Primitive roots.

  • Thread starter nidak
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  • #1
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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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  • #2
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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.
 
Last edited:
  • #3
Look up Henzel's lemma in Lang or Milne's online course note. It is relavant.
 
  • #4
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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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