The Byzantine scholar and teacher Manuel Moschopoulos (whose name incidentally means *little calf*) is well known to classicists for his work in editing and paraphrasing classical Greek texts (see [13], [17]), thereby aiding in their preservation and transmission. He is also remembered by historians of mathematics for his tract on magic squares, since it was here, for the first time in western thought, that the subject was discussed. It is now over 100 years since the Greek text of Moschopoulos' tract has been edited and translated. This was done by Tannery in 1886 ([16]), and the translation was into French. John Calvin McCoy translated this work from French into English in 1941 (see [9]). This paper presents a fairly literal translation of Moschopoulos' tract from the Greek into English, along with some brief biographical and mathematical details. The translation itself begins on page 7 and continues through page 12, with endnotes on page 13 and references on page 14.

The Byzantine scholar and teacher Manuel Moschopoulos (whose name incidentally means *little calf*) is well known to classicists for his work in editing and paraphrasing classical Greek texts (see [13], [17]), thereby aiding in their preservation and transmission. He is also remembered by historians of mathematics for his tract on magic squares, since it was here, for the first time in western thought, that the subject was discussed. It is now over 100 years since the Greek text of Moschopoulos' tract has been edited and translated. This was done by Tannery in 1886 ([16]), and the translation was into French. John Calvin McCoy translated this work from French into English in 1941 (see [9]). This paper presents a fairly literal translation of Moschopoulos' tract from the Greek into English, along with some brief biographical and mathematical details. The translation itself begins on page 7 and continues through page 12, with endnotes on page 13 and references on page 14.

Manuel Moschopoulos is generally thought of as a writer and philologist, whose main interests were in drama and poetry. As a student of Maximus Planudes, whose learning and interests were very wide ranging indeed, he appears to have also taken an interest in some aspects of science as well as letters. He seems to have taken both his scholarship and his teaching very seriously, and is praised in both areas in a letter written to his uncle by Planudes.

Moschopoulos was probably born about 1265 (see [17], p. 244 n.1), and was the nephew of the bibliophile, Nicephoros Moschopoulos, who became bishop of Crete during the reign of Andronikos II. He tells us in one of his letters ([13], p. 134) that his uncle required four horses to transport his library, thus indicating its size, and its quality may be inferred from the fact that there was an attempt made to steal it. Manuel was obviously influenced and encouraged by his uncle in his studies.

In 1305 or 1306 Moschopoulos was involved in some kind of political plot resulting in his incarceration and disgrace. Several letters survive written by him from the jail, in which he complains about his treatment and the lack of food. He admits that he has acted incorrectly, but claims to have acted naively rather than with malicious intent. Byzantium had, at this time, entered a period of turmoil and slow decline. Internal strifes combined with external pressures made it a place wherein criticism of the hierarchy could be looked upon with great suspicion if not as an act of disloyalty, and Moschopoulos appears to have been at best imprudent in his public remarks. He may also have been held responsible for the behaviour of his student Matarangides, a foreigner (possibly Albanian), for whom he had staked personal responsibility. The exact details are not known. (See [13].)

Moschopoulos' work on magic squares was probably written after this difficult period of his life, perhaps about 1315.

While I have no intention of giving a detailed history of the development of magic squares, a summary of what is known prior to the writing of Moschopoulos would be pertinent. (See [3], [9], [10], and [4] ad loc. for more detailed accounts of the history of magic squares.) \[\begin{array} {| c | c | c |} \hline 4 & 9 & 2 \\ \hline 3 & 5 & 7 \\ \hline 8 & 1 & 6 \\ \hline \end{array}\]

**Figure 1**

The *Lo* *Shu* of the Chinese is a pictorial representation of the magic square shown in Figure 1, which makes its first appearance in the first century A.D. It is undoubtedly much older but there is no direct evidence as to how far back it goes, and claims of 2000 B.C. are probably far too extreme (see [6] and [18]). No earlier magic square is known and Tannery's amazing claim that such squares were known to, or even anticipated by the ancient Greeks is to be rejected outright (see [16] and [5]). Andrewes' ([1], pp. 148ff.) equally absurd comments regarding magic squares in relation to Plato's *Republic Book 9*, and *Timaeus 35* should also be ignored.

