## Re: sufficent proof for primes ?

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• ... First, there was a small error in the program, i wrote for 4. p-1 instead of p+1, i corrected it. You are right, that my program does not detect any
Message 1 of 50 , Aug 9, 2010
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> Dear David,
> >
> > > > What is your "strong witness" A for the probable primality of
> > > > p = 170141183460469231731687303715884105727 ?

First, there was a small error in the program, i wrote for 4.
p-1 instead of p+1, i corrected it.

You are right, that my program does not detect any witness for this prime, because p+1=2^n.

I have to exclude this special case.

Nevertheless for this special kind of primes, there is a sufficent
prime algorithm as you know.

Greetings from the primes
Bernhard
• ... [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
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"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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