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Primitive roots.

  1. Mar 6, 2007 #1
    If a is a perfect cube, a= n^3, for some integer n, and p is a prime with p is congreunt to 1 mod 3, then show that a cannot be a primitive root mod p, tat is ep(a) is not equal to p - 1
  2. jcsd
  3. Mar 9, 2007 #2
    Given any primitive root, it follows that r^(p-1)==1 Mod p, but not for any lesser power. But (p-l)/3 =u, is an integer less than p-1, and it follows that:

    (a^3)^u ==1 Mod p.
    Last edited: Mar 9, 2007
  4. Jul 3, 2007 #3
    Look up Henzel's lemma in Lang or Milne's online course note. It is relavant.
  5. Jul 3, 2007 #4
    No need to use Hensel's lemma, what robert did is perfect.
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