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 anonymously by Martin LaBar in 1984, before being included in Richard Guy's Unsolved problems in number theory (2nd edition) in 1994. [1]
The problem is a popular choice for recreational mathematicians, and multiple prizes have been offered for the first solution. [2]
A magic square is a square array of integer numbers in which each row, column and diagonal sums to the same number. [3] The order of the square refers to the number of integers along each side. [4] 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.
The problem asks whether it is possible to construct a third-order magic square such that every element is itself a square number. [5] A square which solves the problem would thus be of the form
and satisfy the following equations [6]
It has been shown that the problem is equivalent to several other problems. [1]
Brute force searches for solutions have been unsuccessful, and suggest that if a solution exists, it would consist of numbers greater than at least . [7]
Rice University professor of mathematics Anthony Várilly-Alvarado has expressed his doubt as to the existence of the magic square of squares. [6]
There have been a number of attempts to construct a magic square of squares by recreational mathematicians.
Recreational mathematician Martin Gardner attempted to produce a solution to the problem, creating a non-trivial semimagic square of squares. In his solution, the diagonal 1272 + 1132 + 972 sums to 38307, not 21609 as for all the other rows and columns, and the other diagonal. [8] [9] [10]
1272 | 462 | 582 | 21609 | |
22 | 1132 | 942 | 21609 | |
742 | 822 | 972 | 21609 | |
21609 | 21609 | 21609 | 21609 | 38307 |
The Parker square [11] is an attempt by Matt Parker to solve the problem. His solution is a trivial, semimagic square of squares, as and both appear twice, and the diagonal sums to 4107, instead of 3051. [12]
292 | 12 | 472 | 3051 | |
412 | 372 | 12 | 3051 | |
232 | 412 | 292 | 3051 | |
4107 | 3051 | 3051 | 3051 | 3051 |
Magic squares of squares of orders greater than 3 have been known since as early as 1770, when Leonard Euler sent a letter to Joseph-Louis Lagrange detailing a fourth-order magic square. [10]
682 | 292 | 412 | 372 |
172 | 312 | 792 | 322 |
592 | 282 | 232 | 612 |
112 | 772 | 82 | 492 |
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. [13]
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 |
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: CS1 maint: DOI inactive as of July 2025 (link)Some 'near misses' have been found such as the Parker Square [2]