Jud McCranie <[EMAIL PROTECTED]> writes:
> At 10:47 AM 7/6/99 +0200, Benny.VanHoudt wrote:
> >Now lets only focus on the set 2^p - 1 with p prime, i.e., the set
> >of numbers that we are checking out at GIMPS. Has anyone proven that
> >an infinite number these are NOT prime (this is VERY likely to be
> >true)? If so, how can one prove this easily (it is probably not
> >possible to indentify a subset that is NOT prime as above
^
> >because then we would not consider all of them for GIMPS)?
> >
> If 2p+1 is prime then it divides 2^p-1. If it has been proven that there are
> in infinite number of prime pairs p and 2p+1 then this proves that there are an
> infinite number of 2^p-1 that is not prime when p is prime. These are called
> Sophie Germain primes, and it has been proven that there are an infinite number
> of them, therefore there are an infinite number of composites of the form 2^p-1
> when p is prime.
Please leave adequate white space in your right margin.
Benny's twice-indented (indentified?) text twice still reads well, but
three of Jud's indented lines wrap around on my 80-character screen.
It is _conjectured_ that p and 2p+1 are simultaneously
prime infinitely often, but I have seen no proof.
This is related to the twin prime conjecture, in which p and p+2
are simultaneously prime. More generally, if f1(x) and f2(x) are
irreducible polynomials with integer coefficients such that
i) f1(x) and f2(x) approach +infinity as x -> +infinity
[This excludes constant polynomials, and 3 - 7*x.];
ii) For each prime q there exists an integer n such that
q does not divide the product f1(n)*f2(n)
[This excludes f1(x) = x and f2(x) = x + 1.];
then the conjecture predicts infinitely many integers n
for which f1(n) and f2(n) are simultaneously prime.
A variation of this conjecture extends the result to more
than two polynomials. Even the one-polynomial result
is unproven when its degree exceeds 1: are there
infinitely many primes of the form x^2 + 1?
Peter Montgomery
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