The smallest possible bimagic square using
distinct integers
See
also the smallest possible bimagic square using
consecutive integers
One of the most intriguing problems on multimagic squares: what is the smallest bimagic square using distinct integers? This problem is the Open problem 3 of my article published in The Mathematical Intelligencer (see also the table).
It is impossible to construct bimagic squares smaller than 8x8 and using consecutive integers. But if we allow the use of various but distinct integers, instead of only consecutive integers, then the freedom to choose the integers used should permit the easy construction of smaller bimagic squares (and there is no blockage owing to the squared integers, because it is possible to construct magic squares of squares smaller than 8x8). But these small bimagic squares are incredibly difficult to construct:
Smallest possible bimagic squares: 5x5 or 6x6?
Order 
Smallest 
Smallest 
Smallest 
Remarks 
3x3 
Impossible 
First proved by Edouard Lucas, 1891 

4x4 
Impossible 
First proved (independently)
by Luke Pebody 

5x5 
Unknown! 
The most wanted 

6x6 
72 
219 
10,663 
First known 6x6 squares 
7x7 
67 
238 
10,400 

8x8 
64 
260 
11,180 
They are the best possible
8x8 characteristics, 
Below are the best known squares. All of them (excluding the second 5x5 example) are at least semibimagic: all their rows and all their columns are bimagic. The difficulty is to succeed to get also two bimagic diagonals... If you have interesting results, send me a message! I will be pleased to add your results to this page.
3rd or 4thorder bimagic square using distinct integers?
Edouard Lucas proved in 1891 that a 3rdorder bimagic square using distinct integers cannot exist (see here for more details). I have published a very short proof in my M.I. article that even a 3rdorder semibimagic square is not possible, "semi" meaning that only sums of rows and columns are needed, not taking account of the diagonals.
Luke Pebody and JeanClaude Rosa proved in 2004 that it is impossible to construct a 4thorder bimagic square using distinct integers (see here for more details). This means that it is impossible to construct a 4thorder square having 20 correct sums. But it is interesting to know what the best possible 4thorder square is. My best result is the following square with 17 correct sums out of 20: only 3 bad sums. It is the CB8 square published in the M.I. article.
4x4 square... 
=205 
>>> 
...squared 
=13939 

9 
55 
105 
36 
=205 
9² 
55² 
105² 
36² 
=15427 

69 
100 
21 
15 
=205 
69² 
100² 
21² 
15² 
=15427 

28 
49 
19 
109 
=205 
28² 
49² 
19² 
109² 
=15427 

99 
1 
60 
45 
=205 
99² 
1² 
60² 
45² 
=15427 

=205 
=205 
=205 
=205 
=173 
=15427 
=15427 
=15427 
=15427 
=12467 
See also this other 4thorder square: the smallest possible semibimagic square using distinct integers. It has 16 correct sums out of 16, because in this case only 204=16 sums are required, "semi" meaning that the 4 sums of the 2 diagonals are not needed.
As seen above, it is impossible to obtain 20 correct sums. But is it possible to construct a 4thorder nearly bimagic square with 18 or 19 correct sums?
In August 2009, Lee Morgenstern proved mathematically that a 4thorder magic square can't be semibimagic:
This proof means that it is impossible to have 18 correct sums combined as follows: 10 correct sums S1, and 8 correct sums S2 (4 rows + 4 columns). But 4thorder nearlybimagic squares with 18 correct sums, using a different combination, are possible. Here is an example constructed by Lee, again in August 2009:
4x4 square... 
=1765 
>>> 
...squared 
=906271 

225 
157 
681 
606 
=1669 
225² 
157² 
681² 
606² 
=906271 

801 
222 
265 
381 
=1669 
801² 
222² 
265² 
381² 
=906271 

382 
633 
597 
57 
=1669 
382² 
633² 
597² 
57² 
=906271 

261 
657 
126 
625 
=1669 
261² 
657² 
126² 
625² 
=906271 

=1669 
=1669 
=1669 
=1669 
=1669 
=906271 
=906271 
=906271 
=906271 
=846943 
The last part of my question above remained open: is it possible to construct a 4thorder nearly bimagic square with 19 correct sums?
In AugustSeptember 2011, Lee tried to answer with these complete parametric solutions, but did not find any example with 19 correct sums:
and finally proved in June 2012 that 4thorder nearly bimagic square with 19 correct sums are impossible (text file).
5thorder bimagic square using distinct integers?
My best result is the following square with 22 correct sums out of 24: only 2 bad sums. It is the CB9 square published in the M.I. article.
5x5 magic square... 
=120 
>>> 
...squared 
=3296 

