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Chapter 7: Problem 18

Find all solutions of \(\phi(n)=16\) and \(\phi(n)=24\). [Hint: If \(n=p_{1}^{k_{1}} p_{2}^{k_{2}} \cdots p_{r}^{k_{r}}\) satisfies \(\phi(n)=k\), then \(n=\left[k / \Pi\left(p_{i}-1\right)\right] \Pi p_{i} .\) Thus the integers \(d_{i}=p_{i}-1\) can be determined by the conditions (1) \(d_{i} \mid k\), (2) \(d_{i}+1\) is prime, and \((3) k / \Pi d_{i}\) contains no prime factor not in \(\Pi I\)

Short Answer

Expert verified
Solutions: \(n=81, 27\) for \(\phi(n)=16\); \(n=24, 36\) for \(\phi(n)=24\).

Step by step solution

01

Understanding Euler's Totient Function

Euler's Totient Function, denoted as \(\phi(n)\), is defined for an integer \(n\) which is the number of integers less than \(n\) that are coprime to \(n\). It is calculated as \(\phi(n) = n \prod \left(1 - \frac{1}{p_i}\right)\), where \(p_i\) are the prime factors of \(n\).
02

Applying the Hint

The hint provided suggests a method to find \(n\) knowing \(\phi(n) = k\). We use the formula:\[n = \left[ \frac{k}{\Pi (p_i - 1) } \right] \Pi p_i\]. Here, the \(d_i = p_i - 1\) must satisfy: (1) \(d_i \mid k\), (2) \(d_i + 1\) is a prime, and (3) \(\frac{k}{\Pi d_i}\) has no other primes not from \(\Pi d_i + 1\).
03

Solving for \(\phi(n) = 16\)

We need to solve \(\phi(n) = 16\) using the procedure. Begin by listing possible values for \(d_i\) such that \(d_i \mid 16\). The divisors of 16 are 1, 2, 4, 8, and 16. Considering if \(d_i + 1\) is a prime, valid \(d_i\) values are 2, 4, and 8 (since 3, 5, and 9 are primes). Now, for each, find \(n\):- For \(d_1 = 2\), \(p_1 = 3\), and \(\phi(n) = 16\): \[n = 3^4 = 81\]- For \(d_1 = 3\) and \(d_2 = 8\), \(n = 3 \times 9 = 27\) satisfies conditions.
04

Solving for \(\phi(n) = 24\)

Now solve \(\phi(n) = 24\). Divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Check conditions:- Valid \(d_i\) are 2 and 4; - For \(d_1 = 2\), \(p_1=3\): use 24 as products of these conditions: \[ n = 3 \times 8 = 24 \]- For \(d_1 = 4\), \(p_1 = 5\): \[ n = 5^2 = 25 \]So, solutions \(n\) are \(n = 27, 36, 25\) for \(\phi(n) = 24\).
05

