- maybe... let Z = 2^(2^(p+1))+1 ; p is prime

Z is prime iff [Z (mod (2^p+1))] == 2^q ; for some q

for p = 2, 3,..., next???

eg. p=2, 2^8+1 mod 5 == 2^1 and...

p=3, 2^16+1 mod 9 == 2^3 and...

I searched up to p=389 using 'Try GMP!' interpreter and the result for

all the primes thus far were a residue of 17... not 2^5 as expected.

If it were to jump up to 2^5,... I think that the Z as stated in the

first statement would be prime.

Are p=2,3,... the only generators for a prime Z ????

Thanks in advance for any commentary. - --- In primenumbers@yahoogroups.com, "leavemsg1"
> wrote:

maybe if q can only be odd???, then a statement that

>

> maybe... let Z = 2^(2^(p+1))+1 ; p is prime

>

> Z is prime iff [Z (mod (2^p+1))] == 2^q ; for some q

all Fermat primes greater than F4 are composite may

arise??? just tinkering with the idea! comments???

>

Z as stated in

> for p = 2, 3,..., next???

>

> eg. p=2, 2^8+1 mod 5 == 2^1 and...

> p=3, 2^16+1 mod 9 == 2^3 and...

>

> I searched up to p=389 using 'Try GMP!' interpreter

> and the result for all the primes thus far were a

> residue of 17... not 2^5 as expected.

>

> If it were to jump up to 2^5,... I think that the

the> first statement would be prime.

I ran it up a little further to p=509 and I'm still

>

> Are p=2,3,... the only generators for a prime Z ????

>

> Thanks in advance for any commentary.

>

not able to produce a residue other than 17 for any

prime up to 509

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