Problem 8 Show that for each integer $m$... [FREE SOLUTION]

91影视

Elementary Number Theory

Gareth A. Jones, Josephine M. Jones

$Math Studyset 91影视 Explanations$ Math

1998 Edition

Chapter 5: Problem 8

Show that for each integer $m$, there are only finitely many integers $n$ such that $\phi(n)=m$.

Short Answer

Expert verified

For each integer $ m $, there are finitely many integers $ n $ with $ \phi(n)=m $ because $ n \leq 2m $.

Step by step solution

Understand Euler's Totient Function

The totient function $ \phi(n) $ counts the number of integers up to $ n $ that are coprime with $ n $. For example, if $ n = 1 $, $ \phi(n) = 1 $, if $ n = 2 $, $ \phi(n) = 1 $, and so on.

Express Euler's Totient Function Formula

Euler's totient function for $ n $, where $ n $ has prime factorization $ n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} $, is given by: \[ \phi(n) = n \left(1 - \frac{1}{p_1}\right) \left(1 - \frac{1}{p_2}\right) \cdots \left(1 - \frac{1}{p_r}\right). \] This expresses $ \phi(n) $ in terms of $ n $'s prime factors.

Set the Condition $ \phi(n) = m $

To solve $ \phi(n) = m $, the equation implies finding values of $ n $ so that when substituted in $ \phi(n) $, results in $ m $.

Bound $ n $ with Inequality

Rearranging the inequality from $ \phi(n) = n \left(1 - \frac{1}{p_1}\right) \cdots \left(1 - \frac{1}{p_r}\right) = m $ shows that $ n > m $. More specifically, $ n/\phi(n) = \prod_i (1 - \frac{1}{p_i})^{-1} > 1.5 $, thus $ n \leq 2m $.

Conclude there are Finite $ n $ for $ m $

Since $ n $ must satisfy that it is less than or equal to $ 2m $, and greater than $ m $, there are only finitely many $ n $ that satisfy this inequality. This range puts a maximum cap on possible values of $ n $, showing that only finitely many integers $ n $ satisfy $ \phi(n) = m $ for each integer $ m $.

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Key Concepts

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

Integer Solutions

An **integer solution** refers to finding whole number values that satisfy an equation. In the context of Euler's Totient Function, we are looking for integer values of $ n $ such that $ \phi(n) = m $. Here are some key points to understand this concept better:

An integer solution is sought when the equation has to be satisfied with integers.
Finding integer solutions often involves strategic manipulation of equations and inequalities.
In our exercise, the integer solution comes from limiting the range of possible values of $ n $ using inequalities and prime factorization of the integer $ n $.

Thus, understanding how to establish and solve for integer solutions becomes fundamental to answering such problems.

Integer Inequalities

**Integer inequalities** involve expressions where we compare integers using inequality symbols like <, >, 鈮�, and 鈮�. In our exercise, dealing with inequalities helps to determine the possible range for $ n $ where $ \phi(n) = m $. Let's break this down:

We start by rearranging the expression for Euler's Totient Function, finding that $ n > m $.
Further manipulation leads to $ n \leq 2m $. This implies our integer solution must satisfy this dual inequality $ m < n \leq 2m $.
Inequalities like these help to limit the search space, ensuring exploration is only within the feasible values of $ n $.

By utilizing integer inequalities, it's clear that there is a finite number of solutions for $ n $, as you're working between a defined upper and lower bound.

Prime Factorization

The concept of **prime factorization** involves expressing an integer as a product of its prime factors. Euler's Totient Function critically depends on the prime factorization of $ n $. Here's how it plays a role:

Euler's Totient Function $ \phi(n) $ is defined using the prime factors of $ n $, expressed as $ n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} $.
With prime factorization, $ \phi(n) $ can be expressed as \[ \phi(n) = n \left(1 - \frac{1}{p_1}\right) \left(1 - \frac{1}{p_2}\right) \cdots \left(1 - \frac{1}{p_r}\right). \]
This factorization allows us to see how $ n $ is structured, which facilitates assessing its range using inequalities.

Understanding prime factorization aids in evaluating $ \phi(n) $ efficiently and comprehending why there are only finitely many integers that satisfy $ \phi(n) = m $ for any given $ m $.

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91影视

Show that for each integer \(m\), there are only finitely many integers \(n\) such that \(\phi(n)=m\).

Short Answer

Step by step solution

Understand Euler's Totient Function

Express Euler's Totient Function Formula

Set the Condition \( \phi(n) = m \)

Bound \( n \) with Inequality

Conclude there are Finite \( n \) for \( m \)

Key Concepts

Integer Solutions

Integer Inequalities

Prime Factorization

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