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 that you 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 not being a square. For example:
7656² 
14543² 
16764² 
18127² 
14916² 
264² 
12804² 
10824² 
16433² 
Interesting, because most of the 3x3 squares with 7 correct sums come from the Lucas family, in which the magic sum is a square. The first known example with a nonsquare 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 nonsquare magic sum, generating an infinite number of 3x3 squares. With such a new parametric solution, it should be easy to check mathematically 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:
"I wore out the meccah high speed calculation cluster at Harvard
University trying, but no success. Spent weeks running the code, but nothing
more turned up than the 1 example that Dr Andrew Bremner found.
Quite
discouraging, actually"
Several years before this problem 2, Randall already worked on Gardner's challenge (=our open problem 1). It was in 1999, look at:
Received November 5th, 13th, 23rd, 2006
JeanClaude Rosa,
France, constructed this 3x3 semimagic square (with 6 correct sums) using only
odd numbers. Interesting, because most of the 3x3 semimagic squares use both odd
and even numbers. Strange: in this smallest possible example, all the numbers
used are squares of primes.
11² 
23² 
71² 
43² 
59² 
19² 
61² 
41² 
17² 
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.
5521² 
10337² 
19069² 
20399² 
9109² 
1367² 
7373² 
17639² 
11639² 
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.
14393² 
2171² 
13507² 
12181² 
10517² 
11633² 
6227² 
16703² 
8749² 
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,
JeanClaude 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, school year 20052006, chose the "3x3 magic square
of squares" as the subject for his TIPE: Travaux d'Initiative Personnelle
Encadrés = supervised personal search. He searched for a magic hourglass;
such a solution would solve the Open problem 2, since it uses 7 squares. He
used 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 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 for
a
3x3 magic square having at least 7 squared integers. Using a similar method as
my study of 2004 [8], he didn't find
any new example different from the only
known example.
You can download and use his Windows application with any other range of center cells, even though 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 semimagic squares of squares (better
than the Lucas formula producing some, but not
all, 3x3 semimagic squares of squares), and a list of 3x3 semimagic 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 fullymagic square must be three
times the central entry. Are there any 3x3 semimagic 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 nonsquare
magic sums that you published on your previous update weren't three times a
square. I searched for 3x3 semimagic 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
semimagic square, any of the entries can become the central one. So here is
my 3x3 semimagic square of squares having a magic sum which is three times
the central entry (S = 3 x 1105²). This is the only known solution."
1751² 
155² 
757² 
595² 
1105² 
1445² 
493² 
1555² 
1001² 
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:
2105² 
25² 
265² 
235² 
1877² 
961² 
125² 
989² 
1873² 
Received from May 17th to July 27th, 2009.
Frank Rubin (http://contestcen.com/)
worked
on 3x3 semimagic 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 nonmagic diagonal, Frank searched for squares having
the best possible ratio Sd1/Sd2 ~ 1.
Using Lee Morgenstern's method for finding 3x3 nearlymagic squares of squares with close sums, here is one of the impressive squares computed by Frank Rubin:
32004859810489663461722097429913882² 
6566314229570234516482551556735538² 
28680635830907835745130420028727601² 
20694684230850857980455894108812542² 
25099843330956651396728662500908321² 
28839804352015135412123656375760542² 
20914717277840113725279028899813359² 
34883918758013437522259135104857422² 
15352303377279966158185135232591402² 
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!
Noticing that the magic sum is a square (S = Sd1 = 43474203911235768609981537098048163²),
I thought that Rubin's square was simply a member of the Lucas
family.
January 22nd 2010, mentioning my
feeling to Randall Rathbun,
he confirmed, finding (p, q, r, s) = (51498645679307420, 68881590670955891,
80839778471595961, 171878882029570731).
It is known that unfortunately this
family can't produce a 3x3 magic square of squares.
Received January 26th, 2010.
Here is Lee
Morgenstern's method, above used by Frank Rubin in MayJuly 2009
for 3x3 nearlymagic squares
of squares with close sums. This method uses Hillyer's formula of Pythagorean
Triangles having equal areas, and Newton's method of finding a root of a polynomial:
Received March 2nd3rd, 2010.
Randall Rathbun made a serious attempt to solve the enigma
#1, searched for a 3x3 magic square having 7 squared entries, different from
the only known example. He created more than 116,000,000 magic squares
having 6 squared entries, looking to see if their 7th, 8th or 9th entries
was a squared integer. But alas, no new solution!
Received April 8th, 2010.
Interesting remark by Lee Sallows
on the enigma #1: it is possible to construct a 3x3 magic square having 7 squared
entries, if we allow... allow from SallowS ? ;)... Gaussian
integers! Here is his nice example having a null magic sum, the two nonsquared
entries being 6 and 6:
(1+2i)² 
(2+2i)² 
(2+i)² 
= 
3+4i 
8i 
3+4i 
6 
0² 
6 
6 
0 
6 

