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January 30th, 2018, 07:03 AM  #1 
Member Joined: Jan 2015 From: italy Posts: 39 Thanks: 1  Semiprime factorization conjecture
Let N be a semiprimo then N=a^2b^2 will have two solutions a1=(N+1)/2 and b1=(N1)/2 a2=? b2=? bruteforce on b2 starting from 0 A+B=a1 , AB=a2 , C+D=b1 ,CD=b2 gcd(A,C)=p_a gcd(B,D)=p_b p=(p_a^2p_b^2) gcd(A,D)=q_a gcd(B,C)=q_b q=(q_b^2q_a^2) Example 15=a^2b^2 a=4 b=1 a=8 b=7 A+B=8 , AB=4 , C+D=7 , CD=1 A=6 ,B=2 ,C=4 ,D=3 gcd(A,C)=2 gcd(B,D)=1 p=(2^21^2)=3 gcd(A,D)=3 gcd(B,C)=2 q=(3^22^2)=5 What do you think about it? 
January 30th, 2018, 09:02 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 276 Thanks: 141 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, 09:21 AM  #3 
Member Joined: Jan 2015 From: italy Posts: 39 Thanks: 1 
thank you


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conjecture, factorization, semiprime 
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