### Proving the Non-Existence of Primitive Roots Modulo 954271**Problem Statement:**Prove that there are no primitive roots modulo $ 954271 = 691 \cdot 1381 $ (691 and 1381 are prime). Do this directly (i.e. without referring to the Primitive Root Theorem or any related results).**Detailed Explanation:**To tackle this problem, a direct proof method needs to be applied. This involves showing that no integer can generate all the units in the multiplicative group modulo $ 954271 $. Let's break down the essential steps required for this proof:1. **Structure of the Multiplicative Group:** - The modulus $ 954271 $ is the product of two primes, 691 and 1381. - The prime factorization of $ 954271 $ is important because it helps in understanding the structure of the group of units modulo $ 954271 $. - The multiplicative group modulo $ 954271 $ has order $(691 - 1)(1381 - 1) = 690 \cdot 1380 = 952200$, which is derived from Euler's totient function $ \phi(954271) $.2. **Properties of Primitive Roots:** - A primitive root modulo $ n $ is an integer $ g $ such that the smallest positive power of $ g $ that is congruent to 1 modulo $ n $ is exactly $ \phi(n) $. - In other words, $ g $ must be a generator of the entire multiplicative group of units.3. **Direct Computation:** - Consider any candidate integer $ g $ for being a primitive root modulo $ 954271 $. - Calculate the order of $ g $ modulo 691 and modulo 1381 separately. - Use the Chinese Remainder Theorem to combine these results and determine if $ g $ can generate $ 952200 $ units.4. **Verification:** - For $ g $ to be a primitive root modulo $ 954271 $, its order modulo 691 and 1381 must be $ 690 $ and $ 1380 $ respectively. - Because $ 954271 = 691 \times 1381 $, if there exists a primitive root $ g $ modulo \(

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Prove that there are no primitive roots modulo 954271 = 691 · 1381 (691 and 1381 are prime). Do this directly (i.e. without referring to the Primitive Root Theorem or any related results).

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

### Proving the Non-Existence of Primitive Roots Modulo 954271

**Problem Statement:**
Prove that there are no primitive roots modulo $ 954271 = 691 \cdot 1381 $ (691 and 1381 are prime). Do this directly (i.e. without referring to the Primitive Root Theorem or any related results).

**Detailed Explanation:**

To tackle this problem, a direct proof method needs to be applied. This involves showing that no integer can generate all the units in the multiplicative group modulo $ 954271 $. Let's break down the essential steps required for this proof:

1. **Structure of the Multiplicative Group:**
- The modulus $ 954271 $ is the product of two primes, 691 and 1381.
- The prime factorization of $ 954271 $ is important because it helps in understanding the structure of the group of units modulo $ 954271 $.
- The multiplicative group modulo $ 954271 $ has order $(691 - 1)(1381 - 1) = 690 \cdot 1380 = 952200$, which is derived from Euler's totient function $ \phi(954271) $.

2. **Properties of Primitive Roots:**
- A primitive root modulo $ n $ is an integer $ g $ such that the smallest positive power of $ g $ that is congruent to 1 modulo $ n $ is exactly $ \phi(n) $.
- In other words, $ g $ must be a generator of the entire multiplicative group of units.

3. **Direct Computation:**
- Consider any candidate integer $ g $ for being a primitive root modulo $ 954271 $.
- Calculate the order of $ g $ modulo 691 and modulo 1381 separately.
- Use the Chinese Remainder Theorem to combine these results and determine if $ g $ can generate $ 952200 $ units.

4. **Verification:**
- For $ g $ to be a primitive root modulo $ 954271 $, its order modulo 691 and 1381 must be $ 690 $ and $ 1380 $ respectively.
- Because $ 954271 = 691 \times 1381 $, if there exists a primitive root $ g $ modulo \(

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