The smallest possible multiplicative magic squares
What are the smallest possible multiplicative squares?
In September 2005, Ed Pegg Jr., author of the wellknown 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.
Order 
Semimagic 
Magic 
Smallest 
3 
120 
(a) 216 
(a) 36 
4 
(c) 4 320 
(b) 5 040 
(b) 28 
5 
277 200 
302 400 
45 
6 
25 945 920 
66 (!) 

7 
3 632 428 800 
91 

8 
(d) 670 442 572 800 
≤ (e) 117 (!) 

9 
(d) 140 792 940 288 000 
≤ (e) 153 (!) 

10 
≤ 2.77E+17 
≤ 290 (!) 

11 
≤ (e) 4.68E+19 
≤ (e) 238 (!) 

12 
≤ (e) 3.23E+22 
≤ (e) 276 (!) 

13 
≤ (e) 1.76E+25 
≤ (e) 350 (!) 

14 
≤ (e) 1.02E+28 
≤ (e) 406 (!) 

15 
≤ (e) 1.12E+31 
≤ (e) 465 (!) 

16 
≤ 1.03E+35 
≤ 848 (!) 

17 
≤ (e) 1.70E+37 
≤ (e) 627 (!) 
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:
Additive 
=12 
>> 
(powering step) 
>> 
Multiplicative 
=4096 

3 
8 
1 
=12 
2^{3} 
2^{8} 
2^{1} 
8 
256 
2 
=4096 

2 
4 
6 
=12 
2^{2} 
2^{4} 
2^{6} 
4 
16 
64 
=4096 

7 
0 
5 
=12 
2^{7} 
2^{0} 
2^{5} 
128 
1 
32 
=4096 

=12 
=12 
=12 
=12 

=4096 
=4096 
=4096 
=4096 
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 3rdorder 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 283294, 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 3rdorder 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:
18 
1 
12 

50 
1 
20 
4 
6 
9 
4 
10 
25 

3 
36 
2 
5 
100 
2 
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:
ab² 
1 
a²b 
a² 
ab 
b² 
b 
a²b² 
a 
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:
"There's a
standard proof that the center of a 3x3 addition magic square is K/3, where K is the row sum (Add up the four lines through the center,
subtract the whole square). Of course
this works for multiplication too, so the magic product is always the cube
of the center.
Minimality of
K = 216 for 3x3 mulgic square, Proof sketch:
K must be a
cube, with at least 9 divisors.
1, 8, 27, 64, 125
have 1, 4, 4, 7, 4 divisors. Done!"
If we do not consider the diagonals, here is the smallest possible semimagic square. It was not required, but one diagonal (not the other) does have the correct product P:
1 
20 
6 
12 
2 
5 
10 
3 
4 
And now, here is the list of the smallest possible magic products requested by Ed:
# 
P 
= 2^ 
· 3^ 
· 5^ 
· 7^ 
· 11^ 
· 13^ 
1 
216 
3 
3 
0 
0 
0 
0 
2 
1000 
3 
0 
3 
0 
0 
0 
3 
1728 
6 
3 
0 
0 
0 
0 
4 
2744 
3 
0 
0 
3 
0 
0 
5 
3375 
0 
3 
3 
0 
0 
0 
6 
4096 
12 
0 
0 
0 
0 
0 
7 
5832 
3 
6 
0 
0 
0 
0 
8 
8000 
6 
0 
3 
0 
0 
0 
9 
9261 
0 
3 
0 
3 
0 
0 
10 
10648 
3 
0 
0 
0 
3 
0 
For more terms: see the 3x3 list referenced in Oct. 2005 under the number A111029 in the OnLine 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 = a^{3}b^{3} is the complete formulation in normalized rational numbers. Here is the nice complete allinteger 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 middleside entries are always squares.
bc 
a² 
bd 
d² 
ab 
c² 
ac 
b² 
ad 
Smallest 4thorder multiplicative magic squares
In 1893, G. Pfeffermann published a 4thorder 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.




