mersenneforum.org Letzte attempt to factor RSA
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2020-08-10, 18:39   #12
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

19×227 Posts

Quote:
 Originally Posted by Alberico Lepore it was meant in chronological order
Right, that's how we mean it too- "last" means "there will not be one after this one," or "chronologically final."

Your equations don't factor anything, let alone RSA. You don't know how to solve the equations you write, and can't explain exceptions that your equations don't work for. You have spent years writing useless equations, yet have never progressed to factor anything larger than a ten year old can solve (and solve faster than you can even write your first equation).

If you do not make this your actual last attempt, some moderator may choose to ban you for a length of time. Stop wasting our time, please.

2020-08-13, 13:13   #13
Alberico Lepore

May 2017
ITALY

1110100012 Posts

Quote:
 Originally Posted by Alberico Lepore Given N in the form N=(6*a+1)*(6*b+1) then if this is solvable solve 2*((N+6*(6*a+1)-1)/6)^2+(N+6*(6*a+1)-1)/6 mod ((6*a+1)^2) =(( (2*((N-1)/6)^2+(N-1)/6) mod (6*a+1)^2)-2*a*(6*a+1)) the factoring is solved Example N=91 solve 2*((91+6*(6*a+1)-1)/6)^2+(91+6*(6*a+1)-1)/6 mod ((6*a+1)^2) =((465 mod (6*a+1)^2)-2*a*(6*a+1)) wolfram solves it https://www.wolframalpha.com/input/?...86*a%2B1%29%29 but how do you solve it?
As this is true

[2*((N+6*(6*a+1)-1)/6)^2+(N+6*(6*a+1)-1)/6 - [2*((N-1)/6)^2+(N-1)/6] +2*a*(6*a+1)]=(4*b+3)*(6*a+1)^2

then we can find the solution in logarithmic times

suppose N=403=13*31

we give the lowest value a=1

[2*((N+6*(6*a+1)-1)/6)^2+(N+6*(6*a+1)-1)/6]/(6*a+1)^2=225

[2*((N-1)/6)^2+(N-1)/6]/(6*a+1)^2=184

a=1 , (225-184)=4*b+3 -> (6*a+1)*(6*b+1)=406

we give correct value a=2

[2*((N+6*(6*a+1)-1)/6)^2+(N+6*(6*a+1)-1)/6]/(6*a+1)^2=76

[2*((N-1)/6)^2+(N-1)/6]/(6*a+1)^2=53

a=2 , (76-53)=4*b+3 -> (6*a+1)*(6*b+1)=403

we give higher value a=3

[2*((N+6*(6*a+1)-1)/6)^2+(N+6*(6*a+1)-1)/6]/(6*a+1)^2=41

[2*((N-1)/6)^2+(N-1)/6]/(6*a+1)^2=25

a=2 , (41-25)=4*b+3 -> (6*a+1)*(6*b+1)=389,...

P.s. Special cases must be taken into account

JOUEZ À VOTRE JEU

Last fiddled with by Alberico Lepore on 2020-08-13 at 13:50 Reason: (41-25)

 2020-08-13, 20:23 #14 fdicarlo   Aug 2020 1 Posts Alberico, as I mentioned in the facebook group where you also bother other people with your posts, here some suggestions: write your math in a proper language, written like this, they just look like garbage, start to use LaTeX, markdown or whatever it is making easier to write/read, the time you use to write here or in facebook (or wherever) is better spent using Matlab, Octave or Mathematica to create algos trying to prove your theories for larger numbers, this was already suggested in other post but it seems you don't like to listen this feedback; it seems is years trying to tackle this problem, did you think about applying your stubbornness (capatosta or testardagine in Italian language) to something else, like kitchen?
 2020-08-14, 02:51 #15 mathwiz   Mar 2019 10101112 Posts This guy (crank? capatost? whatever) has also been told, repeatedly, to test his own algorithms -- preferably on numbers with >=20 digits. He consistently refuses, and just scribbles more equations, with "what do you think?" What we think is that this is all a waste of our time, and a moderator should lock the thread already.
 2020-08-14, 10:33 #16 Alberico Lepore     May 2017 ITALY 3·5·31 Posts DELETE Last fiddled with by Alberico Lepore on 2020-08-14 at 14:01
2020-08-14, 13:15   #17
Dr Sardonicus

Feb 2017
Nowhere

22×5×173 Posts

Quote:
 Originally Posted by mathwiz What we think is that this is all a waste of our time, and a moderator should lock the thread already.
Done.