(1+2i)² 
(2+2i)² 
(2i)² 
34i 
8i 
34i 
Can somebody construct a 3x3 magic square having 8 or 9 distinct squared Gaussian integers?
Received October 15th, 2010
Here is Lee
Morgenstern's method for finding 3x3 magic squares with 7 distinct square entries.
This method would be (perhaps?) able to solve the enigma
1.
Received May 27th, 2011
Ant King
(www.mathstutoring.co.uk) constructed
this nice parametric solution of 3x3 semimagic squares of squares using a single
variable, and only one diagonal is not magic. Because its magic sum is a squared
integer [3(1+ 3k + 3k²)²]², this parametric solution unfortunately
can't produce magic squares of squares (needing a magic sum equaling three times
a squared integer, or more precisely three times its center).
(2 +14k + 37k^{2} + 42k^{3} +18k^{4})^{2} 
(−2 −8k − 7k^{2} + 6k^{3} + 9k^{4})^{2} 
(1+10k + 29k^{2} + 36k^{3} +18k^{4})^{2} 
(2 +12k + 29k^{2} + 30k^{3} + 9k^{4})^{2} 
(1+ 6k +19k^{2} + 30k^{3} +18k^{4})^{2} 
(2 +12k + 29k^{2} + 36k^{3} +18k^{4})^{2} 
(−1− 2k + 7k^{2} + 24k^{3} +18k^{4})^{2} 
(2 +16k + 43k^{2} + 48k^{3} +18k^{4})^{2} 
(2 +10k +19k^{2} +18k^{3} + 9k^{4})^{2} 
Received from November 15th 2011 to January 27th 2012
Lee
Morgenstern sent various interesting remarks and results:
For example, in the last but one document, he extended the above remark of
Lee Sallows on Gaussian integers.
And in the last document, he constructed a magic square of squares
modulo 2^90, therefore better than previous magic hourglasses of squares by Duncan Buell
and Lucien Pech (respectivelly mod 2^46 and mod 2^52).
Received March 2nd, 2012, updated May 6th, 2012,
updated again August 21st, 2012
Mike Winkler
(www.mikewinkler.co.nf),
using Morgenstern's above remark of Feb. 2009, searched
for 3x3 semimagic squares having a magic sum which is three times
a square.
His search with Delphi was done with magic sum < 3 x 320000², but in a nonexhaustive way. He found 7 out of the 20 solutions previously found by Morgenstern < 3 x 5000², and also these new bigger solutions. Sums = 3 x 6115², 7395², 8905², 9345², 9565², 9995², 10195², 12725², 13175², 13765², 14825², 15225², 15525², 16185², 18085², 18115², 20085², 22425², 25075², 26571², 28135², 32625², 37635², 41905², 41925², 42415², 44353², 45025², 45435², 49045², 63495², 76075², 77435², 87135², 117725², 123335², 124845², 140675², 141865², 157675², 195925², 196775², 227035², 264265², 319685². But no new solution with sum three times the central entry.
Received January 21st, 2013
Lee Morgenstern
reused his method sent in April 2008 concerning the
list of 3x3 nearlymagic squares of squares having all odd entries and 7 correct
sums. This is equivalent to three 3square arithmetic progressions having equal
step values. The idea was to look for instances that satisfy the requirements
of a 3x3 fullymagic square of squares. The 2008 results were based on two searches:
His new 2013 searches extended the above:
Received July 2nd and August 8th, 2013
Tim S. Roberts, Bundaberg, east coast of Australia, author of http://unsolvedproblems.org, reminds us that a 3x3 magic square of squares must have this parametric form of any 3x3 magic square:
a² 
b² 
c² 
= 
x+y 
xyz 
x+z 
d² 
e² 
f² 
xy+z 
x 
x+yz 

