I believe Fermat "discovered" this centuries ago (or possibly even

Euclid?). This is simply a Fermat-2 PRP test. Yes, all primes

certainly ARE shown by k being a whole number (i.e. ((2^n)-2)%n == 0)

This can be extened for any "base". So ((2^n)-2)%n==0 implies

Fermat PRP in base 2, and ((b^n)-b)%n==0 implies Fermat PRP in base

b. NOTE these are still just "probably" prime. For your PRP-2

below, 341 is the smallest composite that "passes". However, as

you pointed out, there will be NO primes that fail this test.

This method is most "commonly" used in modular space. Computing

whether ((2^329018309280127)-2)/329018309280127 is a whole number

(without using modular methods), is a VERY hard thing to do, since

2^329018309280127 is a number larger than any "current" computer can

handle. However, computing (2^329018309280127)%329018309280127

then subtracting 2, and checking if the result is zero, can be done

on binary computers in a fraction of a second, by continually

reducing the intermediate results mod n (or 329018309280127 in this

case).

It is always fun to "discover" something, however, when I have done

so in the past, I usually (always in my case) find out that someone,

(like Euclid, Fermat, Erdos, Euler, Newton, Pascal, ...) beat me to

it long long ago. However, I still keep trying ;)

Jim.

--- In primenumbers@yahoogroups.com, "edmorrey" <edmorrey@y...> wrote:

> I have found this new prime conjecture

>

> n is a prime number only when k is a hole number

>

> ((2^n)-2)/n = k

>

>

>

> Discovered by:

> Eduardo Mourey López Negrete

> Torreón Coah. Mexico

> Torreón Jardin

> Lirios 333

> 27200