20200527, 09:16  #1 
May 2017
ITALY
110010011_{2} 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
2^{2}×17 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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