Leonhard Euler, 1707-1783.

Magic squares have been known and studied for many centuries, but there are still surprisingly many unanswered questions about them. In an effort to make progress on these unsolved problems, twelve prizes totalling €8,000 and twelve bottles of champagne have now been offered for the solutions to twelve magic square enigmas.

A magic square consists of whole numbers arranged in a square, so that all rows, all columns and the two diagonals sum to the same number. An example is the following 4×4 magic square, consisting entirely of square numbers, which the mathematician Leonhard Euler sent to Joseph-Louis Lagrange in 1770:

68^{2} |
29^{2} | 41^{2} | 37^{2} |

17^{2} | 31^{2} | 79^{2} |
32^{2} |

59^{2} | 28^{2} | 23^{2} |
61^{2} |

11^{2} | 77^{2} | 8^{2} |
49^{2} |

It's still not known whether a 3×3 magic squares consisting entirely of squares is possible.

The prize money and champagne will be divided between the people who send in first solutions to one of the six main enigmas or the six smaller enigmas listed below. Solutions should be sent to Christian Boyer. His website gives more information about every enigma, and will contain regular updates regarding received progress and prizes won.

Note that the enigmas can be mathematically rewritten as sets of Diophantine equations: for example, a 3×3 magic square is a set of eight equations (corresponding to the three rows, three columns and two diagonals) in ten unknowns (the nine entries and the magic constant to which each line sums).

Here are the six main and six small enigmas:

### How big are the smallest possible magic squares of squares: 3×3 or 4×4?

In 1770 Leonhard Euler was the first to construct 4×4 magic squares of squares, as mentioned above. But nobody has yet succeeded in building a 3×3 magic square of squares or proving that it is impossible. Edouard Lucas worked on the subject in 1876. Then, in 1996, Martin Gardner offered $100 to the first person who could build one. Since this problem — despite its very simple appearance — is incredibly difficult to solve with nine distinct squared integers, here is a question which should be easier:

**Main Enigma 1 (€1000 and 1 bottle):**Construct a 3×3 magic square using seven (or eight, or nine) distinct squared integers different from the only known example and its rotations, symmetries and*k*^{2}multiples. Or prove that it is impossible.

373^{2} | 289^{2} | 565^{2} |

360721 | 425^{2} |
23^{2} |

205^{2} | 527^{2} |
222121 |

### How big are the smallest possible bimagic squares: 5×5 or 6×6?

A *bimagic square* is a magic square which stays magic after squaring its integers. The first known were
constructed by the Frenchman G. Pfeffermann in 1890 (8×8) and 1891 (9×9). It has been proved that 3×3 and 4×4 bimagics
are impossible. The smallest bimagics currently known are 6×6, the first one of
which was built in 2006 by Jaroslaw Wroblewski, a mathematician at Wroclaw University, Poland.

17 | 36 | 55 | 124 | 62 | 114 |

58 | 40 | 129 | 50 | 111 | 20 |

108 | 135 | 34 | 44 | 38 | 49 |

87 | 98 | 92 | 102 | 1 | 28 |

116 | 25 | 86 | 7 | 96 | 78 |

22 | 74 | 12 | 81 | 100 | 119 |

**Main Enigma 2 (€1000 and 1 bottle):**construct a 5×5 bimagic square using distinct positive integers, or prove that it is impossible.

### How big are the smallest possible semi-magic squares of cubes: 3×3 or 4×4?

An *n*×*n semi-magic square* is a square whose *n* rows and *n* columns have the same sum, but whose
diagonals can have any sum. The smallest semi-magic squares of cubes currently known are 4×4
constructed in 2006 by Lee Morgenstern, an American mathematician. We also know 5×5 and 6×6 squares,
then 8×8 and 9×9, but not yet 7×7.

