Thread: Best case Fermat Factors View Single Post
2012-12-11, 18:39   #5
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

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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^2-b^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^2-y^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.

Last fiddled with by science_man_88 on 2012-12-11 at 18:39