**First known 4x4 to 7x7 magic squares of squares**

Can a 3x3 magic square be constructed with nine
distinct square numbers? The answer is unknown today: nobody has succeeded
in constructing a 3x3 magic square of squares, and nobody has proved
that it is impossible to construct such a square. See my article in *The Mathematical
Intelligencer, *and the Powerpoint file of the lecture.

But it is possible to construct other magic squares of squares:

- Well-known 8x8 (and greater) bimagic squares are of course magic squares of squares because, by definition, they remain magic when their numbers are squared. And, as required, all their numbers are distinct because they use consecutive integers. Unfortunately, it is impossible to construct 7x7 (and smaller) bimagic squares using consecutive integers.
- Between the unknown 3x3 and the well-known bimagic 8x8 (and greater), magic squares of squares from 4x4 to 7x7 are known today as I will show in this page.

On this subject, read also
the MathTrek article
written in June 2005 by **Ivars Peterson**:

- ...at... http://www.sciencenews.org/view/generic/id/6308/title/Math_Trek__Magic_Squares_of_Squares
- or ...at... http://www.maa.org/mathland/mathtrek_06_27_05.html

The first known magic square
of squares was sent in 1770 by **Leonhard Euler** to **Joseph Lagrange**. This is the
square LE2 fully explained and described in the *M.I. *article (and lecture slides 22 and 23).

68² |
29² |
41² |
37² |

17² |
31² |
79² |
32² |

59² |
28² |
23² |
61² |

11² |
77² |
8² |
49² |

This square can be constructed with his wonderful 4x4
parametric solution and (a, b, c, d, p, q, r, s) = (5, 5, 9, 0, 6, 4,
2, -3). Or also (5, 5, 9, 0, 2, 3, 6, –4),
giving the same square but permuted. From my *M.I.* paper published in 2005:

"The work of Euler is linked to the theory of quaternions [2] [15] [36] [37] developed later in 1843 by William Hamilton. In his (LE3) square, Euler reused an identity that he found and sent to Christian Goldbach in 1748 [21]:

(a² + b² + c² + d²)(p² + q² + r² + s²) = (ap + bq + cr + ds)² + (aq - bp - cs + dr)² + (ar + bs - cp - dq)² + (as - br + cq - dp)²

This identity also follows from the fact that the norm
of the product of two quaternions is the product of the norms. Euler first
used this identity in 1754 [17] in a partial proof that *every
positive integer is the sum or at most four square integers*,
an old conjecture announced by Diophantus, Bachet, and Fermat. Using as
a basis these partial results of Euler's, Lagrange published in 1770 [55]
the first complete proof of this four square theorem, the same year as the
letter he received with the first 4x4 magic square of squares".

In this article and its supplement of 2005, I gave two sub-families (CB2) and (CB15) of Euler's parametric solution, but using only one parameter k instead of the eight parameters (a, b, c, d, p, q, r, s) from Euler. With k = 5, my sub-family (CB15) produces the above Euler-Lagrange example.

In
November 2018, **Seiji Tomita**, Japan, found 11 more
parametric solutions also using only one parameter k, an excellent set of
solutions. As it was with my two parametric solutions, for each k, we also need
to check that the 16 produced cells are distinct. And as it was with my
two parametric solutions, I remark that his 11 solutions are also sub-families
of Euler's parametric square (a, b, c, d, p, q, r, s), setting b = k and d = 0:

- (9, k, 16, 0, 1, 2, 6, -3)
- (5, k, 16, 0, 2, 3, 9, -6)
- (7, k, 18, 0, 3, 4, 8, -6)
- (5/2, k, 9/2, 0, 1, 3, 15, -5)
- (5, k, 7, 0, 2, 4, 10, -5)
- (2, k, 5, 0, 3, 5, 15, -9)
- (5/2, k, 7/2, 0, 3, 7, 21, -9)
- (5, k, 16, 0, 1, 2, 10, -5)
- (2, k, 9, 0, 2, 3, 12, -8)
- (1, k, 7, 0, 2, 3, 18, -12)
- (7, k, 10, 0, 6, 8, 12, -9)

In 2004, I constructed the first known 5x5 magic squares of squares.
Squares CB4 and CB5 published in the *M.I.* article (and lecture slide
17). The smallest possible is CB4:

1² |
2² |
31² |
3² |
20² |

22² |
16² |
13² |
5² |
21² |

11² |
23² |
10² |
24² |
7² |

12² |
15² |
9² |
27² |
14² |

25² |
19² |
8² |
6² |
17² |

Unfortunately to late to be
published in the *M.I.* article, I constructed, in June 2005, the first
6x6 magic squares of squares.

If I am right, 6x6 magic squares of squares using squared consecutive integers (0² to 35², or 1² to 36²) are impossible. My 6x6 magic square of squares does NOT use squared consecutive integers... but it is interesting to see the used numbers:

- from 0² to 36²
**only excluding**30².

It is impossible to construct a 6x6 magic square of squares with a smaller magic sum. But it is possible to construct other samples with the same magic sum S2 = 2551, or with other bigger sums.

2² |
1² |
36² |
5² |
0² |
35² |

6² |
33² |
20² |
29² |
4² |
13² |

25² |
7² |
14² |
24² |
31² |
12² |

21² |
32² |
11² |
15² |
22² |
16² |

34² |
18² |
23² |
10² |
19² |
9² |

17² |
8² |
3² |
28² |
27² |
26² |

An interesting supplemental characteristics of this sample: the 3 smallest integers (0², 1², 2²) and the 2 biggest (35², 36²) are used together in the first row.

Unfortunately to late to be
published in the *M.I.* article, I constructed, in June 2005, the first
7x7 magic squares of squares.

The smallest order allowing magic squares of squares using squared
**consecutive** integers is the order 7. An indirect
consequence: the impossibility of 7x7 bimagic squares is not coming from a
problem with its squared numbers!

Here is my sample using the squared integers from 0² to 48²:

25² |
45² |
15² |
14² |
44² |
5² |
20² |

16² |
10² |
22² |
6² |
46² |
26² |
42² |

48² |
9² |
18² |
41² |
27² |
13² |
12² |

34² |
37² |
31² |
33² |
0² |
29² |
4² |

19² |
7² |
35² |
30² |
1² |
36² |
40² |

21² |
32² |
2² |
39² |
23² |
43² |
8² |

17² |
28² |
47² |
3² |
11² |
24² |
38² |

An interesting supplemental characteristics added in this sample: the 7 rows are magic (S1=168) when the integers are not squared, meaning that the 7 rows are bimagic!

**Conclusion of this page**: because
4x4 and above are now solved, it
means that 3x3 is
the only remaining open problem
(but the most difficult...) on magic squares of squares!

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