Saturday, June 10, 2006


SF: More equations to check, no claims

Well, as usual, after giving up on surrogate factoring, I thought I'd try just one more idea:

T = (x + y - (2x+1)/2)*(x + y + (2xy+1)/2)


16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T + 5)^2 - (4T - 3)^2

and I used the identity

x^2 + 2xy + y^2 = z^2 + (x+y-z)(x+y+z)

with the condition that

2xy = 2z - 1

so that I could get something weird, hoping it might work.

Oddly enough, in general, it'll give non-rational x and y, so if it did work, it'd be a solution using radicals that behave just enough where it counts.

Just thought up this latest. Really just doodling. Don't want to let SF die.

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