--- In

primenumbers@yahoogroups.com,

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

> i make sure that a is a quadratic residuum by

> calculating a^[(p-1)/2]=1 with p=3 mod 4

Your web page makes the following claim:

> you get a certificate (g, a, p)

> In order to prove the certificate only one Selfridge

> is needed because a^((p-1)/2) = 1 mod p has to verify

> and (gA)^2=1 mod p has to verify.

(where A^2 = a mod p)

You make no reference to the Lucas test in your

final "certificate". So all that you seem to

require for a final "proof" is that

1) p = 3 mod 4

2) kronecker(a,p) = 1

3) a^((p-1)/2) = 1 mod p

4) g^2*a = 1 mod p

or in Pari-GP the test

{cert(g,a,p)=p%4==3&&kronecker(a,p)==1

&&Mod(a,p)^((p-1)/2)==1&&Mod(g,p)^2*a==1;}

\\ Example:

g=3588;a=16;p=14351;

if(cert(g,a,p)==1,print("Helmes says "p" is prime"));

Helmes says 14351 is prime

So you seem to have a problem, since 14351 is not prime.

It is thus quite wrong to claim a 1-selfridge certificate.

Rather I believe that you have a 3-selfridge PRP test

that proves compositeness when it fails, yet does

not prove primality when it succeeds.

David