Magic square

China

While ancient references to the pattern of even and odd numbers in the 3×3 magic square appears in the I Ching, the first unequivocal instance of this magic square appears in a 1st century book Da Dai Liji (Record of Rites by the Elder Dai).^{[7] [8][9]} These numbers also occur in a possibly earlier mathematical text called Shushu jiyi (Memoir on Some Traditions of Mathematical Art), said to be written in 190 BCE. This is the earliest appearance of a magic square on record; and it was mainly used for divination and astrology.^[7] The 3×3 magic square was referred to as the "Nine Halls" by earlier Chinese mathematicians.^[9] The identification of the 3×3 magic square to the legendary Luoshu chart was only made in the 12th century, after which it was referred to as the Luoshu square.^[7][9] The oldest surviving Chinese treatise on the systematic methods for constructing larger magic squares is Yang Hui's Xugu zheqi suanfa (Continuation of Ancient Mathematical Methods for Elucidating the Strange) written in 1275.^[7][9] The contents of Yang Hui's treatise were collected from older works, both native and foreign; and he only explains the construction of third and fourth order magic squares, while merely passing on the finished diagrams of larger squares.^[9] He gives a magic square of order 3, two squares for each order of 4 to 8, one of order nine, and one semi-magic square of order 10. He also gives six magic circles of varying complexity.^[10]

The above magic squares of orders 3 to 9 are taken from Yang Hui's treatise, in which the Luo Shu principle is clearly evident.^[9][11] The order 5 square is a bordered magic square, with central 3×3 square formed according to Luo Shu principle. The order 9 square is a composite magic square, in which the nine 3×3 sub squares are also magic.^[9] After Yang Hui, magic squares frequently occur in Chinese mathematics such as in Ding Yidong's Dayan suoyin (circa 1300), Chen Dawei's Suanfa tongzong (1593), Fang Zhongtong's Shuduyan (1661) which contains magic circles, cubes and spheres, Zhang Chao's Xinzhai zazu (circa 1650), who published China's the first magic square of order ten, and lastly Bao Qishou's Binaishanfang ji (circa 1880), who gave various three dimensional magic configurations.^[7][11] However, despite being the first to discover the magic squares and getting a head start by several centuries, the Chinese development of the magic squares are much inferior compared to the Islamic, the Indian, or the European developments. The high point of Chinese mathematics that deals with the magic squares seems to be contained in the work of Yang Hui; but even as a collection of older methods, this work is much more primitive, lacking general methods for constructing magic squares of any order, compared to a similar collection written around the same time by the Byzantine scholar Manuel Moschopoulos.^[9] This is possibly because of the Chinese scholars' enthralment with the Lo Shu principle, which they tried to adapt to solve higher squares; and after Yang Hui and the fall of Yuan dynasty, their systematic purging of the foreign influences in Chinese mathematics.^[9]

Middle East: Persia, Arabia, North Africa, Muslim Iberia

Original script from the Shams al-Ma'arif.

Printed version of the previous manuscript. Eastern Arabic numerals were used.

Although the early history of magic squares in Persia and Arabia is not known, it has been suggested that they were known in pre-Islamic times.^[12] It is clear, however, that the study of magic squares was common in medieval Islam, and it was thought to have begun after the introduction of chess into the region.^[13][14][15] The first datable appearance of magic square of order 3 occur in the alchemical works of Jābir ibn Hayyān (fl. c. 721– c. 815). While it is known that treatises on magic squares were written in the 9th century, the earliest extant treaties we have date from the 10th-century: one by Abu'l-Wafa al-Buzjani (circa 998) and another by Ali b. Ahmad al-Antaki (circa 987).^[14][16][17] These early treatise were purely mathematical, and the Arabic designation for magic squares is wafq al-a'dad which translates as harmonious disposition of the numbers.^[15] By the end of 10th century, the two treatises by Buzjani and Antaki makes it clear that the Islamic mathematicians had understood how to construct bordered squares of any order as well as simple magic squares of small orders (n ≤ 6) which were used to make composite magic squares.^[14][16] A specimen of magic squares of orders 3 to 9 devised by Islamic mathematicians appear in an encyclopedia from Baghdad circa 983, the Rasa'il Ikhwan al-Safa (the Encyclopedia of the Brethren of Purity).^[18] The squares of order 3 to 7 from Rasa'il are given below:^[18]

The 11th century saw the finding of several ways to construct simple magic squares for odd and evenly-even orders; the more difficult case of evenly-odd case (n = 4k + 2) was solved by Ibn al-Haytham with k even (circa 1040), and completely by the beginning of 12th century, if not already in the latter half of the 11th century.^[14] Around the same time, pandiagonal squares were being constructed. Treaties on magic squares were numerous in the 11th and 12th century. These later developments tended to be improvements on or simplifications of existing methods. From the 13th century on wards, magic squares were increasingly put to occult purposes.^[14] However, much of these later texts written for occult purposes merely depict certain magic squares and mention their attributes, without describing their principle of construction, with only some authors keeping the general theory alive.^[14] One such occultist was the Egyptian Ahmad al-Buni (circa 1225), who gave general methods on constructing bordered magic squares; some others were the 17th century Egyptian Shabramallisi and the 18th century Nigerian al-Kishnawi.^[19]

The magic square of order three was described as a child-bearing charm^[20][21] since its first literary appearances in the alchemical works of Jābir ibn Hayyān (fl. c. 721– c. 815)^[21][22] and al-Ghazālī (1058–1111)^[23] and it was preserved in the tradition of the planetary tables. The earliest occurrence of the association of seven magic squares to the virtues of the seven heavenly bodies appear in Andalusian scholar Ibn Zarkali's (known as Azarquiel in Europe) (1029–1087) Kitāb tadbīrāt al-kawākib (Book on the Influences of the Planets).^[24] A century later, the Egyptian scholar Ahmad al-Buni attributed mystical properties to magic squares in his highly influential book Shams al-Ma'arif (The Book of the Sun of Gnosis and the Subtleties of Elevated Things), which also describes their construction. This tradition about a series of magic squares from order three to nine, which are associated with the seven planets, survives in Greek, Arabic, and Latin versions.^[25] There are also references to the use of magic squares in astrological calculations, a practice that seems to have originated with the Arabs.^[26][27]

India

The 3×3 magic square has been a part of rituals in India since ancient times, and still is today. For instance, the Kubera-Kolam, a magic square of order three, is commonly painted on floors in India. It is essentially the same as the Lo Shu Square, but with 19 added to each number, giving a magic constant of 72 (below, square on the left). The 3×3 magic square first appears in India in Gargasamhita by Garga, who recommends its use to pacify the nine planets (navagraha). The oldest version of this text dates from 100 CE; however passage on planets could not have been written earlier than 400 CE. The first datable instance of 3×3 magic square in India occur in a medical text Siddhayog (ca. 900 CE) by Vrnda, which was prescribed to women in labor in order to have easy delivery.^[28]

The earliest unequivocal occurrence of magic square is found in a work called Kaksaputa, composed by the alchemist Nagarjuna around 1st century CE. All of the squares given by Nagarjuna are 4×4 magic squares, and one of them is called Nagarjuniya after him. Nagarjuna gave a method of constructing 4×4 magic square using a primary skeleton square, given an odd or even magic sum. Incidentally, the special Nagarjuniya square cannot be constructed from the method he expounds.^[29] The Nagarjuniya square is given below, and has the sum total of 100.

The Nagarjuniya square is a pan-diagonal magic square. The Nagarjuniya square is made up of two arithmetic progressions starting from 6 and 16 with eight terms each, with a common difference between successive terms as 4. When these two progressions are reduced to the normal progression of 1 to 8, we obtain the adjacent square.

