[Prime] primes of the form q^2-q+1 with q=2^p-1 for prime p

[Max]

Hans,

If I get you correctly, the "covering congruences" will not help.

I performed sieving of candidate primes p and filtered out all such
that P(p) has a small prime factor less than 1M. The sieving ended up
with still quite a large number of candidates.
C++ source sieve.cpp and its output sieved.txt can be downloaded at
http://www.mytempdir.com/459551

For every prime p from sieved.txt it is guaranteed that P(p) does not
have prime divisors smaller than 1M. It may be useful if somebody
wants to continue search for p beyond 150000.

Max

On 2/18/06, Hans Riesel <riesel at nada.kth.se> wrote:

> Hello Max,
>
>   I think your problem could be solved with the technique of
> "covering congruences". It is easy to prove that for the case
> p=6k-1 the number P(p) is always divisible by 7. The remaining
> numbers p=6k+1 can be split into two classes, p=12k+1 and
> p=12k+7. For the case p=12k+7 it is easy to prove that P(p) is
> aiways divisible by 13.---I have unfortunately not been able to
> do more research on the remaining case p=12k+1, because I have
> been given a new computer, which has not yet had all systems
> installed. But if you could send me factorizations of the
> numbers P(p) for p=13, 25, 37, 49, and 61, say, I'll give it a
> try.
>
> Sincerely yours,
>
> Hans Riesel
>
>
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