## prime observation/question

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• 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...
Message 1 of 2 , Jul 7, 2007
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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.
• ... maybe if q can only be odd???, then a statement that all Fermat primes greater than F4 are composite may arise??? just tinkering with the idea!
Message 2 of 2 , Jul 8, 2007
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> wrote:
>
> 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

maybe if q can only be odd???, then a statement that
all Fermat primes greater than F4 are composite may
arise??? just tinkering with the idea! comments???

>
> 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.
>
I ran it up a little further to p=509 and I'm still
not able to produce a residue other than 17 for any
prime up to 509

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