Latest research on the "3x3 magic square of squares" problem

This page presents the most interesting studies on the Open problems 1 and 2 done after the publication of my article "Some notes on the magic squares of squares problem" in 2005. The two corresponding prizes are still to be won!

If you want to study these two problems, I STRONGLY recommend to first:

Send me your new results, they will be added if interesting.

Received March 6, 2006
Ajai Choudhry, India, constructed some 3x3 squares with 7 correct sums, their magic sums being not a square. For example:

Interesting, because most of the 3x3 squares with 7 correct sums are coming from the Lucas family, in which the magic sum is a square. The first known example with a non-square magic sum was constructed by Michael Schweitzer (Fig MS4 of the M.I. article). It would be very interesting to find a parametric solution with a non-square magic sum, generating an infinite number of 3x3 squares. With such a new parametric solution, it should be easy to mathematically check if this family can or cannot produce a solution with 8 magic sums. The Lucas family is unfortunately proved unable to produce 8 magic sums.

Received October 10th, 2006
Message from Randall Rathbun, USA, on the Open problem 2:

Several years before this problem 2, Randall already worked on the Gardner's challenge (=our open problem 1). It was in 1999, look at:

Received November 5th, 13th, 23rd, 2006
Jean-Claude Rosa, France, constructed this 3x3 semi-magic square (with 6 correct sums) using only odd numbers. Interesting, because most of the 3x3 semi-magic squares useg both odd and even numbers. Strange: in this smallest possible example, all the numbers used are squares of primes.

Using these 3 primitive pythagorean triangles having the same area:

he constructed this 3x3 square (with 7 correct sums) using only odd numbers. 6 of its 9 numbers are squares of primes.

And using other pythagorean triangles having the same area, he produced this other square (with 7 correct sums) using only odd numbers. The magic sum is smaller than the previous square.

Received November 30th, 2006
Theorem from Lee Morgenstern, USA: "In a 3x3 magic square of distinct squares, the smallest square cannot be 1."

Received December 10th, 2006
Only some days after Lee Morgenstern, Jean-Claude Rosa independently proved that a 3x3 magic square can't be constructed using 1 and eight squares of odd numbers.

Received December 19th, 2006
From Lee Morgenstern, USA, far extending his previous proof excluding 1: "if there is a 3x3 magic square of distinct squares, then all entries must be above 10^14."

Received December 23rd, 2006
Lucien Pech, when he was a "Mathématiques Spéciales" student, scholar year 2005-2006, choosed the "3x3 magic square of squares" as the subject for his TIPE: Travaux d'Initiative Personnelle Encadrés = supervised personal search. He searched a magic hourglass, such a solution would solve the Open problem 2, because using 7 squares. He used the Duncan Buell's method (see reference [12], top of this page), but didn't find any solution to this very difficult problem. His best result is an excellent modulo 2^52 example, better than the Duncan Buell's modulo 2^46 example.

Lucien Pech, now a student at the ENS Paris (Ecole Normale Supérieure, rue d'Ulm), sent this revised version of his TIPE:

Received July 21st, 2007. Updated results received October 18th, 2007.
After his paper published in 2003 [49] on the magic squares of squares problem, Landon Rabern, USA, searched a 3x3 magic square having at least 7 squared integers. Using a similar method that my study of 2004 [8], he didn't find any new example different of the only known example.

You can download and use his Windows application with any other range of center cells, even if he doesn't provide any explanation on its use (unfortunately): http://landon314.brinkster.net/MagicSearcher.zip. This application needs the Microsoft .NET Framework. The code sources are provided.

Received April 8th and 12th, 2008.
From Lee Morgenstern, USA, the complete formula for all 3x3 semi-magic squares of squares (better than the Lucas formula producing some, but not all, 3x3 semi-magic squares of squares), and a list of 3x3 semi-magic squares with 7 correct sums and using odd entries (including the two first smallest squares given above by J.-C. Rosa in 2006):

Received February 6th, 2009.
From Lee Morgenstern, USA: "We know that the magic sum of a 3x3 fully-magic square must be three times the central entry. Are there any 3x3 semi-magic squares of squares with a magic sum of three times any square? Looking over the published results, such as with the Lucas formula, all the magic sums are squares.The new non-square magic sums that you published on your previous update weren't three times a square. I searched 3x3 semi-magic squares of squares and found 20 of them with a magic sum that is three times a square (searched up to a magic sum of 3 x 5000²). One of them, the smallest, had a magic sum which was three times one of the actual entries. Since you can rearrange rows and columns to make another semi-magic square, any of the entries can become the central one. So here is my 3x3 semi-magic square of squares having a magic sum which is three times the central entry (S = 3 x 1105²). This is the only known solution."

For his other 19 solutions, the magic sum was three times a square different from any of the entries. Their sums (< 3 x 5000²) are 3 x 1225², 1275², 1533², 1955², 1989², 2125², 2265², 2335², 2345², 2675², 3395², 3485², 3515², 3575², 3655², 3765², 3885², 3995², 4193². Here is the example S = 3 x 1225²:

Received from May 17th to July 27th, 2009.
Frank Rubin worked on 3x3 semi-magic squares of squares having 7 magic sums, only one diagonal being not magic. With Sd1 = magic sum = sum of the magic diagonal, and Sd2 = sum of the non-magic diagonal, Frank searched squares having the best possible ratio Sd1/Sd2 ~ 1.

Using a Lee Morgenstern's method for finding 3x3 nearly-magic squares of squares with close sums, here is one of the impressive squares computed by Frank Rubin:

Sd1 = 1890006405715707401173356334328966702876471825085875343146504267674569
Sd2 = 1890006405715707267778636165444057741201927206436686602706450141117123

Sd1/Sd2 = 1.0000000000000000705... Very close to 1... And Sd1/Sd2 = 1 would be a solution of this 3x3 magic square of squares problem!

Remarking that the magic sum is a square (S = Sd1 = 43474203911235768609981537098048163²), I thought that the Rubin's square was simply a member of the Lucas's family.
January 22nd 2010, mentioning my feeling to Randall Rathbun, he confirmed, finding (p, q, r, s) = (51498645679307420, 68881590670955891, 80839778471595961, 171878882029570731).
It is unfortunately known that this family can't produce a 3x3 magic square of squares.

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