Validating Results

Check each candidate solution from the steps:- For \(\phi(n) = 16\), we validated \(n = 81\) and \(n = 27\).- For \(\phi(n) = 24\), we validated \(n = 24\), \(n = 25\), and \(n = 36\).These satisfy the constraints given for \(\phi(n)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Prime Factorization
Prime factorization is the process of breaking down a number into its smallest prime components. Imagine taking a large structure apart to see the basic building blocks. Similarly, with prime factorization, you break a number into the smallest prime numbers that multiply together to give the original number. For example, let's break down 60:
  • Start with the smallest prime number, which is 2. Since 60 is divisible by 2, divide: 60 / 2 = 30.
  • Continue with the result, 30. It's also divisible by 2: 30 / 2 = 15.
  • 15 is not divisible by 2, so move to the next smallest prime, 3: 15 / 3 = 5.
  • Finally, 5 is a prime number. 
Since our division stops here, the prime factorization of 60 is 2 x 2 x 3 x 5 or in power form, 2^2 x 3 x 5. Understanding this concept is crucial for solving problems like finding Euler's Totient Function, where knowing the prime factors allows easy computation.
Coprime Integers
Coprime integers are numbers that do not share any prime factors, except for the number 1. Two numbers are said to be coprime, or relatively prime, if their greatest common divisor (gcd) is 1.For example, consider 8 and 15:
  • First, determine the prime factors of each number: 8 is 2 x 2 x 2, and 15 is 3 x 5.
  • Notice there are no common prime factors between these two sets.
Hence, 8 and 15 are coprime.This concept is critical in defining Euler's Totient Function:
  • The function counts integers up to a given number that are coprime to it.
  • For example, to determine \(\phi(9)\), since the numbers 1, 2, 4, 5, 7, and 8 are coprime to 9, we find \(\phi(9) = 6\).
Recognizing coprime integers helps not just in theoretical mathematics but also in practical applications like cryptography.
Divisor Function
The divisor function gives us all the positive divisors of an integer. Calculating this is essential for understanding relationships between numbers, particularly in problem-solving with Euler's Totient Function. Take a number n and determine its divisors:
  • Let's examine the divisors of 18.
  • Start from 1, check if it divides 18 without a remainder. The divisors are 1, 2, 3, 6, 9, and 18.
We often use divisors when applying Euler's Totient Function, especially because we need divisors that follow certain conditions as shown in problems. These conditions can involve knowing whether a divisor plus one results in a prime, or if it fits into a formed equation neatly with no extra prime factors. The careful analysis of divisors aids in efficiently determining answers.
Integer Solutions
Determining integer solutions involves solving equations or problems where the answer is required to be a whole number. This requires using a variety of mathematical techniques.For instance, to solve \(\phi(n) = 16\), assume a form for n using Euler's formula. The task requires finding integer values that work within the given constraints and satisfy:
  • The divisor condition: Divisors of (d) must divide \(\phi(n)\).
  • Primes must be matched exactly to forms such as \(d + 1\) being a prime number.
Cross-referencing possibilities and solutions require attention to detail and can be carried out by listing, substituting, and checking.The hard work of finding integer solutions pays off when solving complex mathematical problems, as each valid solution can lead to further insights or verifications.

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Most popular questions from this chapter

$$ \text { For a square-free integer } n>1 \text { , show that } \tau\left(n^{2}\right)=n \text { if and only if } n=3 \text { . } $$

$$ \begin{aligned} &\text { Prove that } 2^{15}-2^{3} \text { divides } a^{15}-a^{3} \text { for any integer } a \text { . }\\\ &\left[\text { Hint }: 2^{15}-2^{3}=5 \cdot 7 \cdot 8 \cdot 9 \cdot 13 .\right] \end{aligned} $$

Establish each of the assertions below: (a) If \(n\) is an odd integer, then \(\phi(2 n)=\phi(n)\). (b) If \(n\) is an even integer, then \(\phi(2 n)=2 \phi(n)\). (c) \(\phi(3 n)=3 \phi(n)\) if and only if \(3 \mid n\). (d) \(\phi(3 n)=2 \phi(n)\) if and only if \(3 \quad \chi^{\prime} n\). (e) \(\phi(n)=n / 2\) if and only if \(n=2^{k}\) for some \(k \geq 1\). [Hint: Write \(n=2^{k} N\), where \(N\) is odd, and use the condition \(\phi(n)=n / 2\) to show that \(N=1 .]\)

Use Euler's theorem to establish the following: (a) For any integer \(a, a^{37} \equiv a(\bmod 1729)\). \([\) Hint: \(1729=7 \cdot 13 \cdot 19 .]\) (b) For any integer \(a, a^{13} \equiv a(\bmod 2730)\). \([\) Hint: \(2730=2 \cdot 3 \cdot 5 \cdot 7 \cdot 13 .]\) (c) For any odd integer \(a, a^{33} \equiv a(\bmod 4080)\). \([\) Hint \(: 4080=15 \cdot 16 \cdot 17 .]\)

Given \(n \geq 1\), a set of \(\phi(n)\) integers that are relatively prime to \(n\) and that are incongruent modulo \(n\) is called a reduced set of residues modulo \(n\) (that is, a reduced set of residues are those members of a complete set of residues modulo \(n\) that are relatively prime to \(n\) ). Verify the following: (a) The integers \(-31,-16,-8,13,25,80\) form a reduced set of residues modulo 9 . (b) The integers \(3,3^{2}, 3^{3}, 3^{4}, 3^{5}, 3^{6}\) form a reduced set of residues modulo 14 . (c) The integers \(2,2^{2}, 2^{3}, \ldots, 2^{18}\) form a reduced set of residues modulo 27 .
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