The idea of the magic square was transmitted to the Arabs from the Chinese, probably through India, in the eighth century and is discussed by Thabit ibn Qurra (known for his formula for amicable numbers) in the early ninth century. A list of squares of all orders from 3 to 9 are displayed in the *Encyclopaedia* (the *Rasa*`*il*) compiled about 990 by a group of Arabic scholars known as the "brethren of purity" (the *Ikhwan al-safa*) (see [11], Vol. 1, pp. 660f.). Despite all this, no general constructive methods appeared until slightly later. In 1225, Ahmed al-Buni showed how to construct magic squares using a simple bordering technique, but he may not have discovered the method himself. Biggs ([3], p. 120), referring to a paper by Camman ([5]), suggests that the methods explained by Moschopoulos may have been of Persian origin and be linked to those expounded by al-Buni. Camman indeed claims that the two methods given by Moschopoulos for constructing odd magic squares were known to the Persians, citing an anonymous Persian manuscript (Garrett Collection no. 1057, Princeton University). Even so, this document contains examples and not explicit methods.

It appears that magic squares were introduced to Europe through Spain. Indeed, Abraham ben Meir ibn Ezra (c. 1090-1167), an Hispano-Jewish philosopher and astrologer, translated many Arabic works into Hebrew and had a deep interest in magic squares and numerology in general. He travelled widely throughout Italy and beyond, and may have been the one of the people responsible for the introduction of magic squares into Europe.

Moschopoulos' mathematical writings on the other hand, seem to have had little influence at the time. They were in fact "lost" until the geometer Philippe de la Hire (1640-1718) found them in the Royal Library in Paris and produced a translation.

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.

**b. The Methods for Evenly-Even Squares**

Moschopoulos' first method for constructing evenly-even squares also works for *any* square whose side length is a multiple of 4. His description of the placement of dots is unnecessarily complicated, since an easy alternative is to divide the given square into 4×4 subsquares and put dots along the two diagonals of each such subsquare. This is particularly perplexing given that the second method given uses the underlying idea of subdivision into 4×4 subsquares. The rule of placement of numbers given by Moschopoulos is equivalent to writing the numbers from \(1\) to \(n\) from right to left beginning at the right-hand side at the bottom. This is shown for \(n=8\) in Figures 4a and 4b below.

\[\begin{array} {| c | c | c | c | c | c | c | c |} \hline \bullet & & & \bullet & \bullet & & & \bullet \\ \hline & \bullet & \bullet & & & \bullet & \bullet & \\ \hline & \bullet & \bullet & & & \bullet & \bullet & \\ \hline \bullet & & & \bullet & \bullet & & & \bullet \\ \hline \bullet & & & \bullet & \bullet & & & \bullet \\ \hline & \bullet & \bullet & & & \bullet & \bullet & \\ \hline & \bullet & \bullet & & & \bullet & \bullet & \\ \hline \bullet & & & \bullet & \bullet & & & \bullet \\ \hline \end{array}\]

**Figure 4a**

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

**Figure 4b**

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

**Figure 4c**

Now each number \(x\) in a cell containing a dot is replaced by its *complement* \(n^2+1-x\) as in the example in Figure 4c. Thus, if we choose a column with no dot in the \(i\)th position (from the right) in the bottom row, adding up that column we have \[\left[i+(i+3n)+(i+4n)+(i+7n)+(i+8n)+\cdots +(i+n(n-1))\right]\] \[+\left[\left[(n^2+1)-(i+n)\right]+ \left[(n^2+1)-(i+2n)\right]+ \left[(n^2+1)-(i+5n)\right]\\ + \left[(n^2+1)-(i+6n)\right] + \cdots + \left[(n^2+1)-(i+n(n-2))\right]\right]\] \[=\left[\frac{n}{2}i+n(n-1)+\frac{n}{4}(n^2-n)-n(n-1)\right]\\+\left[ \frac{1}{2}n(n^2+1)-\left[\frac{n}{2}i+\frac{n}{4}(n^2-n)\right]\right],\]

which equals \(\frac{n}{2}\cdot (n^2+1),\) as expected. One can also check the sum of the columns which contain a dot in the \(i\)th position (from the right) in the bottom row. The rows and diagonals are dealt with similarly. It is reasonable to surmise that magic squares of this type were originally built up from the basic 4×4 square using the technique described. From such examples, Moschopoulos' method was derived.

