i.e. what the famed "Parker Square" was hoping to be: a 3x3 square of distinct perfect square integers such that each row, column, and diagonal sums to the same number. Two such squares are considered "different" for this market if one cannot be obtained from the other by rotating, reflecting, and/or scaling all numbers by a constant.
On Numberphile (https://www.youtube.com/watch?v=U9dtpycbFSY), it is conjectured that such a square is impossible. However, a potential step in the proof is proving that there's only a finite number, which is a weaker result. Will this weaker result be proven?
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.
1,000@Conflux I believe “integer” in your definition should instead say “positive integer”, which would allow each positive integer n to be used a second time in the square by also including -n. I don’t believe the recent proof addresses this, since they use the standard definition. However, since you also reference Parker Squares, I believe the definition of a Parker Square should supersede the definition you provided
https://www.youtube.com/watch?v=stpiBy6gWOA
Resolves YES I believe. You should confirm I didn't misunderstand what's been proven though.
@IsaacKing hmm I don't see why finitely many elliptic curves where solutions may lie upon implies finitely many squares up the the transformations stated in this question.
@TotalVerb Ok yeah I think I'm misunderstanding. Why are there only a finite number of rational points? There must be some additional restriction on the curve that was unstated.
And what does it mean to say there are X curves on a surface? There are an uncountably infinite number, you can take any bisecting plane and that forms a curve.