Dear David,

thank you for your friendly help.

> Here is a counterexample with

> prime g, prime a, and composite p,

> for which every part of your 3-selfridge test

> is applied, including your recent desperate "wriggle".

1. You are right with your counterexample that

the certificate is not right.

I hope you will enjoy this statement.

2. You did not verify the point 4.

http://109.90.3.58/devalco/suf_helmes.htm#1
if i am right in understanding your program

[1+sqrt (2969)]^(p-1)=14278+18966*sqrt (2969)=/=1 mod 29539

which indicate that p is not a prime.

Last but not least, there remains a claim that

the test 1.-4. is a sufficient test for p=3 mod 4, slowly with in average 8 selfridges. This would be an improvement versus the AKS-test.

If you have fun and time, try to find one counterexample

which destroy the remaining test :-)

I think i have 40 years in the future in order to find some nice

algorithms and i hope that you will participate in some.

Greetings from the primes

Bernhard

>

> g = 21283;

> a = 2969;

> p = 29539;

>

> {if(p%4 == 3 &&

> !issquare(a) &&

> kronecker(a,p) == 1 &&

> Mod(a,p)^((p-1)/2) == 1 &&

> Mod((1+x)*Mod(1,p),x^2-a)^((p-1)/2) == g*x &&

> Mod(g,p)^2*a == 1 &&

> !isprime(p), print(" Counterexample!"));}

>

> Counterexample!

>

> David

>