i.e. what the famed "Parker Square" was hoping to be. It must be a 3x3 square of distinct perfect square integers such that each row, column, and diagonal sums to the same number.
Inspired by https://www.youtube.com/watch?v=U9dtpycbFSY, where the latest Numberphile guest conjectures this is impossible.
If an example is found, resolves NO immediately.
See also
General policy for my markets: In the rare event of a conflict between my resolution criteria and the agreed-upon common-sense spirit of the market, I may resolve it according to the market's spirit or N/A, probably after discussion.
Related questions
@Conflux I think the probability of proving this is very small (unless we have AGI by then). Basically, these kind of problems are usually very hard to prove, I don't think that many people work on it and also the base rate of a concrete theorem being proven in 2.5 years is just very small. Although there is a new result in this direction, as far as I know it is only up to constantly many exceptions. I do not know their technique but, again, a priori it's not clear how to push techniques that prove stuff "up to finitely many exceptions" to prove that in fact there are none.
This paper shows that the set of real 3x3 magic squares are parametrized by x,y,z and are of the form (this is what the "equations" of the video reduce to)
[[x+y, x+z, x-y-z],[x-2y-z, x, x+2y+z],[x+y+z, x-z, x-y]]
Since x occurs in the middle, x must be an integer, and from the upper left and upper center, y and z must likewise be integers, so all magic squares of integers are exactly squares of this form for x,y,z integers.
Letting w = y+z, we can rewrite the entries as
[[x+y, x+w-y, x-w],[x-w-y, x, x+w+y],[x+w, x-w+y, x-y]]
So finding a magic square of squares amounts to finding x,w,y such that x + n w + m y is a square for n,m in {-1,0,1}. This makes several sets of arithmetic progressions of length 3 among the values.
This is all presumably still subject to the math he was talking about, but I think it clarifies the problem somewhat.