Show Primitive Roots Cannot be Perfect Cubes Modulo Prime p

In summary, the conversation discusses the relationship between perfect cubes and primitive roots mod p, where p is a prime congruent to 1 mod 3. It is shown that a cannot be a primitive root mod p by proving that (a^3)^u is congruent to 1 mod p, where u is an integer less than p-1. The use of Henzel's lemma, which is discussed in Lang and Milne's course notes, is also mentioned as being relevant to the topic. However, it is stated that the method used by "robert" is sufficient without needing to use Hensel's lemma.
  • #1
nidak
1
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
  • #2
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
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.
 

1. What does it mean for a primitive root to be a perfect cube modulo a prime number?

A primitive root is a number that, when taken to different powers, generates all possible remainders modulo a prime number. A perfect cube is a number that can be written as the cube of another number. Therefore, a show primitive root cannot be a perfect cube modulo a prime number means that there is no number that, when taken to different powers, can generate all possible remainders as cubes modulo that prime number.

2. Why is it important to prove that show primitive roots cannot be perfect cubes modulo a prime number?

Proving that show primitive roots cannot be perfect cubes modulo a prime number is important because it helps us understand the properties of primitive roots and the behavior of numbers modulo a prime. It also has implications in fields such as cryptography, where the use of prime numbers is crucial.

3. Can a non-prime number have a primitive root that is a perfect cube modulo that number?

No, a non-prime number cannot have a primitive root that is a perfect cube modulo that number. This is because a non-prime number has multiple factors, and a perfect cube can only have one possible factorization. Therefore, it is not possible for a non-prime number to have a primitive root that is a perfect cube modulo that number.

4. How does the proof that show primitive roots cannot be perfect cubes modulo a prime number work?

The proof for this statement uses a theorem from number theory called the index theorem. This theorem states that if a number is a perfect kth power modulo a prime, then its index (the smallest number that when raised to the kth power gives the number) is also a perfect kth power modulo that prime. By assuming that a show primitive root is a perfect cube modulo a prime, and using the index theorem, it can be shown that the index must also be a perfect cube modulo that prime, leading to a contradiction.

5. Are there any exceptions to the statement that show primitive roots cannot be perfect cubes modulo a prime number?

No, there are no exceptions to this statement. It has been proven that for any prime number, there is no primitive root that is a perfect cube modulo that prime. However, there may be numbers that are not primitive roots but are perfect cubes modulo a prime. These numbers are known as pseudoprimes.

Similar threads

  • Linear and Abstract Algebra
Replies
2
Views
792
Replies
5
Views
895
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
961
  • Linear and Abstract Algebra
Replies
5
Views
1K
  • Precalculus Mathematics Homework Help
Replies
17
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
654
  • Precalculus Mathematics Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
16
Views
2K
Back
Top