Show Primitive Roots Cannot be Perfect Cubes Modulo Prime p

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SUMMARY

Primitive roots cannot be perfect cubes modulo a prime \( p \) when \( p \equiv 1 \mod 3 \). Specifically, if \( a = n^3 \) for some integer \( n \), then \( e_p(a) \) does not equal \( p - 1 \). This conclusion arises from the fact that for any primitive root \( r \), \( r^{(p-1)} \equiv 1 \mod p \), but \( (p-1)/3 = u \) is an integer less than \( p - 1 \), leading to \( (a^3)^u \equiv 1 \mod p \). Hensel's lemma is referenced as relevant, but its application is deemed unnecessary in this context.

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  • Understanding of primitive roots in modular arithmetic
  • Familiarity with properties of prime numbers, specifically \( p \equiv 1 \mod 3 \)
  • Knowledge of modular exponentiation and its implications
  • Basic comprehension of Hensel's lemma and its applications
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  • Study the properties of primitive roots in modular arithmetic
  • Research the implications of \( p \equiv 1 \mod 3 \) on number theory
  • Learn about Hensel's lemma and its applications in number theory
  • Explore examples of perfect cubes and their behavior in modular systems
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Mathematicians, number theorists, and students studying modular arithmetic and primitive roots, particularly those interested in the properties of primes and their implications in number theory.

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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
 
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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.
 
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Look up Henzel's lemma in Lang or Milne's online course note. It is relavant.
 
LorenzoMath said:
Look up Henzel's lemma in Lang or Milne's online course note. It is relavant.

No need to use Hensel's lemma, what robert did is perfect.
 

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