Magic squares of cubes
Magic squares of fourth powers
Magic squares of fifth powers
See also the Magic squares of squares page

In this page:

In other pages:

Summary and tables

Excerpt from the article Some Notes on the Magic Squares of Squares Problem, Part 6
by Christian Boyer, in The Mathematical Intelligencer, Volume 27, Number 2, Spring 2005, pages 52-64

"It’s implicit in the work of Carmichael that
there can be no 3x3 magic squares with entries
which are cubes or are fourth powers."

Richard K. Guy,
Unsolved Problems in Number Theory, Third Edition, 2004, page 270

The work of Euler implies already that there can be no 3x3 magic square with entries which are cubes. If z3 is the number in the centre cell, then any line going through the centre should have x3 + y3 = 2z3. Euler and Legendre[39] demonstrated that x3 + y3 = kz3 is impossible with distinct integers, for k = 1, 2, 3, 4, 5. Adrien-Marie Legendre mistakenly announced that k = 6 is also impossible: Edouard Lucas published the general solution for k = 6 in the American Journal of Mathematics Pure and Applied of J.J. Sylvester[41], and gave the example 173 + 373 = 6×213. The equation x3 + y3 = 7z3 has been known to be possible since Fermat, one of his examples being 43 + 53 = 7×33.

Legendre showed also that x4 + y4 = 2z2 is impossible if x ¹ y. Because z4 = (z2)2, this implies that there can be no 3x3 magic square with entries which are fourth powers. It’s also implicit in the later work of Carmichael[13] that there can be no 3x3 magic square with entries which are cubes, or are fourth powers or 4k-th powers. Noam Elkies[26] points that with the Andrew Wiles’s proof of Fermat Last Theorem it can be shown that an + bn = 2cn has no solution for n greater than 2, and thus that there can be no 3x3 magic square with entries which are powers greater than 2.

And as said in the D2 problem, “The Fermat problem”, page 219 of the Guy’s book[30]: “It follows from the work of Ribet via Mazur & Kamienny and Darmon & Merel that the equation xn + yn = 2zn has no solution for n > 2 apart from the trivial x = y = z.”

So, 3x3 magic squares of cubes are impossible. I think that 4x4 are also impossible with distinct positive cubes. The 12x12 (WT1) trimagic square of part 7 below, when its numbers are cubed, is a magic square of cubes.

If we accept negative integers, and using the interesting but obvious remark that n3 and (–n)3 are not equal (the rule in a magic square is to use “distinct” integers, and the trick is that they are distinct!), (CB10) and (CB11) are magic squares of cubes having a null magic sum. They seem to be the first published 4x4 and 5x5 magic squares of cubes.

If you do not like the terminological trick I used, then Open problem 5 is for you! And the (CB12) square is a first step.

Open problem 5. Construct the smallest possible magic square of cubes: 5a) using integers having different absolute values, 5b) using only positive integers.
Open problem 6. Construct a magic square of cubes of prime numbers[9].

193

(-3)3

(-10)3

(-18)3

(-42)3

213

283

353

423

(-21)3

(-28)3

(-35)3

(-19)3

33

103

183

 

113

(-20)3

123

133

143

(-15)3

213

33

(-10)3

(-17)3

(-5)3

(-4)3

03

43

53

173

103

(-3)3

(-21)3

153

(-14)3

(-13)3

(-12)3

203

(-11)3

 

93

473

543

643

963

233

973

63

483

723

103

143

673

1013

423

1103

363

213

33

283

403

703

983

183

383

References from the article

[7] Christian Boyer, Multimagic squares, cubes and hypercubes web site, www.multimagie.com/indexengl.htm

[9] Christian Boyer, Supplement to the article “Some notes on the magic squares of squares problem” article, downloadable from [7], 2005:
Download the PDF file (31Kb) or See HTML page at www.multimagie.com/English/Supplement.htm

[13] Robert D. Carmichael, Impossibility of the equation x3 + y3 = 2mz3, and On the equation ax4 + by4 = cz2, Diophantine Analysis, John Wiley and Sons, New-York, 1915, 67-72 and 77-79 (reprint by Dover Publications, New York, in 1959 and 2004)

[26] Martin Gardner, The latest magic, Quantum 6(1996), n°4, 60

[30] Richard K. Guy, Problem D15 – Numbers whose sums in pairs make squares, Unsolved Problems in Number Theory, Third edition, Springer, New-York, 2004, 268-271

