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
