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April 16th, 2009, 09:13 AM  #1 
Senior Member Joined: Nov 2007 Posts: 633 Thanks: 0  Pythagorean primes and Wilson
Hi, Here is a new statement. p>3 
April 16th, 2009, 04:31 PM  #2 
Senior Member Joined: Oct 2008 Posts: 215 Thanks: 0  Re: Pythagorean primes and Wilson
Yes. It is easy to get the solution. But it is still difficult to use it for primality test

April 16th, 2009, 04:41 PM  #3 
Senior Member Joined: Nov 2007 Posts: 633 Thanks: 0  Re: Pythagorean primes and Wilson
Step by step you can solve the problem. The 2 statements have some consequences anyway. Example : If p is pythagorean it divide abs(P(o)P(e)) If p is not pythagorean then P(e) which is equal to m!*2^m will give you a solution for the diophantine equation : a*p = m!*2m + 1 (or p1) and so on 
April 16th, 2009, 04:51 PM  #4 
Senior Member Joined: Oct 2008 Posts: 215 Thanks: 0  Re: Pythagorean primes and Wilson
I think P(o)=P(e) (mod p) for all p in the format of 4k+1 not only Pythagorean primes

April 16th, 2009, 04:57 PM  #5  
Senior Member Joined: Nov 2007 Posts: 633 Thanks: 0  Re: Pythagorean primes and Wilson Quote:
http://www.research.att.com/~njas/seque ... o=Chercher  
April 16th, 2009, 05:03 PM  #6 
Senior Member Joined: Oct 2008 Posts: 215 Thanks: 0  Re: Pythagorean primes and Wilson
Yes. I know the Pythagorean primes are prime of form 4k+1. But the problem is that the first equivalence holds for all integers in the form 4k+1 (so it holds for pythagorean primes too). So the result is so trival 
April 16th, 2009, 05:15 PM  #7  
Senior Member Joined: Nov 2007 Posts: 633 Thanks: 0  Re: Pythagorean primes and Wilson Quote:
I did not say that P(e) mod p is different from zero. It is trivial to signal it. Example 93 = 4*23 +1 =3*31 P(o)=3*5*7*..*31*.....*45 mod (93)=0 P(e)=2*4*6*...........*46 = different from zero because 2*31=61>46.  

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primes, pythagorean, wilson 
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