The existence or not of a non-trivial integer 3x3 magic square of squares is STILL an unsolved problem. As far as I know, the issue was researched for a set of all integers below 10^14.
Let's go for a lesser challenge: the 3 columns, 3 rows and only one of the diagonals has to sum up to the same value.
It is still a D4 problem, if I do not release any hints.
Go for it!
Elements of magic squares of 3x3 have to be in arithmetic progression when they are ordered like below
a12,a33,a21
a13,a22,a31
a23,a11,a32
In our case this elements are squares, so it is necessary to have series of three squares in arithmetic progression
Then a33-a12=a21-a33 => a21+a12=2a33
We can try to find squares so that their double are the sum of two squares.
Anyway, this have been studied and there are formulas to generate series of squares in arithmetic progression, in a similar way as there are formulas to generate pythagorean triples.
Using a version of these formulas I get (for m=2 n=3)
617 526 718
146 802 713
878 503 386
The squares of these numbers sum in each row and column 1172889. And also the main diagonal sum that, the other diagonal has instead a quite different figure.
Edited on March 16, 2018, 7:05 pm
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Posted by armando
on 2018-03-16 18:59:52 |
one solution