g² 
h² 
i² 
xz 
x+y+z 
xy 
As an example, we will look at the modulo 13. Any squared integer can only have one of these 7 "authorized" values, the quadratic residues mod 13: 0, 1, 3, 4, 9, 10, 12. If we check by program all the possible values (x, y, z) mod 13, the nine cells (x+y), (xyz), ...(xy) can be nine squares (meaning that none of them is equal to another value than the 7 authorized values) only when (x, y, z) = (x, y, 0) or (x, 0, z): we conclude that y or z must be divisible by 13. Also equivalent is to say that y*z must be divisible by 13. Here are the final astonishing results that I have checked and slightly extended (5 > 5^2, and 7 > 7^2):
Yes, x*y*z must be divisible by every prime < 40, with the sole exception of 23.
Received from August 10th to September 3rd, 2013
Tim S. Roberts, again using the above parametric form of any 3x3 magic square, remarked that:
Astonishing! An incredibly long list, where 25 is the first integer > 1 not present. We can extend and finish his list, and say:
This should be the full list, a total of 106 integers. All these integers
can be factorized with these primes only (or powers of them): 2 (4, 8,
16),
3 (9), 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 61, 67.
That
is to say, every prime < 70, with the sole exception of 59. We can also
say that:
Received from December 18th 2013 to January 8th, 2014
Mark Underwood, Canada, sent to Tim S. Roberts an interesting remark improving the above results: y and z must both be divisible by 24, then so too must be y+z and yz. So the product of all four must be divisible by 24^4, which is 2^12 * 3^4. And at least one of the items must be divisible by 16, and at least one by 9. So that brings it up to 2^13 * 3^5.
And if I add my previous remark of JulyAugust 2013 that y*z must be divisible by 5^2 * 7^2, then:
Received February 7th and 22nd, 2014
Lee Morgenstern extended again his previous searches of January 2013 and April 2008. No solution :
Received June 30, 2014
Lee Morgenstern completed three new searches for a magic hourglass, but found no solution.
Received September 16, 2014
Not a search... but a call to search for a 3x3 magic square of squares in an article in Scientific American's website. Thanks to Ricki Rusting, Managing Editor, for mentionning my name, linking to multimagie.com http://www.scientificamerican.com/article/canyousolveapuzzleunsolvedsince1996 
Received from February 19 to March 13, 2015
Paul Zimmermann, researcher (INRIA, LORIA, http://www.loria.fr/~zimmerma/), with two young students, Paul Pierrat and François Thiriet, proved these modular properties: the magic sum of any primitive 3x3 magic square of squares must be 3 mod 72, and the squared entries must be 1 mod 24. And extending Buell's and Pech's previous searches, they found modulo 2^n solutions for 3x3 magic squares of 7 squares, up to modulo 2^59.
Their paper is available at http://www.loria.fr/~zimmerma/papers/squares.pdf
Received July 25, 2015
Terry Moriarty, Northern Ireland, found some properties of 3x3 magic squares of squares. For example, their magic sums are 3 mod 72. Look at http://magicsqr.byethost8.com/
Allowing imaginary numbers (see also above), Eddie Gutierrez, New Jersey, USA, found 3x3 magic squares having 7 squared entries. For example, this square with magic sum = 15552:
113² 
(121i)² 
132² 
= 
12769 
14641 
17424 
9839 
72² 
23² 
9839 
5184 
529 

