20051221, 20:08  #144  
Aug 2002
3×37 Posts 
Quote:
Guillermo Last fiddled with by akruppa on 20051221 at 21:07 Reason: by request 

20051221, 23:35  #145 
Dec 2002
33C_{16} Posts 
I was trying to find the encoded residue, but as I was adding bits to one side of my 14" screen they fell of at the other side. So I gave up.

20051222, 00:21  #146  
Jul 2004
Potsdam, Germany
831_{10} Posts 
Quote:
It wouldn't be timeefficient and such, but I don't see a real problem in generating a residue that will be 0x0 in 5000 iterations for a particular exponent. It's only highly unlikely that a simple computer error is responsible. 

20051222, 00:33  #147  
Aug 2002
2^{6}×5 Posts 
Quote:


20051222, 01:14  #148  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Quote:
Therefore, although the inverse iteration is not unique, you need take only one of the possible values at each step. 

20051222, 01:18  #149  
Aug 2002
320_{10} Posts 
Quote:


20051222, 01:44  #150 
"Phil"
Sep 2002
Tracktown, U.S.A.
3·373 Posts 
But isn't just one squareroot computation of the same overall complexity as a complete LL test?

20051222, 01:47  #151  
May 2004
28_{10} Posts 
Quote:


20051222, 04:25  #152 
Dec 2005
2^{2} Posts 
My Encripted Guess
Hi guys,
new to the forums. Do not have much time to spend with prime numbers any more but still follow the news (used to be obsessed). Here is my encrypted guess: 0x36C9248137FFFE Those who know how the false residues are calculated may figure out my method for encryption. Regards, Mike Eaton 
20051222, 05:41  #153 
Oct 2004
2×3^{3} Posts 
I meant about the new Merseene number like everyone is talking about encryption this and encryption that but none of it is a help for me and i'm completely lost about what pacionet and jinydu, t.rex and moo are talking about

20051222, 05:57  #154  
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Quote:


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