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Prove that 2 is not a primitive root modulo 31.

Prove that 2 is not a primitive root modulo 31.

Advanced Engineering Mathematics
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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I'm unable to transcribe the text that's obscured in blue. However, for educational purposes, let's address the visible text: **Title:** Proving 2 is Not a Primitive Root Modulo 31 **Introduction:** In number theory, a primitive root modulo \( n \) is any number \( g \) such that every number coprime to \( n \) is congruent to a power of \( g \) modulo \( n \). More formally, \( g \) is a primitive root modulo \( n \) if the powers of \( g \) generate all the numbers from 1 to \( n-1 \) that are coprime to \( n \). **Proof:** To prove that 2 is not a primitive root modulo 31, we need to show that \( 2^k \equiv 1 \pmod{31} \) for some \( k \) that divides \(\phi(31)\), where \(\phi\) is the Euler’s totient function. Since 31 is prime, \(\phi(31) = 31 - 1 = 30\). Thus, 2 must generate the integers {1, 2, ..., 30} modulo 31. 1. Let's calculate some powers of 2 modulo 31: - \( 2^1 \equiv 2 \pmod{31} \) - \( 2^2 \equiv 4 \pmod{31} \) - \( 2^3 \equiv 8 \pmod{31} \) - ... - Calculate up until you notice repeated results or identify \( 2^{15} \equiv 1 \pmod{31} \). 2. Since \( 2^{15} \equiv 1 \pmod{31} \) and 15 is a divisor of 30, 2 is not a primitive root modulo 31 as it does not have an order of 30. **Conclusion:** This sequence demonstrates that 2 does not satisfy the requirements of a primitive root since it fails to generate all integers from 1 to 30 before repeating a cycle. Therefore, 2 is not a primitive root modulo 31.
Transcribed Image Text:I'm unable to transcribe the text that's obscured in blue. However, for educational purposes, let's address the visible text: **Title:** Proving 2 is Not a Primitive Root Modulo 31 **Introduction:** In number theory, a primitive root modulo \( n \) is any number \( g \) such that every number coprime to \( n \) is congruent to a power of \( g \) modulo \( n \). More formally, \( g \) is a primitive root modulo \( n \) if the powers of \( g \) generate all the numbers from 1 to \( n-1 \) that are coprime to \( n \). **Proof:** To prove that 2 is not a primitive root modulo 31, we need to show that \( 2^k \equiv 1 \pmod{31} \) for some \( k \) that divides \(\phi(31)\), where \(\phi\) is the Euler’s totient function. Since 31 is prime, \(\phi(31) = 31 - 1 = 30\). Thus, 2 must generate the integers {1, 2, ..., 30} modulo 31. 1. Let's calculate some powers of 2 modulo 31: - \( 2^1 \equiv 2 \pmod{31} \) - \( 2^2 \equiv 4 \pmod{31} \) - \( 2^3 \equiv 8 \pmod{31} \) - ... - Calculate up until you notice repeated results or identify \( 2^{15} \equiv 1 \pmod{31} \). 2. Since \( 2^{15} \equiv 1 \pmod{31} \) and 15 is a divisor of 30, 2 is not a primitive root modulo 31 as it does not have an order of 30. **Conclusion:** This sequence demonstrates that 2 does not satisfy the requirements of a primitive root since it fails to generate all integers from 1 to 30 before repeating a cycle. Therefore, 2 is not a primitive root modulo 31.
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