(84i)² 
25009 
(49i)² 
7056 
25009 
2401 
Look at http://www.oddwheel.com/, with some other studies. The above magic square can be found from his table of contents 0B, 15th page (or directly http://www.oddwheel.com/Image_SquareA.html)
Received April 18, 2016
F U N !!! Matt Parker, an Australian living in UK, former maths teacher (https://en.wikipedia.org/wiki/Matt_Parker and http://standupmaths.com/), tried to solve the enigma #1 (3x3 magic square of squares) at (YouTube Numberphile videos by Brady Haran). In his video, 1:021:36, he also presented the solution of the enigma #4c (7x7 magic square of cubes) found in 2015 by Sébastien Miquel:
Invalid solution..., so no, I did not send any prize in euros or any bottle of champagne to Matt! :)
Received February 6 and March 1, 2017
Ben Asselstine, Canada, announced his "set of tools for finding and managing 3x3 magic squares of squares" freely available at http://fituvalu.nongnu.org/. This software uses libgmp (GNU Multiprecision Arithmetic Library) for computing on large integers, and is designed to be easily parallelizable with GNU Parallel.
He added this javascript program http://fituvalu.nongnu.org/checker.html: "It lets people check their 3x3 magic squares with 6 or more (?) perfect squares against a list of 100,000 known squares."
Received February 27, 2017
Lee Morgenstern sent PDFs of two articles written a few years ago, updated versions of the broken links of 2006 :
Received March 10, 2017
Lee Morgenstern produced this 3x3 square, using the new 6th triple of primitive Pythagorean triangles having equal areas found by Duncan Moore a few days earlier:
19720769947309² 
6757561171393² 
11290071470263² 
10987237357337² 
9483582546853² 
18745169816089² 
7239541562993² 
20650330341071² 
9120965347253² 
S = 562039114103450691451191099 for rows, columns, and one diagonal. Unfortunately the other diagonal has a different sum.
Received from March 11 to 13, 2017
Randall Rathbun, USA, found the elliptic curve of rank 4
giving the only previously known example with 7 squared entries. An excellent work, because it is very difficult to find this curve. But unfortunately, thousands of points on this curve (producing numbers with thousands of digits!) did not produce any other magic square with 7 squared entries...
Details in the emails sent by Randall to Andrew Bremner and me.
Christian Woll, California, USA, sent an interesting paper entitled "The Magic Hourglass of Squares related to the Gaussian Integers"
Received September 22, 2017
Joseph Hurban, when he was student tutor at TCNJ (College of New Jersey, USA), sent this nice formula producing 3x3 magic squares of 5 squares:
(10z² + 20z + 5)² 
b = 7²(2z² + 1)²  4z(2z² + 101z + 1) 
c = (2z² + 1)² + 8z(26z² + 38z + 13) 
(2z² + 2z + 1)² 
(10z² + 10z + 5)² 
(14z² + 14z + 7)² 
g = 7²(2z² + 1)² + 8z(24z²  13z + 12) 
h = (2z² + 1)² + 4z(102z² + 151z + 51) 
(10z²  5)² 
Is it possible to obtain supplemental squared integers in cells b, c, g, and h? With a MATLAB program, he tried to find solutions, and from z = 0 to 1,000,000, found five solutions:
Examples with z = 1, 2, 3:
35² 
5² 
25² 