3 
37 
20 
44 
16 
=120 
3² 
37² 
20² 
44² 
16² 
=3970 

34 
35 
1 
12 
38 
=120 
34² 
35² 
1² 
12² 
38² 
=3970 

41 
8 
24 
40 
7 
=120 
41² 
8² 
24² 
40² 
7² 
=3970 

10 
36 
47 
13 
14 
=120 
10² 
36² 
47² 
13² 
14² 
=3970 

32 
4 
28 
11 
45 
=120 
32² 
4² 
28² 
11² 
45² 
=3970 

=120 
=120 
=120 
=120 
=120 
=120 
=3970 
=3970 
=3970 
=3970 
=3970 
=4004 
The above diagonals are not bimagic. If we construct magic squares with their four lines through the centre being bimagic (= 2 bimagic diagonals + bimagic central row + bimagic central column), and with all their rows bimagic, then it seems impossible to get any other bimagic column! An example of such a square with 20 correct sums out of 24, 4 bad sums:
5x5 magic square... 
=120 
>>> 
...squared 
=3856 

1 
17 
37 
26 
39 
=120 
1² 
17² 
37² 
26² 
39² 
=3856 

50 
25 
21 
11 
13 
=120 
50² 
25² 
21² 
11² 
13² 
=3856 

33 
31 
41 
10 
5 
=120 
33² 
31² 
41² 
10² 
5² 
=3856 

14 
7 
19 
35 
45 
=120 
14² 
7² 
19² 
35² 
45² 
=3856 

22 
40 
2 
38 
18 
=120 
22² 
40² 
2² 
38² 
18² 
=3856 

=120 
=120 
=120 
=120 
=120 
=120 
=4270 
=3524 
=3856 
=3566 
=4064 
=3856 
My feeling is that 5thorder bimagic squares do not exist. If yes, who will produce a mathematical proof?
Is it possible to construct a 5thorder bimagic square (= 24 correct sums)? Or at least a 5thorder nearly bimagic square with 23 correct sums?
If we accept nonmagic squares the two examples above are magic, then it is possible to construct a 5thorder nearly bimagic square with 23 correct sums, as remarked by Lee Morgenstern in August 2006, giving this example with only one nonmagic diagonal.
This 5x5 nonmagic square... 
=1505 
>>> 
...becomes magic 
=456929 

296 
388 
217 
268 
316 
=1485 
296² 
388² 
217² 
268² 
316² 
=456929 

220 
328 
350 
349 
238 
=1485 
220² 
328² 
350² 
349² 
238² 
=456929 

394 
265 
286 
314 
226 
=1485 
394² 
265² 
286² 
314² 
226² 
=456929 

301 
280 
370 
202 
332 
=1485 
301² 
280² 
370² 
202² 
332² 
=456929 

274 
224 
262 
352 
373 
=1485 
274² 
224² 
262² 
352² 
373² 
=456929 

=1485 
=1485 
=1485 
=1485 
=1485 
=1485 
=456929 
=456929 
=456929 
=456929 
=456929 
=456929 
In October 2006, Lee Morgenstern searched a 5thorder "associative" bimagic square. In such a square, the sum of two symmetrical numbers around the centre is constant, as it is in the CB9 square. His conclusion: there is no 5x5 associative bimagic square using numbers all smaller than 600,000.
In July 2009, Lee succeeded in finding a 5thorder nearly bimagic square with 23 correct sums, and this time his square is magic!
5x5 magic square... 
=1030 
>>> 
...squared 
=300094 