The oldest datable magic square in the world is found in an encyclopaedic work written by Varahamihira around 587 CE called Brhat Samhita. The magic square is constructed for the purpose of making perfumes using 4 substances selected from 16 different substances. Each cell of the square represents a particular ingredient, while the number in the cell represents the proportion of the associated ingredient, such that the mixture of any four combination of ingredients along the columns, rows, diagonals, and so on, gives the total volume of the mixture to be 18. Although the book is mostly about divination, the magic square is given as a matter of combinatorial design, and no magical properties are attributed to it.^[29][28]

The square of Varahamihira as given above has sum of 18. Here the numbers 1 to 8 appear twice in the square. It is a pan-diagonal magic square. It is also an instance of most perfect magic square. Four different magic squares can be obtained by adding 8 to one of the two sets of 1 to 8 sequence. The sequence is selected such that the number 8 is added exactly twice in each row, each column and each of the main diagonals. One of the possible magic squares shown in the right side. This magic square is remarkable in that it is a 90 degree rotation of a magic square that appears in the 13th century Islamic world as one of the most popular magic squares.^[30]

Around 12th-century, a 4×4 magic square was inscribed on the wall of Parshvanath temple in Khajuraho, India. Several Jain hyms teach how to make magic squares, although they are undatable.^[28]

As far as is known, the first systematic study of magic squares in India was conducted by Thakkar Pheru, a Jain scholar, in his Ganitasara Kaumudi (ca. 1315). This work contains a small section on magic squares which consists of nine verses. Here he gives a square of order four, and alludes to its rearrangement; classifies magic squares into three (odd, evenly even, and oddly even) according to its order; gives a square of order six; and prescribes one method each for constructing even and odd squares. For the even squares, Pheru divides the square into component squares of order four, and puts the numbers into cells according to the pattern of a standard square of order four. For odd squares, Pheru gives the method using horse move or knight's move. Although algorithmically different, it gives the same square as the De la Loubere's method.^[28]

The next comprehensive work on magic squares was taken up by Narayana Pandit, who in the fourteenth chapter of his Ganita Kaumudi (1356) gives general methods for their construction, along with the principles governing such constructions. It consists of 55 verses for rules and 17 verses for examples. Narayana gives a method to construct all the pan-magic squares of fourth order using knight's move; enumerates the number of pan-diagonal magic squares of order four, 384, including every variation made by rotation and reflection; three general methods for squares having any order and constant sum when a standard square of the same order is known; two methods each for constructing evenly even, oddly even, and odd squares when the sum is given. While Narayana describes one older method for each species of square, he claims the method of superposition for evenly even and odd squares and a method of interchange for oddly even squares to be his own invention. The superposition method was later re-discovered by De la Hire in Europe. In the last section, he conceives of other figures, such as circles, rectangles, and hexagons, in which the numbers may be arranged to possess properties similar to those of magic squares.^[29][28] Below are some of the magic squares constructed by Narayana:^[29]

The order 8 square is interesting in itself since it is an instance of the most-perfect magic square. Incidentally, Narayana states that the purpose of studying magic squares is to construct yantra, to destroy the ego of bad mathematicians, and for the pleasure of good mathematicians. The subject of magic squares is referred to as bhadraganita and Narayana states that it was first taught to men by god Shiva.^[28]

Latin Europe

This page from Athanasius Kircher's Oedipus Aegyptiacus (1653) belongs to a treatise on magic squares and shows the Sigillum Iovis associated with Jupiter

Unlike in Persia and Arabia, we have better documentation of how the magic squares were transmitted to Europe. Around 1315, influenced by Islamic sources, the Greek Byzantine scholar Manuel Moschopoulos wrote a mathematical treatise on the subject of magic squares, leaving out the mysticism of his Muslim predecessors, where he gave two methods for odd squares and two methods for evenly even squares. Moschopoulos was essentially unknown to the Latin Europe until the late 17th century, when Philippe de la Hire rediscovered his treatise in the Royal Library of Paris.^[31] However, he was not the first European to have written on magic squares; and the magic squares were disseminated to rest of Europe through Spain and Italy as occult objects. The early occult treaties that displayed the squares did not describe how they were constructed. Thus the entire theory had to be rediscovered.

Magic squares had first appeared in Europe in Kitāb tadbīrāt al-kawākib (Book on the Influences of the Planets) written by Ibn Zarkali of Toledo, Al-Andalus, as planetary squares by 11th century.^[24] The magic square of three was discussed in numerological manner in early 12th century by Jewish scholar Abraham ibn Ezra of Toledo, which influenced later Kabbalists.^[32] Ibn Zarkali's work was translated as Libro de Astromagia in the 1280s,^[33] due to Alfonso X of Castille.^[34][24] In the Alfonsine text, magic squares of different orders are assigned to the respective planets, as in the Islamic literature; unfortunately, of all the squares discussed, the Mars magic square of order five is the only square exhibited in the manuscript.^[35][24]

Magic squares surface again in Florence, Italy in the 14th century. A 6×6 and a 9×9 square are exhibited in a manuscript of the Trattato d'Abbaco (Treatise of the Abacus) by Paolo Dagomari.^[36][37] It is interesting to observe that Paolo Dagomari, like Pacioli after him, refers to the squares as a useful basis for inventing mathematical questions and games, and does not mention any magical use. Incidentally, though, he also refers to them as being respectively the Sun's and the Moon's squares, and mentions that they enter astrological calculations that are not better specified. As said, the same point of view seems to motivate the fellow Florentine Luca Pacioli, who describes 3×3 to 9×9 squares in his work De Viribus Quantitatis by the end of 15th century.^[38][39]

Europe after 15th century

The planetary squares had disseminated into northern Europe by the end of 15th century. For instance, the Cracow manuscript of Picatrix from Poland displays magic squares of orders 3 to 9. The same set of squares as in the Cracow manuscript later appears in the writings of Paracelsus in Archidoxa Magica (1567), although in highly garbled form. In 1514 Albrecht Dürer immortalized a 4×4 square in his famous engraving Melencolia I. Paracelsus' contemporary Heinrich Cornelius Agrippa von Nettesheim published his famous book De occulta philosophia in 1531, where he devoted a chapter to the planetary squares shown below. The same set of squares given by Agrippa reappear in 1539 in Practica Arithmetice by Girolamo Cardano. The tradition of planetary squares was continued into the 17th century by Athanasius Kircher in Oedipi Aegyptici (1653). In Germany, mathematical treaties concerning magic squares were written in 1544 by Michael Stifel in Arithmetica Integra, who rediscovered the bordered squares, and Adam Riese, who rediscovered the continuous numbering method to construct odd ordered squares published by Agrippa. However, due to the religious upheavals of that time, these work were unknown to the rest of Europe.^[32]

In 1624 France, Claude Gaspard Bachet described the "diamond method" for constructing Agrippa's odd ordered squares in his book Problèmes Plaisants. In 1691, Simon de la Loubère described the Indian continuous method of constructing odd ordered magic squares in his book Du Royaume de Siam, which he had learned while returning from a diplomatic mission to Siam, which was faster than Bachet's method. In an attempt to explain its working, de la Loubere used the primary numbers and root numbers, and rediscovered the method of adding two preliminary squares. This method was further investigated by Abbe Poignard in Traité des quarrés sublimes (1704), and then later by Philippe de La Hire in Mémoires de l’Académie des Sciences for the Royal Academy (1705), and by Joseph Sauveur in Construction des quarrés magiques (1710). In Divers ouvrages de mathematique et de physique published posthumously in 1693, Bernard Frenicle de Bessy demonstrated that there were exactly 880 distinct magic squares of order four. De la Hire also introduced concentric bordered square in 1705, while Sauveur introduced magic cubes and lettered squares, which was taken up later by Euler in 1776, who is often credited for devising them. In 1750 d'Ons-le-Bray rediscovered the method of constructing doubly even and singly even squares using bordering technique. By this time the earlier mysticism attached to the magic squares had completely vanished, and the subject was treated as a part of recreational mathematics.^[32][40]

