# The Magic Squares of Manuel Moschopoulos - The First Method for Evenly-Even Squares

Author(s):
Peter G. Brown

As before, two different methods for those squares formed from evenly-even numbers have been found. The first of these is as follows: We draw up the cells for such a square, and then we place (certain) dots17  as shown. In the first such square we place the dots only in the cells along the diagonals, thus. (Fig. 9).

$\begin{array} {| c | c | c | c |} \hline \bullet & & & \bullet \\ \hline & \bullet & \bullet & \\ \hline & \bullet & \bullet & \\ \hline \bullet & & & \bullet \\ \hline \end{array}$

Figure 9

For each square in turn, [we place the dots] firstly along the diagonals and then proceed as follows. (Figs. 10, 11). We count four places in turn to the right from the first of the cells in the topmost row, including the first one and three others, and we place a dot in the fourth cell, and another one in the cell next in turn directly on the right. We again count from this cell 4 places and we place a dot in the 4th cell and another one in the cell directly on the right immediately after it, and so on as far as possible. We continue this procedure also along the other sides of the square in a circle. We then place a series18 of dots from the 4th cell of the topmost (row of cells), counting from left to right, obliquely to the 4th cell of the left side (of the square), counting downwards, so that the dots meet to form an isosceles triangle with the corner of the square. (Likewise, we place a series of dots) from the 5th cell to the 5th cell on the right hand side, counting up (from the bottom). Then counting the 5th as the first, we join the 4th cell (to the 4th) cell at an angle, sloping to the left, then from the 5th cell (sloping) to the right. We continue this until we reach the end of the [remaining] 19 cells on the top row, and then we turn the square (upside down) putting the top side to the bottom, and connect dots from that edge in a similar fashion, as one can see in the diagram.

$\begin{array} {| c | c | c | c | c | c | c | c |} \hline \bullet & & & \bullet & \bullet & & & \bullet \\ \hline & \bullet & \bullet & & & \bullet & \bullet & \\ \hline & \bullet & \bullet & & & \bullet & \bullet & \\ \hline \bullet & & & \bullet & \bullet & & & \bullet \\ \hline \bullet & & & \bullet & \bullet & & & \bullet \\ \hline & \bullet & \bullet & & & \bullet & \bullet & \\ \hline & \bullet & \bullet & & & \bullet & \bullet & \\ \hline \bullet & & & \bullet & \bullet & & & \bullet \\ \hline \end{array}$

Figure 10

$\begin{array} {| c | c | c | c | c | c | c | c | c | c | c | c | c | c | c | c |} \hline \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet \\ \hline & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & \\ \hline & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & \\ \hline \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet \\ \hline \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet \\ \hline & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & \\ \hline & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & \\ \hline \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet \\ \hline \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet \\ \hline & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & \\ \hline & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & \\ \hline \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet \\ \hline \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet \\ \hline & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & \\ \hline & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & \\ \hline \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet & \bullet & & & \bullet \\ \hline \end{array}$

Figure 11

After the placement of the dots as shown, we go through the numbers in turn, starting from the number one, and likewise through the (corresponding) cells in the given square, starting from the first of the cells in the topmost row from left to right. In those cells which have dots, we place the numbers corresponding to the cells, but where there are no dots, we pass over those cells and their corresponding numbers. We continue this process 'till the last of the cells of the entire square. Then again, beginning from the number one, we go through the numbers in turn and the cells of the square, beginning from the first cell in the bottom row from right to left. In those cells which are empty we place the corresponding numbers, but we pass over those cells which have numbers already and their corresponding numbers. We do this running through all the cells as far as the first cell in the topmost row, from which we began our descent.

So that this might become clearer, let us practise (the method) on one such square. In particular then, suppose we are given the square with side 4, which we draw up and place the dots in the cells along the diameters, thus: (Figs. 9, 12).

$\begin{array} {| c | c | c | c |} \hline 1 & 15 & 14 & 4 \\ \hline 12 & 6 & 7 & 9 \\ \hline 8 & 10 & 11 & 5 \\ \hline 13 & 3 & 2 & 16 \\ \hline \end{array}$

Figure 12

We then begin from the number one and from the 1st of the cells along the top, and we place straightaway the number one in that same 1st cell, since it contains a dot. Since there is no dot in the 2nd cell we pass over it and with it the number two which corresponds to it. Similarly we pass over the 3rd cell and with it the number three, but we place 4 in the 4th cell since it contains a dot. We pass over the 5th cell and with it the number five, and we place 6 in the 6th cell, and 7 in the 7th. We pass over 8 and the 8th cell and likewise 9 and the 9th, but we place 10 in the 10th and 11 in the 11th cell. We pass over the 12th and 12, but place 13 in the 13th cell. We pass over 14 and the 14th, and likewise 15 and the 15th cell, but place 16 in the 16th cell. Then we begin again from the number one, and take the first cell on the lowest row [from right to left] as the first cell in the square, counting to the left. We immediately pass over this square and with it the number one which corresponds to it, since it has a number in it. In the 2nd cell we place 2 since there is no number in it, and in the 3rd cell we place 3. We pass over 4 and the 4th cell and place 5 in the 5th. We pass over the 6th cell and 6, and likewise the 7th and 7, but we place 8 in the 8th and 9 in the 9th. The 10th and 10 we pass over and likewise 11 and the 11th, but we place 12 in the 12th and pass over 13 and the 13th. We place 14 in the 14th and 15 in the 15th, but pass over 16 and the 16th cell. All this is clear to see in the diagram. We use this procedure for squares of the same type. Thus the first method has been explained.