187 
289 
3 
109 
442 
=1030 
187² 
289² 
3² 
109² 
442² 
=325744 

111 
457 
205 
250 
7 
=1030 
111² 
457² 
205² 
250² 
7² 
=325744 

493 
103 
130 
219 
85 
=1030 
493² 
103² 
130² 
219² 
85² 
=325744 

178 
147 
463 
1 
241 
=1030 
178² 
147² 
463² 
1² 
241² 
=325744 

61 
34 
229 
451 
255 
=1030 
61² 
34² 
229² 
451² 
255² 
=325744 

=1030 
=1030 
=1030 
=1030 
=1030 
=1030 
=325744 
=325744 
=325744 
=325744 
=325744 
=325744 
We can see that the maximum integer used in the above square is 493. In AugustSeptember 2009, Gildas Guillemot, Ecole Nationale Supérieure d'Arts et Métiers (ENSAM), France, computed that there is no 5x5 bimagic square using 25 distinct integers < 500. His program ran on a PC for 14 days.
In July 2010, Michael Quist found a new 5thorder nearly bimagic square with 23 correct sums, also magic square, but with smaller integers (maximum integer = 340) and smaller magic sums than the previous square of Lee.
5x5 magic square... 
=844 
>>> 
...squared 
=153340 

25 
129 
200 
295 
195 
=844 
25² 
129² 
200² 
295² 
195² 
=182316 

257 
165 
1 
225 
196 
=844 
257² 
165² 
1² 
225² 
196² 
=182316 

127 
340 
171 
111 
95 
=844 
127² 
340² 
171² 
111² 
95² 
=182316 

267 
85 
265 
176 
51 
=844 
267² 
85² 
265² 
176² 
51² 
=182316 

168 
125 
207 
37 
307 
=844 
168² 
125² 
207² 
37² 
307² 
=182316 

=844 
=844 
=844 
=844 
=844 
=844 
=182316 
=182316 
=182316 
=182316 
=182316 
=182316 
In July 2011, trying to find a smaller example, Gildas Guillemot computed that there is no 5x5 nearly bimagic square with 23 correct sums having all its integers < 300. He found these 5719 different squares having 22 correct sums: zipped text file of 336Kb. In his search, the squares are magic, and the bad sums are the diagonals of squared numbers. In January 2014, he recomputed these squares, found the same results, and produced a list of the 5719 squares sorted by their S1: zipped text file of 329Kb.
In November 2014, Lee Morgenstern announced that there is no 5x5 bimagic square with distinct entries using integers in the range 1 through 1500. Using his 5x5 bimagic search technique of January 2014, it took about 2 months of computer time on 4 processors to reach this result.
And what about 5thorder bimagic squares if we allow some repeated numbers? In November 2007, Michael Cohen (Washington DC, USA) submitted a paper entitled "The Order 5 General Bimagic Square" to appear in the Journal of Recreational Mathematics. In this paper now published Vol 34(2) pp107111, 20052006 (but issue printed only in the second half of 2008), he poses this interesting problem:
He gives a square having 5 sets of numbers. I sent him a slightly better square having 6 sets of numbers. In his square, this set {1, 5, 9, 9, 16} uses the number 9 twice. In my alternative square, the numbers are distinct within each set: {4, 9, 18, 26, 28}, {4, 10, 17, 24, 30}, {1, 12, 22, 24, 26}, {9, 10, 12, 20, 34}, {1, 16, 18, 20, 30}, {4, 9, 20, 22, 30}. Who will construct a bimagic square using more than 6 sets of numbers (distinct numbers within each set)?
1 
3 
12 
13 
11 

4 
18 
26 
28 
9 
16 
9 
5 
1 
9 
17 
30 
24 
4 
10 

5 
7 
15 
12 
1 
24 
1 
22 
26 
12 

9 
4 
1 
11 
15 
10 
20 
12 
9 
34 

9 
17 
7 
3 
4 
30 
16 
1 
18 
20 
6thorder bimagic square using distinct integers?
 1) 1894. 6thorder nearly bimagic square by G. Pfeffermann, 26 correct sums out of 28, two bad sums.
 2) 2005. I constructed a better 6thorder square, with 27 correct sums out of 28, only one bad sum:
6x6 magic square... 
=168 
>>> 
...squared 
=6524 