In the 19th century, Bernard Violle gave a comprehensive treatment of magic squares in his three volume Traité complet des carrés magiques (1837—1838), which also described magic cubes, parallelograms, parallelopipeds, and circles. Pandiagonal squares were extensively studied by Andrew Hollingworth Frost, who learned it while in the town of Nasik, India, (thus calling them Nasik squares) in a series of articles: On the knight's path (1877), On the General Properties of Nasik Squares (1878), On the General Properties of Nasik Cubes (1878), On the construction of Nasik Squares of any order (1896). He showed that it is impossible to have normal singly-even pandiagonal magic square. Frederick A.P. Barnard constructed inlaid magic squares and other three dimensional magic figures like magic spheres and magic cylinders in Theory of magic squares and of magic cubes (1888).^[40] In 1897, Emroy McClintock published On the most perfect form of magic squares, coining the words pandiagonal square and most perfect square, which had previously been referred to as perfect, or diabolic, or Nasik.

Luo Shu magic square

Legends dating from as early as 650 BC tell the story of the Lo Shu (洛书) or "scroll of the river Lo".^[11] According to the legend, there was at one time in ancient China a huge flood. While the great king Yu was trying to channel the water out to sea, a turtle emerged from it with a curious pattern on its shell: a 3×3 grid in which circular dots of numbers were arranged, such that the sum of the numbers in each row, column and diagonal was the same: 15. According to the legend, thereafter people were able to use this pattern in a certain way to control the river and protect themselves from floods. The Lo Shu Square, as the magic square on the turtle shell is called, is the unique normal magic square of order three in which 1 is at the bottom and 2 is in the upper right corner. Every normal magic square of order three is obtained from the Lo Shu by rotation or reflection.

Magic square in Parshavnath temple

Magic Square at the Parshvanatha temple, in Khajuraho India

There is a well-known 12th-century 4×4 magic square inscribed on the wall of the Parshvanath temple in Khajuraho, India.^[29][28][41]

This is known as the Chautisa Yantra since its magic sum is 34. It is one of the three 4×4 pandiagonal magic squares and is also an instance of the most-perfect magic square. The study of this square led to the appreciation of pandiagonal squares by European mathematicians in the late 19th century. Pandiagonal squares were referred to as Nasik squares or Jain squares in older English literature.

Albrecht Dürer's magic square

Detail of Melencolia I

The order four magic square Albrecht Dürer immortalized in his 1514 engraving Melencolia I,, referred to above, is believed to be the first seen in European art. The square associated with Jupiter appears as a talisman used to drive away melancholy. It is very similar to Yang Hui's square, which was created in China about 250 years before Dürer's time. The sum 34 can be found in the rows, columns, diagonals, each of the quadrants, the center four squares, and the corner squares (of the 4×4 as well as the four contained 3×3 grids). This sum can also be found in the four outer numbers clockwise from the corners (3+8+14+9) and likewise the four counter-clockwise (the locations of four queens in the two solutions of the 4 queens puzzle^[42]), the two sets of four symmetrical numbers (2+8+9+15 and 3+5+12+14), the sum of the middle two entries of the two outer columns and rows (5+9+8+12 and 3+2+15+14), and in four kite or cross shaped quartets (3+5+11+15, 2+10+8+14, 3+9+7+15, and 2+6+12+14). The two numbers in the middle of the bottom row give the date of the engraving: 1514. The numbers 1 and 4 at either side of the date correspond respectively to the letters "A" and "D," which are the initials of the artist.

Dürer's magic square can also be extended to a magic cube.^[43]

Sagrada Família magic square

A magic square on the Sagrada Família church façade

The Passion façade of the Sagrada Família church in Barcelona, conceptualized by Antoni Gaudí and designed by sculptor Josep Subirachs, features a 4×4 magic square: The magic constant of the square is 33, the age of Jesus at the time of the Passion.^[44] Structurally, it is very similar to the Melancholia magic square, but it has had the numbers in four of the cells reduced by 1.

While having the same pattern of summation, this is not a normal magic square because two numbers (10 and 14) are duplicated and two (12 and 16) are absent, failing the rule that an n x n magic square must contain each of the positive integers 1 through n ². Such squares are not generally mathematically interesting and may be described, in a slightly deprecative sense, as trivial. Lee Sallows has pointed out that, due to Subirachs's ignorance of magic square theory, the renowned sculptor made a needless blunder, and supports this assertion by giving several examples of non-trivial 4 x 4 magic squares showing the desired magic constant of 33.^[45]

Similarly to Dürer's magic square, the Sagrada Familia's magic square can also be extended to a magic cube.^[46]

Magic constant

The constant that is the sum of every row, column and diagonal is called the magic constant or magic sum, M. Every normal magic square has a constant dependent on the order n, calculated by the formula , since the sum of is which when divided by the order n is the magic constant. For normal magic squares of orders n = 3, 4, 5, 6, 7, and 8, the magic constants are, respectively: 15, 34, 65, 111, 175, and 260 (sequence A006003 in the OEIS).

Magic square of order 1 is trivial

The 1×1 magic square, with only one cell containing the number 1, is called trivial, because it is typically not under consideration when discussing magic squares; but it is indeed a magic square by definition, if we regard a single cell as a square of order one.

Magic square of order 2 cannot be constructed

Normal magic squares of all sizes can be constructed except 2×2 (that is, where order n = 2).^[47]

Number of magic squares of a given order

Excluding rotations and reflections, there is exactly one 3×3 magic square, exactly 880 4×4 magic squares, and exactly 275,305,224 5×5 magic squares. For the 6×6 case, there are estimated to be approximately 1.8 × 10¹⁹ squares.^[48]

Center of mass

If we think of the numbers in the magic square as masses located in various cells, then the center of mass of a magic square coincides with its geometric center.

Moment of inertia

The moment of inertia of a magic square has been defined as the sum over all cells of the number in the cell times the squared distance from the center of the cell to the center of the square; here the unit of measurement is the width of one cell.^[48] (Thus for example a corner cell of a 3×3 square has a distance of a non-corner edge cell has a distance of 1, and the center cell has a distance of 0.) Then all magic squares of a given order have the same moment of inertia as each other. For the order-3 case the moment of inertia is always 60, while for the order-4 case the moment of inertia is always 340. In general, for the n×n case the moment of inertia is ^[48]

Birkhoff–von Neumann decomposition

Dividing each number of the magic square by the magic constant will yield a doubly stochastic matrix, whose row sums and column sums equal to unity. However, unlike the doubly stochastic matrix, the diagonal sums of such matrices will also equal to unity. Thus, such matrices constitute a subset of doubly stochastic matrix. The Birkhoff–von Neumann theorem states that for any doubly stochastic matrix , there exists real numbers , where and permutation matrices such that

This representation may not be unique in general. By Marcus-Ree theorem, however, there need not be more than terms in any decomposition.^[49] Clearly, this decomposition carries over to magic squares as well, since we can recover a magic square from a doubly stochastic matrix by multiplying it by the magic constant.