[39] Adrien-Marie Legendre, Théorie des Nombres, 3rd edition, Firmin-Didot, Paris, 2(1830) 4-5, 9-11, and 144-145 (reprint by Albert Blanchard, Paris, in 1955)

[41] Edouard Lucas, Sur l’analyse indéterminée du troisième degré – Démonstration de plusieurs théorèmes de M. Sylvester, American Journal of Mathematics Pure and Applied 2(1879) 178-185

3x3 magic squares of cubes
3x3 magic squares of fourth powers

As seen above, 3x3 magic squares are proved impossible. But the status of 3x3 semi-magic squares of cubes and of fourth powers is still unknown. My best result is:

In October 2006, Frank Rubin searched a 3x3 semi-magic square of cubes, and concludes that there is no solution using numbers all smaller than 300,0003.

In March 2008, Lee Morgenstern proposed an interesting method. If we find two numbers x and y being differerences of two cubes in 3 different ways and having 3 terms in common:

then we have found a 3x3 semi-magic square of cubes:

In May 2008, Uwe Hollerbach worked on this method immediately after his confirmation that my upper bound 933528127886302221000 is the real Cabtaxi(10) number (the smallest number sum or difference of two cubes in 10 different ways). Using his long list of 10,597,218 primitive solutions ≤ 950000519472444752221 which are sums or differences of two cubes in 3-or-more ways, he did not find any solution to the system (3.1) (3.2), that is to say, for any x and y < 9.5 *1020.

Who will be the first to construct a 3x3 semi-magic square of cubes, using distinct positive integers? Or prove that it is impossible? In 2007, I also asked this question in the website of Carlos Rivera: http://www.primepuzzles.net/puzzles/puzz_412.htm

Who will be the first to construct a 3x3 semi-magic square of fourth powers? Or prove that it is impossible?

4x4 magic squares of cubes
4x4 magic squares of fourth powers

In June 2006, after his 6x6 and 7x7 bimagic squares using distinct integers, Lee Morgenstern studied magic squares of cubes. He found a very nice method to construct 4x4 magic squares of cubes (or of any nth-powers):

Such a square of nth-powers is magic if the three equations are true:

Its magic sum is Sn = uv. If only the two equations (4.1) and (4.2) are true, then the square is only a semi-magic square.

Definition of standard Taxicab numbers: integers which can be expressed as the sum of 2 cubes, in 2 different ways.
Definition of Taxicab(j) numbers: integers which can be expressed as the sum of 2 cubes, in j different ways.
 (look at http://www.research.att.com/~njas/sequences/A001235
or http://mathworld.wolfram.com/TaxicabNumber.html
or http://euler.free.fr/taxicab.htm):

Using for example the famous smallest Taxicab number 1729 of Ramanujan, previously known by Bernard Frénicle de Bessy as early as 1657, and the second smallest Taxicab number 4104

we can directly construct the square:

After an exhaustive research done by Lee Morgenstern (meaning not only with his above method), this is THE smallest possible 4x4 semi-magic square of cubes. Not counting the obvious 23 multiple of the previous square, his method generates also the second smallest possible: using again 1729, but together with the third smallest primitive Taxicab number 20683 = 103 + 273 = 193 + 243, giving the second smallest S3 = 1729 * 20683 = 35,760,907. The third smallest square can't be produced by his method:

He confirmed also that my CB10 square allowing negative integers is the best possible.

Remarks on the Morgenstern's 4x4 method

I worked on this powerful method which would be able to generate 4x4 magic square of cubes, with two magic diagonals, if we can find at least one solution of the three above equations (4.1) (4.2) (4.3) when power n=3. For example, this system of 3 equations is possible when n=2, generating a magic square of squares S2 = 125*8357:

but I was unable to find at least one solution when n=3, or 4, or more. Using lists kindly provided by Jaroslaw Wroblewski, and if my computation is correct, I can say that there is no solution to the 3 equations when n=3:

If the Morgenstern's method can generate 4x4 fully magic squares of cubes, then its numbers are huge!