16^{3} | 20^{3} | 18^{3} | 192^{3} |

180^{3} | 81^{3} | 90^{3} | 15^{3} |

108^{3} | 135^{3} | 150^{3} | 9^{3} |

2^{3} | 160^{3} | 144^{3} | 24^{3} |

**Main Enigma 3 (€1000 and 1 bottle):**Construct a 3×3 semi-magic square using positive distinct cubed integers, or prove that it is impossible.**Small Enigma 3a (€100 and 1 bottle):**Construct a 7×7 semi-magic square using positive distinct cubed integers, or prove that it is impossible.

### How big are the smallest possible magic squares of cubes: 4×4, 5×5, 6×6, 7×7 or 8×8?

The first known magic square of cubes was constructed by the Frenchman Gaston Tarry in 1905, thanks to a large 128×128 trimagic square (magic up to the third power). The smallest currently known magic squares of cubes are 8×8 squares constructed in 2008 by Walter Trump, a German teacher of mathematics. We do not know any 4×4, 5×5, 6×6 or 7×7 squares. It has been proved that 3×3 magic squares of cubes are impossible.

11^{3} | 9^{3} | 15^{3} | 61^{3} | 18^{3} |
40^{3} | 27^{3} | 68^{3} |

21^{3} | 34^{3} | 64^{3} | 57^{3} | 32^{3} |
24^{3} | 45^{3} | 14^{3} |

38^{3} | 3^{3} | 58^{3} | 8^{3} | 66^{3} |
2^{3} | 46^{3} | 10^{3} |

63^{3} | 31^{3} | 41^{3} | 30^{3} | 13^{3} |
42^{3} | 39^{3} | 50^{3} |

37^{3} | 51^{3} | 12^{3} | 6^{3} | 54^{3} |
65^{3} | 23^{3} | 19^{3} |

47^{3} | 36^{3} | 43^{3} | 33^{3} | 29^{3} |
59^{3} | 52^{3} | 4^{3} |

55^{3} | 53^{3} | 20^{3} | 49^{3} | 25^{3} |
16^{3} | 5^{3} | 56^{3} |

1^{3} | 62^{3} | 26^{3} | 35^{3} | 48^{3} |
7^{3} | 60^{3} | 22^{3} |

**Main Enigma 4 (€1000 and 1 bottle):**Construct a 4×4 magic square using distinct positive cubed integers, or prove that it is impossible.**Small Enigma 4a (€500 and 1 bottle):**Construct a 5×5 magic square using distinct positive cubed integers, or prove that it is impossible.**Small Enigma 4b (€500 and 1 bottle):**Construct a 6×6 magic square using distinct positive cubed integers, or prove that it is impossible.**Small Enigma 4c (€200 and 1 bottle):**Construct a 7×7 magic square using distinct positive cubed integers, or prove that it is impossible.*(When such a square is constructed, if small enigma 3a of the 7×7 semi-magic is not yet solved, then the person will win both prizes — that is to say a total of €300 and 2 bottles.)*

### How big are the smallest integers allowing the construction of a multiplicative magic cube?

Contrary to all other enigmas which concern the magic squares, this one concerns magic cubes. An
*n×n×n* multiplicative magic cube is a cube whose *n*^{2} rows, *n*^{2} columns, *n*^{2} pillars, and 4 main diagonals
have the same product *P*. Today the best multiplicative magic cubes known are 4×4×4 cubes in which the largest
used number among their 64 integers is equal to 364. We do not know if it is possible to construct a cube
with smaller numbers.

A 4×4×4 multiplicative magic cube by Christian Boyer. Max number=364. *P*=17,297,280.

**Main Enigma 5 (€1000 and 1 bottle):**Construct a multiplicative magic cube in which the distinct positive integers are all strictly lower than 364. The size is free: 3×3×3, 4×4×4, 5×5×5,... . Or prove that it is impossible.

