[math-fun] Artin's conjecture
The absence of concrete results on Artin's conjecture is amazing. According to Wikipedia there is not even one integer N for which it is known that N is a primitive root modulo p for infinitely many primes p. Not even N = 2. Amazing. --Dan << And that is devastating because (pathetically!) it is not even known that 10 is a prim-root modulo more than a finite set of primes: http://en.wikipedia.org/wiki/Artin's_conjecture_on_primitive_roots
________________________________________________________________________________________ It goes without saying that .
Although the "Proof Attempts" section is interesting. It says that Artin's conjecture is true for all primes except possibly at most two exceptional primes p. It's not clear to me whether this is the weak conjecture (p is a primitive root for an infinite number of moduli) or the strong conjecture (p is a primitive root for Artin's conjectured proportion of moduli). On 1/31/2012 7:53 PM, Dan Asimov wrote:
The absence of concrete results on Artin's conjecture is amazing.
According to Wikipedia there is not even one integer N for which it is known that N is a primitive root modulo p for infinitely many primes p. Not even N = 2. Amazing.
--Dan
<< And that is devastating because (pathetically!) it is not even known that 10 is a prim-root modulo more than a finite set of primes: http://en.wikipedia.org/wiki/Artin's_conjecture_on_primitive_roots
And while I was exploring Artin's conjecture, I started looking at "full reptend primes" (primes p with decimal repeating block of maximal length p-1). In the MathWorld article on Full Reptend Primes, we find: A necessary (but not sufficient) condition that p be a full reptend prime is that the number 9R_(p-1) (where R_p is a repunit) is divisible by p, which is equivalent to 10^(p-1)-1 being divisible by p. For example, values of n such that 10^(n-1)-1 is divisible by n are given by 1, 3, 7, 9, 11, 13, 17, 19, 23, 29, 31, 33, 37, ... (Sloane's A104381). A104381 is titled "Numbers n such that 10^(n-1) == 1 (mod n)." By Fermat's Little Theorem, A104381 includes every prime except 2 and 5. Additionally, it contains the base-10 Fermat pseudoprimes, A005939. So the comment "Superset of full reptend primes" is not particularly noteworthy, it amounts to saying 2 and 5 are not full reptend primes.
On Tue, Jan 31, 2012 at 10:53 PM, David Wilson <davidwwilson@comcast.net> wrote:
Although the "Proof Attempts" section is interesting. It says that Artin's conjecture is true for all primes except possibly at most two exceptional primes p. It's not clear to me whether this is the weak conjecture (p is a primitive root for an infinite number of moduli) or the strong conjecture (p is a primitive root for Artin's conjectured proportion of moduli).
It's the weak conjecture. Towards an unconditional proof of the strong conjecture, the best unconditional result (i.e. without assuming extended Riemann Hypotheses) is an old result of Morris (now known as Dorian) Goldfeld: If a is an integers which isn't 0, +/- 1 or a square, let N_a(x) be the number of primes p <= x for which a is a primitive root, and c_a be the proportion of primes for which Hooley's theorem says that a is a primitive root. Set R_a(x) = N_a(x) - c_a*pi(x) be the remainder (i.e. the difference between the actual and predicted number. Hooley's conditional theorem says that R_a(x) = O(sqrt(x)*log(x)^c) for some constant c. Goldfeld's theorem says that if you look at the average of R_a(x) for all a in some interval that you get essentially what you'd get by plugging in Hooley's remainder (with perhaps an extra log(x) factor or two). What this shows is that if even if the ERH weren't true, that you would have a hard time finding an exceptional a by looking at counts. That is, it's conceivable that there would be an a for which there are only a finite number of primes having it as a primitive root, but even that a might look like it's behaving as Hooley's theorem predicts for large finite chunks of integers. Victor
On 1/31/2012 7:53 PM, Dan Asimov wrote:
The absence of concrete results on Artin's conjecture is amazing.
According to Wikipedia there is not even one integer N for which it is known that N is a primitive root modulo p for infinitely many primes p. Not even N = 2. Amazing.
--Dan
<< And that is devastating because (pathetically!) it is not even known that 10 is a prim-root modulo more than a finite set of primes: http://en.wikipedia.org/wiki/Artin's_conjecture_on_primitive_roots
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participants (3)
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Dan Asimov -
David Wilson -
Victor Miller