--- In

primenumbers@yahoogroups.com,

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

> > If p is an odd prime, then the group (Z/Zp)*

> > of integers from 1 to p-1 is cyclic under

> > multiplication modulo p.

>

> ThatÂ´s right, but does not consider the

> ring with an adjoined square root.

I invited you to consider that simpler case

in order to impress on you the importance

of p being a prime.

> If p is an odd prime, then the group of natural numbers with an

> adjoined square root is cyclic with the order of p+1 under

> multiplication modulo p.

In this case you also begin with the crucial caveat:

"if p is an odd prime".

That means that you have no hope of proving the primality

of a probable prime, say n, that passes your test.

Any "cyclicity" that you use must be modulo a *prime* divisor

of n. All the classical proofs begin with the wording:

"Let p be a prime divisor of n" and then attempt to show that

(for example) p > sqrt(n), as in Morrison's theorem.

But in your "proof" of primality you considered the case

n = f*g where you claim that you do not need f and g

to be prime. That's simply crazy, to my mind.

David