20181217, 13:17  #1 
May 2017
ITALY
5^{2}×19 Posts 
was this formula known relative to primality and factorization that you know?
Good morning and happy holidays I wanted to ask you:
was this formula known relative to primality and factorization that you know? Any natural number(N) n odd can be written in the form 6*x^2+5*y*x+y^2=n with x natural number(N) even and with y integer(Z) odd number Let p and q be two factors of n such that n=p*q then x=(qp) y=(p2*(qp)) n=6*x^2+5*y*x+y^2=6*(qp)^2+5*(p2*(qp))*(qp)+(p2*(qp))^2=p*q it should be shown that this is the only way in which we can represent n through this formula in its positive and negative factors and above all to solve it thank tou 
20181218, 08:02  #2 
Dec 2012
The Netherlands
2×3×5^{2}×11 Posts 
Try writing \(6x^2+5yx+y^2\) as \((3x+y)(2x+y)\) and you will probably understand the pattern better!

20181219, 14:39  #3  
May 2017
ITALY
5^{2}·19 Posts 
Quote:
Hey, thank you Nick you gave me an idea You could try such a way look for z near 0 Example N = 67586227 N=p*q=(z+m)*(z+2*m)=2*m^2+3*m*z+z^2=N 2*m^2+3*m*z+z^2=N 8*(2*m^2+3*m*z+z^2)=8*N 16*m^2+24*m*z+8*z^2=8*N (4*m+3*z)^2z^2=8*N sqrt(8*N)=23252,..... sqrt(8*N+z^2)=23529 138 steps recalling that z is odd N=p*q=(z+2*m)*(z+3*m)=6*m^2+5*m*z+z^2=N 6*m^2+5*m*z+z^2=N 24*(6*m^2+5*m*z+z^2)=24*N (12*m+5*z)^2z^2=24*N sqrt(24*N)=40274,..... sqrt(24*N+z^2)=40277 2 steps recalling that z is odd N=p*q=(z+3*m)*(z+4*m)=12*m^2+7*m*z+z^2=N 12*m^2+7*m*z+z^2=N 48*(12*m^2+7*m*z+z^2)=N (24*m+7*z)^2z^2=48*N sqrt(48*N)=56957,..... sqrt(48*N+z^2)=57025 33 steps recalling that z is odd what do you think about it? 

20181219, 17:28  #4 
May 2017
ITALY
5^{2}·19 Posts 
I thought of
N = p * q = (z + A * m) * (z + (A + 1) * m) = (A + 1) * A * m ^ 2 + (A + A + 1) * m * z + z ^ 2 with A ranging from 1 to 100 to cover 2> q / p> 1.01 multiplies by 4 * A * (A + 1) if p mod (qp) = z or [p mod (qp)]  (qp)=z is near to zero compared to 4 * A * (A + 1) * N then N is factorizable 
20190102, 15:08  #5  
May 2017
ITALY
5^{2}×19 Posts 
Quote:
from what you explained to me I tried to create a symmetric cryptographic code https://www.academia.edu/38048144/Le..._Vernam_cipher I would like to understand where I was wrong? Would you like to take a look? 

20190103, 08:50  #6 
Dec 2012
The Netherlands
2×3×5^{2}×11 Posts 
In practice, designing your own cryptographic algorithm is a bad idea  it is much safer to use one of the standard algorithms that have been tested by many people all over the world.
If you want to do it anyway, as an academic exercise, you need to work out not just how messages are encrypted and decrypted but also what someone trying to break it could try. 
20190103, 15:49  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,347 Posts 

20190103, 23:42  #8 
Aug 2006
3^{2}·5·7·19 Posts 
Designing your own cryptosystem, as Nick mentioned, is probably an example of a wrong turn.
Another would be basing it off an antiquated, broken cipher. Another would be designing a cryptosystem without an understanding of the fundamentals of cryptography and cryptanalysis. You don't give a reduction to factoring, you don't discuss diffusion or nonlinearity, you don't give an analysis of the avalanche properties, you don't discuss resistance to any sort of attack model (distinguishing, chosenplaintext, linear, differential, brute force, quantum, etc.). 
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