20170202, 06:22  #1 
May 2016
7×23 Posts 
Can I decrease the factorization time with this formula ?
Formula ( 3 Steps ) :
Note : Step 1 : Step 2 : Step 3 : First Way It work always (for prime factors similar value or distant) Second Way It work only for prime factors distant value Example First Way : Step 1 : Step 2 : Step 3 : First Way It work always (for prime factors similar value or distant) Second Way It work only for prime factors distant value EDIT Can I decrease the factorization time with this formula for big numbers? . Last fiddled with by Godzilla on 20170202 at 06:55 
20170202, 10:09  #2 
Feb 2016
! North America
5×13 Posts 
What is the purpose of the "Second way" if the first one will "It work always (for prime factors similar value or distant)"?
And by looking at it and simplifying it, sqrt(product) < N3 < product. Always. [ N3 = ( sqrt(3)+2 )/ sqrt(3) * sqrt(product) ] which is ~2,1547 *sqrt(product) N4 = product / (~2,15*sqrt(product) )* 3,5 < which is like (sqrt(product)/~2,15 ) * 3,5 = sqrt prod * ~1,6 < rough rounding ^ It might be wrong, and a mess, I'm little tired. Also it's not scientific in any way. Ok, so product = 1852927 sqrt is 1361,222612213 N3 by using sqrt(3)s is 2933,027095 N4 by using ...*3,5 is 2211,109645 So what are the factors? (yafu: 0.0090 sec) Last fiddled with by thyw on 20170202 at 10:11 
20170202, 12:31  #3  
May 2016
10100001_{2} Posts 
Quote:
First Way Second Way Now trying First Way and Second Way , but only First Way is correct , than ..... ...continue until I think that for large numbers ( 300 digits ), could to be efficient ...... . 

20170202, 12:43  #4 
May 2016
7×23 Posts 

20170202, 17:42  #5  
Aug 2006
13357_{8} Posts 
Quote:
Let the product be N and its square root be S. Then N1 = sqrt(3330)/50 * S = 1.154...S N2 = S N3 = (1 + sqrt(3330)/50)*S = 2.154...S N4 = 35(3*sqrt(370)  50)*S/166 = 1.62479...S So basically you have four numbers, all of which are fixed algebraic multiples of the square root of the number you're trying to factor. How does this help? 

20170202, 19:21  #6  
May 2016
7·23 Posts 
Quote:
and First Way and Second Way this is the minimum distance of the two prime factors for First Way ( ) and First Way and Second Way and First Way and Second Way and First Way and Second Way this is the maximum distance of the two prime factors for First Way and First Way and Second Way Now Second Way is correct and First Way and Second Way Now Second way is correct You say it does not work in proportion also for large numbers? P.S. Should be tried with a program . . Last fiddled with by Godzilla on 20170202 at 19:22 

20170202, 19:35  #7 
Aug 2006
5871_{10} Posts 
Let's say I give you a number:
N = 67646988554084448110460372160440713890061829425106624449473216837898618103759093968747 Then I get (rounding down): N1 = 9492406899940630172687271283182766199836725 N2 = 8224778936487256622736720394209992127637443 N3 = 17717185836427886795423991677392758327474168 N4 = 13363547807490383809719662986158762809329777 Knowing this, how do we factor the number N? Or what can be said about the factors of this number? Note that N is not particularly large, and most of the people who post here could factor it without difficulty. 
20170202, 19:55  #8  
May 2016
7×23 Posts 
Quote:
But , I begin by "(N4)" N42 , N44 , N46...(First Way and Second Way) and not by"(N)" N2 , N4 , N6.....! And I bet it will be faster. . 

20170202, 20:18  #9 
Aug 2006
3·19·103 Posts 
A 64bit integer division takes 73 clocks, and a 286bit division won't be faster than that. Assuming your clock speed is less than 100 GHz and you have no more than a million CPU cores it would take you more than 10^20 years to factor the number in this fashion. But it should only take a few minutes for SIQS.
Last fiddled with by CRGreathouse on 20170202 at 20:18 
20170202, 20:30  #10  
May 2016
241_{8} Posts 
Quote:
I was thinking in pencil, for example. I do not know how to work the clock. 

20170202, 21:50  #11 
Aug 2006
3·19·103 Posts 
OK, let me rephrase. Your method, starting from N4 and working down through the odds, would take more than 10^40 divisions, and at one division per second would take 10^25 times the age of the universe to factor the number.

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