### How big are the smallest possible additive-multiplicative magic squares: 5×5, 6×6, 7×7 or 8×8?

An *n×n* additive-multiplicative magic square is a square in which the *n* rows, *n *columns and two diagonals have
the same sum *S*, and also the same product *P*. The smallest known are 8×8 squares, the first one of which
was constructed in 1955 by Walter Horner, an American teacher of mathematics. We do not know any 5×5,
6×6 or 7×7 squares. It has been proved that 3×3 and 4×4 additive-multiplicative magic squares are impossible.

162 | 207 | 51 | 26 | 133 | 120 | 116 | 25 |

105 | 152 | 100 | 29 | 138 | 243 | 39 | 34 |

92 | 27 | 91 | 136 | 45 | 38 | 150 | 261 |

57 | 30 | 174 | 225 | 108 | 23 | 119 | 104 |

58 | 75 | 171 | 90 | 17 | 52 | 216 | 161 |

13 | 68 | 184 | 189 | 50 | 87 | 135 | 114 |

200 | 203 | 15 | 76 | 117 | 102 | 46 | 81 |

153 | 78 | 54 | 69 | 232 | 175 | 19 | 60 |

**Main Enigma 6 (€1000 and 1 bottle):**Construct a 5×5 additive-multiplicative magic square using distinct positive integers, or prove that it is impossible.**Small Enigma 6a (€500 and 1 bottle):**Construct a 6×6 additive-multiplicative magic square using distinct positive integers, or prove that it is impossible.**Small enigma 6b (€200 and 1 bottle):**Construct a 7×7 additive-multiplicative magic square using distinct positive integers, or prove that it is impossible.

## Comments

## 3 x 3 magic square of squares

Here is an "almost" magic square of squares, where all the rows and columns, and one of the diagonals sum to 21609. Sadly, the diagonal from top left to bottom right sums to 14358

94^2, 2^2 , 113^2

97^2, 74^2, 82^2

58^2, 127^2, 46^2

Arthur Vause

arthur dot vause at gmail.com

## This is just the

This is just the transformation of Lee Sallows's magic square (L. Sal lows, The lost theorem, Math. Intelligencer 19 (1997), no. 4, 51-54).

## your solution

is that a parker square?

## 3 x 3 magic square of squares

Here is an "almost" magic square.

All rows and columns, and one diagonal sum to 21609. Sadly the diagonal from top left to bottom right sums to 14358

94^2, 2^2, 113^2

97^2, 74^2, 82^2

58^2, 127^2, 46^2

## searching for 3x3

I've pieced together a small script in Excel searching for a perfect 3x3 magic square of squares.

The first version was limited to search for integers between 1 and 1000. But I can't be the first one trying to brute force the main enigma 1 problem here. Not when there is a bottle at stake. But how far up in the integers have we searched for a solution?

## re: searching for 3x3

It has been mathematically proven that the smallest number in a complete 3x3 magic square of squares – if it even exists – would have to be larger than 10^14.

I would definitely recommend starting with the algebra before moving on to brute force. Personally, I've found a parametric formula which turns 4 variables of almost any arbitrary value into a 5/9 square of squares (the 4 corners and the center), and I've brute forced about 150 specific values which create 6/9 squares (the 4 corners, the middle, and one of the sides).

Out of all of the 6/9 squares that I have found myself, my favorite (per my love of horrifyingly large numbers) is the one with

17-digit square, 14-digit square, 18-digit square

18-digit non-square, 17-digit square, 15-digit non-square

14-digit square, 18-digit non-square, 17-digit square

However, I have not found a general parametric formula for generating 6/9 squares from any arbitrary value of variables, nor have I found any 7/9 squares by applying brute force to my 5/9 formula.

## excel brute force

well i reached 20 million on excel and gave up. good luck :)

## About Main Enigma1

I've find another magic square using 7 distinct squares. Is it worthy?

## Yes.

Yes.