85² 
2281 
3169 

155² 
13825 
95² 
5² 
25² 
35² 
13² 
65² 
91² 
25² 
125² 
175² 

25² 
35² 
5² 
5281 
6169 
35² 
22225 
17425 
85² 
I checked that there is no other solution with more than 5 squared integers for any z < 10^10.
Received November 15, 2017
Adrian Suter, Switzerland, working on Moriarty's results, sent a paper unfortunately producing a 3x3 magic square of nondistinct squares...
Received November 28, 2017
Benjamin Bartsch sent this parametric solution, similar to King's solutions, producing 3x3 semimagic squares:
(2  6x  7x^{2}  2x^{3}  x^{4})^{2} 
(2 + 5x^{2} + 4x^{3} + 2x^{4})^{2} 
(1  6x  5x^{2}  4x^{3}  2x^{4})^{2} 
(2  4x  5x^{2} + 2x^{4})^{2} 
(1  2x  7x^{2}  6x^{3}  2x^{4})^{2} 
(2  4x  5x^{2}  6x^{3}  x^{4})^{2} 
(1  2x  7x^{2}  8x^{3}  2x^{4})^{2} 
(2  8x  7x^{2}  2x^{3} + x^{4})^{2} 
(2 + 2x + x^{2} + 2x^{3} + 2x^{4})^{2} 
Seven magic lines with S = 9 + 36x + 90x^{2} + 144x^{3} + 171x^{4} + 144x^{5} + 90x^{6} + 36x^{7} + 9x^{8}, one diagonal being not magic with a different sum.
I remarked that with x = 8.6131187, we have a fully magic square with eight magic lines.... but of course using nonintegers :)
Two years after his magic squares of 7 squares allowing imaginary numbers, Eddie Gutierrez found a new 3x3 magic square of 7 squares, this time linked to Bremner's square (www.oddwheel.com/special squares.html), S = 780300:
806425 
(697i)² 
678² 
= 
806425 
485809 
459684 
86641 
510² 
779² 
86641 
260100 
606841 

246² 
1003² 
(535i)² 
60516 
1006009 
286225 
Thesis, May 2018
Giancarlo Labruna, Montclair State University (New Jersey, USA), submitted a thesis entitled "Magic Squares of Squares of Order Three over Finite Fields". Giancarlo is now lecturer, School of General Studies, Kean University (New Jersey).
Published from September to November 2018
Christian Woll, after his above paper of June 2017, published two new papers on arXiv:
Two examples of his Lunar magic squares of squares. On the left in base 10, S = 24². On the right in base 2, S = 1011111.
22² 
0² 
14² 

11² 
101² 
1001² 
1² 
24² 
2² 
110² 
1011² 
1² 

4² 
3² 
23² 
1010² 
0² 
111² 
On Lunar arithmetic, and primes on the Moon, Numberphile video featuring Neil Sloane (OEIS): https://youtu.be/cZkGeR9CWbk, Nov. 2018.
Lunar
addition, Lunar multiplication, and Neil Sloane
Received November 18, 2018
Vlad Volosatov, Russia, cleverly remarked that 3x3 magic squares of squares are possible using quaternions (i² = j² = k² = ijk = 1). For example, with S = 75 :
(5i)² 
(7k)² 
(j)² 
= 
25 
49 
1 
(k)² 
(5j)² 
(7i)² 
1 
25 
49 

(7j)² 
(i)² 
(5k)² 
49 
1 
25 
I add a remark: there is already a link between quaternions and 4x4 magic squares of squares!
Received December 30, 2018
Arkadiusz Wesolowski, Poland, found two parametric solutions of 3x3 magic squares of 5 squared integers. With n >= 1, let
Or equivalent, the value of x can also be obtained with this recurrence formula:
Then we obtain this magic square, the integers in black being squared integers:
17x^{4} + 44x^{3} + 34x^{2} + 10x + 1 
9x^{2} + 6x + 1 
10x^{4} + 28x^{3} + 23x^{2} + 8x + 1 
2x^{4} + 8x^{3} + 11x^{2} + 6x + 1 
9x^{4} + 24x^{3} + 22x^{2} + 8x + 1 
16x^{4} + 40x^{3} + 33x^{2} + 10x + 1 
8x^{4} + 20x^{3} + 21x^{2} + 8x + 1 
18x^{4} + 48x^{3} + 35x^{2} + 10x + 1 
x^{4} + 4x^{3} + 10x^{2} + 6x + 1 
First examples with n = 1 (> x = 4) and n = 2 (> x = 28):
7753 
13² 
4753 

