- 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

http://beablue.selfip.net/devalco/symmetrie_adjoined.htm

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 - --- In primenumbers@yahoogroups.com,

"bhelmes_1" <bhelmes@...> wrote:

> 1. let a jacobi (a, p)=-1

[4] is meaningless, as it stands.

> 2. let a^(p-1)/2 = -1 mod p

> 3. a^6 =/= 1 mod p

> 4. (1+sqrt (a))^p = 1-sqrt (a)

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....

The group of units (Z/nZ)* is /not/ cyclic

> there is a cyclic order ...

> 3....

> there is a cyclic order ...

if n has at least two distinct odd prime fators.

David