- ...

if p= 4*k +1, and q= 2*p +3 are both prime, then if

[(Mp)^p -p] mod q== N, and q mod N== +/-1, then (Mp),

the base... is prime.

(someone would have to prove this conjecture.)

...

let 2^127 -1 = 170141183460469231731687303715884105727

...

C5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*

77158673929+1) -1

...

let p= 5, q= 13 such that [(C5)^5 -5] mod 13 = N; and

2^54 mod 13 == 12 == (-1) (by chance!!!), and then...

...

(-1)^(odd power)*2 -1 ==

(-1)^49*19*43*73*127*337*5419*92737*649657*

77158673929*2^1 -1 ==

(-1)*2 = -2 -1 and (-3)^5 -5 = -248 and [(C5)^5 -5] mod

13 == 12

...

thus, if [(C5)^5 -5] mod 13 == 12, and 13 mod 12 == 1,

then C5 must be prime!

...

Rewards,

...

Bill Bouris - the 4 equations for a Mersenne number (Mr) where r is prime.

(either one or the other is true)

p=4k+1, q=2p+3 (both prime) [(Mr)^p-p] mod q == -1, or

p=4k+3, q=2p+3 (again) [(Mr)^p-p] mod q == -1 (can't be +1)

(but, both of the following must be true)

p=4k+1, q=2p+1 (again) [(Mr)^p-p] mod q == p

p=4k+3, q=2p+1 (again) [(Mr)^p-p] mod q == p+2

look!!!> let 2^127 -1 = 170141183460469231731687303715884105727

C5 is definitely prime, if my study is correct.

> ...

> C5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1

> ...

> let p= 3, q= 7 such that [(C5)^3 -3] mod 7 = N; and 2^27 mod 7

> == 1 (again, by chance!!!) and then...

> ...

> (1)^(left over exponent)*2 -1 ==

> (1)^2*49*19*43*73*127*337*5419*92737*649657*77158673929*2^1 -1 ==

> (1)*2 -1 = 2 -1 and (1)^3 -3 = 1 -3 = -2 and [(C5)^3 -3] mod 7 == 5

> ...

> thus, if [(C5)^3 -3] mod 7 == 5, or p +2 = 3 +2 = 5, then C5 must

> be prime!

> ...

> --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:

Bill Bouris

>

> so many times, there are typos when using e-mail. in math, Mp

> is often used to describe a Mersenne number with a prime expo-

> nent. I have corrected it to (Mr). this conjecture cannot be

> found in print. it's my own new idea which may need tweaking.

> ...

> (conjecture)

> if p= 4*k +1, and q= 2*p +3 are both prime, then if [(Mr)^p -p]

> mod q == N, & q mod N == +/-1, then (Mr), the base... is prime.

> also, if (Mr) mod p = 1, then choose a different 'p', or if N is

> a square, then that (Mr) is prime. finally, the exponent cannot

> be such that q mod r == 1. there may be other small restrictions.

> (someone would have to prove this conjecture.)

> ...

> Now, I think that it is stated IN FULL. It's a brand new idea.

> ...

> let 2^127 -1 = 170141183460469231731687303715884105727

> ...

> C5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1

> ...

> let p= 5, q= 13 such that [(C5)^5 -5] mod 13 = N; and 2^54 mod 13

> == 12 == (-1) (by chance!!!) and then...

> ...

> (-1)^(odd power)*2 -1 ==

> (-1)^49*19*43*73*127*337*5419*92737*649657*77158673929*2^1 -1 ==

> (-1)*2 -1 = -2 -1 and (-3)^5 -5 = -248 and [(C5)^5 -5] mod 13 == 12

> ...

> thus, if [(C5)^5 -5] mod 13 == 12, and 13 mod 12 == 1, then C5 must

> be prime!

> ...