11441977 
85² 
6779473 
35² 
65² 
85² 
1189² 
2465² 
3277² 

3697 
91² 
697 
5372977 
3485² 
710473 
The other parametric solution is a derivative of the first one above. Again with n >= 1:
17x^{4}  44x^{3} + 34x^{2}  10x + 1 
9x^{2}  6x + 1 
10x^{4}  28x^{3} + 23x^{2}  8x + 1 
2x^{4}  8x^{3} + 11x^{2}  6x + 1 
9x^{4}  24x^{3} + 22x^{2}  8x + 1 
16x^{4}  40x^{3} + 33x^{2}  10x + 1 
8x^{4}  20x^{3} + 21x^{2}  8x + 1 
18x^{4}  48x^{3} + 35x^{2}  10x + 1 
x^{4}  4x^{3} + 10x^{2}  6x + 1 
First examples with n = 1 (> x = 6) and n = 2 (> x = 30):
13693 
17² 
7693 
12612301 
89² 
7364461 

35² 
85² 
115² 
1189² 
2581² 
3451² 

6757 
119² 
757 
5958661 
3649² 
710821 
In the two parametric solutions, the numbers increase quickly: i.e. with n = 100, the magic sums are as big as ~2.38*10^307... However maybe, for some values of n, cells in red can become (huge) squared integers, giving 3x3 magic squares with more than 5 squared integers?
Onno M. Cain, USA, worked on a paper, later published in August 2019 on arXiv, entitled "Gaussian Integers, Rings, Finite Fields, and the Magic Square of Squares".
And, after Woll's paper on 3x3 magic squares of squares on the Moon, Onno constructed this 3x3 magic square of squared Lunar primes:
1001101² 
110101² 
1010011² 
111001² 
1010111² 
1011001² 
1001011² 
1011101² 
1001111² 
His papers (including this square) are available at https://sites.google.com/view/onnomc/papers, and his software for Lunar Arithmetic calculations in Python is also available at https://github.com/onnomc/lunararithmeticwrapper and https://repl.it/@onnomc/LunarArithmeticPlayground
Received June 24, 2019
Ben Asselstine published a PDF of 41 pages, looking as a PowerPoint file, on the 3x3 magic square of squares problem, including his analysis on magic squares of 6 squares.
Received September 16, 2019
Sunil Kumar, Tamil Nadu, India, 16 years old, found this nearmiss of a 3x3 magic square of 7 squares, S = 108329642031:
65118629677 
49565² 
201877² 
108377² 
190026² + 1 
245915² 
177385² 
264127² 
7101131677 
Received December 20 and 22, 2019
One year after his above parametric solutions of 3x3 magic squares of 5 squared integers, one more squared integer! Arkadiusz Wesolowski found this marvelous parametric solution of 3x3 magic squares of 6 squared integers:
(xy  z)² 
(xz + y)² 
x² + y²z² 
(x + yz)² 
(x² + y²)(z² + 1)/2 
(xz  y)² 
x²z² + y² 
(yz  x)² 
(xy + z)² 
The three first examples of 3x3 magic squares, constructed with n = 1 (> (x, y, z) = (11, 3, 4)), n = 2 (> 41, 11, 15), and n = 3 (> 153, 41, 56):
29² 
47² 
265 

436² 
626² 
28906 

6217² 
8609² 
5295025 
23² 
1105 
41² 
206² 
203626 
604² 
2449² 
39353665 
8527² 

