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Primitive roots - annoying problem

  1. May 11, 2008 #1
    Let [tex]r[/tex] be a primitive root of a prime number [tex]p \geq 3[/tex]. Prove that if [tex]p \equiv 1 (mod 4)[/tex], then [tex]-r[/tex] is also a primitive root of [tex]p[/tex].

    I've been told it's quite easy, but I can't see why it's true for the life of me :frown:
  2. jcsd
  3. May 11, 2008 #2
    If r is a primitive root, then r^[(p-1)/2] is not equal to 1. The question is about -r.
  4. May 11, 2008 #3
    thanks for the reply :smile:
    if r is a primitve root then r^[(p-1)/2] is -1 (mod p). However, (-r)^[(p-1)/2] is the same since (p-1)/2 is even, am I right?

    Where do we go from here? Maybe I could say that the order of (-r) couldn't be lower than (p-1), because if it was so, ord(-r)|(p-1)/2 and then (-r)^[(p-1)/2]=1(mod p) contradicting the fact that p [tex]\neq[/tex] 2?
    Am I on the right track?

    Thanks again! :smile:
  5. May 11, 2008 #4
    Richard Cypher: Am I on the right track?

    Sounds O.K. to me. Take, (-r)^2 = r^2, well then what power is necessry to raise r^2 to 1?
  6. May 11, 2008 #5
    OK, I got it!
    Bunch of thanks for your help!:biggrin:
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