The second method is very poorly described by Moschopoulos and would be impossible to follow without an example. Moreover, he gives no clues as to where or how he obtains the basic 4×4 square which is used in the construction. It is not one that arises from the first method, nor from any row or column permutation of such a square and I have been unable to find it anywhere else. It is however very cleverly constructed since the square given in Fig. 14a (in the translation) contains precisely the numbers from \(1\) to \(32\) thus allowing for the placement of the remaining 32 numbers following the order of the basic square. Each subsquare is also a magic square, and this enables us to see why the method works.

The translation follows on pages 7-12, with notes on page 13 and references on page 14.

I have attempted to give as literal a translation as possible. Where I was forced to add in words or phrases not in the Greek, I have used round brackets to enclose additions which are designed to assist the reader but are not technically essential, and square brackets for additions without which the sense would be obscure or misleading.

I have throughout used the Greek text edited by Tannery [16].

I would like to thank Professors V. Katz and F. Swetz for their extremely useful comments and references.

**MAGIC SQUARES**

**by Manuel Moschopoulos - 13th century**

An exposition by the most learned and most blessed master Manuel Moschopoulos with regard to the invention of number squares, which he has made at the insistence of Nicolaos Artabasdos of Smyrna, arithmetician and geometer, the Rhabdas^{1}.

**Introduction**

Among the set of numbers, some are *odd* and some are *even*. Furthermore, among the even numbers, some are *evenly-even*^{2} - those capable of repeated division into two equal parts, finally resulting in the number one^{3}; and some are *evenly-odd*^{4}* * - those which are not capable of repeated division into two equal parts resulting in the number one.

Every number when multiplied by itself, makes a square of equal sides^{5}. For example, the number^{6} 3 , multiplied by itself makes 9, and 9 is a square, its side^{7} being the number 3. For the side of each square is the number that when multiplied by itself, produces that square. It is possible for [the sum of the numbers along] this side^{8} to be the same in all directions even along the diagonals, but so that this may become clearer by a diagram, let a square be drawn (Fig. 1) and in it let the cells^{9} of the number square be contained by lines, thus:\[\begin{array} {| c | c | c |} \hline 1 & 1 & 1 \\ \hline 1 & 1 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{array}\quad{Side}=3\]

**Figure 1**

Further, let a one be placed in each of the cells, then it is altogether obvious that the addition of all these ones amounts to 9, but that the addition of (the numbers on) each of the sides in any direction gives 3, even along the diagonals. This is easy to understand. But if a square were drawn, and cells are drawn in it equal to the (side) number squared, and furthermore, there were placed not ones, but the number one, and those numbers one after another from the number one, then the *Sides* will no longer be equal in all directions, since the numbers in succession have been placed in the cells. Now if an arrangement were sought which could make the *Sides* equal in all directions, including the diagonals, this would not be found very easily. Even if after some difficulty such an arrangement were found for one square, there is no reason to expect one might be found in a different (size) square. However, if one were guided by a method, one would easily have the arrangement which produces this effect, in whatever square one likes.

There is no single method for this, but one method which works for squares arising from odd numbers, another for those arising from evenly-even numbers, and yet another for those arising from evenly-odd numbers. Concerning these things^{10} we now propose to speak.

