## sufficent proof for primes ?

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• A beautifull evening I have tried to make a sufficent proof for all odd primes: http://beablue.selfip.net/devalco/suf_prime.html There remains two question: 1.
Message 1 of 50 , Aug 5, 2010
A beautifull evening

I have tried to make a sufficent proof for all odd primes:
http://beablue.selfip.net/devalco/suf_prime.html

There remains two question:
1. if (1+sqrt (a))^d mod p is a natural number,
does this reveal a group with order d in the ring of natural numbers with an adjoined square ?

2. if p=f*g and there exist a group in p with order d in the ring of natural numbers with an adjoined square,
does that means, that f and g have also a group with order d.

The test bases on the symmetry of primes shown by

I hope that there are some people who are interested in the proof
and who can help me to finish the proof.

The best greetings from the primes
Bernhard

http://devalco.de
• ... [4] is meaningless, as it stands. You should write a double mod: 4. (1+x)^p = 1-x mod(x^2-a,p) ... There is no reason whatsoever to believe that [1] to [4]
Message 50 of 50 , Sep 29, 2011
"bhelmes_1" <bhelmes@...> wrote:

> 1. let a jacobi (a, p)=-1
> 2. let a^(p-1)/2 = -1 mod p
> 3. a^6 =/= 1 mod p
> 4. (1+sqrt (a))^p = 1-sqrt (a)

[4] is meaningless, as it stands.
You should write a double mod:

4. (1+x)^p = 1-x mod(x^2-a,p)

> 1. Is it possible that there are other exceptions

There is no reason whatsoever to believe that
[1] to [4] establish that p is prime. Morevoer,
some folk believe that, for every epsilon > 0,
the number of pseudoprimes less than x may
exceed x^(1-epsilon), for /sufficiently/ large x.

> 2....
> there is a cyclic order ...

> 3....
> there is a cyclic order ...

The group of units (Z/nZ)* is /not/ cyclic
if n has at least two distinct odd prime fators.

David
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