20121211, 06:15  #1 
Dec 2012
3_{8} Posts 
Best case Fermat Factors
The logic that it is hard to do Fermat factorization for odd composite which is a product of two large prime factors with big difference is wrong. Reason when those primes are best Fermat factors then Fermat factorization will be very easy. For more details please follow the link http://kadinumberprops.blogspot.in

20121211, 07:51  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×11×19×23 Posts 
Then it would be no problem for you to factor
Code:
4872694181406339617512781250710256288128420426749870494701352170485888238522036701839697408990043865362740060996930806408048841117542674271031589079075642908938171217283398153697602454775549091739003927335892645964656077739143953748851155087308230066486278985637829660170144978240247037951 ? 
20121211, 09:45  #3 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Straight to Misc. Math.

20121211, 10:06  #4 
Dec 2012
3 Posts 
Please understand i didn't mean it. Please understand i gave the fresh analysis of Fermat factorization and i coined the term Best Fermat factor. Please reply whether you understood the article thoroughly. I said Fermat factorization complexity is easy when the number's factors are Best Fermat factors or Nearer to Best Fermat factors.

20121211, 18:39  #5  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
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. Last fiddled with by science_man_88 on 20121211 at 18:39 

20121212, 04:18  #6 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}×613 Posts 
That is too heavy!
Next number is a C170, whose two prime factors ratio is very close to "best fermat factor" ratio, the distance to such a split is lower than 1/10^40 (i.e. 0.000(about35zeroes)0001). There should be no problem to factor it, and if this is possible, it should be nice, because there is no other method (beside long taking gnfs) to factor it: both factors are P1 and ECM "tough"  I did not make them so on pupose, they just come out like that from the first trial of nextprime(random(perfectsplit)), hehe. So, it should resist p1 and ECM trials. You only have to try the other primes in the ~10^30 interval that remained. Code:
54553746256351348704340853049775196132111315873148539674440243786571807634659955885688337948063058916001903005889468533003195598599269527286081791979809069615167325135781 Last fiddled with by LaurV on 20121212 at 04:42 Reason: forgot the number, grrr... 
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