My Math Forum Semi-prime factorization conjecture

 Number Theory Number Theory Math Forum

 January 30th, 2018, 06:03 AM #1 Member   Joined: Jan 2015 From: italy Posts: 39 Thanks: 1 Semi-prime factorization conjecture Let N be a semiprimo then N=a^2-b^2 will have two solutions a1=(N+1)/2 and b1=(N-1)/2 a2=? b2=? bruteforce on b2 starting from 0 A+B=a1 , A-B=a2 , C+D=b1 ,C-D=b2 gcd(A,C)=p_a gcd(B,D)=p_b p=(p_a^2-p_b^2) gcd(A,D)=q_a gcd(B,C)=q_b q=(q_b^2-q_a^2) Example 15=a^2-b^2 a=4 b=1 a=8 b=7 A+B=8 , A-B=4 , C+D=7 , C-D=1 A=6 ,B=2 ,C=4 ,D=3 gcd(A,C)=2 gcd(B,D)=1 p=(2^2-1^2)=3 gcd(A,D)=3 gcd(B,C)=2 q=(3^2-2^2)=5 What do you think about it?
 January 30th, 2018, 08:02 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 413 Thanks: 227 Math Focus: Dynamical systems, analytic function theory, numerics This is known as the quadratic sieve factoring algorithm. Its slightly souped up big brother (the number field sieve) is the currently the fastest known factoring algorithm for integers which aren't the product of "a lot" of smaller factors. In this case, the ellliptic curve factoring algorithm is extremely fast which leads to the common strategy. 1. Given a number to be factored, use EC factoring to split off small divisors until the remaining product is the product of only a few large factors. It is easy to check primality so you can tell when this happens because EC stops giving you results but primality tests still fail. 2. Switch to number field sieve (or quadratic sieve) and try to split off larger factors.
 January 31st, 2018, 08:21 AM #3 Member   Joined: Jan 2015 From: italy Posts: 39 Thanks: 1 thank you

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post mobel Number Theory 32 August 21st, 2016 11:32 AM cm318 Number Theory 20 April 4th, 2015 04:17 PM mobel Number Theory 9 July 1st, 2014 09:58 AM Bogauss Number Theory 58 November 28th, 2011 11:19 AM momo Number Theory 19 June 23rd, 2008 09:18 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top