It is necessary first to speak of the *Side* produced by the numbers from one to the [area of the] square we seek. This value we find as follows: We add the numbers from one to the square and then divide the total of this addition by the number which when multiplied by itself gives the square. This quotient^{11} we consider to be the *Side* of the numbers from one to the square we seek. For example, suppose we seek the *Side* of the numbers from one to 9. We therefore add 2 to one, giving 3; then add 3 to 3 which makes 6, then 4 to 6, which makes 10, and 5 to 10, which makes 15 and so on as far as 9. The total then becomes 45. This we divide by 3, since that number, multiplied by itself, gives 9. The quotient then is 15. These are the *Sides* of the numbers from one to 9. The method is the same in regard to other squares.

But lest, if the number happens to be very large, we grow weary of adding the numbers from one, having made inquiry, we found a method where we can easily arrive at the total of the addition of the numbers from one to as far as we wish. This method is as follows: We keep in mind the number to which the addition will proceed, and we multiply it by itself. We divide the number obtained from the multiplication into two equal parts, and then to one of these parts we add half of the number we multiplied by itself. It is a fact^{12} that the result of the addition of half of the number obtained from the multiplication and [half of]^{13} the number being multiplied by itself, is equal to that obtained from the addition of the numbers from one to the number which was multiplied by itself.

This would become clearer when applied to some specific numbers, for example: Suppose again that the result of the sum of the numbers from one to 9 is sought. We multiply this number then by itself and it becomes 81. This we divide into two equal parts, so there corresponds 40 and a half to each part. Then we divide 9 into two equal parts, and there corresponds to each part 4 and a half. We add this to half of the result of the multiplication, that is to say, to 40 and a half, and together they make 45. The result of the addition of the numbers from 1 to 9 was in fact also 45. The same thing happens also on all other occasions.

That part concluded, it is now time to speak about the placement (of the numbers). Let us commence with the simplest case. One can form first of all a square based on the number 3 (Fig. 2) and we will speak firstly about this, however the method about to be expounded in regard to this square, will apply to all squares of similar type^{14} . One can make an arrangement giving (the sum) equal in each direction, using [the method of] two's and three's and it is possible also to use [the method of] three's and five's.

**The Method of Two's and Three's**

The [method of] two's and three's is as follows:

Suppose the cells of the smallest possible square are drawn up, that is to say, of the (square corresponding to the) number 9, thus. We place the number one in the middle cell of the three along the bottom row, and we count two cells, including that which holds the number one and then the next below this in a straight line. The general rule is always to go to the lower one. Since however, we do not find (such a cell), we return again to the top (of the square) in a straight line moving through the cells in a cyclic fashion. We count that cell as the second one. Then we place 2 in the cell to the right of this in a straight line.

\[\begin{array} {| c | c | c |} \hline 4 & 9 & 2 \\ \hline 3 & 5 & 7 \\ \hline 8 & 1 & 6 \\ \hline \end{array}\quad{Side}=15\]

**Figure 2**

We again count two cells, the one which holds the 2 and another below it and we look to the cell on the right in a straight line so that we might place the number 3; the general rule is always to move to the right. Since we do not find (such a cell), we return to the left (hand side of the square) in a straight line. For whenever a row of the square ends you return to the beginning (of that row)^{15} . We place 3 in the last cell as we cycle back, that is, in the first cell as we move to the right, the one which we come to as we count the cells from the beginning in circular movement. Now since we have come to the number 3, which when multiplied by itself gives the (area of the) square, that is to say, it is the side of the number 9, we no longer count two places in order to put 4 in the cell on the right, but three places, as follows. We count the first as that which holds 3, the cell below this as the second as we take as third the cell directly below, and since we do not find (a cell there), we return to the top moving in a straight line, and count that one as the third. In that very cell we place 4, not turning aside (to the right). Following from this and taking that cell as our starting point we count again two cells and place the next number in turn (in the cell) on the right, following the previously explained pattern, and we continue with this until we again come to [a multiple of] the side of 9, that is to say, to 6, which is twice 3. Again, having come to this number, we count three cells and place the next number in turn in the third cell, not turning aside (to the right). Once again we count two and we make the placement in the cell on the right. We continue doing this to the end. As before, we count through two cells for all the other numbers, but through three cells precisely when we come upon a multiple of the side^{16}. This is the procedure for all squares of the same form, for we count according to this pattern; by two's until we reach the cells containing (a multiple of) the side of the given square, and by three's whenever we come upon these numbers in turn, and we continue thus till the end, cycling back through the cells, as above.