Classification of magic squares

Euler diagram of requirements of some types of 4×4 magic squares. Cells of the same colour sum to the magic constant. * In 4×4 most-perfect magic squares, any 2 cells that are 2 cells diagonally apart (including wraparound) sum to half the magic constant, hence any 2 such pairs also sum to the magic constant.

While the classification of magic squares can be done in many way, some useful categories are given below. An n×n square array of integers 1, 2, ..., n² is called:

• Semi-magic square when its rows and columns sum to give the magic constant.
• Simple magic square when its rows, columns, and two diagonals sum to give magic constant and no more. They are also known as ordinary magic squares.
• Associated magic square when it is a magic square with a further property that every number added to the number equidistant, in a straight line, from the center gives n² + 1. They are also called symmetric magic squares. Associated magic squares do not exist for squares of singly even order.
• Pan-diagonal magic square when it is a magic square with a further property that the broken diagonals sum to the magic constant. They are also called panmagic squares, perfect squares, diabolic squares, Jain squares, or Nasik squares. Panmagic squares do not exist for singly even orders. However, singly even non-normal squares can be panmagic.
• Ultra magic square when it is a both an associated as well as pan-diagonal magic square. Ultra magic square exist only for orders n ≥ 5.
• Bordered magic square when it is a magic square and it remains magic when the rows and columns at the outer edge is removed. They are also called concentric bordered magic squares.
• Composite magic square when it is a magic square that can be partitioned into smaller magic squares, which may or may not overlap with each other.
• Most perfect magic square when it is a pandiagonal magic square with two further properties (i) each 2×2 subsquare add to 1/k of the magic constant where n = 4k, and (ii) all pairs of integers distant n/2 along any diagonal (major or broken) are complementary (i.e. they sum to n² + 1). The first property is referred to as compactness, while the second property is referred to as completeness. Most perfect magic squares exist only for squares of doubly even order. All the pan-diagonal squares of order 4 are also most perfect.

Transformations that preserve the magic property

• A magic square remains magic when its numbers are multiplied by any fixed number.^[50]
• A magic square remains magic when its numbers are added to or subtracted from any fixed number. In particular, if every element in a normal magic square is subtracted from n² + 1, we obtain the complement of the original square.^[50] In the example below, a elements of 4×4 square on the left is subtracted from 17 to obtain the complement of the square on the right.

• Any magic square can be rotated and reflected to produce 8 trivially distinct squares. In magic square theory, all of these are generally deemed equivalent and the eight such squares are said to make up a single equivalence class.^[51][50] In discussing magic squares, equivalent squares are usually not considered as distinct. The 8 equivalent squares is given for the 3×3 magic square below:

• Given any magic square, another magic square of the same order can be formed by interchanging the row and the column which intersect in a cell on a diagonal with the row and the column which intersect in the complementary cell (i.e. cell symmetrically opposite from the center) of the same diagonal.^[50][40] For an even square, there are n/2 pairs of rows and columns that can be interchanged; thus we can obtain 2^n/2 equivalent magic squares by combining such interchanges. For odd square, there are (n - 1)/2 pairs of rows and columns that can be interchanged; and 2^(n-1)/2 equivalent magic squares obtained by combining such interchanges. Interchanging all the rows and columns rotates the square by 180 degree. In the example using a 4×4 magic square, the left square is the original square, while the right square is the new square obtained by interchanging the 1st and 4th rows and columns.

• Given any magic square, another magic square of the same order can be formed by interchanging two rows on one side of the center line, and then interchanging the corresponding two rows on the other side of the center line; then interchanging like columns. For an even square, since there are n/2 same sided rows and columns, there are n(n - 2)/8 pairs of such rows and columns that can be interchanged. Thus we can obtain 2^n(n-2)/8 equivalent magic squares by combining such interchanges. For odd square, since there are (n - 1)/2 same sided rows and columns, there are (n - 1)(n - 3)/8 pairs of such rows and columns that can be interchanged. Thus, there are 2^{(n - 1)(n - 3)/8} equivalent magic squares obtained by combining such interchanges. Interchanging every possible pairs of rows and columns rotates each quadrant of the square by 180 degree. In the example using a 4×4 magic square, the left square is the original square, while the right square is the new square obtained by this transformation. In the middle square, row 1 has been interchanged with row 2; and row 3 and 4 has been interchanged. Note that the middle square is also a magic square, since the original square is an associative magic square.

• A magic square remains magic when any of its non-central rows x and y are interchanged, along with the interchange of their complementary rows n - x + 1 and n - y + 1; and then interchanging like columns. This is a generalization of the above two transforms. When y = n - x + 1, this transform reduces to the first of the above two transforms. When x and y are on the same side of the center line, this transform reduces to the second of the above two transforms. In the example below, the original square is on the left side, while the final square on the right. The middle square has been obtained by interchanging rows 1 and 3, and rows 2 and 4 of the original square. The final square on the right is obtained by interchanging columns 1 and 3, and columns 2 and 4 of the middle square. In this example, this transform amounts to interchanging the quadrants diagonally. Since the original square is associative, the middle square also happens to be magic.

• A magic square remains magic when its quadrants are diagonally interchanged. This is exact for even ordered squares. For odd ordered square, the halves of the central row and central column also needs to be interchanged.^[50] Examples for even and odd squares are given below:

• An associated magic square remains an associated magic square when two rows or columns equidistant from the center are interchanged. For an even square, there are n/2 pairs of rows or columns that can be interchanged; thus we can obtain 2^n/2 × 2^n/2 = 2ⁿ equivalent magic squares by combining such interchanges. For odd square, there are (n - 1)/2 pairs of rows or columns that can be interchanged; and 2^n-1 equivalent magic squares obtained by combining such interchanges. Interchanging all the rows flips the square vertically (i.e. reflected along the horizontal axis), while interchanging all the columns flips the square horizontally (i.e. reflected along the vertical axis). In the example below, a 4×4 associated magic square on the left is transformed into a square on the right by interchanging the 2nd and 3rd row, yielding the famous Durer's magic square.

• An associated magic square remains an associated magic square when two same sided rows (or columns) are interchanged along with corresponding other sided rows (or columns). For an even square, since there are n/2 same sided rows (or columns), there are n(n - 2)/8 pairs of such rows (or columns) that can be interchanged. Thus, we can obtain 2^n(n-2)/8 × 2^n(n-2)/8 = 2^n(n-2)/4 equivalent magic squares by combining such interchanges. For odd square, since there are (n - 1)/2 same sided rows or columns, there are (n - 1)(n - 3)/8 pairs of such rows or columns that can be interchanged. Thus, there are 2^{(n - 1)(n - 3)/8} × 2^{(n - 1)(n - 3)/8} = 2^{(n - 1)(n - 3)/4} equivalent magic squares obtained by combining such interchanges. Interchanging all the same sided rows flips each quadrants of the square vertically, while interchanging all the same sided columns flips each quadrant of the square horizontally. In the example below, the original square is on the left, whose rows 1 and 2 are interchanged with each other, along with rows 3 and 4, to obtain the transformed square on the right.

• A pan-diagonal magic square remains a pan-diagonal magic square under cyclic shifting of rows or of columns or both.^[50] This allows us to position a given number in any one of the n² cells of an n order square. Thus, for a given pan-magic square, there are n² equivalent pan-magic squares. In the example below, the original square on the left is transformed by shifting the 1st row to the bottom to obtain a new pan-magic square in the middle. Next, the 1st and 2nd column of the middle pan-magic square is circularly shifted to the right to obtain a new pan-magic square on the right.

