Show Primitive Roots Cannot be Perfect Cubes Modulo Prime p

[nidak]

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

 

[robert Ihnot]

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.

 

[LorenzoMath]

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

 

[Kummer]

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.