41 
35 
6 
2 
37 
47 
=168 
41² 
35² 
6² 
2² 
37² 
47² 
=6524 

5 
42 
33 
39 
46 
3 
=168 
5² 
42² 
33² 
39² 
46² 
3² 
=6524 

38 
7 
45 
43 
1 
34 
=168 
38² 
7² 
45² 
43² 
1² 
34² 
=6524 

22 
55 
13 
11 
49 
18 
=168 
22² 
55² 
13² 
11² 
49² 
18² 
=6524 

53 
10 
17 
23 
14 
51 
=168 
53² 
10² 
17² 
23² 
14² 
51² 
=6524 

9 
19 
54 
50 
21 
15 
=168 
9² 
19² 
54² 
50² 
21² 
15² 
=6524 

=168 
=168 
=168 
=168 
=168 
=168 
=168 
=6524 
=6524 
=6524 
=6524 
=6524 
=6524 
=6012 
 3) Is it possible to construct a 6thorder bimagic square (= 28 correct sums)? YES!
Jaroslaw Wroblewski (Wroclaw 1962  )
Photo during
the "LIII Olimpiada Matematyczna" of Poland, in 2002. Click on the image to enlarge it.
2006, February 4th: Jaroslaw Wroblewski, University of Wroclaw, Poland, was the first to construct a 6thorder bimagic square, 28 correct sums out of 28. An important result: before him, nobody had succeeded in constructing a bimagic square smaller than the classical 8thorder bimagic squares first built in 1890!
6x6 magic square... 
=408 
>>> 
...squared 
=36826 

17 
36 
55 
124 
62 
114 
=408 
17² 
36² 
55² 
124² 
62² 
114² 
=36826 

58 
40 
129 
50 
111 
20 
=408 
58² 
40² 
129² 
50² 
111² 
20² 
=36826 

108 
135 
34 
44 
38 
49 
=408 
108² 
135² 
34² 
44² 
38² 
49² 
=36826 

87 
98 
92 
102 
1 
28 
=408 
87² 
98² 
92² 
102² 
1² 
28² 
=36826 

116 
25 
86 
7 
96 
78 
=408 
116² 
25² 
86² 
7² 
96² 
78² 
=36826 

22 
74 
12 
81 
100 
119 
=408 
22² 
74² 
12² 
81² 
100² 
119² 
=36826 

=408 
=408 
=408 
=408 
=408 
=408 
=408 
=36826 
=36826 
=36826 
=36826 
=36826 
=36826 
=36826 
Structured as most of the other squares of this page, Wroblewski's square is "associative": the sum of two symmetrical numbers around the centre is constant. His square is announced as the smallest possible 6thorder associative bimagic square.
After that square with (MaxNb, S1, S2) = (135, 408, 36826), he found three bigger 6thorder associative bimagic squares: (205, 618, 86978), (215, 648, 86684), (237, 714, 111074).
 4) A next possible step? Because Jaroslaw Wroblewski tried to find associative squares, his above smallest square was perhaps not the smallest possible 6thorder bimagic square. Hence, this new question: is it possible to construct a smaller 6thorder bimagic square, with MaxNb<135, or S1<408, or S2<36826? (with another structure, or without any structure)
Walter Trump searched for 6thorder symmetrical left/right bimagic squares: no solution with any MaxNb < 192.
In May 2006, immediately after his 7thorder bimagic squares, Lee Morgenstern found two better 6thorder squares than Wroblewski's squares: (72, 219, 10663) and (109, 330, 26432). He used unusual structures. Look at his best example below: in the two columns on the left, two cells of the same colour have the same sum 73. Not coloured below, but exactly the same structure in the two central columns, and in the two right columns.
6x6 magic square... 
=219 
>>> 
...squared 
=10663 

72 
18 
17 
16 
49 
47 
=219 
72² 
18² 
17² 
16² 
49² 
47² 
=10663 

13 
52 
36 
5 
50 
63 
=219 
13² 
52² 
36² 
5² 
50² 
63² 
=10663 

38 
35 
7 
66 
15 
58 
=219 
38² 
35² 
7² 
66² 
15² 
58² 
=10663 

20 
53 
34 
39 
69 
4 
=219 
20² 
53² 
34² 
39² 
69² 
4² 
=10663 

55 
1 
57 
56 
26 
24 
=219 
55² 
1² 
57² 
56² 
26² 
24² 
=10663 

21 
60 
68 
37 
10 
23 
=219 
21² 
60² 
68² 
37² 
10² 
23² 
=10663 

=219 
=219 
=219 
=219 
=219 
=219 
=219 
=10663 
=10663 
=10663 
=10663 
=10663 
=10663 
=10663 
Lee Morgenstern used another unusual structure in his other 6thorder bimagic square.
Is it possible to construct a smaller 6thorder bimagic square, with MaxNb<72, or S1<219, or S2<10663?
In October 2008, Lee Morgenstern finished his exhaustive search: there is no solution with MaxNb<72, which means that his above square can't be beaten! But the minimum S1 and S2 are still open questions.
And in July 2009, after several months of computation using up to 8 processors, Lee Morgenstern concluded that his solution has also the smallest S1 and the smallest S2. His square above is THE smallest possible 6thorder bimagic square. Thus, the answer to the above question is no!
7thorder bimagic square using distinct integers?
 1) 2005. My best result was the following square with 31 correct sums out of 32: only one bad sum.
7x7 magic square... 
=196 
>>> 
...squared 
=7244 

