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Old 2018-12-17, 13:17   #1
Alberico Lepore
 
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Default 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
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Old 2018-12-18, 08:02   #2
Nick
 
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Try writing \(6x^2+5yx+y^2\) as \((3x+y)(2x+y)\) and you will probably understand the pattern better!
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Old 2018-12-19, 14:39   #3
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Quote:
Originally Posted by Nick View Post
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?
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Old 2018-12-19, 17:28   #4
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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
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Old 2019-01-02, 15:08   #5
Alberico Lepore
 
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Quote:
Originally Posted by Nick View Post
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

https://www.academia.edu/38048144/Le..._Vernam_cipher

I would like to understand where I was wrong?
Would you like to take a look?
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Old 2019-01-03, 08:50   #6
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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.
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Old 2019-01-03, 15:49   #7
Batalov
 
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Quote:
Originally Posted by Alberico Lepore View Post
would like to understand where I was wrong?
https://online.stanford.edu/courses/...cryptography-i
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Old 2019-01-03, 23:42   #8
CRGreathouse
 
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Quote:
Originally Posted by Alberico Lepore View Post
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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