20200515, 15:23  #12 
May 2017
ITALY
388_{10} Posts 

20200515, 15:25  #13 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{5}·173 Posts 

20200515, 15:45  #14 
May 2017
ITALY
2^{2}·97 Posts 

20200515, 15:59  #15 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
15A0_{16} Posts 

20200515, 16:08  #16 
May 2017
ITALY
2^{2}·97 Posts 

20200515, 16:11  #17 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
1010110100000_{2} Posts 
There is no best solution. You're posting stuff that doesn't work on anything except your cherry picked values like 65 and 187.
Last fiddled with by retina on 20200517 at 02:44 
20200515, 20:53  #18  
May 2017
ITALY
2^{2}×97 Posts 
Quote:
Quote:
I would like to implement this algorithm but I still can't implement it N^2*x*(x+2+4*h)[8*((96 k² + 24 k + 1)3*a*(a1)/2)+1]/3=0 [N^2*x^2+(4*h+2)*N^2*x][256*k^2+64*k]+[4*a^24*a]3=0 [[N*x+(2*h+1)*N]^2(2*h+1)^2*N^2][[16*k+2]^24]+[[2*a1]^21]3=0 [N*x+(2*h+1)*N]^2[16*k+2]^2=(2*h+1)^2*N^2[2*a1]^24+3+1 now I put a semiprime in place of N to continue the explanation solve [N*x+(2*h+1)*N]^2[16*k+2]^2=(2*h+1)^2*N^2[2*a1]^24+3+1 , [N*x+(2*h+1)*N][16*k+2] =2*12*N,x=3,N=66390187 ,a > 8*k66390187*h+663901871=0 apply the Generalized Euclidean algorithm > k= 24896320 + 66390187 t h = 13 + 8 t but h > 0 little bruteforce from 1 to ..... for t =1 k=41493867 h=5 > a=165975468 > k=41493867 ,a=165975468 , sqrt[(8*(96 k² + 24 k + 1)+1)/3+1](2*a1)=P > P=99915 GCD(66390187,99915)=6661 what do you think? 

20200516, 08:29  #19 
May 2017
ITALY
2^{2}·97 Posts 
Generalized algorithm of [1]
V=1 While(GCD(N,P)!= p && GCD(N,P)!=q){ solve [N*x+(2*h+1)*N]^2[16*k+2]^2=(2*h+1)^2*N^2[2*a1]^24+3+1 , [N*x+(2*h+1)*N][16*k+2] =2*12*V^2*N,x=3 ,a,k > a=1/2 (1 + sqrt(N^2 (1 + 4 h + 4 h^2  192 V^2  96 h V^2 + 576 V^4))) 8*kN*h+(12*V^22)*N+1=0 apply the Generalized Euclidean algorithm 8*kN*h+(12*V^22)*N+1=0 nel paramether t vary t in h with h>=0 until a has complex roots in the meantime, calculate sqrt[(8*(96 k² + 24 k + 1)+1)/3+1](2*a1)=P GCD(N,P)=p  q  N if GCD == p  q > the algorithm ends otherwise V=V+1 } someone please could try it for pleasure Last fiddled with by Alberico Lepore on 20200516 at 09:01 
20200517, 00:08  #20 
May 2017
ITALY
2^{2}·97 Posts 
[187*x+(2*h+1)*187]^2[16*k+2]^2=(2*h+1)^2*187^2[2*a1]^24+3+1
, x=3 , h=1 , sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]+(2*a1)=27*187*[2*(3*z+1(zy+1))+1(4*y2)] , sqrt[(8*(96 k² + 24 k + 1)+1)/3+1](2*a1)=[2*(3*z+1(zy+1))+1] , 70=3*z*(z+1)/23*y*(y1)/2+(3*z+1)*(3*z+2)/2 Last fiddled with by Alberico Lepore on 20200517 at 00:18 
20200517, 03:28  #21 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{5}·173 Posts 

20200517, 10:33  #22  
May 2017
ITALY
2^{2}·97 Posts 
Quote:
give values to a and apply Lenstra ellipticcurve factorization hoping that the factors of the transformed number are less than 40 digits [1] N^2*x*(x+2+4*h)=[8*((96 k² + 24 k + 1)3*a*(a1)/2)+1]/3 sqrt[(8*(96 k² + 24 k + 1)+1)/3+1](2*a1)=P GCD(N,P)= p  q  N [2] N^2*x*(x+2+4*h)=[8*((96 k² + 72 k + 13)3*a*(a1)/2)+1]/3 sqrt[(8*(96 k² + 72 k + 13)+1)/3+1](2*a1)=P GCD(N,P)= p  q  N [3] N^2*x*(x+2+4*h)=[8*((96 k² + 120 k + 37)3*a*(a1)/2)+1]/3 sqrt[(8*(96 k² + 120 k + 37)+1)/3+1](2*a1)=P GCD(N,P)= p  q  N [4] N^2*x*(x+2+4*h)=[8*((96 k² + 168 k + 73)3*a*(a1)/2)+1]/3 sqrt[(8*(96 k² + 168 k + 73)+1)/3+1](2*a1)=P GCD(N,P)= p  q  N 

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