Primitive roots - annoying problem

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The discussion centers on proving that if r is a primitive root of a prime number p ≥ 3 and p ≡ 1 (mod 4), then -r is also a primitive root of p. Participants confirm that if r is a primitive root, then r^[(p-1)/2] is -1 (mod p), leading to the conclusion that (-r)^[(p-1)/2] remains consistent due to the even exponent. The conversation emphasizes the importance of understanding the order of elements in modular arithmetic to validate the properties of primitive roots.

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  • Understanding of primitive roots in number theory
  • Familiarity with modular arithmetic and congruences
  • Knowledge of the properties of prime numbers
  • Basic concepts of group theory related to orders of elements
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  • Study the properties of primitive roots and their applications in cryptography
  • Learn about modular exponentiation and its computational techniques
  • Explore the implications of the Legendre symbol in number theory
  • Investigate the structure of multiplicative groups of integers modulo p
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This discussion is beneficial for mathematicians, number theorists, and students studying abstract algebra, particularly those interested in the properties of primitive roots and modular arithmetic.

RichardCypher
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Let r be a primitive root of a prime number p \geq 3. Prove that if p \equiv 1 (mod 4), then -r is also a primitive root of p.

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 \neq 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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