Here are the values of p, factorization of n = q^2 - q + 1
for small p == 1 (mod 12). They were found through Maple's
ifactor(n, easy) function.
The value of n is a probable prime when n = 25 or 901.
If we pick a large p == 1 (mod 12), then n will be about 4^p.
We estimate n will be prime with probability
1/log(4^p) = (1/log(4))/p. p will be prime with probability about
1/log(p), Summing
1/log(4) 1
-------- -------
p log(p)
over p = 13, 25, 37, 49, ..., the series diverges,
so we expect infinitely many cases where both are prime.
But they can be hard to find, just as 1093 and 3511 are
the only known primes p for which 2^p == 2 (mod p^2).
1, 1
13, (193) (347587)
25, (1125899806179331)
37, (367) (379) (3234619) (41984721013)
49, (254110928791681607803) (48337) (25801)
61, (937) _c34_1
73, (43) (21002580922364503755525539978253220069) (98773)
85, (223) _c44_1 (222553)
97, _c47_1 (865209213469)
Peter Montgomery
>
> On Fri, 17 Feb 2006, Max wrote:
>
>> Hello!
>> This problem is somewhat related to Mersenne numbers.
>> Let p be a prime number and q = 2^p - 1 (so q is a Mersenne number but
not necessary prime). Can the number q^2 - q + 1 be prime?
>> Substituting the value for q, we can formulate the same problem as
follows:
>> Can the number P(p) = 2^(2*p) - 3*2^p + 3 be prime for prime p? What is
known:
>> If p is not limited to primes, then there are many primes P(p). There
are also many primes P(p) if p is limited to powers of primes. But no
prime P(p) found for primes p < 150000.
>> Can anybody prove that P(p) can never be prime for prime p or
>> find prime P(p) for some prime p (in which case p must be > 150000) ?
Thanks,
>> Max
>> P.S. The origin of this problem (in russian):
>> http://www.nsu.ru/phorum/read.php?f=29&i=5095&t=5095
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>
>
> 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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