All the squares (of this type) simply follow the same pattern except in regard to the placing of the number one, for this is not always put in the same place but assumes a different position in each square. In the first square which can be constructed from an odd number, it is placed in the middle of the cells along the bottom. In the second such square, (it is placed) in the middle cell of the row one up from the bottom, and in the third, in the middle of the row one further up. Clearly, as the numbers increase, so the number one rises up through the cells. It always transpires that it is placed in the cell which lies directly below the cell in the middle of the given square of that type. One can see all these things more clearly in the diagram. (Figs. 3,4,5).

\[\begin{array} {| c | c | c | c | c |} \hline 11 & 24 & 7 & 20 & 3 \\ \hline 4 & 12 & 25 & 8 & 16 \\ \hline 17 & 5 & 13 & 21 & 9 \\ \hline 10 & 18 & 1 & 14 & 22 \\ \hline 23 & 6 & 19 & 2 & 15 \\ \hline \end{array}\quad{Side}=65\]

**Figure 3**

\[\begin{array} {| c | c | c | c | c | c | c |} \hline 22 & 47 & 16 & 41 & 10 & 35 & 4 \\ \hline 5 & 23 & 48 & 17 & 42 & 11 & 29 \\ \hline 30 & 6 & 24 & 49 & 18 & 36 & 12 \\ \hline 13 & 31 & 7 & 25 & 43 & 19 & 37 \\ \hline 38 & 14 & 32 & 1 & 26 & 44 & 20 \\ \hline 21 & 39 & 8 & 33 & 2 & 27 & 45 \\ \hline 46 & 15 & 40 & 9 & 34 & 3 & 28 \\ \hline \end{array}\quad{Side}=175\]

**Figure 4**

\[\begin{array} {| c | c | c | c | c | c | c |} \hline 37 & 78 & 29 & 70 & 21 & 62 & 13 & 54 & 5 \\ \hline 6 & 38 & 79 & 30 & 71 & 22 & 63 & 14 & 46 \\ \hline 47 & 7 & 39 & 80 & 31 & 72 & 23 & 55 & 15 \\ \hline 16 & 48 & 8 & 40 & 81 & 32 & 64 & 24 & 56 \\ \hline 57 & 17 & 49 & 9 & 41 & 73 & 33 & 65 & 25 \\ \hline 26 & 58 & 18 & 50 & 1 & 42 & 74 & 34 & 66 \\ \hline 67 & 27 & 59 & 10 & 51 & 2 & 43 & 75 & 35 \\ \hline 36 & 68 & 19 & 60 & 11 & 52 & 3 & 44 & 76 \\ \hline 77 & 28 & 69 & 20 & 61 & 12 & 53 & 4 & 45 \\ \hline \end{array}\quad{Side}=369\]

**Figure 5**

(The method of counting by) three's and five's is as follows: We draw the square and we draw in it the cells of the number square. We then place the number one always in the middle of the topmost (row of) cells. We count three cells, including the one which holds the number one, and two below this in turn, and we place 2 in the cell directly on the right of the third cell. We again count from there three cells in a similar fashion and in the cell to the right we place 3. If there is no cell on the right, we return to the left side in a straight line, as in the previous method, and we place it in the last cell as we cycle back, that is, in the first cell as we move to the right. We continue to do this until we come to the side of the given square, and, after we have reached this number, we count five cells, including the cell having the side length and four below it. Then we place in the fifth cell the next number in turn after the side length, not turning (to the right), and again we count by three's until we reach (a multiple of) the side, cycling through the cells as in the previous method. We continue this to the end. This method is in all respects similar to the previous one except that there, the number one was placed in a different position for each different square, but here it is always (placed) in the middle of the top (row of) cells. There, moreover, we counted by two's and three's, while here by three's and five's. One can see these things in the diagram. (Figs. 6, 7, 8)

\[\begin{array} {| c | c | c |} \hline 8 & 1 & 6 \\ \hline 3 & 5 & 7 \\ \hline 4 & 9 & 2 \\ \hline \end{array}\quad{Side}=15\]

