- Membership
- Publications
- Meetings
- Competitions
- Community
- Programs
- Students
- High School Teachers
- Faculty and Departments
- Underrepresented Groups
- MAA Awards
- MAA Grants

- News
- About MAA

There are two obvious questions to be asked in relation to Moschopoulos' methods.

i) Why do they work?

ii) How were these methods arrived at?

These questions are, of course, intimately related. By the second question, I mean: Are there some "simple" techniques which generate the magic squares obtained by following the "recipes" in Moschopoulos, which were then replaced by the "recipes" themselves?

**a. The Methods for Odd Squares**

The *Lo Shu* magic square was created according to the scheme shown below in Figure 2 ([18], p. 97). Essentially, the numbers from \(1\) to \(9\) are written in order in a diagonal fashion and the central square is removed, giving the basis of the magic square. The remaining numbers are then transposed from top to bottom and side to side to complete the square. \[{\begin{array} {| c | c | c | c | c |} \hline & & 1 & & \\ \hline & 4 & & 2 & \\ \hline 7 & & 5 & & 3 \\ \hline & 8 & & 6 & \\ \hline & & 9 & & \\ \hline \end{array}}\rightarrow\begin{array} {| c | c | c |} \hline 4 & & 2 \\ \hline & 5 & \\ \hline 8 & & 6 \\ \hline \end{array}\rightarrow\begin{array} { c | c | c | c | c } & & 9 & & \\ \hline & 4 & & 2 & \\ \hline 3 & & 5 & & 7 \\ \hline & 8 & & 6 & \\ \hline & & 1 & & \\ \end{array}\rightarrow\begin{array} {| c | c | c |} \hline 4 & 9 & 2 \\ \hline 3 & 5 & 7 \\ \hline 8 & 1 & 6 \\ \hline \end{array}\]

**Figure 2**

The idea of this construction generalises to produce precisely the odd-side magic squares one obtains using Moschopoulos' *Method of Two's and Three's*. I will illustrate for the magic square of side 5.

Draw up the 9-sided square grid as shown in Figure 3a, and place the numbers \(1\) to \(25\) in standard order as shown. From this array we extract the central square with entries and spacings as shown in Figure 3b.

\[\begin{array} {| c | c | c | c | c | c | c | c | c |} \hline & & & & 1 & & & &{\phantom{25}} \\ \hline & & & 6 & & 2 & & & \\ \hline & & 11 & & 7 & & 3 & & \\ \hline & 16 & & 12 & & 8 & & 4 & \\ \hline 21 & & 17 & & 13 & & 9 & & 5 \\ \hline & 22 & & 18 & & 14 & & 10 & \\ \hline & & 23 & & 19 & & 15 & & \\ \hline & & & 24 & & 20 & & & \\ \hline & & & & 25 & & & & \\ \hline \end{array}\]

**Figure 3a**

\[\begin{array} {| c | c | c | c | c |} \hline 11 & & 7 & & 3 \\ \hline & 12 & & 8 & \\ \hline 17 & & 13 & & 9 \\ \hline & 18 & & 14 & \\ \hline 23 & & 19 & & 15 \\ \hline \end{array}\rightarrow\begin{array} {| c | c | c | c | c |} \hline 11 & 24 & 7 & 20 & 3 \\ \hline 4 & 12 & 25 & 8 & 16 \\ \hline 17 & 5 & 13 & 21 & 9 \\ \hline 10 & 18 & 1 & 14 & 22 \\ \hline 23 & 6 & 19 & 2 & 15 \\ \hline \end{array}\]

**Figure 3b**

The remaining numbers are transposed from the top to the bottom and side to side to obtain the final square in Figure 3b.

As to the mathematical correctness of the configuration, it is easy to see that in general, the sum of the main diagonal of an \(n\)-sided square will be the sum of the \(n\) consecutive integers starting from \(\frac{n^2-n+2}{2},\) which gives the expected total \(\frac{n}{2}\cdot (n^2+1)\) and the back diagonal sum will be \[\sum_{k=0}^{n-1}\left[\frac{n+1}{2}+nk\right],\] which again equals \(\frac{n}{2}\cdot (n^2+1).\) For the columns, observe that we are simply adding the numbers down the \(j\)th column of the \((2n-1)\)-sided square, \(1\le j\le n,\) along with the numbers in the \((j+n)\)th column. (We regard the \((2n)\)th column as empty.) For example, in the 5-sided square, if we take \(j=2\) we add \(16\) and \(22\) with the numbers \(3,\) \(9\) and \(15,\) and if we take \(j=4\) we add \(6,\) \(12,\) \(18\) and \(24\) with the number \(5.\) Doing this in general, for each \(j\) from \(1\) to \(n\) we have to evaluate \[\sum_{k=0}^{j-1}\,[{1+(n-j)n+(n+1)k}]+\sum_{k=0}^{n-j-1}[{(j+1)+(n+1)k}],\] which, as expected, simplifies to \(\frac{n}{2}\cdot (n^2+1).\) The row sum can be similarly shown to be \(\frac{n}{2}\cdot (n^2+1).\)

It would seem that this diamond to square method must have been the technique used to generalise the *Lo Shu* square. Once a number of such squares were created, a "simpler" algorithm was devised to produce them, which appears in Moschopoulos as the *Method of Two's and Three's*.

Those squares obtained using the *Method of Three's and Five's* can be obtained by permuting the rows of the squares of the same size which arise using the *Method of Two's and Three's*. Observe that in each case row 2 is left fixed. The *Method* was probably devised on the basis of the smaller examples.

Peter G. Brown, "The Magic Squares of Manuel Moschopoulos - The Mathematics of the Methods: Odd Squares," *Loci* (July 2012)