1 
24 








168 
7 
In 1913, Sayles published various samples of 4thorder 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.
1 
15 
24 
14 

1 
14 
12 
30 
12 
28 
3 
5 
15 
24 
7 
2 

21 
6 
10 
4 
42 
3 
10 
4 

20 
2 
7 
18 
8 
5 
6 
21 
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:
A 
C 
D 
B 
X 
a 
b 
c 
d 
= 
Aa 
Cb 
Dc 
Bd 
B 
D 
C 
A 
c 
d 
a 
b 
Bc 
Dd 
Ca 
Ab 

C 
A 
B 
D 
d 
c 
b 
a 
Cd 
Ac 
Bb 
Da 

D 
B 
A 
C 
b 
a 
d 
c 
Db 
Ba 
Ad 
Cc 
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 mostperfect magic square: all 2x2 subsquares have the same magic product.
1 
24 
10 
60 
30 
20 
3 
8 
12 
2 
120 
5 
40 
15 
4 
6 
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.

98 
102 
51 
64 
79 
44 

46 
39 
231 
11 

64 
93 
132 
11 

58 
46 
39 
231 
11 
53 
121 
21 
26 
69 
121 
12 
62 
96 
95 
121 
21 
26 
69 
106 

13 
253 
33 
42 
31 
352 
33 
24 
50 
13 
253 
33 
42 
47 

63 
22 
23 
143 
36 
22 
32 
341 
96 
63 
22 
23 
143 
91 


41 
93 
52 
61 
94 
97 
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) = (x_{1}, x_{2}, x_{3}, x_{4}) = (23, 11, 13, 21) and (a, b, c, d) = (x_{5}, x_{6}, x_{7}, x_{8}) = (2, 3, 11, 1). Moreover, if we write each of those eight numbers x_{i} = 10u_{i} + v_{i}, with 0 ≤ u_{i} ≤ 9 and 1 ≤ v_{i} ≤ 9, since x_{i}*x_{j} = (10u_{i} + v_{i})*(10u_{j} + v_{j}) = 100u_{i}u_{j} + 10(u_{i}v_{j} + u_{j}v_{i}) + v_{i}v_{j}, 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:
11 
264 
672 
451 

11 
462 
276 
154 
352 
861 
132 
22 
253 
168 
231 
22 

492 
32 
242 
231 
294 
23 
242 
132 

462 
121 
41 
384 
264 
121 
14 
483 
The main results of my research done in 2005 are:
1 
bc 
ab^{3} 
a^{3} 
>> 
1 
15 
54 
8 
a^{3}b 
ab^{2} 
c 
b 
24 
18 
5 
3 

ac 
ab 
a^{2}b 
b^{2} 
10 
6 
12 
9 

b^{3} 
a^{2} 
a 
abc 
27 
4 
2 
30 
16 
1 
10 
27 
5 
24 
18 
2 
6 
12 
3 
20 
9 
15 
8 
4 
# 
P 
= 2^ 
· 3^ 
· 5^ 
· 7^ 
· 11^ 
· 13^ 
1 
5040 
4 
2 
1 
1 
0 
0 
2 
6480 
4 
4 
1 
0 
0 
0 
3 
6720 
6 
1 
1 
1 
0 
0 
4 
7200 
5 
2 
2 
0 
0 
0 
5 
7560 
3 
3 
1 
1 
0 
0 
6 
7920 
4 
2 
1 
0 
1 
0 
7 
8400 
4 
1 
2 
1 
0 
0 
8 
8640 
6 
3 
1 
0 
0 
0 
9 
9072 
4 
4 
0 
1 
0 
0 
10 
9240 
3 
1 
1 
1 
1 
0 
For more terms: see the 4x4 list referenced in Oct. 2005 under the number A111030 in the OnLine 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).
1 
ce 
adf 
abg 
bf 
adg 
c 
ae 
acg 
f 
abe 
d 
ade 
ab 
g 
cf 
From this first formulation, here is his reasoning that produces the next formulation in normalized rational numbers of 4x4 pandiagonal multiplicative magic squares:
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
1 
cfg 
df 
cdg 
cdf 
dg 
c 
fg 
cg 
f 
cdfg 
d 
dfg 
cd 
g 
cf 
Smallest 5thorder multiplicative magic squares
In 1893, G. Pfeffermann published two samples of 5thorder 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?


