- Hi all,

(1) Proth's theorem states:

Let n = h.2k+1 with 2k > h. If there is an integer a such that a(n-1)/2 = -1 (mod n), then n is prime.

(2) Now, is there a theorem which states?:

Let n = h.2k - 1 with 2k > h. If there is an integer a such that a(n-1)/2 = 1 (mod n), then n is prime.

If there is no theorem for (2), can anyone prove or disprove it?.....

---Cletus Emmanuel

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[Non-text portions of this message have been removed] - --- Cletus Emmanuel <cemmanu@y...> wrote:
> (2) Now, is there a theorem which states?:

a(n-1)/2 = 1 (mod n), then n is prime.

> Let n = h.2k - 1 with 2k > h. If there is an integer a such that

>

As I indicated to the poster in a private response,

> If there is no theorem for (2), can anyone prove or disprove it?.....

disprove this statement with n=175, h=11, k=4, a=51. - --- In primenumbers@yahoogroups.com, "jbrennen" <jack@b...> wrote:
> --- Cletus Emmanuel <cemmanu@y...> wrote:

it?.....

> > (2) Now, is there a theorem which states?:

> > Let n = h.2k - 1 with 2k > h. If there is an integer a such that

> a(n-1)/2 = 1 (mod n), then n is prime.

> >

> > If there is no theorem for (2), can anyone prove or disprove

>

Or indeed any composite n of the above form with a=1!

> As I indicated to the poster in a private response,

> disprove this statement with n=175, h=11, k=4, a=51.

Rick