20160720, 09:21  #45  
May 2016
7·23 Posts 
Quote:
I found this, now it works with any two numbers, but the product must be three digits. n=101⋅3=303 n1=303++303=303303 303303/11=n2 sqrt(n2⋅(n−99))/22=n3=( maximum limit) the 9 in proportion of the product ex. 99 (two digits) because the product is 303 (three digits) sqrt(n2⋅(n−99−99))/22=n4=(minimum limit) just that the margin of error is always high. just that. Last fiddled with by Godzilla on 20160720 at 09:23 

20160813, 19:06  #46 
May 2016
10100001_{2} Posts 
I have this now :
P1 greater than P2 ( 2178 * ( P1 * P2 ) ) / 22 = N1 99 / (P1 * P2) = N2 N1 / N2 = N3 N3 / (prime number random but only P1 is correct because is the bigger and only it) = N4 N4 / N1 = P2 Example : 10007 * 3 = 30021 ( 2178 * ( 10007 * 3 ) ) / 22 = 2972079 99 / 30021 = 0,00329769161 2972079 / 0,003297 ( I have take only six digits ) = 901449499,545 901449499 / 10007 (considered trying, Random P1 ) = 90081,89 90081 / 30021 = 3,000599 and this is P2 = 3 Now it is always difficult but possible. 
20160813, 19:44  #47 
May 2016
7×23 Posts 
I can't cancel , there is an error , sorry , the operations are too much
Last fiddled with by Godzilla on 20160813 at 20:00 Reason: error 
20160813, 19:59  #48  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
99*(P1*P2)=N1 99/(P1*P2)=N2 N1/N2 = N3 = (99*(P1*P2))*((P1*P2)/99) = (P1 ^2)*(P2 ^2) N4 = P1*(P2 ^2) P2 = P2/99 ? what you have in the example is: y = P1*P2; 99*y = N1; 99/y = N2; N1/trunc(N2,x)= N3 floor(N3)/P1 =N4 floor(floor(N4)/y) = P2 Last fiddled with by science_man_88 on 20160813 at 20:11 

20160813, 20:32  #49 
May 2016
7·23 Posts 
But how many steps do you to find the product of 3 * 5 = 15? Starting at the highest number?
13 * 11 13 * 7 13 * 5 13 * 3 11 * 7 11 * 5 11 * 3 7 * 5 7 * 3 5 * 7 5 * 3 and stop Instead I : (N3 / 13) / 15 (N3 / 11) / 15 (N3 / 7) / 15 (N3 / 5) / 15 and stop but I have to check with a multiplication but not all operations Last fiddled with by Godzilla on 20160813 at 20:48 
20160813, 20:50  #50 
May 2016
7·23 Posts 
I have wrong to write before, this is correct N4 / (P1*P2) = P2
not this N4 / N1 = P2 P1 greater than P2 ( 2178 * ( P1 * P2 ) ) / 22 = N1 99 / (P1 * P2) = N2 N1 / N2 = N3 N3 / (prime number random but only P1 is correct because is the bigger and only it) = N4 N4 / (P1 *P2) = P2 Example : 10007 * 3 = 30021 ( 2178 * ( 10007 * 3 ) ) / 22 = 2972079 99 / 30021 = 0,00329769161 2972079 / 0,003297 ( I have take only six digits ) = 901449499,545 901449499 / 10007 (considered trying, Random P1 ) = 90081,89 90081 / 30021 = 3,000599 and this is P2 = 3 Last fiddled with by Godzilla on 20160813 at 21:02 
20160813, 21:02  #51  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:


20160813, 21:08  #52  
May 2016
7·23 Posts 
Quote:
I know this, but do you think the time is lower or higher? P.S. But only one factor and the higher factor and test it with multiplication. Last fiddled with by Godzilla on 20160813 at 21:16 

20160813, 21:51  #53 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
the problem with that PS is if we know the higher factor of a semiprime we also know the lower factor and therefore can do an addition instead of this convoluted procedure.

20160827, 18:41  #54 
May 2016
161_{10} Posts 
I conclude and resume with the first algorithm of this thread saying that:
Limit (max) and Limit (min) The algorithm is a conclusion rather trivial because it is found that the sum of two prime numbers similar is "around" the double of their average : with for and . is equal to . P.S. I replaced 22 with 22,5 Last fiddled with by Godzilla on 20160827 at 18:44 
20160827, 19:06  #55  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
(2*k+1)^2+2*(2*k+1)*(2*j+1)+(2*j+1)^2 = (4*k^2+4*k+1)+(8*j*k+4*j+4*k+2) + (4*j^2+4*j+1) = 4*(k^2+k+2*j*k+j+k+j^2+j+1) which when divided by four gives (k^2+k+2*j*k+j+k+j^2+j) so your second statement is equivalent to 4*j*k+2*j+2*k+1 ~ = (k^2+k+2*j*k+j+k+j^2+j) lets reduce this a bit we can get 2*j*k+1 ~=(k^2+j^2) which uses less variables overall. let's see what you can get out of that first I want to work on other things right now ( like being lazy). Last fiddled with by science_man_88 on 20160827 at 19:21 

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