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 2018-12-17, 13:17 #1 Alberico Lepore     May 2017 ITALY 52×19 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=(q-p) y=(p-2*(q-p)) n=6*x^2+5*y*x+y^2=6*(q-p)^2+5*(p-2*(q-p))*(q-p)+(p-2*(q-p))^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
 2018-12-18, 08:02 #2 Nick     Dec 2012 The Netherlands 2×3×52×11 Posts Try writing $$6x^2+5yx+y^2$$ as $$(3x+y)(2x+y)$$ and you will probably understand the pattern better!
2018-12-19, 14:39   #3
Alberico Lepore

May 2017
ITALY

52·19 Posts

Quote:
 Originally Posted by Nick Try writing $$6x^2+5yx+y^2$$ as $$(3x+y)(2x+y)$$ and you will probably understand the pattern better!

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)^2-z^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)^2-z^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)^2-z^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?

 2018-12-19, 17:28 #4 Alberico Lepore     May 2017 ITALY 52·19 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 (q-p) = z or [p mod (q-p)] - (q-p)=z is near to zero compared to 4 * A * (A + 1) * N then N is factorizable
2019-01-02, 15:08   #5
Alberico Lepore

May 2017
ITALY

52×19 Posts

Quote:
 Originally Posted by Nick Try writing $$6x^2+5yx+y^2$$ as $$(3x+y)(2x+y)$$ and you will probably understand the pattern better!

from what you explained to me I tried to create a symmetric cryptographic code

I would like to understand where I was wrong?
Would you like to take a look?

 2019-01-03, 08:50 #6 Nick     Dec 2012 The Netherlands 2×3×52×11 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.
2019-01-03, 15:49   #7
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

22×2,347 Posts

Quote:
 Originally Posted by Alberico Lepore would like to understand where I was wrong?
https://online.stanford.edu/courses/...cryptography-i

2019-01-03, 23:42   #8
CRGreathouse

Aug 2006

32·5·7·19 Posts

Quote:
 Originally Posted by Alberico Lepore I would like to understand where I was wrong?
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, chosen-plaintext, linear, differential, brute force, quantum, etc.).

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