Euler, who as I am sure you are aware was a great 18th century mathematician, came up with a 4x4 magic square that just had square numbers. Yet no one has been able to do the same with a 3x3 magic square and no one has proved that it is not possible.
I am sure that I am missing something but when can you not just say the following to prove it true. In a normal 3x3 magic square there will always be a diagonal, row or column with a constant difference between all the numbers in the row. Therefore as this cannot be the case if the numbers are square numbers then a 3x3 magic square cant exist as this can never be the case with square numbers.
I know that this is too simple for such a mathematical mystery but I can’t find the flaw in the argument. If you could tell me where the flaw is I would be grateful.
No, it’s a reply to matt12’s claim that it’s not possible to have a row in which there are three numbers, all squares, in which the first one is more than the second one by the same amount that the third one is more than the second one. In naita’s and septimus’s posts are four examples to show that this claim is wrong:
205^2 425^2 565^2, where the difference is 136,800
Wasn’t supposed to be. matt12 is apparently right that there is no known 3x3 magic square of squares, but as shown, his approach at proving there can’t be is flawed.
where s is the square of the sum of two squares in four different ways, i.e.
s = (xx+yy)^2 = (zz+ww)^2 = (aa+bb)^2 = (cc+dd)^2
and t, u depend on the decomposition of s
t = 2xy (x+y) (x-y)
u = 2zw (z+w) (z-w)
t+u = 4ab(a+b)(a-b)
t-u = 4cd(c+d)(c-d)
(That the cells are squares if and only if the above holds follows from a theorem due to Leonardo of Pisa (c. 1170 – c. 1250) !)
So all we need, to form the desired magic square, is to find an s expressible as the square of the sum of two squares in four different waysand with the constraints on t+u and t-u holding. We can achieve three different ways, e.g.
Sorry my internet was down for a while. The approach works if you can prove that it must always be the case for a 3x3 magic square which I can’t do. I decided to attack the proof differently using permutations and some intuition which will work I think. I would post some of the proof but I have completed 1/3 of the proof outline and it is 15 pages. So I am sure that you will understand when i say that it will be far too long and that is assuming that it will work and there is no guarantee of that either. Thanks for your help though really quite glad that a proof doesn’t already exist or I would’ve wasted the last week of my life.
Recreating a proof whose outline will take about 6 weeks and when done properly is easily 100 pages long. May even be 200-300 pages long. I can’t do it full time I’m at university I do it in gaps so it will take 12-18 months. If no one has written one then it’s exciting and a bit of fun. If someone else has done it then I feel that it is a bit of a waste.
I don’t know much about this stuff, but here’s a decent reference on the current state of the 3x3 magic square of squares that may be worthwhile for your research, if you haven’t come across it already. There are updates going up through 2015, so it looks like it is being maintained and updated. I don’t see any proof that a 3x3 square of squares does not exist on that page yet. I do see this, though, which apparently has a proof: “if there is a 3x3 magic square of distinct squares, then all entries must be above 10^14.”