The smallest possible bimagic square using distinct 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:

• no 5x5 bimagic square currently known
• first 6x6 bimagic squares constructed only in February 2006, by Jaroslaw Wroblewski, more than one century after the first 6x6 studies
• first 7x7 bimagic squares constructed only in May 2006, by Lee Morgenstern

Smallest possible bimagic squares: 5x5 or 6x6?

 Order SmallestMax nb SmallestS1 SmallestS2 Remarks 3x3 Impossible First proved by Edouard Lucas, 1891 4x4 Impossible First proved (independently) by Luke Pebodyand Jean-Claude Rosa, 2004 5x5 Unknown! The most wantedbimagic problem !!!!!! 6x6 72 219 10,663 First known 6x6 squaresfound by Jaroslaw Wroblewski, 2006.But best possible 6x6 square and first known 7x7 squares found by Lee Morgenstern, 2006:is his best 7x7 the best possible? 7x7 67 238 10,400 8x8 64 260 11,180 They are the best possible 8x8 characteristics,because using consecutive integers from 1 to 64.First found by G. Pfeffermann, 1890.

Below are the best known squares. All of them (excluding the second 5x5 example) are at least semi-bimagic: 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 4th-order bimagic square using distinct integers?

Edouard Lucas proved in 1891 that a 3rd-order 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 3rd-order semi-bimagic square is not possible, "semi" meaning that only sums of rows and columns are needed, not taking account of the diagonals.

Luke Pebody and Jean-Claude Rosa proved in 2004 that it is impossible to construct a 4th-order bimagic square using distinct integers (see here for more details). This means that it is impossible to construct a 4th-order square having 20 correct sums. But it is interesting to know what the best possible 4th-order 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 4th-order square: the smallest possible semi-bimagic square using distinct integers. It has 16 correct sums out of 16, because in this case only 20-4=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 4th-order nearly bimagic square with 18 or 19 correct sums?

In August 2009, Lee Morgenstern proved mathematically that a 4th-order magic square can't be semi-bimagic:

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 4th-order nearly-bimagic 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 4th-order nearly bimagic square with 19 correct sums?

In August-September 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 4th-order nearly bimagic square with 19 correct sums are impossible (text file).

5th-order 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 5th-order bimagic squares do not exist. If yes, who will produce a mathematical proof?

Is it possible to construct a 5th-order bimagic square (= 24 correct sums)? Or at least a 5th-order nearly bimagic square with 23 correct sums?

If we accept non-magic squares -the two examples above are magic-, then it is possible to construct a 5th-order nearly bimagic square with 23 correct sums, as remarked by Lee Morgenstern in August 2006, giving this example with only one non-magic diagonal.

 This 5x5 non-magic square... =1505 >>> ...becomes magicwhen squared =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 5th-order "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 5th-order 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 August-September 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 5th-order 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.

March 2018: first 5x5 semi-bimagic square of primes, by Nicolas Rouanet.

And what about 5th-order 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) pp107-111, 2005-2006 (but issue printed only in the second half of 2008), he poses this interesting problem:

• Find an order 5 bimagic square in which all 12 row-column-diagonal sets of numbers are distinct sets

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

In March 2017, Frank Rubin, author of http://contestcen.com, did not find more than 6 sets, but found another example with 6 sets. His example can be obtained from my above square, replacing 1->35, 4->28, 9->47, 10->21, 12->24, 16->12, 17->29, 18->43, 20->42, 22(center)->15, 24->54, 26->7, 28->10, 30->3, 34->1:

 28 43 7 10 47 29 3 54 28 21 54 35 15 7 24 21 42 24 47 1 3 12 35 43 42

And also in March 2017, just for fun because not useful in N, I constructed this 5x5 fully bimagic square example in Z/41Z (or if you prefer modulo 41) using 25 distinct integers!

 1 6 13 8 32 7 29 3 37 25 20 22 12 2 4 40 28 21 36 17 33 16 11 18 23

6th-order bimagic square using distinct integers?

--- 1) 1894. 6th-order nearly bimagic square by G. Pfeffermann, 26 correct sums out of 28, two bad sums.

--- 2) 2005. I constructed a better 6th-order 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 6th-order 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 6th-order 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 8th-order 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 6th-order associative bimagic square.

After that square with (MaxNb, S1, S2) = (135, 408, 36826), he found three bigger 6th-order associative bimagic squares: (205, 618, 86978), (215, 648, 86684), (237, 714, 111074).

• Download the four 6th-order bimagic squares of Jaroslaw Wroblewski (Excel file, 27Kb)

--- 4) A next possible step? Because Jaroslaw Wroblewski tried to find associative squares, his above smallest square was perhaps not the smallest possible 6th-order bimagic square. Hence, this new question: is it possible to construct a smaller 6th-order bimagic square, with MaxNb<135, or S1<408, or S2<36826? (with another structure, or without any structure)

Walter Trump searched for 6th-order symmetrical left/right bimagic squares: no solution with any MaxNb < 192.

In May 2006, immediately after his 7th-order bimagic squares, Lee Morgenstern found two better 6th-order 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 6th-order bimagic square.

• Download the two 6th-order bimagic squares of Lee Morgenstern (Excel file, 25Kb)

Is it possible to construct a smaller 6th-order 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 6th-order bimagic square. Thus, the answer to the above question is no!

March 2018: first 6th-order semi-bimagic square of primes, by Nicolas Rouanet.

7th-order 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 7th-order 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 7th-order 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 Tangles Town USA Puzzle Contest organized by the Los Angeles Times.

 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).

• Download the eight 7th-order bimagic squares of Lee Morgenstern (Excel file, 55Kb)

Is it possible to construct a smaller 7th-order bimagic square, with MaxNb<67, or S1<238, or S2<10400?

March 2018: first 7th-order semi-bimagic square of primes, with one bimagic diagonal, by Nicolas Rouanet.