• A bordered magic square remains a bordered magic square after permuting the border cells in the rows or columns, together with their corresponding complementary terms, keeping the corner cells fixed. Since the cells in each row and column of every concentric border can be permuted independently, when the order n ≥ 5 is odd, there are ((n-2)! × (n-4)! × ··· × 3!)² equivalent bordered squares. When n ≥ 6 is even, there are ((n-2)! × (n-4)! × ··· × 4!)² equivalent bordered squares. In the example below, a square of order 5 is given whose border row has been permuted. We can obtain (3!)² = 36 such equivalent squares.

• A bordered magic square remains a bordered magic square after each of its concentric borders are independently rotated or reflected with respect to the central core magic square. If there are b borders, then this transform will yield 8^b equivalent squares. In the example below of the 5×5 magic square, the border has been rotated 90 degrees anti-clockwise.

• A composite magic square remains a composite magic square when the embedded magic squares undergo transformations that do not disturb the magic property (e.g. rotation, reflection, shifting of rows and columns, and so on).

Method for constructing a magic square of order 3

In the 19th century, Édouard Lucas devised the general formula for order 3 magic squares. Consider the following table made up of positive integers a, b and c:

These 9 numbers will be distinct positive integers forming a magic square so long as 0 < a < b < c − a and b ≠ 2a. Moreover, every 3×3 square of distinct positive integers is of this form.

In 1997 Lee Sallows discovered that leaving aside rotations and reflections, then every distinct parallelogram drawn on the Argand diagram defines a unique 3×3 magic square, and vice versa, a result that had never previously been noted.^[51]

Method for constructing a magic square of odd order

Yang Hui's construction method

A method for constructing magic squares of odd order was published by the French diplomat de la Loubère in his book, A new historical relation of the kingdom of Siam (Du Royaume de Siam, 1693), in the chapter entitled The problem of the magical square according to the Indians.^[55] The method operates as follows:

The method prescribes starting in the central column of the first row with the number 1. After that, the fundamental movement for filling the squares is diagonally up and right, one step at a time. If a filled square is encountered, one moves vertically down one square instead, then continues as before. When an "up and to the right" move would leave the square, it is wrapped around to the last row or first column, respectively.

Starting from other squares rather than the central column of the first row is possible, but then only the row and column sums will be identical and result in a magic sum, whereas the diagonal sums will differ. The result will thus be a semimagic square and not a true magic square. Moving in directions other than north east can also result in magic squares.

A method of constructing a magic square of doubly even order

Doubly even means that n is an even multiple of an even integer; or 4p (e.g. 4, 8, 12), where p is an integer.

Generic pattern All the numbers are written in order from left to right across each row in turn, starting from the top left hand corner. Numbers are then either retained in the same place or interchanged with their diametrically opposite numbers in a certain regular pattern. In the magic square of order four, the numbers in the four central squares and one square at each corner are retained in the same place and the others are interchanged with their diametrically opposite numbers.

A construction of a magic square of order 4 Go left to right through the square counting and filling in on the diagonals only. Then continue by going left to right from the top left of the table and fill in counting down from 16 to 1. As shown below.

An extension of the above example for Orders 8 and 12 First generate a pattern table, where a '1' indicates selecting from the square where the numbers are written in order 1 to n² (left-to-right, top-to-bottom), and a '0' indicates selecting from the square where the numbers are written in reverse order n² to 1. For M = 4, the pattern table is as shown below (third matrix from left). When we shade the unaltered cells (cells with '1'), we get a criss-cross pattern.

Note that a) there are equal number of '1's and '0's in each row and column; b) each row and each column are "palindromic"; c) the left- and right-halves are mirror images; and d) the top- and bottom-halves are mirror images (c & d imply b). The pattern table can be denoted using hexadecimals as (9, 6, 6, 9) for simplicity (1-nibble per row, 4 rows). The simplest method of generating the required pattern for higher ordered doubly even squares is to copy the generic pattern for the fourth order square in each four-by-four sub-squares.

For M = 8, possible choices for the pattern are (99, 66, 66, 99, 99, 66, 66, 99); (3C, 3C, C3, C3, C3, C3, 3C, 3C); (A5, 5A, A5, 5A, 5A, A5, 5A, A5) (2-nibbles per row, 8 rows).

For M = 12, the pattern table (E07, E07, E07, 1F8, 1F8, 1F8, 1F8, 1F8, 1F8, E07, E07, E07) yields a magic square (3-nibbles per row, 12 rows.) It is possible to count the number of choices one has based on the pattern table, taking rotational symmetries into account.

Euler's method

This method consists of first constructing two preliminary squares, which when added gives the magic square. As a running example, we will consider a 3×3 magic square. We can uniquely label each number of the 3×3 square by a pair of numbers as

where every pair of Greek and Latin alphabets, e.g. αa, are meant to be added together, i.e. αa = α + a. Here, (α, β, γ) = (0, 3, 6) and (a, b, c) = (1, 2, 3). The numbers 0, 3, and 6 are referred to as the root numbers while the numbers 1, 2, and 3 are referred to as the primary numbers. An important general constraint to note here is that

• a Greek letter is paired with a Latin letter only once.

Thus, the original square can now be split into two simpler squares:

The lettered squares are referred to as Greek square or Latin square if the they are filled with Greek or Latin letters, respectively. A magic square can be constructed by ensuring that the Greek and Latin squares are magic squares too, provided that the Greek and Latin alphabets are paired with each other only once. The converse of this statement is also often, but not always (e.g. bordered magic squares), true: A magic square can be decomposed into a Greek and a Latin square, which are themselves magic squares. Thus the method is useful for both synthesis as well as analysis of a magic square. Lastly, by examining the pattern in which the numbers are laid out in the finished square, it is often possible to come up with a faster algorithm to construct higher order squares that replicate the given pattern, without the necessity of creating the preliminary Greek and Latin squares.

During the construction of the 3×3 magic square, the Greek and Latin squares with just three unique terms are much easier to deal with than the original square with nine different terms. The row sum and the column sum of the Greek square will be the same, α + β + γ, if

• each letter appears exactly once in a given column or a row.

This can be achieved by cyclic permutation of α, β, and γ. Satisfaction of these two conditions ensures that the resulting square is a semi-magic square; and such Greek and Latin squares are said to be mutually orthogonal to each other. To construct a magic square, we should also ensure that the diagonals sum to magic constant. For this, we have a third condition:

• either all the letters should appear exactly once in both the diagonals; or in case of odd ordered squares, one of the diagonals should consist entirely of the middle term, while the other diagonal should have all the letters exactly once.

The mutually orthogonal Greek and Latin squares that satisfy the first part of the third condition (that all letters appear in both the diagonals) are said to be mutually orthogonal doubly diagonal Graeco-Latin squares.

Odd squares: For the odd square, since α, β, and γ are in arithmetic progression, their sum is equal to the product of the square's order and the middle term, i.e. α + β + γ = 3 β. Thus, the diagonal sums will be equal if we have βs in the main diagonal and α, β, γ in the skew diagonal. Similarly, for the Latin square. The resulting Greek and Latin squares and their combination will be as below. Note that the Latin square is just a rotation of the Greek square with the corresponding letters interchanged. Substituting the values of the Greek and Latin letters will give the 3×3 magic square.

For the odd squares, this method explains why the Siamese method (method of De la Loubere) and its variants work. This basic method can be used to construct odd ordered magic squares of higher orders. To summarise:

• For odd ordered squares, to construct Greek square, place the middle term along the main diagonal, and place the rest of the terms along the skew diagonal. The remaining empty cells are filled by diagonal moves. The Latin square can be constructed by rotating or flipping the Greek square, and replacing the corresponding alphabets. The magic square is obtained by adding the Greek and Latin squares.

