 mersenneforum.org was this formula known relative to primality and factorization that you know?
 Register FAQ Search Today's Posts Mark Forums Read 2018-12-17, 13:17 #1 Alberico Lepore   May 2017 ITALY 2·32·29 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 5×192 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

2×32×29 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 2×32×29 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

2·32·29 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 180510 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

23·3·419 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

5,987 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.).  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post carpetpool Miscellaneous Math 5 2018-02-05 05:20 Alberico Lepore Alberico Lepore 43 2018-01-17 15:55 Alberico Lepore Alberico Lepore 48 2017-12-30 09:43 Alberico Lepore Alberico Lepore 26 2017-12-17 18:44 Godzilla Miscellaneous Math 53 2017-02-07 07:58

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