Magic square of squares

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Template:Short description

Template:Unsolved The magic square of squares is an unsolved problem in mathematics which asks whether it is possible to construct a three-by-three magic square, the elements of which are all square numbers. The problem was first posed by Martin LaBar in the January 1984 College Mathematics Journal, before being included in Richard Guy's Unsolved problems in number theory (2nd edition) in 1994.[1][2]

The problem has been a popular choice for recreational mathematicians following two articles Martin Gardner published in Quantum Magazine on the problem, offering a prize of Template:Currency in 1996.[3][4] Other prizes have subsequently been offered for the first solution.[5]

Background

Template:Main

The smallest (and unique up to rotation and reflection) non-trivial case of a magic square, order 3

A magic square is a square array of integer numbers in which each row, column, and diagonal sums to the same number.[6] The order of the square refers to the number of integers along each side.[7] A trivial magic square is a magic square which has at least one repeated element, and a semimagic square is a magic square in which the rows and columns, but not both diagonals, sum to the same number.

Problem

The problem asks whether it is possible to construct a third-order magic square such that every element is itself a square number. A square which solves the problem would thus be of the form

x12 x22 x32
x42 x52 x62
x72 x82 x92

and satisfy the following equations[8]

x12+x22+x32=x42+x52+x62=x72+x82+x92=x12+x52+x92=x32+x52+x72=x12+x42+x72=x22+x52+x82=x32+x62+x92

Current research

It has been shown that the problem is equivalent to several other problems.[2]

  1. Do there exist three arithmetic progressions such that each has three terms, each has the same difference between terms as the other two, the terms are all perfect squares, and the middle terms of the three arithmetic progressions themselves form an arithmetic progression?
  2. Do there exist three rational right triangles with the same area, such that the squares of the hypotenuses are in arithmetic progression?
  3. Does there exist an elliptic curve, y2=x3−n2x, where n is a congruent number, with three rational points on the curve, (x1,y1), (x2,y2), (x3,y3), such that each point is "double" another rational point on the curve ("double" in the sense of the group structure for points on an elliptic curve), and x1, x2 and x3 are in arithmetic progression?

Properties of a magic square of squares

Let S denote the sum of each row, column or main diagonals of a primitive (the greatest common divisor of all entries is 1) third-order magic square of squares. If such a magic square of squares exists, it must satisfy the following properties.

  • All elements must be odd.[9]
  • More specifically, they must be of the form 24k+1.[10]
  • S=3a2 where a2is the middle element.[11]
  • S must thus be of the form S=72k+3.[10]
  • No element can have a prime divisor p of the form p=8k+3.[9]
  • All prime divisors p of the middle element must be of the form p=4k+1.[9]
  • A prime p of the form p=8k+5 cannot divide a non-corner edge element.[9]
  • If a prime p of the form p=4k+3 divides a corner element then it must also divide the two non-corner edge elements that are not adjacent to that corner.[9]
  • If a prime p of the form p=8k+5 divides a corner element then it must also divide the center element and the opposite corner element.[12]

Brute force searches for solutions have been unsuccessful, and suggest that if a solution exists, it would consist of numbers greater than at least 1014.[13]

Rice University professor of mathematics Anthony Várilly-Alvarado has expressed his doubt as to the existence of the magic square of squares.[8]

Notable attempts

There have been a number of attempts to construct a magic square of squares by recreational mathematicians.

Sallows' Square

Following Gardner's prize offer for anyone who could find a magic square of squares in 1996, Lee Sallows published his attempt in The Mathematical Intelligencer. His attempt is unique in that all of the rows and columns, and one of the diagonals, all sum to the same square number.[14][8]

Sallows' Square[14]
127Template:Sup 46Template:Sup 58Template:Sup 147Template:Sup
2Template:Sup 113Template:Sup 94Template:Sup 147Template:Sup
74Template:Sup 82Template:Sup 97Template:Sup 147Template:Sup
147Template:Sup 147Template:Sup 147Template:Sup 147Template:Sup 38307

Bremner Square

In 1999, Andrew Bremner published his attempt at the problem, and further research surrounding magic squares of squares.[15] Bremner's attempt differs from others in that not all elements of the square are square numbers, while all the rows, columns and diagonals sum to the same number.[8]

Bremner Square[15]
373Template:Sup 289Template:Sup 565Template:Sup 541875
360721 425Template:Sup 23Template:Sup 541875
205Template:Sup 527Template:Sup 222121 541875
541875 541875 541875 541875 541875

Parker square

The Parker square[16] is an attempt by Matt Parker to solve the problem. His solution is a trivial, semimagic square of squares, as 412, 292 and 12 all appear twice, and the diagonal 232+372+472 sums to 4107 instead of 3051.[17][13]

The Parker Square, with sums shown in bold.
29Template:Sup 1Template:Sup 47Template:Sup 3051
41Template:Sup 37Template:Sup 1Template:Sup 3051
23Template:Sup 41Template:Sup 29Template:Sup 3051
4107 3051 3051 3051 3051

Non third-order magic squares of squares

Magic squares of squares of orders greater than 3 have been known since as early as 1770, when Leonhard Euler sent a letter to Joseph-Louis Lagrange detailing a fourth-order magic square.[18]

Euler's magic square of squares
68Template:Sup 29Template:Sup 41Template:Sup 37Template:Sup
17Template:Sup 31Template:Sup 79Template:Sup 32Template:Sup
59Template:Sup 28Template:Sup 23Template:Sup 61Template:Sup
11Template:Sup 77Template:Sup 8Template:Sup 49Template:Sup

Multimagic squares are magic squares which remain magic after raising every element to some power. In 1890, Georges Pfeffermann published a solution to a problem he posed involving the construction of an eighth-order 2-multimagic square.[19]

Pfeffermann's eighth order 2-multimagic square[20]
56 34 8 57 18 47 9 31 260
33 20 54 48 7 29 59 10 260
26 43 13 23 64 38 4 49 260
19 5 35 30 53 12 46 60 260
15 25 63 2 41 24 50 40 260
6 55 17 11 36 58 32 45 260
61 16 42 52 27 1 39 22 260
44 62 28 37 14 51 21 3 260
260 260 260 260 260 260 260 260 260 260

References

Template:Reflist