Primitive roots - annoying problem

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    Primitive Roots
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Discussion Overview

The discussion revolves around the properties of primitive roots, specifically examining the claim that if \( p \equiv 1 \mod 4 \), then \(-r\) is also a primitive root of a prime number \( p \geq 3\) given that \( r \) is a primitive root of \( p \). The scope includes mathematical reasoning and exploration of the properties of primitive roots.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant states that if \( r \) is a primitive root, then \( r^{(p-1)/2} \) is not equal to 1, raising the question of the behavior of \(-r\).
  • Another participant suggests that since \( r^{(p-1)/2} \equiv -1 \mod p \), it follows that \((-r)^{(p-1)/2}\) must also yield the same result due to the even exponent.
  • A further exploration is made regarding the order of \(-r\), proposing that if the order were lower than \( p-1 \), it would lead to a contradiction with \( (-r)^{(p-1)/2} \equiv 1 \mod p \).
  • A participant confirms that the reasoning appears sound and prompts further investigation into the necessary power to raise \( r^2 \) to 1.

Areas of Agreement / Disagreement

Participants express varying levels of confidence in their reasoning, but there is no explicit consensus on the proof or resolution of the original claim regarding \(-r\) being a primitive root.

Contextual Notes

There are unresolved assumptions regarding the properties of orders of elements in modular arithmetic and the implications of the conditions set by the prime \( p \).

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

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