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Old 2020-05-27, 09:16   #1
Alberico Lepore
 
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May 2017
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Default 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*(a-1)/2)+1]/3
,
sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]-(2*a-1)=[2*(3*b+1-(b-a+1))+1-(4*a-2)]=x*[2*(3*z+1-(z-y+1))+1]^2
,
sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]+(2*a-1)=[2*(3*b+1-(b-a+1))+1]=(x+2+4*u)*[2*(3*z+1-(z-y+1))+1-(4*y-2)]^2
,
(96 k² + 24 k + 1-1)/3=2*b*(b+1)
,
N^2*w*(w+2+4*v)=[8*((96 m² + 24 m + 1)-3*c*(c-1)/2)+1]/3
,
sqrt[(8*(96 m² + 24 m + 1)+1)/3+1]-(2*c-1)=[2*(3*d+1-(d-c+1))+1-(4*c-2)]=w*[2*(3*z+1-(z-y+1))+1]^2
,
sqrt[(8*(96 m² + 24 m + 1)+1)/3+1]+(2*c-1)=[2*(3*d+1-(d-c+1))+1]=(w+2+4*v)*[2*(3*z+1-(z-y+1))+1-(4*y-2)]^2
,
(96 m² + 24 m + 1-1)/3=2*d*(d+1)
,
N^2*g*(g+2+4*h)=[8*((96 n² + 24 n + 1)-3*f*(f-1)/2)+1]/3
,
sqrt[(8*(96 n² + 24 n + 1)+1)/3+1]-(2*f-1)=[2*(3*r+1-(r-f+1))+1-(4*f-2)]=g*[2*(3*z+1-(z-y+1))+1]^2
,
sqrt[(8*(96 n² + 24 n + 1)+1)/3+1]+(2*f-1)=[2*(3*r+1-(r-f+1))+1]=(g+2+4*h)*[2*(3*z+1-(z-y+1))+1-(4*y-2)]^2
,
(96 n² + 24 n + 1-1)/3=2*r*(r+1)
,
(3*N-1)/8=3*z*(z+1)/2-3*y*(y-1)/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
Attached Files
File Type: pdf phantom factorization.pdf (46.2 KB, 10 views)
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Old 2020-05-27, 12:20   #2
mathwiz
 
Mar 2019

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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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