Journal of Integer Sequences

Who will be the first to construct a 4x4 magic square of cubes, using distinct positive integers? Or prove that it is impossible? This problem is also given in part 8.3 "Who can construct of 4x4 magic square of cubes?" of my paper "New Upper Bounds for Taxicab and Cabtaxi Numbers" published in Journal of Integer Sequences, 2008, Volume 11, Issue 1.
See http://www.christianboyer.com/taxicab and http://www.cs.uwaterloo.ca/journals/JIS/vol11.html

And what about magic squares of fourth powers? With n=4, I generated easilly the smallest solution to (4.1)+(4.2), using the 635,318,657 number first found by Euler:

(see the list of sums of two 4th powers in more than one way at http://www.research.att.com/~njas/sequences/A003824).

But mixing the 1420 solutions of:

(list from Jaroslaw Wroblewski, available at http://www.math.uni.wroc.pl/~jwr/422/index.htm), and if my computation is correct, I can say that there is no solution to the system of 3 equations when n=4, if a,b,c,d,e,f,g,h < 10,000,000.

Who will be the first to construct a 4x4 magic square of fourth powers? Or prove that it is impossible?

And what about magic squares of fifth powers? With n=5, nobody knows any solution to the equation:

After an exhaustive search done in September 2002, Stuart Gascoigne did not find any solution u < 3.26 * 10^32. And Jaroslaw Wroblewski did not find any solution in 2006 for special forms up to 2.43 * 10^37. And, more generally, nobody knows a solution to:

However, today, no known mathematical proof that the equation is impossible. Look at the "Computing Minimal Equal Sums Of Like Powers" website http://euler.free.fr/ of Jean-Claude Meyrignac, where this equation is called (n, 2, 2): nth powers, 2 positive terms on the left side, 2 positive terms on the right side.

It means that we are unable to construct a semi-magic square of fifth powers, or more, using the Morgenstern's method.

Who will be the first to construct a 4x4 (at least semi-) magic square of fifth powers? Or prove that it is impossible?

5x5 magic squares of cubes
5x5 magic squares of fourth powers

Lee Morgenstern confirmed that my CB12 square is the smallest possible 5x5 semi-magic square of cubes, with S3 = 1,408,896. He found the second smallest solution:

Who will be the first to construct a 5x5 magic square of cubes, using distinct positive integers? Or prove that it is impossible?

Who will be the first to construct a 5x5 magic square of fourth powers? Or prove that it is impossible?

He confirmed also that my CB11 square of cubes allowing negative integers is the best possible.

6x6 magic squares of cubes
6x6 magic squares of fourth powers

Lee Morgenstern found a method to construct 6x6 semi-magic squares of cubes (or of any nth-powers). If the two equations (6.1) (6.2) are true:

then this square is a semi-magic square of nth-powers, with magic sum Sn = uv:

The solutions of these equations are well-known. When n=3, we can find for example using:

that their smallest solutions are respectively:

generating the square:

Lee Morgenstern constructed also a 6x6 magic square of cubes allowing negative integers with S3=0, as my CB10 and CB11 squares. All integers are used, from -18 to 18, excluding 0:

Remarks on the Morgenstern's 6x6 method

The above square S3 = 88,327,172,871 is the smallest possible using the method, but it is probably not the smallest possible of all the 6x6 semi-magic square of cubes. And nobody knows a 6x6 magic square of cubes, with two magic diagonals.

Who will be the first to construct a 6x6 magic square of cubes, using distinct positive integers? Or prove that it is impossible?

I add that this Morgenstern's method cannot generate a 6x6 semi-magic square of fourth powers with the current status of the research on Taxicab numbers:

Definition of generalized Taxicab(n, i, j) numbers:
x=Taxicab(n, i, j) is an integer x which can be expressed as the sum of i nth-powers, in j different ways. For n=3 and i=2, they coincide with Taxicab(j) numbers.

Who will be the first to construct a 6x6 magic square of fourth powers? Or prove that it is impossible?

7x7 magic squares of cubes
7x7 magic squares of fourth powers

According Lee Morgenstern's computation done in May 2008, there is no 7x7 semi-magic square of cubes using any possible set of 49 cubes between 13 and 553. He extended his computation in August 2008: also impossible up to 583.

Who will be the first to construct a 7x7 magic square of cubes, using distinct positive integers? Or prove that it is impossible?

Who will be the first to construct a 7x7 magic square of fourth powers? Or prove that it is impossible?

In this magic square of cubes allowing negative integers, all integers are used, from -24 to 24, including 0:

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