1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Primitive root !

  1. Mar 12, 2008 #1
    If a is a perfect square then a is not a primitive root modulo p (p is an odd prime). (from Artin's conjecture on primitive roots) http://en.wikipedia.org/wiki/Artin's_conjecture_on_primitive_roots

    This is what I know: suppose a = b^2
    a is a primitive root mod p when , a^(p-1) congruent to 1 (mod p)
    that means b^2^(p-1) congruent to 1 (mod p)..,... I got stuck from here. Someone kindly gives me a hint ?

    thank you
     
  2. jcsd
  3. Mar 13, 2008 #2
    Actually the test you give is not correct. The test is that every number [tex]n[/tex] such that [tex]n^{p-1} = 1 \mod p[/tex] must equal some power of [tex]a[/tex] modulus [tex]p[/tex]. In other words if p is prime and a is a primitive root then the set of [tex]a^i \mod p[/tex] equals the congruence set {1,2,3 ... p-1}. But a = b^2 = a primitive root implies b is also a primitive root. Could b and b^2 both be primitive roots?
     
    Last edited: Mar 13, 2008
  4. Mar 13, 2008 #3
    The matter is easy enough. Every element is such that a^(p-1)==1 Mod p, for p prime.

    But to be a primitive root, a must be such that a, a^2, a^3....a^(p-1) all generate
    different elements.

    However should a ==b^2 Mod p, then a^(p-1)/2 ==b^(p-1) ==1 Mod p. So that a is then capable of generating no more than half the multiplicative group.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Primitive root !
  1. Primitive roots (Replies: 10)

  2. Primitive roots (Replies: 5)

  3. Primitive roots. (Replies: 3)

Loading...