The smallest possible bimagic square
See also the smallest possible bimagic square using distinct integers
See also the smallest possible bimagic cube

The smallest multimagic square must be a bimagic square (2-multimagic). What is the smallest bimagic square possible? The smallest known bimagic squares are 8th-ordered, which is the same order as the first bimagic discovered in 1890 by Pfeffermann. Is the mythic 7th-order bimagic square possible?

3rd or 4th-order bimagic square?

The famous Edouard Lucas, who also wrote in Les Tablettes du Chercheur, proved easily that 3rd-order bimagic squares can't exist even if composed with non-consecutive numbers (unequals). He also reported that 4th-order bimagic squares can't exist if composed with consecutive numbers. This demonstration was published on March 1st 1891, just some weeks after the publishing of the first Pfeffermann 8th-order square. A short time later Edouard Lucas died accidentally, alas, in this same year 1891.

But it is however possible to construct 4th-order squares which are at least semi-bimagic (neither magic nor bimagic for the 2 diagonals) and non-normal (= using different but non-consecutive positive integers), like the following sample that I have constructed in 2003. For the 4 rows and the 4 columns, the magic sum is 143 and the bimagic sum is 7063. It is impossible to construct a square having the same characteristics (different integers, higher or equal to 1) with smaller integers. But it is possible to construct other squares with bigger integers.

Answering the puzzle 287 "Multimagic prime squares" proposed by Carlos Rivera in his site http://www.primepuzzles.net/, Dr. Luke Pebody, of Trinity College, Cambridge, England, and Jean-Claude Rosa, France, prove independantly in October 2004 that a non-normal 4th-order bimagic square, = using distinct integers, is impossible. Read here the nice and short demonstration by Luke Pebody.

See also the problem of the 4th-order nearly bimagic square using distinct integers.

5th-order bimagic square?

For 5th-order squares, there exists only 8 series of 5 distinct integers, contained between 1 and 25 (=5²), having a sum of 65 (magic sum = 5x(5² + 1)/2) and having a squares sum of 1105 (bimagic sum = 5x(5² + 1)*(2x5² + 1)/6):

It is evident that these 8 series are not sufficient to construct a 5th-order bimagic square since at least 12 series (5 rows, 5 columns and 2 diagonals) are needed. And, for example, the number 3 can't ever be present.

Here is a magic square, that we can find in a lot of books, which is partially bimagic. Two rows, two columns and the two diagonals are bimagic. In fact, the diagonals are even trimagic.

See also the problem of the 5th-order bimagic square using distinct integers.

6th-order bimagic square?

It is more difficult, but feasible, to demonstrate that a 6th-order bimagic square cannot exist. Therefore, we may have a nearly bimagic square like this magic square where all the columns are bimagic, but alas not the rows and the diagonals:

And here is another 6th-order square also by Pfeffermann, but this time non-normal semi-bimagic. It uses non consecutive integers between 1 and 49. The square is magic. The 6 rows and 6 columns are bimagic. Only the diagonals are not bimagic.

See also the problem of the 6th-order bimagic square using distinct integers.

7th-order bimagic square?

In 1891, G. Pfeffermann published in the daily newspaper L'Echo de Paris a 7th-order partially bimagic square. His square is magic, and has 4 bimagic rows and 4 bimagic columns.

In a 7th-order square, there are 49 cells. So we must place numbers from 1 to 49, or 25 odd and 24 even numbers. The rank's S1 sum of such a square must be equal to 7*(49+1)/2 = 175. The rank's S2 sum of the squared elements must be equal to 7*(49+1)*(2*49+1)/6 = 5775, which is of the 4k+3 form.

The squares of even numbers being of the 4k form, and the squares of odd numbers being of the 4k+1 form, a sum of 7 squares can be of the 4k+3 form if and only if the rank contains 3 or 7 odd numbers. So, in order to place 25 odd numbers in 7 ranks, the only solution is to have:

Hence the main question: can a series of 7 distinct odd numbers, from 1 to 49, having 175 for sum and having 5775 for sum of squares, exist?

In 1892, Michel Frolov claimed, in the famous Société Mathématique de France bulletin, that none of these series can exist: so, he concluded that a 7th-order bimagic square cannot exist. Sorry, Monsieur Frolov! But we will forgive him, because computers didn't exist during his period. I have found 60 series having these properties, like for example:

In order to make up a row and a column af such a square, it is sufficient that at least a pair (among these 60 series) having one and only one number in common, exists. If our calculations are correct, there are exactly 424 possible pairs among these series of 7 odd numbers. For example the G2 and G13 series listed above have only the number 49 in common, so the number 49 will be at the row/column intersection of the G2/G13 pair.