A peculiarity of the construction method given above for the odd magic squares is that the middle number (n² + 1)/2 will always appear at the center cell of the magic square. Since there are (n - 1)! ways to arrange the skew diagonal terms, we can obtain (n - 1)! Greek squares this way; same with the Latin squares. Also, since each Greek square can be paired with (n - 1)! Latin squares, and since for each of Greek square the middle term may be arbitrarily placed in the main diagonal or the skew diagonal (and correspondingly along the skew diagonal or the main diagonal for the Latin squares), we can construct a total of 2 × (n - 1)! × (n - 1)! magic squares using this method. For n = 3, 5, and 7, this will give 8, 1152, and 1,036,800 different magic squares, respectively. Dividing by 8 to neglect equivalent squares due to rotation and reflections, we obtain 1, 144, and 129,600 essentially different magic squares, respectively.

As another example, the construction of 5×5 magic square is given. Numbers are directly written in place of alphabets. The numbered squares are referred to as primary square or root square if they are filled with primary numbers or root numbers, respectively. The numbers are placed about the skew diagonal in the root square such that the middle column of the resulting root square has 0, 5, 10, 15, 20 (from bottom to top). The primary square is obtained by rotating the root square counter-clockwise by 90 degrees, and replacing the numbers. The resulting square is an associative magic square, in which every pair of numbers symmetrically opposite to the center sum up to the same value, 26. For e.g., 16+10, 3+23, 6+20, etc. In the finished square, 1 is placed at center cell of bottom row, and successive numbers are placed via elongated knight's move (two cells right, two cells down). When a collision occurs, the break move is to move one cell up. Also note that all the odd numbers occur inside the central diamond formed by 1, 5, 25 and 21, while the even numbers are placed at the corners. We have re-created the so called lozenge method.

A variation of the above example, where the skew diagonal sequence is taken in different order, is given below. The resulting magic square is the famous Agrippa's Mars magic square. It is an associative magic square and is the same as that produced by Moschopoulos's method. Here the resulting square starts with 1 placed in the cell which is to the right of the centre cell, and proceeds as De la Loubere's method, with downwards-right move. When a collision occurs, the break move is to shift two cells to the right.

In the previous examples, for the Greek square, the second row can be obtained from the first row by circularly shifting it to the right by one cell. Similarly, the third row is a circularly shifted version of the second row by one cell to the right; and so on. Likewise, the rows of the Latin square is circularly shifted to the left by one cell. Note that the row shifts for the Greek and Latin squares are in mutually opposite direction. It is possible to circularly shift the rows by more than one cell to create the Greek and Latin square.

• For odd ordered squares, whose order is not divisible by three, we can create the Greek squares by shifting a row by two places to the left or to the right to form the next row. The Latin square is made by flipping the Greek square along the main diagonal and interchanging the corresponding letters. This gives us a Latin square whose rows are created by shifting the row in the direction opposite to that of the Greek square. A Greek square and a Latin square should be paired such that their row shifts are in mutually opposite direction. The magic square is obtained by adding the Greek and Latin squares. When the order also happens to be a prime number, this method always creates pandiagonal magic square.

This essentially re-creates the knight's move. All the letters will appear in both the diagonals, ensuring correct diagonal sum. Since there are n! permutations of the Greek letters by which we can create the first row of the Greek square, there are thus n! Greek squares that can be created by shifting the first row in one direction. Likewise, there are n! such Latin squares created by shifting the first row in the opposite direction. Since a Greek square can be combined with any Latin square with opposite row shifts, there are n! × n! such combinations. Lastly, since the Greek square can be created by shifting the rows either to the left or to the right, there are a total of 2 × n! × n! magic squares that can be formed by this method. For n = 5 and 7, since they are prime numbers, this method creates 28,800 and 50,803,200 pandiagonal magic squares. Dividing by 8 to neglect equivalent squares due to rotation and reflections, we obtain 3,600 and 6,350,400 equivalent squares. Further dividing by n² to neglect equivalent panmagic squares due to cyclic shifting of rows or columns, we obtain 144 and 129,600 essentially different panmagic squares. The condition that the square's order not be divisible by 3 means that we cannot construct squares of orders 9, 15, 21, 27, and so on, by this method.

In the example below, the square has been constructed such that 1 is at the center cell. In the finished square, the numbers can be continuously enumerated by the knight's move (two cells up, one cell right). When collision occurs, the break move is to move one cell up, one cell left.The resulting square is a pandiagonal magic square. The central 3×3 square is also a magic square with constant 33. This square also has a further diabolical property that any five cells in quincunx pattern formed by any odd sub-square, including wrap around, sum to the magic constant, 65. For e.g., 13+7+1+20+24, 23+1+9+15+17, 13+21+10+19+2 etc. Also the four corners of any 5×5 square and the central cell, as well as the middle cells of each side together with the central cell, including wrap around, give the magic sum: 13+10+19+22+1 and 20+24+12+8+1. Lastly the four rhomboids that form elongated crosses also give the magic sum: 23+1+9+24+8, 15+1+17+20+12, 14+1+18+13+19, 7+1+25+22+10.

We can also combine the Greek and Latin squares constructed by different methods. In the example below, the primary square is made using knight's move. We have re-created the magic square obtained by De la Loubere's method. As before, we can form 4 × (n - 1)! × n! magic squares by this combination.

Even squares: We can also construct even ordered squares in this fashion. Since there is no middle term among the Greek and Latin alphabets for even ordered squares, in addition to the first two constraint, for the diagonal sums to yield the magic constant, all the letters in the alphabet should appear in the main diagonal and the skew diagonal in the Greek and Latin square of even order.

An example of a 4×4 square is given below. For the given diagonal and skew diagonal in the Greek square, the rest of the cells can be filled using the condition that each letter appear only once in a row and a column.

Similarly, an 8×8 magic square can be constructed as below. Here the order of appearance of the numbers is not important; however the quadrants imitate the layout pattern of the 4×4 Graeco-Latin squares.

Euler's method has given rise to the study of Graeco-Latin squares. Euler's method for constructing magic squares is valid for any order except 2 and 6.

Variations: Magic squares constructed from mutually orthogonal doubly diagonal Graeco-Latin squares are interesting in themselves since the magic property emerges from the relative position of the alphabets in the square, and not due to any arithmetic property of the value assigned to them. This means that we can assign any value to the alphabets of such squares and still obtain a magic square. This is the basis for constructing squares that display some information (e.g. birthdays, years, etc) in the square. For example, we can display the number π ≈ 3·141592 at the bottom row of a 4×4 magic square using the Graeco-Latin square given above by assigning (α, β, γ, δ) = (10, 0, 90, 15) and (a, b, c, d) = (0, 2, 3, 4). We will obtain the following non-normal magic square with the magic sum 124:

Narayana-De la Hire's method for even orders

Narayana-De la Hire's method for odd square is the same as that of Euler's. However, for even squares, we drop the second requirement that each Greek and Latin letter appear only once in a given row or column. This allows us to take advantage of the fact that the sum of an arithmetic progression with an even number of terms is equal to the sum of two opposite symmetric terms multiplied by half the total number of terms. Thus, when constructing the Greek or Latin squares,

• for even ordered squares, a letter can appear n/2 times in a column but only once in a row, or vice versa.

