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Jun 8, 2018 at 8:59 comment added Fedor Petrov The second matrix is simply antisymmetric of odd order, but the first is not even if we permute the rows and columns (there are not enough many zero entries)
Jun 8, 2018 at 4:58 comment added Zhi-Wei Sun By (1.16) of my paper arxiv.org/abs/1308.2900, for any prime $p\equiv 3\pmod 4$ we have $\det[(\frac{i^2-j^2}p)]_{1\le i,j\le(p-1)/2}=0$.
Jun 8, 2018 at 4:57 comment added Zhi-Wei Sun @Amdeberhan Have you really proved the equivalence?
Jun 8, 2018 at 4:35 history answered T. Amdeberhan CC BY-SA 4.0