 2020-08-15, 11:51 #18 Dr Sardonicus     Feb 2017 Nowhere 22·5·173 Posts I have re-opened the thread. Upon further consideration, I concluded that in entitling the thread "Last attempt to factor RSA" the user merely made an error in word choice. I believe the meaning the user intended by the first word was "Latest."
2020-08-15, 14:26   #19
Alberico Lepore

May 2017
ITALY

3×5×31 Posts

Quote:
 Originally Posted by Dr Sardonicus I have re-opened the thread. Upon further consideration, I concluded that in entitling the thread "Last attempt to factor RSA" the user merely made an error in word choice. I believe the meaning the user intended by the first word was "Latest."
English

Since you were kind for once I would like to reciprocate by showing you something

It can be easily demonstrated that:

[[[2*((N-1)/6)^2+(N-1)/6] mod N]-2*a^2-a+h*N]/(6*a+1)=a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2
,
(6*a+1)*(6*b+1)=N
,
h=(b-2*a-2)/3 || h=(b-2*a-1)/3 || h=(b-2*a)/3

And now IF in this case (2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N is VERIFIED

then the solution of the factorization is thus found

(2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N
,
a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2=k*a
,
(6*a+1)*(6*b+1)=N
,
h=(b-2*a-2)/3 || h=(b-2*a-1)/3 || h=(b-2*a)/3

Example

(2*a^2+a+(k)*(6*a^2+a))=59+h*133 , a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2=k*a ,h=(b-2*a-1)/3 , (6*a+1)*(6*b+1)=133

In the next few days I will try to analyze the statistics on (2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N is verified

Italian

Visto che per una volta siete stati gentili vorrei ricambiare mostrandovi una cosa

E' facilmente dimostrabile che:

[[[2*((N-1)/6)^2+(N-1)/6] mod N]-2*a^2-a+h*N]/(6*a+1)=a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2
,
(6*a+1)*(6*b+1)=N
,
h=(b-2*a-2)/3 || h=(b-2*a-1)/3 || h=(b-2*a)/3

Ed ora SE nel caso questa (2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N è verififcata

allora la soluzione della fattorizzazione si trova così

(2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N
,
a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2=k*a
,
(6*a+1)*(6*b+1)=N
,
h=(b-2*a-2)/3 || h=(b-2*a-1)/3 || h=(b-2*a)/3

Esempio

(2*a^2+a+(k)*(6*a^2+a))=59+h*133 , a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2=k*a ,h=(b-2*a-1)/3 , (6*a+1)*(6*b+1)=133

Nei prossimi giorni cercherò di analizzare le statistiche su (2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N è verififcata

Ciao.

