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April 16th, 2009, 09:13 AM   #1
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Pythagorean primes and Wilson

Hi,

Here is a new statement.
p>3
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April 16th, 2009, 04:31 PM   #2
duz
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Re: Pythagorean primes and Wilson

Yes. It is easy to get the solution. But it is still difficult to use it for primality test
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April 16th, 2009, 04:41 PM   #3
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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 p-1)

and so on
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April 16th, 2009, 04:51 PM   #4
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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
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April 16th, 2009, 04:57 PM   #5
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Re: Pythagorean primes and Wilson

Quote:
Originally Posted by duz
I think P(o)=P(e) (mod p) for all p in the format of 4k+1 not only Pythagorean primes
But at my knowledge (maybe am I wrong) a Pythagorean prime is a prime of the form of 4k+1.

http://www.research.att.com/~njas/seque ... o=Chercher
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April 16th, 2009, 05:03 PM   #6
duz
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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
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April 16th, 2009, 05:15 PM   #7
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Re: Pythagorean primes and Wilson

Quote:
Originally Posted by duz
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
If an integer is not prime how the statement is going to hold?
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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