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Old 2020-08-10, 18:39   #12
VBCurtis
 
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Quote:
Originally Posted by Alberico Lepore View Post
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.
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Old 2020-08-13, 13:13   #13
Alberico Lepore
 
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Quote:
Originally Posted by Alberico Lepore View Post
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)
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Old 2020-08-13, 20:23   #14
fdicarlo
 
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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?
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Old 2020-08-14, 02:51   #15
mathwiz
 
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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.
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Old 2020-08-14, 10:33   #16
Alberico Lepore
 
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DELETE

Last fiddled with by Alberico Lepore on 2020-08-14 at 14:01
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Old 2020-08-14, 13:15   #17
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Quote:
Originally Posted by mathwiz View Post
What we think is that this is all a waste of our time, and a moderator should lock the thread already.
Done.
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Old 2020-08-15, 11:51   #18
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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."
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Old 2020-08-15, 14:26   #19
Alberico Lepore
 
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Quote:
Originally Posted by Dr Sardonicus View Post
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
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Old 2020-08-15, 15:11   #20
Alberico Lepore
 
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Quote:
Originally Posted by Alberico Lepore View Post
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
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Old 2020-08-16, 05:17   #21
CRGreathouse
 
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Quote:
Originally Posted by Alberico Lepore View Post
(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
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Old 2020-08-17, 18:18   #22
Alberico Lepore
 
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Quote:
Originally Posted by CRGreathouse View Post
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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