Primitive roots - annoying problem

  • #1

Main Question or Discussion Point

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:
 

Answers and Replies

  • #2
1,056
0
If r is a primitive root, then r^[(p-1)/2] is not equal to 1. The question is about -r.
 
  • #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:
 
  • #4
1,056
0
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?
 
  • #5
OK, I got it!
Bunch of thanks for your help!:biggrin:
 

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