1945 
1² 
37² 
378346 
124² 
466² 
73412305 
2143² 
6329² 
When n is an even positive number, we can prove that the cells in red can't be squared integers. His quick search of odd n <= 2*10^5 didn't produce 3x3 magic squares with seven, eight, or nine square numbers. But maybe a larger odd n?
This was not given by Arkadiusz, but x, y, z can be easily computed with these recurrence formulas:
And less than one month later, Arkadiusz Wesolowski found another marvelous parametric solution of 3x3 magic squares of 6 squared integers. The following identity expresses the product of two sums of two squares as a sum of two squares in three different ways.
And since
we can form a magic square with:
(wz + xy)² 
(wy  xz)² 
(2y²  z²)x² + (2z²  y²)w² 
2(x²y² + w²z²)  (wy + xz)² 
x²y² + w²z² 
(wy + xz)² 
x²z² + w²y² 
2(x²y² + w²z²)  (wy  xz)² 
(wz  xy)² 
The three first examples of 3x3 magic squares, constructed with n = 1, 2, 3:
127² 
46² 
20062 

878² 
503² 
905719 

3191² 
2162² 
11588158 
16702 
113² 
94² 
778039 
802² 
713² 
10220638 
2969² 
2722² 

74² 
23422 
97² 
617² 
1033399 
718² 
2458² 
12955678 
2729² 
I checked that the cells in red are not squared integers for any n < 10^10, meaning magic sum < 9.72*10^82. So difficult to obtain a 7th squared integer! But maybe possible for a larger n ?
Received May 18, 2020
Three years after his first parametric solution, Joseph Hurban sent this nice parametric solution of 3x3 magic squares of 4 squared integers:
(qr  ps)² 
[(qr)² + (ps)² + (pr)² + (qs)²]/2 + 4pqrs 
(pr  qs)² 
[3(pr)² + 3(qs)²  (qr)²  (ps)²]/2 
[(qr)² + (ps)² + (pr)² + (qs)²]/2 
[3(qr)² + 3(ps)²  (pr)²  (qs)²]/2 
(qr + ps)² 
[(qr)² + (ps)² + (pr)² + (qs)²]/2  4pqrs 
(pr + qs)² 
We can obtain a lot of solutions with two more squared integers, but again it is very difficult to obtain a 7th... Here are two examples of solutions with 6 squared integers, obtained with (p, q, r, s) = (1, 3, 2, 11) and (1, 9, 5, 8):
5² 
889 
31² 

37² 
5089 
67² 
1561 
25² 
311 
6769 
3649 
23² 

17² 
19² 
35² 
53² 
47² 
77² 
Before sending me your results, don't reinvent the wheel!
Read
the top of this page. For example, yes, the below results are well known
and proved,
do not send them again.
If a 3x3 magic square of 9 squared integers exist (primitive,
with cells relatively prime), then all cells are odd, and all cells are 3k+1
>
all cells are 6k+1.
And no cell multiple
of 3, 11, 19, 43,... because already proved are:
1) the center cell
is a (squared) product of 4k+1 prime numbers = 5, 13, 17, 29, 37, 41,...
2)
the 4 cells middleside are (squared) products of 8k+1 and/or
8k+7 prime numbers = 7, 17, 23, 31, 41, 47,...
3) the 4 corners are (squared)
products of 8k+1, +5, and/or +7 prime numbers = 5, 7, 13, 17, 23, 29,
31, 37, 41, 47,...
In a 3x3 magic square of 9 squared integers, if we accept that
some squared integers can be repeated, then any solution uses only
1 or 3
squared integers.
No other way, impossible with 2, 4, 5, 6, 7, or 8
distinct squared integers. But unknown for 9, that's the problem...
On the
left, the only possible family of solutions with repeated squared integers.
3x3 magic square of squares, S = 3c², 

Obvious 3x3 magic square of squares,
S
= 3, 

3x3 magic square of squares, S = 75, 

a² 
b² 
c² 
1² 
1² 
1² 
1² 
7² 
5² 

b² 
c² 
a² 
1² 
1² 
1² 
7² 
5² 
1² 

c² 
a² 
b² 
1² 
1² 
1² 
5² 
1² 
7² 
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