20201106, 01:13  #45 
Jan 2010
2×43 Posts 
any one can try on their own. since then my task has been to reduce number of samples for given numbers.

20201106, 02:25  #46 
"Curtis"
Feb 2005
Riverside, CA
3×1,559 Posts 
There's nothing to try. You can't get factors yourself. We certainly aren't going to be able to either.
You haven't demonstrated any factoring. Just a bunch of BS. Factor something. 
20201106, 03:51  #47 
Jan 2010
2×43 Posts 
bs? :) curious "synonym" for approximation. actually, i have no time to crack any number for NOTHING + it has no sense. so far here i have met laughable stock w/ deep phobias to open 7zip archive. So it was extremely naive for me to wait adequate dispute

20201106, 04:32  #48 
Mar 2019
2^{4}·3^{2} Posts 
Except you haven't actually demonstrated any sort of method or algorithm whatsoever.

20201106, 05:12  #49 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·3^{2}·13^{2} Posts 

20201106, 20:42  #50  
Jan 2010
126_{8} Posts 
Quote:
So, you want superb algo(s) & want to do NOTHING for :) so, any not magical algo is only bs for you 

20201106, 20:56  #51  
Mar 2019
2^{4}·3^{2} Posts 
Quote:
(1) efficient algorithms to factor large integers, or (2) efficient algorithms for finding large primes (and proving them prime). You've demonstrated no meaningful algorithm for doing either. 

20201106, 21:10  #52 
Jan 2010
56_{16} Posts 
really? how could you have efficient algo, if you have done no research??? perhaps new algos appear out of the blue, right? :)

20201106, 21:10  #53  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
249F_{16} Posts 
Quote:
You need to solve the first one before your next post (and you must post the solution in your next post), or you may be banned (because you fail to show that your method works and just want to be a troll). Then you must solve the next one (and post the answer) to continue posting. After you have posted the answer to the third, then we will believe that you are on to something. First test. Factor this semiprime (using your method): 3870092038884345663779821427477643475136534002402905753076769909311217 Second test Factor this semiprime (using your method) 150996579069406676849328254452885095204370060219736837010494868333342863322970462956706063 Third test. Factor this semiprime (using your method) Code:
4735324369078304459849659757648833978535825054585241886997488631830939162381565490995222405517413444400835669 Last fiddled with by Uncwilly on 20201107 at 00:30 

20201106, 21:23  #54  
Jan 2010
2×43 Posts 
Quote:


20201106, 21:25  #55  
May 2020
2×3×5 Posts 
Quote:
If you wanted any cooperation, it would have been more helpful to ask questions about the topic  "Has there been any work done on factorizing RSA numbers with approximations of the sums of the primes in the product?"  rather than to announce "big results" and then be gleefully sour when people are doubtful. It would have saved us a lot of time, and it's likely there's already been a lot of research into the topic  with a good reason as to why it's failed so far. Last fiddled with by Gelly on 20201106 at 21:26 Reason: formatting is whack 

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