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The other method has the following procedure.

I draw up the cells of the smallest possible square, that is to say, the one having side 4. I fill the cells with numbers as shown. (Fig. 13)

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

**Figure 13**

I then use this square as model and archetype for each square of similar form in turn^{20}. For all such squares in turn can be split into such subdivisions. The next square in turn after this one has double its side. Doubling the side always makes the (area of the) square four times as great as the area of the square whose side was doubled, therefore the square next in turn can be divided into four such squares. The next square again after this has double its side, and four times (the side of) the first square. Its area is four times as great as the second square, but, compared with the first, since its side is four times as great, it has area 16 times as great. It can therefore be divided into 16 such squares. We can easily find what multiple (the area of) one square is of another from the side, for we look to see what multiple one side is of the other, and we take the number by which one is a multiple of the other, and we multiply this number by itself and the number resulting from the multiplication is the ratio of (the area of) one square to the other. For example, suppose one side is four times the other. I take 4 and multiply it by itself and it makes 16. I have thus proven that the (larger) square is 16 times the (smaller) square. The other cases are similar.

We must now come to the positioning of the numbers which is as follows:

We draw up the cells of another such square after the first one, which is already given, and we divide it up, using certain lines^{21}, into as many squares of the first kind as possible. Then we fill half the cells of the squares in turn starting from the top, looking to the first square and placing the numbers in the same order as in the first square. Then, beginning from the bottom, we reverse back to the top, filling the other half of the cells, those remaining in each square, looking again to the first and placing the numbers in the same order as in the first square^{22}.

For greater clarity, let one such square be drawn, and let us show the placing (of the numbers) in it. Let it be, in fact, the square immediately after the first, (Figs. 14a, 14b) which we draw up thus:

\[\begin{array} {| c | c | c | c |} \hline 1 & {\phantom{24}} & & 8 \\ \hline {\phantom{24}} & 7 & 2 & \\ \hline 6 & & {\phantom{24}} & 3 \\ \hline & 4 & 5 & {\phantom{24}} \\ \hline \end{array}\begin{array} {| c | c | c | c |} \hline 9 & & & 16 \\ \hline & 15 & 10 & \\ \hline 14 & & & 11 \\ \hline & 12 & 13 & \\ \hline \end{array}\\\begin{array} {| c | c | c | c |} \hline 17 & & & 24 \\ \hline & 23 & 18 & \\ \hline 22 & & & 19 \\ \hline & 20 & 21 & \\ \hline \end{array}\begin{array} {| c | c | c | c |} \hline 25 & & & 32 \\ \hline & 31 & 26 & \\ \hline 30 & & & 27 \\ \hline & 28 & 29 & \\ \hline \end{array}\]

**Figure 14a**

\[\begin{array} {| c | c | c | c |} \hline 1 & 62 & 59 & 8 \\ \hline 60 & 7 & 2 & 61 \\ \hline 6 & 57 & 64 & 3 \\ \hline 63 & 4 & 5 & 58 \\ \hline \end{array}\begin{array} {| c | c | c | c |} \hline 9 & 54 & 51 & 16 \\ \hline 52 & 15 & 10 & 53 \\ \hline 14 & 49 & 56 & 11 \\ \hline 55 & 12 & 13 & 50 \\ \hline \end{array}\\\begin{array} {| c | c | c | c |} \hline 17 & 46 & 43 & 24 \\ \hline 44 & 23 & 18 & 45 \\ \hline 22 & 41 & 48 & 19 \\ \hline 47 & 20 & 21 & 42 \\ \hline \end{array}\begin{array} {| c | c | c | c |} \hline 25 & 38 & 35 & 32 \\ \hline 36 & 31 & 26 & 37 \\ \hline 30 & 33 & 40 & 27 \\ \hline 39 & 28 & 29 & 34 \\ \hline \end{array}\]

**Figure 14b**

We divide it by lines^{19} into as many squares of the first kind as possible. It can in fact be divided into 4. We fill half the cells as shown, beginning from the top and descending to the bottom^{23}. We then reverse, in turn, beginning from the bottom, back to the top from whence we descended, filling those remaining cells in the same order as in the first square^{24}, and when the whole thing is full, the *Sides* are equal in any direction.

The procedure is the same in all other such squares.

Observe that in this arrangement, whenever you divide up the square into four sub-squares, the *Side* of each part equals that of the first - this was not the case with the previous arrangement^{25}. If the sides are divided in two, each part has the same sum (along the side) - this will happen in all cases except the first. The square has other elegant and graceful features, which the preceeding account does not contain.

The End |

Peter G. Brown, "The Magic Squares of Manuel Moschopoulos - The Second Method for Evenly-Even Squares," *Convergence* (July 2012)