Peter, David Broadhurst provided similar arguments to show that there is likely to infinite number of such primes and that their count grows proportionally to log(log(x)): http://groups.yahoo.com/group/primeform/message/6938
But he still has some doubts (and I share them) if probabilistic assumptions he made hold for this problem: http://groups.yahoo.com/group/primeform/message/6965 But from the point of finding a new prime, these results are interesting but useless. At least I do not see how they can help in the search. Max On 2/18/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote: > 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 > >> _______________________________________________ > >> Prime mailing list > >> [email protected] > >> http://hogranch.com/mailman/listinfo/prime > > > > > > 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 > > > > > > _______________________________________________ > > Prime mailing list > > [email protected] > > http://hogranch.com/mailman/listinfo/prime > > > > > > > > _______________________________________________ > Prime mailing list > [email protected] > http://hogranch.com/mailman/listinfo/prime > _______________________________________________ Prime mailing list [email protected] http://hogranch.com/mailman/listinfo/prime
