20200810, 18:39  #12 
"Curtis"
Feb 2005
Riverside, CA
19×227 Posts 
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. 
20200813, 13:13  #13  
May 2017
ITALY
111010001_{2} Posts 
Quote:
[2*((N+6*(6*a+1)1)/6)^2+(N+6*(6*a+1)1)/6  [2*((N1)/6)^2+(N1)/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*((N1)/6)^2+(N1)/6]/(6*a+1)^2=184 a=1 , (225184)=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*((N1)/6)^2+(N1)/6]/(6*a+1)^2=53 a=2 , (7653)=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*((N1)/6)^2+(N1)/6]/(6*a+1)^2=25 a=2 , (4125)=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 20200813 at 13:50 Reason: (4125) 

20200813, 20:23  #14 
Aug 2020
1 Posts 
Alberico,
as I mentioned in the facebook group where you also bother other people with your posts, here some suggestions:

20200814, 02:51  #15 
Mar 2019
1010111_{2} 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. 
20200814, 10:33  #16 
May 2017
ITALY
3·5·31 Posts 
DELETE
Last fiddled with by Alberico Lepore on 20200814 at 14:01 
20200814, 13:15  #17 
Feb 2017
Nowhere
2^{2}×5×173 Posts 

20200815, 11:51  #18 
Feb 2017
Nowhere
2^{2}·5·173 Posts 
I have reopened 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." 
20200815, 14:26  #19  
May 2017
ITALY
3×5×31 Posts 
Quote:
Since you were kind for once I would like to reciprocate by showing you something It can be easily demonstrated that: [[[2*((N1)/6)^2+(N1)/6] mod N]2*a^2a+h*N]/(6*a+1)=a+((3+4*a)+(3+4*a)+4*(b2*a1))*(b2*a)/2 , (6*a+1)*(6*b+1)=N , h=(b2*a2)/3  h=(b2*a1)/3  h=(b2*a)/3 And now IF in this case (2*a^2+a+(k)*(6*a^2+a))=[[2*((N1)/6)^2+(N1)/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*((N1)/6)^2+(N1)/6] mod N]+h*N , a+((3+4*a)+(3+4*a)+4*(b2*a1))*(b2*a)/2=k*a , (6*a+1)*(6*b+1)=N , h=(b2*a2)/3  h=(b2*a1)/3  h=(b2*a)/3 Example (2*a^2+a+(k)*(6*a^2+a))=59+h*133 , a+((3+4*a)+(3+4*a)+4*(b2*a1))*(b2*a)/2=k*a ,h=(b2*a1)/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*((N1)/6)^2+(N1)/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*((N1)/6)^2+(N1)/6] mod N]2*a^2a+h*N]/(6*a+1)=a+((3+4*a)+(3+4*a)+4*(b2*a1))*(b2*a)/2 , (6*a+1)*(6*b+1)=N , h=(b2*a2)/3  h=(b2*a1)/3  h=(b2*a)/3 Ed ora SE nel caso questa (2*a^2+a+(k)*(6*a^2+a))=[[2*((N1)/6)^2+(N1)/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*((N1)/6)^2+(N1)/6] mod N]+h*N , a+((3+4*a)+(3+4*a)+4*(b2*a1))*(b2*a)/2=k*a , (6*a+1)*(6*b+1)=N , h=(b2*a2)/3  h=(b2*a1)/3  h=(b2*a)/3 Esempio (2*a^2+a+(k)*(6*a^2+a))=59+h*133 , a+((3+4*a)+(3+4*a)+4*(b2*a1))*(b2*a)/2=k*a ,h=(b2*a1)/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*((N1)/6)^2+(N1)/6] mod N]+h*N è verififcata Ciao. Last fiddled with by Alberico Lepore on 20200815 at 14:48 Reason: red edit 

20200815, 15:11  #20  
May 2017
ITALY
3×5×31 Posts 
Quote:
Example 2: (2*a^2+a+(k)*(6*a^2+a))=374+h*13*37, a+((3+4*a)+(3+4*a)+4*(b2*a1))*(b2*a)/2=k*a ,h=(b2*a2)/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 20200815 at 15:38 Reason: EDIT: not good 

20200816, 05:17  #21  
Aug 2006
1011100100001_{2} Posts 
Quote:
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 

20200817, 18:18  #22  
May 2017
ITALY
3·5·31 Posts 
Quote:
as this is true and is easily demonstrable [[2*((N1)/6)^2+(N1)/6] mod N]+N*h2*a^2aa*(6*a+1)=(6*a+1)*((1+b2*a1)*(2*b+1)) ,(6*a+1)*(6*b+1)=N h=(b2*a)/3  (b2*a1)/3  (b2*a2)/3 then solving this [[[2*((N1)/6)^2+(N1)/6] mod N]+N*h2*a^2aa*(6*a+1)]/((1+b2*a1)*(2*b+1))=A ,(6*a+1)*(6*b+1)=N or this [[[2*((N1)/6)^2+(N1)/6] mod N]+N*h2*a^2aa*(6*a+1)]/(1+b2*a1)=B ,(6*a+1)*(6*b+1)=N or this [[[2*((N1)/6)^2+(N1)/6] mod N]+N*h2*a^2aa*(6*a+1)]/(2*b+1)=C ,(6*a+1)*(6*b+1)=N we will have our factorization GCD 

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