20200910, 09:21  #12 
May 2017
ITALY
5^{2}·19 Posts 
before understanding this
I have to understand this would you give me a little clue 
20200910, 09:48  #13  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{2}·653 Posts 
Quote:
It's up to you to show us how you factor things, not the other way around. Your claim, you prove it. 

20200910, 11:38  #14  
May 2017
ITALY
5^{2}×19 Posts 
Quote:
(a + b) mod 3 = 0 in two cases M=[(a+b)/2((a+b)/61)/2]*[2*[(a+b)/2((a+b)/61)/2]3] and M=[(a+b)/2((a+b)/6+1)/2]*[2*[(a+b)/2((a+b)/6+1)/2]+3] so I tried to bring back a generic number (a + b) mod 3 = 0 in M=[(a+b)/2((a+b)/61)/2]*[2*[(a+b)/2((a+b)/61)/2]3] but I didn't get any useful results Example N=161 , 2*(N+(n/2)^2((a+b)/61)^2)+2*a^2+((ba)/2)^2=((3*a+b)/2)^2 , a*b=(N+(n/2)^2((a+b)/61)^2) , 2*(N+(n/2)^2((a+b)/61)^2)+2*1^2+((a+b)/2+1)^2((3*a+b)/2)^2=0 but I will continue to study Last fiddled with by Alberico Lepore on 20200910 at 11:42 

20200910, 12:20  #15 
May 2017
ITALY
5^{2}×19 Posts 
Bruteforce could be attempted for a multiple of 9 :
9*F N=161 , 2*(N*9*F)+2*a^2+((ba)/2)^2=((3*a+b)/2)^2 , a*b=(N*9*F) , 2*(N*9*F)+2*1^2+((a+b)/2+1)^2((3*a+b)/2)^2=0 , F=15 > a=105 GCD(105,161)=7 but I think this is very RANDOM 
20200910, 16:48  #16 
May 2017
ITALY
111011011_{2} Posts 
If we solve F as a function of a and N
solve 2*(N*9*F)+2*a^2+((ba)/2)^2=((3*a+b)/2)^2 , a*b=(N*9*F) , 2*(N*9*F)+2*1^2+((a+b)/2+1)^2((3*a+b)/2)^2=0 ,F,b > 9*N*F=2*a^23*a multiplying by 2 and imposing 2 * a = A we will have 18*N*F=A^23*A A0 < sqrt(18*N) is it possible to apply the Coppersmith method? https://en.wikipedia.org/wiki/Coppersmith_method 
20200910, 21:42  #17 
Mar 2019
2·3^{2}·7 Posts 
Please, stop posting.

20200918, 17:51  #18  
May 2017
ITALY
111011011_{2} Posts 
Quote:
I don't know with what efficiency solve (N*F1)/8=(X^21)/82*((ba)/8)^2 ,a*b=(N*F) , 2*(N*F)+2*1^2+((a+b)/2+1)^2((3*a+b)/2)^2=0 8*X^26*X9=F*N*9 8*X^2+6*X9=F*N*9 multiplying everything by 2 and imposing A = 4 * X and B = 4 * X are obtained A^23*A18=F*N*9*2 B^2+3*B18=F*N*9*2 solve (65*F1)/8=(X^21)/82*((ba)/8)^2 ,a*b=(65*F) , 2*(65*F)+2*1^2+((a+b)/2+1)^2((3*a+b)/2)^2=0 8*X^26*X9=F*65*9 8*X^2+6*X9=F*65*9 multiplying everything by 2 and imposing A = 4 * X and B = 4 * X are obtained A^23*A18=F*65*9*2 B^2+3*B18=F*65*9*2 and these are the first two and then solve (N*F1)/8=x*(x+1)/22*((ba)/8)^2 ,a*b=(N*F) , 2*(N*F)+2*1^2+((a+b)/2+1)^2((3*a+b)/2)^2=0 ,F 32*x^2+20*x7=F*N*9 32*x^2+44*x+5=F*N*9 multiplying everything by 2 and imposing A = 8 * X and B = 8 * X are obtained A^2+5*A14=F*N*9*2 B^2+11*B+10=F*N*9*2 and these are the other 2 

20200918, 17:58  #19 
Aug 2006
1732_{16} Posts 

20200918, 19:28  #20 
May 2017
ITALY
733_{8} Posts 

20200920, 09:11  #21 
May 2017
ITALY
5^{2}×19 Posts 

20200920, 11:52  #22  
Feb 2017
Nowhere
3784_{10} Posts 
I am brought to mind of the following:
Quote:
But an infinity of nonsense! You've got the Professor at the grand Academy of Lagado beat, hands down. 

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