As a running example, if we take a 4×4 square, where the Greek and Latin terms have the values (α, β, γ, δ) = (0, 4, 8, 12) and (a, b, c, d) = (1, 2, 3, 4), respectively, then we have α + β + γ + δ = 2 (α + δ) = 2 (β + γ). Similarly, a + b + c + d = 2 (a + d) = 2 (b + c). This means that the complementary pair α and δ (or β and γ) can appear twice in a column (or a row) and still give the desired magic sum. Thus, we can construct:

• For even ordered squares, the Greek magic square is made by first placing the Greek alphabets along the main diagonal in some order. The skew diagonal is then filled in the same order or by picking the terms that are complementary to the terms in the main diagonal. Finally, the remaining cells are filled column wise. Given a column, we use the complementary terms in the diagonal cells intersected by that column, making sure that they appear only once in a given row but n/2 times in the given column. The Latin square is obtained by flipping or rotating the Greek square and interchanging the corresponding alphabets. The final magic square is obtained by adding the Greek and Latin squares.

In the example given below, the main diagonal (from top left to bottom right) is filled with sequence ordered as α, β, γ, δ, while the skew diagonal (from bottom left to top right) filled in the same order. The remaining cells are then filled column wise such that the complementary letters appears only once within a row, but twice within a column. In the first column, since α appears on the 1st and 4th row, the remaining cells are filled with its complementary term δ. Similarly, the empty cells in the 2nd column are filled with γ; in 3rd column β; and 4th column α. Each Greek letter appears only once along the rows, but twice along the columns. As such, the row sums are α + β + γ + δ while the column sums are either 2 (α + δ) or 2 (β + γ). Likewise for the Latin square, which is obtained by flipping the Greek square along the main diagonal and interchanging the corresponding letters.

The above example explains why the "criss-cross" method for doubly even magic square works. Another possible 4×4 magic square, which is also pan-diagonal as well as most-perfect, is constructed below using the same rule. However, the diagonal sequence is chosen such that all four letters α, β, γ, δ appear inside the central 2×2 sub-square. Remaining cells are filled column wise such that each letter appears only once within a row. In the 1st column, the empty cells need to be filled with one of the letters selected from the complementary pair α and δ. Given the 1st column, the entry in the 2nd row can only be δ since α is already there in the 2nd row; while, in the 3rd row the entry can only be α since δ is already present in the 3rd row. We proceed similarly until all cells are filled. The Latin square given below has been obtained by flipping the Greek square along the main diagonal and replacing the Greek alphabets with corresponding Latin alphabets.

We can use this approach to construct singly even magic squares as well. However, we have to be more careful in this case since the criteria of pairing the Greek and Latin alphabets uniquely is not automatically satisfied. Violation of this condition leads to some missing numbers in the final square, while duplicating others. Thus, here is an important proviso:

• For singly even squares, in the Greek square, check the cells of the columns which is vertically paired to its complement. In such a case, the corresponding cell of the Latin square must contain the same letter as its horizontally paired cell.

Below is a construction of a 6×6 magic square, where the numbers are directly given, rather than the alphabets. The second square is constructed by flipping the first square along the main diagonal. Here in the first column of the root square the 3rd cell is paired with its complement in the 4th cells. Thus, in the primary square, the numbers in the 1st and 6th cell of the 3rd row are same. Likewise, with other columns and rows. In this example the flipped version of the root square satisfies this proviso.

As another example of a 6×6 magic square constructed this way is given below. Here the diagonal entries are arranged differently. The primary square is constructed by flipping the root square about the main diagonal. In the second square the proviso for singly even square is not satisfied, leading to a non-normal magic square (third square) where the numbers 3, 13, 24, and 34 are duplicated while missing the numbers 4, 18, 19, and 33.

The last condition is a bit arbitrary and may not always need to be invoked, as in this example:

As one more example, we have generated an 8×8 magic square. Unlike the criss-cross pattern of the earlier section for evenly even square, here we have a checkered pattern for the altered and unaltered cells. Also, in each quadrant the odd and even numbers appear in alternating columns.

Variations: A number of variations of the basic idea are possible: a complementary pair can appear n/2 times or less in a column. That is, a column of a Greek square can be constructed using more than one complementary pair. This method allows us to imbue the magic square with far richer properties. The idea can also be extended to the diagonals too. An example of an 8×8 magic square is given below. In the finished square each of four quadrants are pan-magic squares as well, each quadrant with same magic constant 130.

Bordering method for order 3

In this method, the objective is to wrap a border around a smaller magic square which serves as a core. Let us consider the 3×3 square for example. Subtracting the middle number 5 from each number 1, 2, ..., 9, we obtain 0, ± 1, ± 2, ± 3, and ± 4, which we will refer to as bone numbers. The magic constant of a magic square, which we will refer to as the skeleton square, made by these bone numbers will be zero since adding all the rows of a magic square will give nM = Σ k = 0; thus M = 0.

It is not difficult to argue that the middle number should be placed at the center cell: let x be the number placed in the middle cell, then the sum of the middle column, middle row, and the two diagonals give Σ k + 3 x = 4 M. Since Σ k = 3 M, we have x = M / 3. Here M = 0, so x = 0.

Putting the middle number 0 in the center cell, we want to construct a border such that the resulting square is magic. Let the border be given by:

Since the sum of each row, column, and diagonals must be a constant (which is zero), we have

a + a* = 0,

b + b* = 0,

u + u* = 0,

v + v* = 0.

Thus, it must be that if we have chosen a, b, u, and v, then we have a* = - a, b* = - b, u* = - u, and v* = - v. But how should we choose a, b, u, and v? We have the sum of the top row and the sum of the right column as

u + a + v = 0,

v + b + u* = 0.

Since 0 is an even number, there are only two ways that the sum of three integers will yield an even number: 1) if all three were even, or 2) if two were odd and one was even. Since in our choice of numbers we only have two even non-zero number (± 2 and ± 4), the first statement is false. Hence, it must be the case that the second statement is true: that two of the numbers are odd and one even.

The only way that both the above two equations can satisfy this parity condition simultaneously, and still be consistent with the set of numbers we have, is when u and v are odd. For on the contrary, if we had assumed u and a to be odd and v to be even in the first equation, then u* = - u will be odd in the second equation, making b odd as well, in order to satisfy the parity condition. But this requires three odd numbers (u, a, and b), contradicting the fact that we only have two odd numbers (± 1 and ± 3) which we can use.

Thus, taking u = 1 and v = 3, we have a = - 4 and b = - 2. Hence, the finished skeleton square will be as in the left. Adding 5 to each number, we get the finished magic square.

Similar argument can be used to construct larger squares.

Bordering method for order 5

Let us consider the fifth order square. For this, we have a 3×3 magic core, around which we will wrap a magic border. The bone numbers to be used will be ± 5, ± 6, ± 7, ± 8, ± 9, ± 10, ± 11, and ± 12. Disregarding the signs, we have 8 bone numbers, 4 of which are even and 4 of which are odd. Let the magic border be given as

As before, we should

• place a bone number and its complement opposite to each other, so that the magic sum will be zero.

It is sufficient to determine the numbers u, v, a, b, c, d, e, f to describe the magic border. As before, we have the two constraint equations for the top row and right column:

u + a + b + c + v = 0

v + d + e + f + u* = 0.

Multiple solutions are possible. The standard procedure is to

• first try to determine the corner cells, after which we will try to determine the rest of the border.

There are 28 ways of choosing two numbers from the set of 8 bone numbers for the corner cells u and v. However, not all pairs are admissible. Among the 28 pairs, 16 pairs are made of an even and an odd number, 6 pairs have both as even numbers, while 6 pairs have them both as odd numbers.

