Thread: January 2019
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Old 2019-03-05, 15:03   #69
Dr Sardonicus
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Feb 2017

2·2,081 Posts

At one point, I had hopes of producing formulaic solutions. And, for the case of two sets of two numbers each, it's actually pretty easy. With the 2x2 matrix of squares [a^2, b^2; c^2, c^2] you have the relation

a^2 - b^2 = c^2 - d^2, or a^2 + d^2 = c^2 + b^2.

OK, just use the usual formula for a product of two sums of two squares.

Unfortunately, I was unable to expand on this idea. It is possible that quaternions etc might be used to satisfy some of the conditions that crop up in certain cases, but I didn't pursue this.

I looked at single-variable polynomials, and did not see any promising lines of attack.

The sort of simultaneous conditions that crop up are curious. The simultaneous equality of several differences of two squares points to numbers with lots of factors. For this reason, using primorials would seem to increase the odds of success...
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