20090109, 18:16  #34 
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
5984_{10} Posts 
does anyone know about a program that can output the full residue from a rabinmiller test and not just the RES64

20090109, 18:23  #35  
Jan 2005
479 Posts 
Quote:
Hmm... Rather interesting :) No more beef for MooMoo ;) (Or else prove that the residue is incorrect....) 

20090109, 18:27  #36 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
7×13×47 Posts 

20090110, 10:40  #37  
May 2007
Kansas; USA
2×23×233 Posts 
Quote:
Oh come on now, you can do better than that! 123*2^67867868711 cannot possibly have a residue of 12BEEF15BAD431DE. It has factors of 7 AND 13!! As I recall, that would mean that it would have a residue of 0000000000000007! (no joke) HARDY HARDY HARDY HA HA HA HA HA !!!!!!!!!!!!!!! Lest anyone wonder how I came up with that: My good old mainstay...Alpertron's site. All I did was calculate the values for all n=1 thru 30 and found the following patterns: For all 123*2^n1, all n==(1 mod 3) has a factor of 7 and all n==(7 mod 12) has a factor of 13. 6786786871 is both == (1 mod 3) and == (7 mod 12). Mike, you can continue eating beef. Sometimes I even amaze myself. Gary 

20090110, 12:08  #38  
Jan 2005
111011111_{2} Posts 
Quote:
I didn't even check for small factors :0 Continue to eat your beef :) 

20090110, 13:24  #39  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
7·13·47 Posts 
Quote:
Quote:
Do you have any mathematical proof (even if I couldn't understand it ) that all 123*2^n1 with n==(1 mod 3) has a factor of 7 or that all n==(7 mod 12) has a factor of 13? Oh, I'm sure he can for now anyway. Is there anything that can actually try to factor something that large to show certainly that it has trivial factors? Oh Lord it's hard to be humble, when you're perfect in every way! 

20090110, 14:13  #40  
"Jacob"
Sep 2006
Brussels, Belgium
3·5·11^{2} Posts 
Quote:
n=3*m+1. 123*2^n1=123*2^(3*m+1)1=123*2*2^(3*m)1 2^3==1 mod 7 => 2^(3*m)=(2^3)^m==1 mod 7 123*2*2^(3*m)1==246*11 mod 7 245==0 mod 7 Use the same path for the other case. Jacob 

20090110, 17:45  #41 
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
2^{5}×11×17 Posts 

20090111, 09:00  #42 
May 2007
Kansas; USA
2·23·233 Posts 

20090111, 14:35  #43  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10B5_{16} Posts 
Quote:
2^3==1 mod 7 => 2^(3*m)=(2^3)^m==1 mod 7 Why do we know that 2^(3*m)==1 mod 7 is equivalent to 2^3==1 mod 7? Is it just a law/thereom/whatever that there's no further explanation to or is there a reason that could be explained? 

20090111, 17:31  #44  
"Jacob"
Sep 2006
Brussels, Belgium
3427_{8} Posts 
Quote:
(2^3)^m=8^m 8^m=(7+1)^m (7+1)^m==(0+1)^m mod 7 1^m=1 so 2^(3*m)==1 mod 7 Jacob 

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