If we don't limit ourselves to this odd/even question, there exists 1844 series of distinct numbers, even or odd, from 1 to 49, having 175 for sum and having 5775 for sum of squares. In 1909, the list of these 1844 series was published by Achille Rilly in Sphinx-Œdipe. Without error, and without a computer. Congratulations for this feat!

It is possible to construct a software program which, from these 424 pairs, and in using the 1844 series, tries to construct a bimagic square. Barring a bug from me, I say that a 7th-order bimagic square can't exist. But… here is a 7th-order square that I have obtained, from the G2 / G13 pair placed in 1st row / 6th column.

Where are its flaws? My challenge: will you find a better square? Is it magic? Is it bimagic? I will let you take your calculator or your spreadsheet, and enter it. So, where are the magic or bimagic flaws of this square?

Because of this challenge, Walter Trump (Germany) has tried to also approach the 7th-order bimagic square. With his computer, he has arrived at the same conclusion as I: a 7th-order bimagic square cannot exist. Because it is proved, as already seen, that a 7th-order bimagic square has to use a pair of bimagic series of 7 odd numbers among the 60 existing series, Walter Trump directly searched -as I did- squares using a pair of these series. Here is one of his nearly bimagic squares (the pair used of series being in bold) which has also 3 wrong bimagic sums but closer to the correct bimagic sum. Sum of the 3 differences = 866, compared to the 928 of my above square.

In April 2004, we can confirm and maintain that there is no 7th-order bimagic square.

Even if a mathematical proof has not yet been found, Bogdan Golunski, Germany, confirms also that it is impossible to construct such a bimagic square. With a third different person, using a third different program, and resulting in the same conclusion, we can now safely maintain the assertion.

In December 2004, a fourth person with a fourth different program has found that it is impossible to construct a 7th-order bimagic square. Pan Fengchu, China, has found that there are 86 nearly bimagic squares having three wrong S2 sums, and that it is impossible to get a better result.

His squares use one or two series of 7 odd numbers. His squares #26 and #86 are his best squares having the smallest sum of the 3 differences = 348, and use only one series of 7 odd numbers (below in bold). We can consider, as remarked by Walter Trump, that these two squares are the same: the square #86 can be constructed from the square #26, complementing it (50-x) and applying the permutation 6-7-3-4-5-1-2 for rows and columns. In fact, the 86 Fengchu squares can be reduced to 43 completely different ones. Among them, we can recognize previously known squares: his square #25 (and #46) is the above Trump square, and his square #76 (and #11) is the above Boyer square.

 

In June 2005, I constructed the first 7x7 magic square of squares. This example uses consecutive squared integers. It means that the impossibility of a 7th-order bimagic square is not coming from an impossibility to arrange its squared integers in magic square.

See also the problem of the 7th-order bimagic square using distinct integers.

In August-September 2012, Francis Gaspalou and Walter Trump proved that the smallest order of an axially symmetric bimagic square is not 8, but 12.

Open their PDF file.

D.N. Lehmer and the smallest bimagic square problem

D.N. Lehmer in 1934 (Somerset, Indiana, 1867 - Berkeley, California, 1938)

The mathematician D.N. Lehmer was interested in magic squares. For example, he published in 1929 an article entitled On the congruences connected with certain magic squares, in the Transactions of the American Mathematical Society (AMS). See some of his works in other fields at:

In an AMS meeting held at the University of California at Los Angeles, December 1, 1934, he presented a paper entitled Bimagic squares. Here is the abstract:

Only this abstract has been published by the AMS, the entire paper having never been published. It seems that this Lehmer paper is today unfortunately lost. Many thanks to:

After having studied the 3-4-5-6th order bimagic squares problems, D.N. Lehmer also studied the 7th-order. Reviewing the book Carrés Magiques au Degré n of General Cazalas, D.N. Lehmer wrote in 1935:

But it seems that the "reviewer" (= D.N. Lehmer) has never answered that question of the order 7. He died three years later, in 1938.

References: Bulletin of the AMS, Volume XLI, 1935, pages 18, 146-147, and 316-317

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