Dear David,

> > with A quadratic residuum

>

> You cannot determine whether this is the case without

> factorizing the target. Your code seems to ask for a

> positive kronecker symbol (which you call "Jacobi").

> But that is not the same as what you assumed.

i changed the algorithm

http://109.90.3.58/devalco/suf_prime_2.html
First of all i limit the search for p=3 mod 4

Instead of a strong probablistic test i make now

a probablistic test for A^[(p-1)/2]=1

As p=3 mod 4 (p-1)/2 is odd and therefore the pretest

makes sure that A is a quadratic residue.

The test is not very fast, because i need two pretest and two

test for the certificate in the average for proving primes,

that makes 8 Selfridges as i guess.

Nevertheless no factorisation of p-1 or p+1 is needed

The certificate could be verified very fast by two multiplication

and a modulo operation.

Perhaps there might be some improvement.

This test seems to be faster than the AKS-test which is an improvement.

Nice greetings from the primes

Bernhard