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The smallest possible multiplicative magic squares

What are the smallest possible multiplicative squares?

In September 2005, Ed Pegg Jr., author of the well-known MathPuzzle web site, asked the question: "Is there a list of smallest multiplicative constants for various NxN multiplicative magic squares somewhere?"

Strangely, it seems that no list has never been published. After the question was asked by Ed, I immediately studied the question and you will find in this website the answer to Ed's question: the 3x3, 4x4, 5x5, 6x6, 7x7 lists, and much more information... A lot of new results: for example, the smallest P=302400 for 5x5 squares is new, found in September 2005. As far as I know, all the examples of 5x5 multiplicative squares published before were bigger.

Thanks to Ed for his very interesting question! After the first answers to his question, Ed wrote an interesting paper in his Math Games column of MAA Online:

Thanks to Dan Asimov, Michael Kleber, Richard Schroeppel, and David Wilson for their ideas. And thanks to Edwin Clark, Don Reble, and Günter Stertenbrink for checking my big multiplicative squares, confirming that they have all the announced properties.

What is a multiplicative magic square?

It is a square which is magic using multiplication instead of addition.

A "multiplicative" magic square is very easy to construct from a standard "additive" magic square, using the numbers of the additive magic square as powers of a fixed integer. For example:

When we add the numbers of any line in the left square, we get always the same number (here 12). When we multiply the numbers of any line in the right square, we get always the same number (here 4096). This 3rd-order multiplicative square on the right was published by Antoine Arnauld in Nouveaux Eléments de Géométrie, Paris, in... 1667... a long time ago!

Some properties on multiplicative magic squares:

I recommend the book Magic squares and cubes, by W.S. Andrews, republished by Dover in 1960: it includes, pages 283-294, an excellent paper on multiplicative squares written by Harry A. Sayles, a paper which was initially published in The Monist in 1913. Various examples of multiplicative magic squares are given. Some of them are used in this page, referenced by Sayles[page in the Andrews book, figure number]. For example, the above square P=4096 can also be found in this book: Sayles[284, 510]. And I recommend the old French magazine Les Tablettes du Chercheur, year 1893, in which G. Pfeffermann published various multiplicative squares as games.

Smallest 3rd-order multiplicative magic squares

Because the construction method given above uses powers, the generated numbers are big. It is possible to construct 3x3 squares with magic products smaller than 4096.

Sayles published two examples in 1913:

I discovered in 2006 that G. Pfeffermann published a lot of 3x3 multiplicative squares 20 years before Sayles: he published 17 squares in 1893! These squares were games, as for his first bimagic squares 8x8 and 9x9. His first nine 3x3 multiplicative squares were the following, including the smallest P=216. Will you succeed in filling them?

In 1917, in his Amusements in Mathematics, Henry E. Dudeney also published the square P=216. We note that these squares can be generated by the same method:

With a=2 and b=3, we get the square P=216. With a=2 and b=5, we get the square P=1000.

As first proved in a paper published in 1983 in Discrete Mathematics by Debra K. Borkovitz (currently at Wheeklock College, Boston, USA) and Frank K.-M. Hwang (currently at National Chiao Tung University, Taiwan), the minimum magic product for 3x3 multiplicative squares is 216. In 2005, just after the above question asked by Ed, here is another -and very short- proof given by Rich Schroeppel:

If we do not consider the diagonals, here is the smallest possible semi-magic square. It was not required, but one diagonal (not the other) does have the correct product P:

And now, here is the list of the smallest possible magic products requested by Ed:

For more terms: see the 3x3 list referenced in Oct. 2005 under the number A111029 in the On-Line Encyclopedia of Integer Sequences, OEIS Foundation.

Within the 17 squares (3x3) published by G. Pfeffermann in 1893, the above first 10 smallest P were already all present. Excellent analysis done more than one century ago!

My formulation above with P = a3b3 is the complete formulation in normalized rational numbers. Here is the nice complete all-integer formulation constructed by Lee Morgenstern in December 2007, where ab = cd. All solutions may be obtained from this by scaling the 9 entries by the same constant. In an unscaled solution, the 4 middle-side entries are always squares.

Smallest 4th-order multiplicative magic squares

In 1893, G. Pfeffermann published a 4th-order pandiagonal square as a game, as usual for him. Complete the next square P=28224 with the numbers 2, 3, 4, 6, 8, 12, 14, 21, 28, 42, 56, 84. The numbers in the four quarters and in the central 2x2 square are also expected to have the same P. Will you succeed? G. Pfeffermann gave 3 solutions, but as remarked by Philippe Deléham in October 2014, his problem has 6 solutions.

In 1913, Sayles published various samples of 4th-order squares: P = 5040, 7560, 11760, 14112, 14400, 21000, 2985984. In their paper Discrete Mathematics of 1983, Borkovitz and Hwang proved that the minimum magic product for 4x4 multiplicative squares is 5040, and produced a different example from the one of Sayles.

These 2 squares can be constructed with a multiplication of cells of two latin squares (the two squares producing an Eulerian square), using two set of numbers (A, B, C, D) and (a, b, c, d). For example, Sayles's square can be built using:

and the Borkovitz & Hwang example can be produced by the same method, but using other Latin squares and other sets: (1, 2, 3, 6) and (1, 4, 5, 7).

