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)
where
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.
T = (x + y - (2x+1)/2)*(x + y + (2xy+1)/2)
where
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.
# posted by James Harris @ 19:40 
