 36 & 43 & 45  & 27 \\
    8 & 76  & 99  & 92  & 10  & 35 & 67 & 49 & 26  & 33
 \end{bmatrix}.
$$
\end{example}



\begin{theorem} \label{ref-sc}
Suppose $A^*$ and $B$ are as defined in the Definitions \ref{asc} and \ref{br} respectively.  Then
$[A^*; B]$ yields a reduced ref-symmetric  magic square of order $2(2m+1)$ when its entries are considered as integers represented in base $(2m+1)^2$.
\begin{proof}
The magicness of  $A^*$ follows from that of  $A$ (which has already been shown in the proof of Theorem \ref{strachey}).

The fact that $A_2 = 3J_{2m+1} - A_1$  and $A_3 = 3J_{2m+1} -A_0$ together with the condition that  $\rho(A_0) = A_0$ and $\rho(A_1) = A_1$ imply that
\begin{equation}  \label{a*}  \rho(A^*) + A^* = 3 J_{2(2m+1)}.
\end{equation}


Clearly the rows of $B$ are magic. We assert that the columns of $B$ are also magic.

 Since  $M$ is ro-symmetrical, we have
 \begin{equation} \label{m13}
  M_1 + \pi(M_3) = ((2m+1)^2 -1) J_{m, 2m+1}
 \end{equation}
\noindent  and
 \begin{equation}  \label{m2}
M_2 + \pi(M_2) = ((2m+1)^2 -1) J_{1, 2m+1}
\end{equation}

\noindent where  $J_{k,l}$ is the $k \times l$ matrix whose entries are equal to $1$. This proves the assertion.


(\ref{m13}) and (\ref{m2}) imply that
  \begin{equation} \label{b1}
B_1 +\rho(B_1^*)  = ((2m+1)^2-1)J_{2m+1}
\end{equation}
\noindent and this implies that
\begin{equation} \label{rob1}
 \pi(B_1) +\rho(\pi(B_1^*))  = ((2m+1)^2-1)J_{2m+1}
\end{equation}

 (\ref{b1}) and (\ref{rob1})  imply that
\begin{equation}  \label{bref}
B +\rho(B)  = ((2m+1)^2-1)J_{2(2m+1)}.
\end{equation}

 To show that both the diagonals are magic, assume that $M = (a_{i,j})$ so that $\{ a_{i, i} \ | \ i=1, 2, \ldots, 2m+1\}$  and
$ \{ a_{i, 2m+2-i} \ | \ i=1, 2, \ldots, 2m+1\}$  are the main diagonal and anti-diagonal of $M$ respectively.


Note that $B_1$ and $B_1^*$ have the same sets of main diagonal and anti-diagonal. The same goes for those of $\pi(B_1)$ and $\pi(B_1^*)$.


Now the main diagonal of $B_1$ consists of the entries
$$\{ a_{i,i} \ | \ i=1, 2, \ldots, m+1\} \cup  \{ a_{i, 2m+2-i} \ | \ i=m+2, m+3, \ldots, 2m+1\}$$
while the main diagonal of $\pi(B_1^*)$ consists of the entries
$$\{ a_{i, 2m+2-i} \ | \ i= 1 \ldots, m\} \cup \{ a_{i,i} \ | \ i= m+1, m+2, \ldots, 2m+1\}. $$
The union of these sets turns out to be the set of entries in the diagonal  and anti-diagonal of $M$.


Likewise, the anti-diagonal of $B$ consists of the anti-diagonals of $\pi(B_1)$ and $B_1^*$ whose entries are
$$  \{ a_{i,i} \ | \ i=1, 2, \ldots, m+1\} \cup  \{ a_{i, 2m+2-i} \ | \ i=m+2, m+3, \ldots, 2m+1\}       $$ and
$$  \{ a_{i, 2m+2-i} \ | \ i= 1 \ldots, m\} \cup \{ a_{i,i} \ | \ i= m+1, m+2, \ldots, 2m+1\}   $$ respectively so their union is the set of entries in the diagonal  and anti-diagonal of $M$.