**Figure 6**

\[\begin{array} {| c | c | c | c | c |} \hline 10 & 18 & 1 & 14 & 22 \\ \hline 4 & 12 & 25 & 8 & 16 \\ \hline 23 & 6 & 19 & 2 & 15 \\ \hline 17 & 5 & 13 & 21 & 9 \\ \hline 11 & 24 & 7 & 20 & 3 \\ \hline \end{array}\quad{Side}=65\]

**Figure 7**

\[\begin{array} {| c | c | c | c | c | c | c |} \hline 38 & 14 & 32 & 1 & 26 & 44 & 20 \\ \hline 5 & 23 & 48 & 17 & 42 & 11 & 29 \\ \hline 21 & 39 & 8 & 33 & 2 & 27 & 45 \\ \hline 30 & 6 & 24 & 49 & 18 & 36 & 12 \\ \hline 46 & 15 & 40 & 9 & 34 & 3 & 28 \\ \hline 13 & 31 & 7 & 25 & 43 & 19 & 37 \\ \hline 22 & 47 & 16 & 41 & 10 & 35 & 4 \\ \hline \end{array}\quad{Side}=175\]

**Figure 8**

These are the methods in regard to squares formed from odd numbers.

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) dots^{17}^{ } 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 series^{18} 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.

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 |

1. *Rhabdas* means *staff* or *stick*. Its meaning here is unknown. Sir Thomas Heath [7] seems to regard it as a man's name. Conceivably it could mean *son of Rhabdas*, a common use of \(\tau\omicron\overline\upsilon\) + genitive

2. An *evenly-even* number is a positive integer of the form \(2^n,\) for some positive integer \(n.\)

3. Literally 'the monad', Greek \(\mu\omicron\nu\acute{\alpha}\varsigma.\) Early Greek arithmetical theory regarded the number 1, the monad, as a special object, from which all numbers emanated, rather than as a number in its own right. I have translated the term as 'the number one' throughout. See [7], Vol. 1, pp. 69f.

4. An *evenly-odd* number is generally taken to refer to a number of the form \(2(2m+1).\) See [7], Vol. 1, p. 72. Moschopoulos may, however, be refering to those numbers traditionally called *oddly-even*, which have the form \(2^n (2m+1),\) for \(n, m\ge 1.\)* *

5. Literally, an 'equal sided quadrangle'. Moschopoulos abbreviates this to 'quadrangle' in what follows. I translate this as 'square' or 'number square'. Square integers are still thought of geometrically, and the 'side' of a square refers, of course, to the square root. This leads to some confusion in the Greek which I have attempted to overcome in translation.

The term 'magic square' does not appear in Moschopoulos. The exact origin of the term is not known but may have been derived from the fact that medieval astrologers engraved such magic squares on charms which supposedly possessed magical properties. There is not the slightest suggestion that Moschopoulos attached any magical or spiritual power to these squares or their construction. See [4], p. 122.

6. Number represented by Greek letters are simply transposed into modern numerals, while those written out in Greek are translated by their equivalent English terms.

7. i.e. square root.

8. Moschopoulos here uses the term 'side', \(\pi\lambda\epsilon\upsilon\rho\acute{\alpha},\) to refer to the sum of the numbers on any 'side' (i.e. row or column). He often confuses this use of the word 'side' with the 'side length' of the square. Thus the 'side' of a three-by-three magic square is 15, but the 'side of the square' is 3. To avoid this confusion I have used the term *Side* for the sum of the numbers along any side and 'side' for the side length.

9. The word translated as 'cell' is \(\tau\acute{\omicron}\pi\omicron\varsigma,\) meaning 'a place' or 'position'.

10. In fact, Moschopoulos only speaks of the construction of squares of odd length sides, and those whose sides are *evenly-even*, that is, a power of 2.

