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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
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
> 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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