51 
8 
29 
21 
26 
11 
50 
=196 
51² 
8² 
29² 
21² 
26² 
11² 
50² 
=7244 

32 
10 
53 
18 
33 
43 
7 
=196 
32² 
10² 
53² 
18² 
33² 
43² 
7² 
=7244 

25 
34 
44 
1 
41 
9 
42 
=196 
25² 
34² 
44² 
1² 
41² 
9² 
42² 
=7244 

19 
39 
2 
28 
54 
17 
37 
=196 
19² 
39² 
2² 
28² 
54² 
17² 
37² 
=7244 

14 
47 
15 
55 
12 
22 
31 
=196 
14² 
47² 
15² 
55² 
12² 
22² 
31² 
=7244 

49 
13 
23 
38 
3 
46 
24 
=196 
49² 
13² 
23² 
38² 
3² 
46² 
24² 
=7244 

6 
45 
30 
35 
27 
48 
5 
=196 
6² 
45² 
30² 
35² 
27² 
48² 
5² 
=7244 

=196 
=196 
=196 
=196 
=196 
=196 
=196 
=196 
=7244 
=7244 
=7244 
=7244 
=7244 
=7244 
=7244 
=7706 
It is interesting to note that this square is very nearly a normal magic square. The 49 integers used are from 1 to 55, which means only six integers missing. This square does not use consecutive integers, but is not far off!
 2) Is it possible to construct a 7thorder bimagic square (=32 correct sums)? YES!
2006, May 9th: competing with Frank Rubin, who had found another square with 31 correct sums, Lee Morgenstern constructed the first known 7thorder bimagic square, 32 correct sums. From May 9th to May 25th, Lee Morgenstern found a total of eight squares! His best one, with the smallest sums and numbers, is the square below having (MaxNb, S1, S2) = (67, 238, 10400). It has very unusual symmetries, the colored squares or rectangles being inside associative: the sum of two opposite numbers is constant = 68 = twice the central number of the square.
Lee Morgenstern is a retired mathematician living in Los Angeles, California, USA. He now solves puzzles as a hobby. He is a winner of some of Frank Rubin's contests at www.contestcen.com, and has solved some of Rubin's harder puzzles. In 1989, when he was living in Sylmar, California, he won the Tanglestown USA Puzzle Contest (see the Los Angeles Times, March 12, 1989, Part II, page 4).
7x7 magic square... 
=238 
>>> 
...squared 
=10400 

26 
50 
51 
21 
19 
10 
61 
=238 
26² 
50² 
51² 
21² 
19² 
10² 
61² 
=10400 

18 
42 
49 
47 
17 
7 
58 
=238 
18² 
42² 
49² 
47² 
17² 
7² 
58² 
=10400 

57 
41 
1 
22 
54 
38 
25 
=238 
57² 
41² 
1² 
22² 
54² 
38² 
25² 
=10400 

15 
53 
31 
34 
37 
62 
6 
=238 
15² 
53² 
31² 
34² 
37² 
62² 
6² 
=10400 

27 
11 
14 
46 
67 
43 
30 
=238 
27² 
11² 
14² 
46² 
67² 
43² 
30² 
=10400 

66 
39 
48 
5 
24 
33 
23 
=238 
66² 
39² 
48² 
5² 
24² 
33² 
23² 
=10400 

29 
2 
44 
63 
20 
45 
35 
=238 
29² 
2² 
44² 
63² 
20² 
45² 
35² 
=10400 

=238 
=238 
=238 
=238 
=238 
=238 
=238 
=238 
=10400 
=10400 
=10400 
=10400 
=10400 
=10400 
=10400 
=10400 
Among its 8 squares: 5 use this symmetry, 2 use another unusual symmetry, and 1 is an associative square (69, 245, 11483).
Is it possible to construct a smaller 7thorder bimagic square, with MaxNb<67, or S1<238, or S2<10400?
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