# Primitive roots.

1. Mar 6, 2007

### 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

2. Mar 9, 2007

### 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.

Last edited: Mar 9, 2007
3. Jul 3, 2007

### LorenzoMath

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

4. Jul 3, 2007

### Kummer

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