- Dear David,

> > 1. p is an odd positive integer

Please give a hint how you produce these impression

> > 2. a is an integer with kronecker(a,p) = -1

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

> > 5. Mod(1+x,x^2-a)^p = 1-x mod p

>

> > Puzzle: Find a pseudoprime, p,

> > with a false witness, a,

> > that fools (1), (2), (3), (5) and has

> > znorder(Mod(a,p)) > 3414.

>

> Comment: The gremlins can generate false witnesses

> for this puzzle at a rate in excess of 50 kHz,

> finding more than 600,000,000 in less than 3 hours.

rate, some mathematical background would be very nice,

perhaps also for others.

I would like to participate in the calculation,

but how do you generate pseudoprimes for which (1), (2), (3) and (5)

are o.k. ?

By the way, is there a mathematical book, about

complex, adjoined squares and complex adjoined squares number,

which somebody could recommand ?

> Hint: All of these false witnesses fail my gremlin-trap:

Pure chance ;-)

>

> > 4. gcd(a^6-1,p) = 1

Nice Greetings from the primes

Bernhard - --- 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