11. Literally, 'the amount corresponding to each monad of the sides'.

12. Literally, 'it occurs by necessity.' Moschopoulos is, of course, using the fact that \[\Sigma_{i=1}^n i = \frac{1}{2}n^2 + \frac{1}{2}n.\] Nowhere in his writings does he attempt to prove any of his claims. This result was well known in antiquity, since for each positive integer \(n\) this formula gives the triangular numbers. See [14], Vol II, p. 499.

13. Tannery adds in the words, \(\tau\omicron\overline{\upsilon}\) '\(\eta\mu\acute{\iota}\sigma\epsilon\omicron\varsigma\) \(\mu\acute{\epsilon}\rho\omicron\upsilon\varsigma.\)

14. That is, odd squares.

15. Literally, 'when the cells are filled it is always necessary to return to their beginning'.

16. Literally, 'when we vary from one side to another side'.

17. Literally 'signs', \(\sigma\eta\mu\epsilon\overline{\iota}\alpha.\)

18. Literally, 'we lead' or 'carry' dots.

19. Omitted in some manuscripts.

20. It is not clear why M. chose this *archetype* rather than the one constructed using the earlier method. It has been suggested that M. is simply copying some earlier material of which he has little understanding.

21. Literally 'signs'.

22. Compared with the verbose descriptions of the earlier methods, this section is remarkably terse, and would be difficult to follow without an example.

23. As shown, we move from left to right and add 8, 16, and 24, respectively, to the entries in the basic square, as we do so.

24. Adding, of course, 24, 32, 40, and 48, respectively, to the basic square as we reverse back.

25. This also holds for the diagonal sum of each sub-square.

[1] W. S. Andrews, *Magic Squares and Cubes*, 1960. Dover reprint of 1917 edition.

[2] William Benson and Oswald Jacoby, *New Recreations with Magic Squares*, 1976, Dover Publications.

[3] N. L. Biggs, Roots of Combinatorics, 1979, in *Historia Mathematica*, Vol. 6, pp. 118-136.

[4] Florian Cajori, *A History of Mathematics*, 1919, Macmillan (Revised Edition).

[5] Schuyler Cammann, The Evolution of Magic Squares in China, 1960, in *J. American Oriental Society* 80, pp. 116-124.

[6] Cheng-Yih Chen (Ed.), *Science and Technology in Chinese Civilisation*, 1987, Singapore, pp. 93ff.

[7] Thomas L. Heath, *A History of Greek Mathematics*, Vols. 1, 2, 1921/1981, Oxford: Clarendon Press. New edition: 1981, Dover, New York.

[8] Maurice Kraitchich, *Mathematical Recreations*, 1942, Dover Publications.

[9] John Calvin McCoy, *Scripta Mathematica*, 8, 1941, pp. 15-26.

[10] Clifford Pickover, *The Zen of Magic Squares, Circles and Stars*, 2002, Princeton University Press.

[11] George Sarton, *Introduction to the History of Science, Vols. 1-3*, 1947, Baltimore.

[12] Jacques Sesiano, Les Carrés magiques de Manuel Moschopoulos, *Arch. Hist. Exact Sciences* 53 (1998), pp. 377-397, Springer Verlag.

[13] Ihor Sevcenko, The Imprisonment of Manuel Moschopoulos in the Year 1305 or 1306, 1952, *Speculum*, Vol. 27, No. 2, pp. 133-157.

[14] David Eugene Smith, *History of Mathematics*, 1958, Dover, New York.

[15] Frank Swetz, *Legacy of the Luoshu*, 2002, Open Court.

[16] Paul Tannery, Le Traité de Manuel Moschopoulos sur Les Carrés Magiques, 1886. In *Ann. de l'Assoc. pour l'Encouragement des Etudes Grecques*, pp. 88-104.

[17] Nigel Wilson, *Scholars of Byzantium*, 1983, Johns Hopkins University Press, Baltimore, Maryland.

[18] Li Yan and Du Shiran, *Chinese Mathematics: A Concise History*, 1987, Clarendon Press, Oxford.