Sayles also published a pandiagonal multiplicative square, pandiagonal meaning that all the broken diagonals give also the same magic product. As it is for Pfeffermann's 4x4 square above, it is also a most-perfect magic square: all 2x2 subsquares have the same magic product.

In 1957, Ronald B. Edwards, Rochester (NY, USA) http://geniimagazine.com/magicpedia/Ronald_B._Edwards, published in Scripta Mathematica (Vol XXII, p.202) an astonishing 4x4 multiplicative square. When we write all its numbers backward, we get again a 4x4 multiplicative square!!! What a surprising square! Its magic product 4558554 is a palindromic number. And we can border the square to obtain a 6x6 additive square.

What was the method used to build it, 50 years ago, without a computer? In February 2015, Claude Bégin, Boucherville (Québec, Canada), points out that Edwards's multiplicative square can be obtained with the construction method already seen above using these two Latin squares: (A, B, C, D) = (x1, x2, x3, x4) = (23, 11, 13, 21) and (a, b, c, d) = (x5, x6, x7, x8) = (2, 3, 11, 1). Moreover, if we write each of those eight numbers xi = 10ui + vi, with 0 ≤ ui ≤ 9 and 1 ≤ vi ≤ 9, since xi*xj = (10ui + vi)*(10uj + vj) = 100uiuj + 10(uivj + ujvi) + vivj, each number of the final multiplicative square can be reversed because, for each 1 ≤ i ≤ 4 and for each 5 ≤ j ≤ 8, we have :

We now can compute some other examples. The smallest product generating 16 distinct integers is obtained with (A, B, C, D) = (1, 2, 11, 12) and (a, b, c, d) = (1, 3, 4, 13), giving P = ABCDabcd = 41184. The largest product is obtained with (2, 11, 12, 21) and (31, 32, 33, 41), giving P = 7441023744. The smallest palindromic product is obtained with (1, 2, 11, 22) and (1, 3, 4, 12), giving P = 69696. Several other palindromic products are possible, the largest palindrome being obtained with (1, 11, 12, 21) and (11, 22, 32, 41), giving this square:

The main results of my research done in 2005 are:

For more terms: see the 4x4 list referenced in Oct. 2005 under the number A111030 in the On-Line Encyclopedia of Integer Sequences, OEIS Foundation.

In December 2007, Lee Morgenstern gives this complete formulation in normalized rational numbers of 4x4 multiplicative magic squares. It requires 7 terms. For example (a, b, c, d, e, f, g) = (1, 2, 3, 4, 5, 6, 7) gives the smallest solution P=5040 of Sayles (see his square above).

From this first formulation, here is his reasoning that produces the next formulation in normalized rational numbers of 4x4 pandiagonal multiplicative magic squares:

To be pan-magic, the 6 pan-diagonals must be magic:
(1) (ce)(c)(d)(ade) = aabcdefg
(2) (adf)(ae)(acg)(ab) = aabcdefg
(3) (abg)(bf)(f)(g) = aabcdefg
(4) (ae)(abe)(ab) = aabcdefg
(5) (d)(g)(ce)(bf) = aabcdefg
(6) (cf)(adf)(adg)(acg) = aabcdefg

Cancelling terms simplifies to
(1a) cde = abfg
(2a) aa = 1
(3a) bfg = acde
(4a) abe = cdfg
(5a) 1 = aa
(6a) acdfg = be

(5a) matches (2a).  After substituting a = 1,
(6a) matches (4a) and (3a) matches (1a).

(1b) cde = bfg
(2b) a = 1
(4b) be = cdfg

Substituting e = cdfg/b from (4b) into (1b) simplifies to
(1c) b = cd

Substituting (1c) into (4b) simplifies to
(4c) e = fg

Therefore, 3 terms are eliminated from the general 4x4 to produce a pan 4x4.
Substituting a = 1, b = cd, e = fg into the formulation

Smallest 5th-order multiplicative magic squares

In 1893, G. Pfeffermann published two samples of 5th-order pandiagonal squares: P=665280 and 1182720. Supplemental info given by Pfeffermann about the square P=1182720: the missing numbers are 1, 2, 3, 4, 5, 6, 7, 8, 10, 11. Will you succeed?

In 1913, Sayles published two other pandiagonal examples: P = 362880 and 720720. And Dudeney a bigger example P=60466176. Here is Sayles's smallest example which can be built with two Latin squares (producing an Eulerian square), and which is pandiagonal:

The main results of my research done in 2005 are:

For more terms: see the 5x5 list referenced in Oct. 2005 under the number A111031 in the On-Line Encyclopedia of Integer Sequences, OEIS Foundation.

In December 2007, Lee Morgenstern gives this complete formulation in normalized rational numbers of 5x5 multiplicative magic squares. It requires 14 terms. For example, setting s = t = u = v = w = x = y = z = 1, then setting e = f = a, we get my previous formulation above where P = a6b3c2d.

In February-March 2015, extending Edwards-Bégin's 4x4 method to 5x5 squares, and using the 5x5 Eulerian construction above, we can construct several squares, for example this one with (A, B, C, D, E) = (1, 2, 3, 11, 12) and (a, b, c, d, e) = (12, 13, 21, 22, 23):

The smallest possible product is obtained with (1, 2, 11, 12, 21) and (1, 3, 4, 13, 14) giving P = 12108096, the largest with (2, 3, 11, 12, 21) and (21, 23, 31, 32, 33) giving P = 262976668416. Contrary to 4x4 squares, it seems impossible to obtain a palindromic product.

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