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New idea about Integer Factorization |
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| Dec26-12, 11:25 AM | #1 |
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New idea about Integer Factorization
The logic that odd composite with least difference will be factored easily and large difference would factored hardly is wrong. B'coz whatever be the difference between the factors their exist Best Fermat Factors to make the Fermat factorization easier. Please follow the link to know more. http://kadinumberprops.blogspot.in
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| Dec31-12, 09:28 AM | #2 |
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Kadhir, what people want to see is simple: take a big enough number ( 50 digits say to start with) and use your method to factorize it in less time than with previous methods. If you do that, people will listen to you.
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| Jan1-13, 08:20 AM | #3 |
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At your website you seem to suspect yourself that what you are doing is wrong- you say "Please don't try any logical proofs on my ideas". If you are sure you are right, why would you object to logical proofs? You seem to be under the impression that if your method works for some examples, then it must always work and that is certainly NOT true!
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| Jan1-13, 08:53 AM | #4 |
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New idea about Integer Factorization |
| Jan1-13, 09:30 AM | #5 |
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Mentor
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What do you mean with "if n is something, then p should be"? What happens if p is not?
It is pointless to generate some numbers based on n - if you know n for your number, the factorization is done anyway. Here is some large digit for you to analyze/factorize: 291025469390636121509355493847053288310414921649815747 It has two factors, with 24 and 31 digits, and conventional methods (but not Fermat) can factorize it on my home computer in 4 seconds. |
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