20200910, 09:21  #12 
May 2017
ITALY
1E2_{16} 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
2^{2}·5·307 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
2×241 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
2×241 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
2×241 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
157 Posts 
Please, stop posting.

20200918, 17:51  #18  
May 2017
ITALY
482_{10} 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
3^{2}×5×7×19 Posts 

20200918, 19:28  #20 
May 2017
ITALY
2·241 Posts 

20200920, 09:11  #21 
May 2017
ITALY
2·241 Posts 

20200920, 11:52  #22  
Feb 2017
Nowhere
10645_{8} 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. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Phantom factorization  subcubic factorization of integers?  Alberico Lepore  Alberico Lepore  1  20200527 12:20 
Best approach for P1?  Mark Rose  PrimeNet  6  20170523 23:58 
Unorthodox approach to primes  Erkan  PrimeNet  7  20170110 03:34 
An Analytic Approach to Subexponential Factoring  akruppa  Math  2  20091211 18:05 
Best approach for 130 digits?  boothby  Factoring  10  20091016 17:24 