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Magic squares

Numbers in a square where every row, column and diagonal has the same total

A magic square of order \(n\) holds the numbers \(1, 2, \dots, n^2\) in an \(n \times n\) grid so that every row, every column and both main diagonals add up to the same number, the magic constant. They are among the oldest puzzles in the world, and for centuries people believed they had special powers.

Try it

Fill a 3 × 3 square with the numbers 1 to 9, each once, so that all three rows, all three columns and both diagonals add up to the same total. First work out what the total must be. Then: which number has to go in the middle? Which numbers can go in the corners?

Answer

The numbers add to \(1 + 2 + \dots + 9 = 45\). The three rows share that equally, so each line adds to 15.

The middle square lies on four lines (one row, one column, two diagonals). Add up those four lines: you get \(4 \times 15 = 60\), and you have counted every number once, except the middle, which you counted four times. So \(45 + 3 \times \text{middle} = 60\), and the middle is 5.

The corners must be even. A corner lies on three lines. If it held 1, each of those lines would need two more numbers adding to 14, and only 9 + 5 and 8 + 6 do: two pairs for three lines. The same happens for 3, 7 and 9. Trying the cases leaves 2, 4, 6, 8 in the corners, and the odd numbers on the edges. Up to turning and flipping, there is only one 3 × 3 magic square:

4	9	2
3	5	7
8	1	6

The Lo Shu

That square is the Lo Shu, from Chinese legend: a turtle climbed out of the Luo River with this pattern of dots on its shell, in the time of the flood-taming emperor Yu. The written record is much later than the legend, but by the first centuries CE the square was well known in China, and it travelled through India and the Islamic world to Europe. Its pattern of odd and even numbers, and the pairs adding to 10 on opposite sides of the centre, appear in Chinese cosmology and in feng shui.

How to think about it: the magic constant

For any order \(n\), the numbers \(1\) to \(n^2\) add up to \(\frac{n^2(n^2+1)}{2}\), and the \(n\) rows share this, so the magic constant is

\[ M(n) = \frac{n(n^2 + 1)}{2}. \]

That gives 15 for \(n = 3\), 34 for \(n = 4\), 65 for \(n = 5\). There is no 2 × 2 magic square: with 1, 2, 3, 4, each line would need to add to 5, but then the top row \(a + b = 5\) and the left column \(a + c = 5\) give \(b = c\), and the numbers must be different.

A recipe for odd squares

For any odd \(n\) there is a simple recipe, described by Simon de la Loubère after a visit to Siam in 1688:

1. Put 1 in the middle of the top row.
2. Put each next number one up and one to the right, wrapping around the edges: off the top you come in at the bottom, off the right you come in at the left.
3. If that square is already full, put the number directly below the last one instead.

For \(n = 3\) this gives 8 1 6 / 3 5 7 / 4 9 2, the Lo Shu flipped upside down. Try it for \(n = 5\) on paper; the constant should be 65.

Dürer's square

In 1514 the artist Albrecht Dürer put a 4 × 4 magic square in his engraving Melencolia I:

16	3	2	13
5	10	11	8
9	6	7	12
4	15	14	1

Every row, column and diagonal adds to 34, and so do the four corners, the four centre squares and each 2 × 2 quarter. The middle of the bottom row reads 1514, the year of the engraving.

Going deeper

How many magic squares are there?

Counting squares that differ by more than a rotation or reflection: order 3 has 1, order 4 has 880 (found by Bernard Frénicle de Bessy in 1693), order 5 has 275,305,224 (computed by Richard Schroeppel in 1973). For order 6 the number is not known exactly; estimates put it near \(1.8 \times 10^{19}\). The numbers grow so fast that counting them is itself a hard research problem.

Squares of order 4n and 4n + 2

De la Loubère's recipe works only for odd orders. For orders divisible by 4 there is a neat trick: write the numbers in order, then replace the numbers on both diagonals of every 4 × 4 block by \(n^2 + 1\) minus themselves. Try it on a 4 × 4 square. Orders 6, 10, 14, ... are the hardest, and need a method such as Ralph Strachey's (1918), which builds them from four odd squares.

Latin squares and Euler's 36 officers

Magic squares are cousins of Latin squares, where each row and column holds each symbol once. Euler asked in 1782 whether 36 officers of 6 ranks and 6 regiments could stand in a 6 × 6 square with each rank and regiment once in every row and column. He guessed not, and Gaston Tarry proved it in 1901. Euler also guessed it was impossible for 10, 14, 18, ...; in 1959 Bose, Shrikhande and Parker proved him wrong. See Sudoku.

Magic squares of squares

Is there a 3 × 3 magic square made of nine different perfect squares (\(1, 4, 9, 16, \dots\))? Nobody knows. Martin Gardner and others popularised the question, and it is still open, though many near-misses are known.

See also: Counting · Sudoku · Eight queens

Sources

• Frank J. Swetz, Legacy of the Luoshu: The 4,000 Year Search for the Meaning of the Magic Square of Order Three, A K Peters, 2008.
• Simon de la Loubère, Du Royaume de Siam, 1691 (English: A New Historical Relation of the Kingdom of Siam, 1693).
• Martin Gardner, Time Travel and Other Mathematical Bewilderments, Freeman, 1988, chapter 17.
• Gaston Tarry, "Le problème des 36 officiers", Compte Rendu de l'Association Française pour l'Avancement des Sciences 1 (1900), 122–123, and 2 (1901), 170–203.
• R. C. Bose, S. S. Shrikhande, E. T. Parker, "Further results on the construction of mutually orthogonal Latin squares and the falsity of Euler's conjecture", Canadian Journal of Mathematics 12 (1960), 189–203.
• OEIS A006052, "Number of magic squares of order n".