Quote:
Originally Posted by Batalov
Then it would be no problem for you to factor
Code:
4872694181406339617512781250710256288128420426749870494701352170485888238522036701839697408990043865362740060996930806408048841117542674271031589079075642908938171217283398153697602454775549091739003927335892645964656077739143953748851155087308230066486278985637829660170144978240247037951 ?
?

here's what I got from reading it:
N=a^2b^2 ~ s^2
d=2n
so say y is the lower factor in a 2 best factor setup y+2n is the other to go along with it y*(y+2n) = y^2+2ny~s^2 this leads to s^2y^2 ~ 2ny since both y^2 and 2ny have a common factor y, so do ~s^2 and y^2 namely y so in closing the lowest of the 2 has a multiple close to the sqrt of the N to be factored. of course how close depends on the amount of rounding I think.
edit: ~ is for approximately.