 mersenneforum.org Formubla-bla-bla to calculate the sum of two Prime numbers just by knowing the product
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Godzilla

May 2016

7·23 Posts Quote:
 Originally Posted by retina Not really. It is purely a guesswork and trial-and-error approach. Any numbers it produces have no mathematical rigour whatsoever. And for numbers of unknown character there is no way to know if there are only two divisors, or more -- so you can't apply the results until you already know all the factors.
retina is right but

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 2016-07-20 at 09:23   2016-08-13, 19:06 #46 Godzilla   May 2016 101000012 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.   2016-08-13, 19:44 #47 Godzilla   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 2016-08-13 at 20:00 Reason: error   2016-08-13, 19:59   #48
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts Quote:
 Originally Posted by Godzilla 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.
so you have ( effectively, before floors, ceilings, and I won't follow my own math operations):
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 2016-08-13 at 20:11   2016-08-13, 20:32 #49 Godzilla   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 2016-08-13 at 20:48   2016-08-13, 20:50 #50 Godzilla   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 2016-08-13 at 21:02   2016-08-13, 21:02   #51
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts Quote:
 Originally Posted by Godzilla I have wrong before to write , 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
You do realize, if you do a bit of algebra within your steps, you can avoid floor functions right ? I just showed you why. also your whole procedure is dependant on knowing one of the prime factors.   2016-08-13, 21:08   #52
Godzilla

May 2016

7·23 Posts Quote:
 Originally Posted by science_man_88 You do realize, if you do a bit of algebra within your steps, you can avoid floor functions right ? I just showed you why. also your whole procedure is dependant on knowing one of the prime factors.

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 2016-08-13 at 21:16   2016-08-13, 21:51   #53
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts Quote:
 Originally Posted by Godzilla 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.
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.   2016-08-27, 18:41 #54 Godzilla   May 2016 16110 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 2016-08-27 at 18:44   2016-08-27, 19:06   #55
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts Quote:
 Originally Posted by Godzilla;440870 [B 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 . [/B]
p+q = (2*(p+q)/2) is true it's not about and then by that math you get n^2 = ((p+q)^2)/4 = (p^2+2*p*q+q^2)/4 assuming p=2*k+1 and q=2*j+1 we then can first expand the numerator to:

(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).

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