Show Primitive Roots Cannot be Perfect Cubes Modulo Prime p

  • Context: Graduate 
  • Thread starter Thread starter nidak
  • Start date Start date
  • Tags Tags
    Primitive Roots
Click For Summary

Discussion Overview

The discussion centers on the question of whether a perfect cube can be a primitive root modulo a prime \( p \) where \( p \equiv 1 \mod 3 \). The scope includes mathematical reasoning and theoretical exploration of number theory concepts.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant posits that if \( a \) is a perfect cube, expressed as \( a = n^3 \) for some integer \( n \), then it cannot be a primitive root modulo \( p \) because \( e_p(a) \) is not equal to \( p - 1 \).
  • Another participant notes that for any primitive root \( r \), it holds that \( r^{p-1} \equiv 1 \mod p \), but this does not apply to any lesser power. They introduce the integer \( u = (p-1)/3 \) and suggest that \( (a^3)^u \equiv 1 \mod p \).
  • Several participants reference Hensel's lemma as relevant to the discussion, with one asserting that it is unnecessary for the current argument, indicating that a previous contribution is sufficient.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of Hensel's lemma and the validity of the claims regarding perfect cubes and primitive roots, indicating that the discussion remains unresolved.

Contextual Notes

There are references to specific mathematical properties and theorems that may require further clarification or assumptions, particularly regarding the implications of \( p \equiv 1 \mod 3 \) and the nature of primitive roots.

nidak
Messages
1
Reaction score
0
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
 
Physics news on Phys.org
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:
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.
 

Similar threads

Undergrad Question on Primitive Roots and GCDs
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad How to show ##p(x)=g(x)x\pm 1\in\Bbb{Q}[x]## is irreducible in ##\Bbb{Q}_{\Bbb{Z}}[x]##?
  • · Replies 48 ·
2
Replies
48
Views
6K
MHB Number of natural numbers that have primitive roots
  • · Replies 5 ·
Replies
5
Views
2K
MHB Is There a Strategy for Finding Primitive Roots of 2p^n If There is One for p^n?
  • · Replies 3 ·
Replies
3
Views
2K
Graduate Relationship between primitive roots of a prime
  • · Replies 4 ·
Replies
4
Views
4K
Graduate Artin's Conjecture on Primitive Roots: Perfect Squares
  • · Replies 2 ·
Replies
2
Views
6K
Undergrad Localising a non integral domain at a prime
  • · Replies 17 ·
Replies
17
Views
3K
Proving Primitive Roots of Odd Numbers Modulo pm
  • · Replies 3 ·
Replies
3
Views
2K
Graduate Is α a Primitive Element Modulo p Given α^q ≡ -1 mod p?
  • · Replies 4 ·
Replies
4
Views
6K
Undergrad I think I've hit upon a very easy proof of impossibility of quintic
  • · Replies 2 ·
Replies
2
Views
2K