## Re: primes of the form q^2-q+1...

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• ... not ... True, it is not divisible by 2,3 and 5 but it is often divisible by 7 and 13. My guess is that there is nothing per se preventing prime generation.
Message 1 of 2 , Feb 28, 2006
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--- In primenumbers@yahoogroups.com, "Werner D. Sand"
<Theo.3.1415@...> wrote:
>
> Hello Max,
>
> Some simple partial results:
>
> Let q=2^p-1; a=q²-q+1 then
> q=0 mod 3, if p is even
> q=1 mod 3, if p is odd (e.g. prime)
> in any case a=1 mod 3
> a is not divisible by 2,3,5; the more amazing that a=prime does
not
> occur (p>3).
> a cannot be factorized within the real numbers such as e.g.
> a²-b²=(a+b)*(a-b).
>
> Werner
>

True, it is not divisible by 2,3 and 5 but it is often divisible by 7
and 13. My guess is that there is nothing per se preventing prime
generation. Afterall the first two numbers generated are prime, and
after that the numbers quickly increase.

Max originally said that there are 'many' primes generated for
q^2-q+1
given that
q = 2^n-1
where n is not prime and n<150000, but it would help to know how many.

If instead of q=2^p-1 we use q=2^(2*p)-1 (p prime)
then q^2-q+1 is prime for p=2,3 and 5. (Are there anymore?)

If instead we use q=2^(4*p)-1 then is q^2-q+1 ever prime?

Along a similar vein:

let q = 3^n + 1

It turns out that q^2 - q + 1 contains a factor of 13 (thirteen) if n
does not contain a factor of 3 (three). (Fancy that!)

Thus for q^2 - q + 1 to be prime then n must contain a factor of 3.
(This is excepting where n=1, where q^2 - q + 1 = 13.)

q^2 - q + 1 is prime when n=1 and n=3, and I can't find anymore. (Not
that I searched that high...)

Mark
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