## Help!!!

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• 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
Message 1 of 3 , Dec 2, 2003
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

---------------------------------
Do you Yahoo!?

[Non-text portions of this message have been removed]
• ... a(n-1)/2 = 1 (mod n), then n is prime. ... As I indicated to the poster in a private response, disprove this statement with n=175, h=11, k=4, a=51.
Message 2 of 3 , Dec 2, 2003
--- Cletus Emmanuel <cemmanu@y...> wrote:
> (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?.....

As I indicated to the poster in a private response,
disprove this statement with n=175, h=11, k=4, a=51.
• ... it?..... ... Or indeed any composite n of the above form with a=1! Rick
Message 3 of 3 , Dec 2, 2003
--- In primenumbers@yahoogroups.com, "jbrennen" <jack@b...> wrote:
> --- Cletus Emmanuel <cemmanu@y...> wrote:
> > (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?.....
>
> As I indicated to the poster in a private response,
> disprove this statement with n=175, h=11, k=4, a=51.

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

Rick
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