## Re: sufficent proof for primes ?

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• ... If p is an odd prime, then the group (Z/Zp)* of integers from 1 to p-1 is cyclic under multiplication modulo p. If n is a square-free odd composite
Message 1 of 50 , Aug 9, 2010
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"bhelmes_1" <bhelmes@...> wrote:

> The description of a cyclic group is borrowed from the
> cyclic group of natural numbers.

If p is an odd prime, then the group (Z/Zp)*
of integers from 1 to p-1 is cyclic under
multiplication modulo p.

If n is a square-free odd composite integer, then
the group (Z/Zn)* of integers from 1 to n-1 that
are coprime to n is _not_ cyclic.

> You are very fast in judging the proof.

That was because it contained obvious "cyclic" nonsense.

> I did not wrote that f and g must be primes

That may be because you do not understand cyclic groups.

David
• ... [4] is meaningless, as it stands. You should write a double mod: 4. (1+x)^p = 1-x mod(x^2-a,p) ... There is no reason whatsoever to believe that [1] to [4]
Message 50 of 50 , Sep 29 4:14 AM
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"bhelmes_1" <bhelmes@...> wrote:

> 1. let a jacobi (a, p)=-1
> 2. let a^(p-1)/2 = -1 mod p
> 3. a^6 =/= 1 mod p
> 4. (1+sqrt (a))^p = 1-sqrt (a)

[4] is meaningless, as it stands.
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....
> there is a cyclic order ...

> 3....
> there is a cyclic order ...

The group of units (Z/nZ)* is /not/ cyclic
if n has at least two distinct odd prime fators.

David
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