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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.
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