Magic square

 ( Matrices ) 

$A\; =\; \backslash theta\_1\; P\_1\; +\; \backslash cdots\; +\; \backslash theta\_k\; P\_k.$

This representation may not be unique in general. By Marcus-Ree theorem, however, there need not be more than $k\; \backslash le\; n^2\; -\; 2n\; +\; 2$ terms in any decomposition. 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.

• 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 or normal magic squares.
• Self-complementary magic square when it is a magic square which when complemented (i.e. each number subtracted from n² + 1) will give a rotated or reflected version of the original magic square.
• Associative 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. All associated magic square are self-complementary magic squares as well.
• Pandiagonal 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 both associative and pandiagonal 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 on the outer edge are removed. They are also called concentric bordered magic squares if removing a border of a square successively gives another smaller bordered magic square. Bordered magic square do not exist for order 4.
• Composite magic square when it is a magic square that is created by "multiplying" (in some sense) smaller magic squares, such that the order of the composite magic square is a multiple of the order of the smaller squares. Such squares can usually be partitioned into smaller non-overlapping magic sub-squares.
• Inlaid magic square when it is a magic square inside which a magic sub-square is embedded, regardless of construction technique. The embedded magic sub-squares are themselves referred to as inlays.
• 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 = 4 k, 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 pandiagonal squares of order 4 are also most perfect.
• Franklin magic square when it is a doubly even magic square with three further properties (i) every bent diagonal adds to the magic constant, (ii) every half row and half column starting at an outside edge adds to half the magic constant, and (iii) the square is compact.
• Multimagic square when it is a magic square that remains magic even if all its numbers are replaced by their k-th power for 1 ≤ k ≤ P. They are also known as P-multimagic square or satanic squares. They are also referred to as bimagic squares, trimagic squares, tetramagic squares, pentamagic squares when the value of P is 2, 3, 4, and 5 respectively.

Low order squares

The number of different n × n magic squares for n from 1 to 6, not counting rotations and reflections is:

1, 0, 1, 880, 275305224, 17753889197660635632.

Magic tori

1, 0, 1, 255, 251449712 .

Higher order squares and tori

The 880 magic squares of order 4 are displayed on 255 magic tori of order 4 and the 275,305,224 squares of order 5 are displayed on 251,449,712 magic tori of order 5. The number of magic tori and distinct normal squares is not yet known for any higher order. Anything but square: from magic squares to Sudoku by Hardeep Aiden, Plus Magazine, March 1, 2006

Algorithms tend to only generate magic squares of a certain type or classification, making counting all possible magic squares quite difficult. Traditional counting methods have proven unsuccessful, statistical analysis using the Monte Carlo method has been applied. The basic principle applied to magic squares is to randomly generate n × n matrices of elements 1 to n² and check if the result is a magic square. The probability that a randomly generated matrix of numbers is a magic square is then used to approximate the number of magic squares.

More intricate versions of the Monte Carlo method, such as the exchange Monte Carlo, and Monte Carlo backtracking have produced even more accurate estimations. Using these methods it has been shown that the probability of magic squares decreases rapidly as n increases. Using fitting functions give the curves seen to the right.

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

• The numbers of a magic square can be substituted with corresponding numbers from a set of s arithmetic progressions with the same common difference among r terms, such that r × s = n², and whose initial terms are also in arithmetic progression, to obtain a non-normal magic square. Here either s or r should be a multiple of n. Let us have s arithmetic progressions given by

:$$

where a is the initial term, c is the common difference of the arithmetic progressions, and d is the common difference among the initial terms of each progression. The new magic constant will be

:$M\; =\; na\; +\; \backslash frac\{n\}\{2\}\; \backslash big.$

If s = r = n, then we have the simplification

:$M\; =\; na\; +\; \backslash frac\{n\}\{2\}(n-1)(c+d).$

If we further have a = c = 1 and d = n, we obtain the usual M = n( n²+1)/2. For given M we can find the required a, c, and d by solving the linear Diophantine equation. In the examples below, we have order 4 normal magic square on the left most side. The second square is a corresponding non-normal magic square with r = 8, s = 2, a = 1, c = 1, and d = 10 such that the new magic constant is M = 38. The third square is an order 5 normal magic square, which is a 90 degree clockwise rotated version of the square generated by De la Loubere method. On the right most side is a corresponding non-normal magic square with a = 4, c = 1, and d = 6 such that the new magic constant is M = 90.

• 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.
  Maurice Kraitchik (2025). ((9780486201634)), Dover Publications, Inc.. . ((9780486201634))
  In discussing magic squares, equivalent squares are usually not considered as distinct. The 8 equivalent squares are 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. 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. The final square on the right is obtained by interchanging columns 1 and 2, and columns 3 and 4 of the middle square. In this particular example, this transform amounts to rotating the quadrants by 180 degree. 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. Examples for even and odd squares are given below:

• An associative magic square remains associative 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 = 2 ⁿ 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 associative magic square on the left is transformed into a square on the right by interchanging the second and third row, yielding the famous Durer's magic square.

• An associative magic square remains associative 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. 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 first 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).