 This completes the proof that $B$ is a magic square.

 By following the same argument as in the proof of Theorem \ref{strachey},  we see that  the entries in $[A^*; B]$ consist of distinct ordered pairs of entries $(x, y)$ where  $x \in A^*$ and $y \in B$.


Since $A^*$ and $B$ are  magic squares with magic sums  $3(2m+1)$ and $(2m+1)((2m+1)^2-1)$ respectively, we conclude (as in the proof of Theorem \ref{strachey}) that $[A^*; B]$ yields a reduced magic square of order $2(2m+1)$. Moreover, by Equations (\ref{a*}) and (\ref{bref}), the resulting reduced magic square is ref-symmetrical.	
\end{proof}
\end{theorem}

\section{Some  remarks} \label{remark}

Some remarks are in order.

\begin{enumerate}

\item By taking the $(m+1)$-th row of $A_0$ to be $(0, 3, \ldots, 3, 0 \ldots, 0)$, and all other rows to be $(3, \ldots, 3, 0, \ldots 0)$ where the numbers of $3$ and $0$ are $m$ and  $m+1$ respectively, and each row in $A_1$ to be $(1,\ldots, 1, 2, \ldots, 2)$ (where the numbers of $1$ and $2$ are $m+2$ and  $m-1$ respectively),  $B_1 = B_2$ to be the reduced De la Loub\`{e}re  magic square of order $2m+1$, the magic  square $[A; B]$ is the reduced Strachey square of order  $2(2m+1)$.

\item By cyclically permuting the columns of $A_1$ as constructed in Remark $1$, we obtain new matrices which satisfy condition (ii) of Definition \ref{a}.


\item By taking the reflection along the central vertical line of the matrix $A_0$ in Remark $1$, we obtain a different matrix satisfying the condition (i) of Definition \ref{a}.


\item How many matrices $A_0$ and $A_1$ are there that satisfy the conditions of Definition \ref{a} (or Definition \ref{asc}) for a given natural number $m$?
\end{enumerate}


\section{Acknowledgments}
The research was supported by the Ministry of Higher Education (MoHE) through Fundamental Research Grant Scheme project FRGS/1/2018/STG06/UTAR/01/1.


\begin{thebibliography}{9}


\bibitem{cc:refer} P. S. Chee and G. L. Chia, Construction of magic squares by compositions, {\em Menemui Mat.\/} {\bf 5} (1983), 51--62.

\bibitem{chia:refer}  G. L. Chia, Self-complementary magic squares of doubly even orders, {\em Discrete Math.\/} {\bf 341} (2018), 1359--1362.

\bibitem{chialee:refer} G. L. Chia and Angeline P. L. Lee, Self-complementary magic squares, {\em Ars Combinat.\/} {\bf 114} (2014), 449--460.

\bibitem{km:refer} J. M. Kudrle and S. B. Menard,
Magic square, in C. J. Colbourn and J. H. Dinitz,
eds., {\em The CRC Handbook of Combinatorial Designs},
2nd ed., CRC Press, 2007, pp. 524--528.

\bibitem{baco:refer} W. W. Rouse Ball and H. S. M. Coxeter, {\em
Mathematical Recreations and Essays}, 12th ed.,
 University of Toronto Press, 1974.

\end{thebibliography}

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\noindent 2020 {\it Mathematics Subject Classification}: 05B15.

\noindent \emph{Keywords:} Self-complementary magic square, singly even order, Strachey method,  De la Loub\`{e}re method.

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\noindent (Concerned with sequence
\seqnum{A006052}.)

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\noindent
Received July 15  2022;
revised versions received  July 17 2022; February 28 2023; 
March 20 2023.
Published in {\it Journal of Integer Sequences}, January 11 2024.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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