Like the famed "Parker Square", but with distinct digits. Examples exist for 4x4, 5x5, 6x6, maybe higher. 3x3 is conjectured impossible. It must be a 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 the case.
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.
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If you choose a sufficiently large row sum and choose arbitrary values for the bottom (n-4) x n submatrix then (ignoring diagonals) the remaining 4 x n values can be made to work out vertically using Lagrange's four squares theorem. It remains to see if you can permute each of these n quadruplets so that the sums work out horizontally. This is a bit similar to IMO 2020 Problem 3 and I wonder if there's a connection there.