5 
2 



28 
48 

88 
10 
4 


7 

40 


112 


8 
14 
20 
9 
44 
16 
14 
24 




11 
3 
16 
56 

20 
176 


1 
6 



80 
22 


12 
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:
Aa 
Bb 
Cc 
Dd 
Ee 
>> 
1 
10 
21 
32 
54 
Dc 
Ed 
Ae 
Ba 
Cb 
28 
48 
9 
2 
15 

Be 
Ca 
Db 
Ec 
Ad 
18 
3 
20 
42 
8 

Eb 
Ac 
Bd 
Ce 
Da 
30 
7 
16 
27 
4 

Cd 
De 
Ea 
Ab 
Bc 
24 
36 
6 
5 
14 
The main results of my research done in 2005 are:
a^{2}b 
cd 
1 
a^{3}c 
ab^{2} 
>> 
12 
35 
1 
40 
18 
a^{2}b^{2} 
a 
a^{3}b 
d 
c^{2} 
36 
2 
24 
7 
25 

ad 
b^{2}c 
bc 
a^{2} 
a^{3} 
14 
45 
15 
4 
8 

c 
a^{4} 
abd 
abc 
b 
5 
16 
42 
30 
3 

ac 
ab 
a^{2}c 
b^{2} 
a^{2}d 
10 
6 
20 
9 
28 
9 
1 
14 
44 
50 
2 
35 
55 
18 
4 
25 
33 
8 
7 
6 
22 
20 
3 
10 
21 
28 
12 
15 
5 
11 
# 
P 
= 2^ 
· 3^ 
· 5^ 
· 7^ 
· 11^ 
· 13^ 
1 
302400 
6 
3 
2 
1 
0 
0 
2 
332640 
5 
3 
1 
1 
1 
0 
3 
362880 
7 
4 
1 
1 
0 
0 
4 
393120 
5 
3 
1 
1 
0 
1 
5 
403200 
8 
2 
2 
1 
0 
0 
6 
415800 
3 
3 
2 
1 
1 
0 
7 
423360 
6 
3 
1 
2 
0 
0 
8 
443520 
7 
2 
1 
1 
1 
0 
9 
453600 
5 
4 
2 
1 
0 
0 
10 
475200 
6 
3 
2 
0 
1 
0 
For more terms: see the 5x5 list referenced in Oct. 2005 under the number A111031 in the OnLine 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 = a^{6}b^{3}c^{2}d.
abfstuvyz 
cd 
1 
a^{2}cftxy 
b^{2}esuv^{2}wy 
ab^{2}fv^{2}w 
a 
abefs^{2}tuy 
d 
c^{2}tuvxy^{2}z 
de 
b^{2}cs^{2}tuwy^{3} 
bcvx 
afuv^{2}z 
a^{2}ft 
cx 
a^{2}f^{2}tz 
abduv^{2}y 
bcestuvwy^{2} 
bs 
acstuy^{2} 
beuv^{3}x 
acftwyz 
b^{2}s 
adf 
In FebruaryMarch 2015, extending EdwardsBégin'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):
12 
26 
63 
242 
276 

21 
62 
36 
242 
672 
231 
264 
23 
24 
39 
132 
462 
32 
42 
93 

46 
36 
143 
252 
22 
64 
63 
341 
252 
22 

156 
21 
44 
69 
132 
651 
12 
44 
96 
231 

66 
253 
144 
13 
42 
66 
352 
441 
31 
24 
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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