Thread: January 2019 View Single Post 2019-03-05, 18:51 #71 Dr Sardonicus   Feb 2017 Nowhere 5×761 Posts Curiously, I completely failed to notice a similar identity to the sum-of-two-squares identity, and one more directly applicable to the problem, for the product of two differences of two squares, namely (a^2 - b^2)*(c^2 - d^2) = (a*c +/- b*d)^2 - (a*d +/- b*c)^2. As with the sum-of-two-squares identity, the product of k factors gives 2^(k-1) formally different expressions of the product as a difference of two squares. Alas, the expressions involve terms of total degree k. There is also a "head-to-tail" property of the various conditions that arise in the problem, which I didn't see how to use efficiently. Taking the example from the solutions A=[9, 28224, 419904, 3968064]; B=[0, 47952, 259072, 2442960] and letting Mij = Ai + Bj, we see that taking the difference of row j and row i of M gives the constant difference Aj - Ai. (Similarly with differences of two columns) Taking row 2 - row 1, row 3 - row 2, and row 4 - row 3 we find the conditions 168^2 - 3^2 = 276^2 - 219^2 = 536^2 - 509^2 = 1572^2 - 1536^2, 648^2 - 168^2 = 684^2 - 276^2 = 824^2 - 536^2 = 1692^2 - 1572^2, and 1992^2 - 648^2 = 2004^2 - 684^2 = 2056^2 - 824^2 = 2532^2 - 1692^2 where the "head" of each expression in one set of conditions becomes the "tail" of an expression in the next.  