We can prove that the corner cells u and v cannot have an even and an odd number. This is because if this were so, then the sums u + v and v + u* will be odd, and since 0 is an even number, the sums a + b + c and d + e + f should be odd as well. The only way that the sum of three integers will result in an odd number is when 1) two of them are even and one is odd, or 2) when all three are odd. Since the corner cells are assumed to be odd and even, neither of these two statements are compatible with the fact that we only have 3 even and 3 odd bone numbers at our disposal. This proves that u and v cannot have different parity. This eliminates 16 possibilities.

Now consider the case when both u and v are even. The 6 possible pairs are: (6, 8), (6, 10), (6, 12), (8, 10), (8, 12), and (10, 12). Since the sums u + v and v + u* are even, the sums a + b + c and d + e + f should be even as well. The only way that the sum of three integers will result in an even number is when 1) two of them are odd and one is even, or 2) when all three are even. The fact that the two corner cells are even means that we have only 2 even numbers at our disposal. Thus, the second statement is not compatible with this fact. Hence, it must be the case that the first statement is true: two of the three numbers should be odd, while one be even.

Let a, b, d, e be odd numbers while c and f be even numbers. Given the odd bone numbers at our disposal: ± 5, ± 7, ± 9, and ± 11, their differences range from D = { ± 2, ± 4, ± 6} while their sums range from S = {± 12, ± 14, ± 16, ± 18, ± 20}. It is also useful to have a table of their sum and differences for later reference. Now, given the corner cells (u, v), we can check its admissibility by checking if the sums u + v + c and v + u* + f fall within the set D or S. The admissibility of the corner numbers is a necessary but not a sufficient condition for the solution to exist.

For example, if we consider the pair (u, v) = (8, 12), then u + v = 20 and v + u* = 6; and we will have ± 6 and ± 10 even bone numbers at our disposal. Taking c = ± 6, we have the sum u + v + c to be 26 and 14, both of which do not fall within the sets D or S. Likewise, taking c = ± 10, we have the sum u + v + c to be 30 and 10, both of which again do not fall within the sets D or S. Thus, the pair (8, 12) is not admissible. By similar process of reasoning, we can also rule out the pair (6, 12).

As another example, if we consider the pair (u, v) = (10, 12), then u + v = 22 and v + u* = 2; and we will have ± 6 and ± 8 even bone numbers at our disposal. Taking c = ± 6, we have the sum u + v + c to be 28 and 16. While 28 does not fall within the sets D or S, 16 falls in set S. By inspection, we find that if (a, b) = (-7, -9), then a + b = -16; and it will satisfy the first constraint equation. Also, taking f = ± 8, we have the sum v + u* + f to be 10 and -6. While 10 does not fall within the sets D or S, -6 falls in set D. Since -7 and -9 have already been assigned to a and b, clearly (d, e) = (-5, 11) so that d + e = 6; and it will satisfy the second constraint equation.

Likewise, taking c = ± 8, we have the sum u + v + c to be 30 and 14. While 30 does not fall within the sets D or S, 14 falls in set S. By inspection, we find that if (a, b) = (-5, -9), then a + b = -14. Also, taking f = ± 6, we have the sum v + u* + f to be 8 and -4. While 8 does not fall within the sets D or S, -4 falls in set D. Clearly, (d, e) = (-7, 11) so that d + e = 4, and the second constraint equation will be satisfied.

Hence the corner pair (u, v) = (10, 12) is admissible; and it admits two solutions: (a, b, c, d, e, f) = (-7, -9, -6, -5, 11, -8) and (a, b, c, d, e, f) = ( -5, -9, -8, -7, 11, -6). The finished skeleton squares are given below. The magic square is obtained by adding 13 to each cells.

Using similar process of reasoning, we can construct the following table for the values of u, v, a, b, c, d, e, f expressed as bone numbers as given below. There are only 6 possible choices for the corner cells, which leads to 10 possible border solutions.

Given this group of 10 borders, we can construct 10×8×(3!)² = 2880 essentially different magic squares. Here the bone numbers ± 5, ..., ± 12 were consecutive. More bordered squares can be constructed if the numbers are not consecutive. Voille lists a total of 594 such magic borders for order 5 square; thus there are 594×8×(3!)² = 171,072 essentially different magic squares.

Continuous enumeration methods

Exhaustive enumeration of all the borders of a magic square of a given order, as done previously, is very tedious. As such a structured solution is often desirable, which allows us to construct a border for a square of any order. Below we give three algorithms for constructing border for odd, evenly even, and evenly odd squares. These continuous enumeration algorithms were discovered in 10th century by Islamic scholars; and their earliest surviving exposition comes from the two treatises by al-Buzjani and al-Antaki, although they themselves were not the discoverers.^[16]

Odd ordered squares: The following is the algorithm given by al-Buzjani to construct a border for odd squares. A peculiarity of this method is that for order n square, the two adjacent corners comprise of numbers n - 1 and n + 1.

Starting from the cell above the lower left corner, we put the numbers alternately in left column and bottom row until we arrive at the middle cell. The next number is written in the middle cell of the bottom row just reached, after which we fill the cell in the upper left corner, then the middle cell of the right column, then the upper right corner. After this, starting from the cell above middle cell of the right column already filled, we resume the alternate placement of the numbers in the right column and the top row. Once half of the border cells are filled, the other half are filled by numbers complementary to opposite cells. The subsequent inner borders is filled in the same manner, until the square of order 3 is filled.^[16]

Below is an example for 9th order square.

Evenly even order: The following is the method given by al-Antaki. Consider an empty border of order n = 4k with k ≥ 3. The peculiarity of this algorithm is that the adjacent corner cells are occupied by numbers n and n - 1.

Starting at the upper left corner cell, we put the successive numbers by groups of four, the first one next to the corner, the second and the third on the bottom, and the fourth at the top, and so on until there remains in the top row (excluding the corners) six empty cells. We then write the next two numbers above and the next four below. We then fill the upper corners, first left then right. We place the next number below the upper right corner in the right column, the next number on the other side in the left column. We then resume placing groups of four consecutive numbers in the two columns as before. Once half of the border cells are filled, the other half are filled by numbers complementary to opposite cells.^[16]

The example below gives the border for order 16 square.

For order 8 square, we just begin directly with the six cells.

Evenly odd order: For evenly odd order, we have the algorithm given by al-Antaki. Here the corner cells are occupied by n and n - 1. Below is an example of 10th order square.

Start by placing 1 at the bottom row next to the left corner cell, then place 2 in the top row. After this, place 3 at the bottom row and turn around the border in anti-clockwise direction placing the next numbers, until n - 2 is reached on the right column. The next two numbers are placed in the upper corners (n - 1 in upper left corner and n in upper right corner). Then, the next two numbers are placed on the left column, then we resume the cyclic placement of the numbers until half of all the border cells are filled. Once half of the border cells are filled, the other half are filled by numbers complementary to opposite cells.^[16]

For squares of order m × n where m, n > 2

There is a method reminiscent of the Kronecker product of two matrices, that builds an nm × nm magic square from an n × n magic square and an m × m magic square.^[56] The "product" of two magic squares creates a magic square of higher order than the two multiplicands. This product is made as follows: Let the two magic squares be of orders m and n. The final square will be of order m × n. Divide the final m × n square into n × n sub-squares, totaling m × m such sub-squares. In the square of order n, reduce by 1 the value of all the numbers. Multiply these reduced values by m × m. The squares of order m are added n × n times to the sub-squares of the final square. The smallest composite magic square of order 9, composed of two order 3 squares is given below.