Because: Main Enigma 1 (€1000 and 1 bottle): Construct a 3×3 magic square using seven (or eight, or nine) distinct squared integers different from the only known example and its rotations, symmetries and k2 multiples. Or prove that it is impossible.

## Smallest 3×3 magic square using six distinct squared integers

I looked into that topic a bit and came to the conclusion that the smallest possible one with six squares is.

17^2; 35^2; 19^2

697 ; 25^2; 553

889 ; 5^2 ; 31^2

Maybe that is already known

## Smallest 3x3 magic square using six distinct squared integers

Hi renes, I found an even smaller one, SUM: 435

5^2; 241; 13^2

17^2; 145; 1^2

11^2; 7^2; 265

## 3 x 3

I checked all solutions up tot S=1,500,000. Did not find another 3 x 3 magic square of 7 or more distinct squared integers. I did however find many with 6. They seem to get more rare the higher you get. The last one I found was:

532^2/980^2/476^2

433552/700^2/546448

868^2/140^2/696976

S = 1,470,000

## Method

I'm interested as to how you check these. Could I possibly have some info on that?

## Yeah that will be a great

Yeah that will be a great help if you tell how to check it using computer.. THANK YOU.

Because applying simple algebra, I have also found a magic square consisting of 6 out of 9 different square numbers which is already shown above in a message.The total sum is 1875.

35^2 (-311) 31^2

19^2 25^2 889

17^2 1561 5^2

## 4×4×4 multiplicative magic cube

I have the code that generates possible P values for it but I'm not sure how given the P value to check if there is a possible 4x4x4 multiplicative magic cube that works (and hopefully without having to try all 64! possibilities. Is there a pretty easy way to do this?

## Formula for 3x3 magic square using squared integers

x and y are triangular numbers, and we know 8* any triangular number = an odd square.

1+8*(2*y); 1+8*(x+(2*y)); 1+8*(2*(x+y))

1+8*(y); 1+8*(x+y); 1+8*(y+(2*x))

1; 1+8*(x); 1+8*(2*x))

So if 2*x and 2*y are triangular numbers, then you automatically have 5 squared integers.

Then if two out of four possible options of: x+y or 2*(x+y) or x+2*y or y+2*x are triangular numbers. Then you have the 7!

## Formula for Magic square to make square numbers

X; 2*Y; Y+2*X

2*(X+Y); X+Y; 0

Y; 2*X; X+2*Y

SUM: 3*(X+Y)

If all entries are triangular numbers, then we know that 8 times any triangular number +1 equals an odd square. (0*8 + 1=1^2)

## Possibly solved

I think I may have solved this but I am afraid the solution I came up with is a few typed pages long. Is there an email or something I can send the proof to? Be warned its a little rough but all of the necessary ideas are there and if it isn't I'd gladly reply to questions.

## Possible solution?

This was the closest I could come but I’m unhappy with the answer. If you want to contact me- nikoliishot@aim.com

X being any positive integer. i being imaginary

(x sqrt(3))^2, ((x2)i)^2, (x)^2

((x sqrt(2))i)^2, 0, (x sqrt(2))^2

(xi)^2, (x2i)^2, ((x sqrt(3)i)^2

I’m formatting on a phone sorry

## 3x3 impossibility proof

has the impossibility of 3x3 magic square with 9 squared integers been prooven?

## 3x3 impossibility proof

if i prove that a 3x3 magic square of all 9 squared integers is impossible, do i win the prize?

## Computational Result of 3x3

Nothing found under S = 1,200,000,000

## 3 x 3 magic square of squares: Computational result

Searched until S=12,000,000,000,000 but could not find a solution for 9 distinct squares.

## 12*10^12 in one day?

How is this even possible? Did you use Fugaku? =)

## 3x3 Magic Square

3,46415849^2, 999,9694993^2, 7^2

999,9469986^2, 5^2, 9^2

9,695377457^2, 6^2, -999,9349979^2

Excel solver - if you need them in integers, just multiply out the decimals...