Special methods are standard and most simple ways to construct a magic square. It follows certain configurations / formulas / algorithm which generates regular patterns of numbers in a square. The correctness of these special methods can be proved using one of the general methods given in later sections. After a magic square has been constructed using a special method, the transformations described in the previous section can be applied to yield further magic squares. Special methods are usually referred to using the name of the author(s) (if known) who described the method, for e.g. De la Loubere's method, Starchey's method, Bachet's method, etc.

Magic squares exist for all values of n, except for order 2. Magic squares can be classified according to their order as odd, doubly even ( n divisible by four), and singly even ( n even, but not divisible by four). This classification is based on the fact that entirely different techniques need to be employed to construct these different species of squares. Odd and doubly even magic squares are easy to generate; the construction of singly even magic squares is more difficult but several methods exist, including the LUX method for magic squares (due to John Horton Conway) and the Strachey method for magic squares.

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.

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.

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 Starting from top left, go left to right through each row of the square, counting each cell from 1 to 16 and filling the cells along the diagonals with its corresponding number. Once the bottom right cell is reached, continue by going right to left, starting from the bottom right of the table through each row, and fill in the non-diagonal cells counting up from 1 to 16 with its corresponding number. 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.

The patterns are 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 and 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.

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 here is

• 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 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. 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. For a given order n, there are at most n - 1 squares in a set of mutually orthogonal squares, not counting the variations due to permutation of the symbols. This upper bound is exact when n is a prime number.

In order 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 3×3 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. The Latin square is just a 90 degree anti-clockwise rotation of the Greek square (or equivalently, flipping about the vertical axis) 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), or equivalently, bishop's move (two cells diagonally down right). When a collision occurs, the break move is to move one cell up. 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. The occurrence of the even numbers can be deduced by copying the square to the adjacent sides. The even numbers from four adjacent squares will form a cross.

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 flipped version of 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. 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. For order 5 squares, these are the only panmagic square there are. 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. 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 8 × ( n - 1)! × n! magic squares by this combination. For n = 5 and 7, this will create 23,040 and 29,030,400 magic squares. After dividing by 8 in order to neglect equivalent squares due to rotation and reflection, we get 2,880 and 3,628,800 squares.

For order 5 squares, these three methods give a complete census of the number of magic squares that can be constructed by the method of superposition. Neglecting the rotation and reflections, the total number of magic squares of order 5 produced by the superposition method is 144 + 3,600 + 2,880 = 6,624.

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 in the skew diagonal.

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.

Using these two Graeco-Latin squares, we can construct 2 × 4! × 4! = 1,152 magic squares. Dividing by 8 to eliminate equivalent squares due to rotation and reflections, we get 144 essentially different magic squares of order 4. These are the only magic squares constructible by the Euler method, since there are only two mutually orthogonal doubly diagonal Graeco-Latin squares of order 4.

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 and for creating "reversible squares". For example, we can display the number π ≈ 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:

• 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, where in the root square each cell is vertically paired with its complement:

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.

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.

Now, if we have chosen a, b, u, and v, then we have a* = - a, b* = - b, u* = - u, and v* = - v. This means that if we assign a given number to a variable, say a = 1, then its complement will be assigned to a*, i.e. a* = - 1. Thus out of eight unknown variables, it is sufficient to specify the value of only four variables. We will consider a, b, u, and v as independent variables, while a*, b*, u*, and v* as dependent variables. This allows us to consider a bone number ± x as a single number regardless of sign because (1) its assignment to a given variable, say a, will automatically imply that the same number of opposite sign will be shared with its complement a*, and (2) two independent variables, say a and b, cannot be assigned the same bone number. 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. This proves that the odd bone numbers occupy the corners cells. When converted to normal numbers by adding 5, this implies that the corners of a 3×3 magic square are all occupied by even numbers.

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. Since there does not exist a 2×2 magic square around which we can wrap a border to construct a 4×4 magic square, the next smallest order for which we can construct bordered square is the order 5.

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.

Using similar type reasoning we can also draw some conclusions about the sets { a, b, c} and { d, e, f}. If u and v are both even, then both the sets should have two odd numbers and one even number. If u and v are both odd, then one of the sets should have three even numbers while the other set should have one even number and two odd numbers.

As a running example, 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.

Now 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, depending on the sign of ± 6 taken, 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 bordered magic squares. Here the bone numbers ± 5, ..., ± 12 were consecutive. More bordered squares can be constructed if the numbers are not consecutive. If non-consecutive bone numbers were also used, then there are a total of 605 magic borders. Thus, the total number of order 5 essentially different bordered magic squares (with consecutive and non-consecutive numbers) is 174,240.http://oz.nthu.edu.tw/~u9621110/IT2010/txt/0929/canterburypuzzle00dudeuoft.pdf The Canterbury Puzzles and Other Curious Problems, Henry Ernest Dudeney, 1907 http://budshaw.ca/howMany.html, Bordered Square Numbers, S. Harry White, 2009 See history.http://www.law05.si/iwms/presentations/Styan.pdf Some illustrated comments on 5×5 golden magic matrices and on 5×5 Stifelsche Quadrate, George P. H. Styan, 2014. It is worth noting that the number of fifth-order magic squares constructible via the bordering method is about 26 times larger than via the superposition method.