
Number Theory Number Theory Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 11th, 2009, 07:24 AM  #1 
Senior Member Joined: Nov 2007 Posts: 633 Thanks: 0  Wilson theorem new formulation?
I have never seen before such formulation that is why I'm willing to know your thoughts. While trying to find some way to express factorial as sequential recurrence (not trivial one) I have found this. If n is prime then (k! n(k+1)!) mod n = 1 or n1 with 0<=k<=n1 Example 10!12! mod 23 = 22 9!13! mod 23 = 1 I know it is not hard to prove but I have little knowledge to do it. Is there some consequences if written this way? I do not know. Thank you for any explanation or comment. 
April 11th, 2009, 10:49 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Wilson theorem new formulation?
This is no easier to test than Wilson's theorem  you're still doing about n multiplications. You can show an even stronger statement, that the result is 0 if n is composite (and > 4?), by checking factors. I'm still a bit too hazy to do the proof at the moment (have been sick). 

Tags 
formulation, theorem, wilson 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
wilson theorem  agustin975  Number Theory  15  April 8th, 2013 06:44 AM 
Relation of Clement's theorem, his corollary and Wilson's th  ibougueye  Number Theory  13  January 15th, 2012 04:58 PM 
Wilson's Theorem  Scooter  Number Theory  9  October 21st, 2010 01:10 PM 
Wilson's theorem proof  Ben92  Number Theory  1  July 15th, 2009 12:01 PM 
Wilson's theorem question  mathsss22  Number Theory  1  November 8th, 2008 08:26 PM 