Last fiddled with by Alberico Lepore on 2020-08-15 at 14:48 Reason: red edit

2020-08-15, 15:11   #20
Alberico Lepore

May 2017
ITALY

3×5×31 Posts

Quote:
 Originally Posted by Alberico Lepore English Since you were kind for once I would like to reciprocate by showing you something It can be easily demonstrated that: [[[2*((N-1)/6)^2+(N-1)/6] mod N]-2*a^2-a+h*N]/(6*a+1)=a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2 , (6*a+1)*(6*b+1)=N , h=(b-2*a-2)/3 || h=(b-2*a-1)/3 || h=(b-2*a)/3 And now IF in this case (2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N is VERIFIED then the solution of the factorization is thus found (2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N , a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2=k*a , (6*a+1)*(6*b+1)=N , h=(b-2*a-2)/3 || h=(b-2*a-1)/3 || h=(b-2*a)/3 Example (2*a^2+a+(k)*(6*a^2+a))=59+h*133 , a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2=k*a ,h=(b-2*a-1)/3 , (6*a+1)*(6*b+1)=133 In the next few days I will try to analyze the statistics on (2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N is verified Italian Visto che per una volta siete stati gentili vorrei ricambiare mostrandovi una cosa E' facilmente dimostrabile che: [[[2*((N-1)/6)^2+(N-1)/6] mod N]-2*a^2-a+h*N]/(6*a+1)=a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2 , (6*a+1)*(6*b+1)=N , h=(b-2*a-2)/3 || h=(b-2*a-1)/3 || h=(b-2*a)/3 Ed ora SE nel caso questa (2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N è verififcata allora la soluzione della fattorizzazione si trova così (2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N , a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2=k*a , (6*a+1)*(6*b+1)=N , h=(b-2*a-2)/3 || h=(b-2*a-1)/3 || h=(b-2*a)/3 Esempio (2*a^2+a+(k)*(6*a^2+a))=59+h*133 , a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2=k*a ,h=(b-2*a-1)/3 , (6*a+1)*(6*b+1)=133 Nei prossimi giorni cercherò di analizzare le statistiche su (2*a^2+a+(k)*(6*a^2+a))=[[2*((N-1)/6)^2+(N-1)/6] mod N]+h*N è verififcata Ciao.
GCD (495 ,59)=a=1

Example 2:

(2*a^2+a+(k)*(6*a^2+a))=374+h*13*37, a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2=k*a ,h=(b-2*a-2)/3 , (6*a+1)*(6*b+1)=13*37

GCD (6440 ,374)=a=2

EDIT:

I analyzed 134217726 cases

number of p without repetitions = 20

total of p with repetitions = 82

I conclude: not good

Last fiddled with by Alberico Lepore on 2020-08-15 at 15:38 Reason: EDIT: not good

2020-08-16, 05:17   #21
CRGreathouse

Aug 2006

10111001000012 Posts

Quote:
 Originally Posted by Alberico Lepore (2*a^2+a+(k)*(6*a^2+a))=59+h*133 , a+((3+4*a)+(3+4*a)+4*(b-2*a-1))*(b-2*a)/2=k*a ,h=(b-2*a-1)/3 , (6*a+1)*(6*b+1)=133
Have you considered re-titling your method, "Finding single-digit prime factors of semiprimes coprime to 6"?

Code:
listAL(lim)=setunion(5*primes([5,lim\5]),7*primes([5,lim\7]))
factorAL(n)=if(n%5, 7, 5) \\ Finds a factor of one of the above numbers

2020-08-17, 18:18   #22
Alberico Lepore

May 2017
ITALY

3·5·31 Posts

Quote:
 Originally Posted by CRGreathouse Have you considered re-titling your method, "Finding single-digit prime factors of semiprimes coprime to 6"? Code: listAL(lim)=setunion(5*primes([5,lim\5]),7*primes([5,lim\7])) factorAL(n)=if(n%5, 7, 5) \\ Finds a factor of one of the above numbers
N=(6*a+1)*(6*b+1)

as this is true and is easily demonstrable

[[2*((N-1)/6)^2+(N-1)/6] mod N]+N*h-2*a^2-a-a*(6*a+1)=(6*a+1)*((1+b-2*a-1)*(2*b+1)) ,(6*a+1)*(6*b+1)=N

h=(b-2*a)/3 || (b-2*a-1)/3 || (b-2*a-2)/3

then solving this

[[[2*((N-1)/6)^2+(N-1)/6] mod N]+N*h-2*a^2-a-a*(6*a+1)]/((1+b-2*a-1)*(2*b+1))=A ,(6*a+1)*(6*b+1)=N

or this

[[[2*((N-1)/6)^2+(N-1)/6] mod N]+N*h-2*a^2-a-a*(6*a+1)]/(1+b-2*a-1)=B ,(6*a+1)*(6*b+1)=N

or this

[[[2*((N-1)/6)^2+(N-1)/6] mod N]+N*h-2*a^2-a-a*(6*a+1)]/(2*b+1)=C ,(6*a+1)*(6*b+1)=N

we will have our factorization

GCD

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