20200527, 09:16  #1 
May 2017
ITALY
2^{2}·97 Posts 
Phantom factorization  subcubic factorization of integers?
There are 48 systems in the paper "phantom factorization" .
Will the combination of three of them lead to factorization? Here is an example of a combination of three times the first system (but it is better to use three different systems) N=187 , x=w=g=3 , u=1 , v=7 ,h=9 N^2*x*(x+2+4*u)=[8*((96 k² + 24 k + 1)3*a*(a1)/2)+1]/3 , sqrt[(8*(96 k² + 24 k + 1)+1)/3+1](2*a1)=[2*(3*b+1(ba+1))+1(4*a2)]=x*[2*(3*z+1(zy+1))+1]^2 , sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]+(2*a1)=[2*(3*b+1(ba+1))+1]=(x+2+4*u)*[2*(3*z+1(zy+1))+1(4*y2)]^2 , (96 k² + 24 k + 11)/3=2*b*(b+1) , N^2*w*(w+2+4*v)=[8*((96 m² + 24 m + 1)3*c*(c1)/2)+1]/3 , sqrt[(8*(96 m² + 24 m + 1)+1)/3+1](2*c1)=[2*(3*d+1(dc+1))+1(4*c2)]=w*[2*(3*z+1(zy+1))+1]^2 , sqrt[(8*(96 m² + 24 m + 1)+1)/3+1]+(2*c1)=[2*(3*d+1(dc+1))+1]=(w+2+4*v)*[2*(3*z+1(zy+1))+1(4*y2)]^2 , (96 m² + 24 m + 11)/3=2*d*(d+1) , N^2*g*(g+2+4*h)=[8*((96 n² + 24 n + 1)3*f*(f1)/2)+1]/3 , sqrt[(8*(96 n² + 24 n + 1)+1)/3+1](2*f1)=[2*(3*r+1(rf+1))+1(4*f2)]=g*[2*(3*z+1(zy+1))+1]^2 , sqrt[(8*(96 n² + 24 n + 1)+1)/3+1]+(2*f1)=[2*(3*r+1(rf+1))+1]=(g+2+4*h)*[2*(3*z+1(zy+1))+1(4*y2)]^2 , (96 n² + 24 n + 11)/3=2*r*(r+1) , (3*N1)/8=3*z*(z+1)/23*y*(y1)/2+(3*z+1)*(3*z+2)/2 Since I don't have CAS and I don't know how to use them I'm not sure, could someone tell me, please, if the system of the three systems is resolved? This is a free copy of https://www.academia.edu/43115308/Phantom_factorization for MersenneForum Friend 
20200527, 12:20  #2 
Mar 2019
67 Posts 
I'd be amazed if anybody on the forum has the patience to put up with your gibberish anymore, after you've been told repeatedly to test your own equations.

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