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Chapter 6: Problem 18

Let \(f\) and \(g\) be multiplicative functions that are not identically zero and such that \(f\left(p^{k}\right)=\) \(g\left(p^{i}\right)\) for each prime \(p\) and \(k \geq 1\). Prove that \(f=g\).

Short Answer

Expert verified
The multiplicative functions \(f\) and \(g\) are equal, \(f = g\).

Step by step solution

01

Understand the Problem

We need to show that two multiplicative functions \(f\) and \(g\) are equal. The given condition states that for every prime \(p\) and integer \(k \geq 1\), the values \(f(p^k) = g(p^i)\) are equal for some prime power \(p^i\).
02

Use the Property of Multiplicative Functions

Given that \(f\) and \(g\) are multiplicative, for any coprime integers \(a\) and \(b\), \(f(ab) = f(a)f(b)\) and \(g(ab) = g(a)g(b)\). This property will help us analyze the functions on arbitrary integers.
03

Analyze Prime Powers

For any prime \(p\), \(f(p^k) = g(p^i)\). By choosing specific prime powers, particularly when \(i = k\), we get \(f(p^k) = g(p^k)\). This equality must hold for all \(k\).
04

Extend to Arbitrary Integers

Since both \(f\) and \(g\) satisfy \(f(p^k) = g(p^k)\) for all \(k \geq 1\) and are multiplicative, we can extend this to any integer \(n\) with prime factorization \(n = p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m}\). Thus, \(f(n) = \prod f(p_i^{k_i}) = \prod g(p_i^{k_i}) = g(n)\).
05

Conclude the Functions are Equal

Therefore, since \(f(n) = g(n)\) for every integer \(n\), the multiplicative functions \(f\) and \(g\) must be equal.

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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 all about expressing a number as a product of prime numbers. It means breaking down any integer into a set of prime numbers that multiply together to give the original number. For example, let's take 30. The prime factorization of 30 is 2, 3, and 5 because:
  • 2 脳 3 脳 5 = 30
Prime numbers like 2, 3, and 5 are the building blocks for all integers. That's why prime factorization is so useful, especially when dealing with multiplicative functions. Once you know the primes and their powers, you can reconstruct numbers easily. This property of numbers is fundamental because it gives us a structured way to look at integers, making it easier to analyze complex problems involving multiplicative functions.
Properties of Multiplicative Functions
Multiplicative functions are special kinds of functions used mainly in number theory. These functions have a unique characteristic: if you have two numbers that are coprime (they have no common factors except 1), the function's value at their product is the product of the function's values at those numbers. In mathematical language, if you have a multiplicative function \( f \) and two coprime integers \( a \) and \( b \), then:
  • \( f(ab) = f(a) \times f(b) \)
This property allows us to 'build up' the function for larger numbers by knowing its values for lower powers of primes. It makes them predictable when dealing with prime factorization. Understanding this property helps you work with multiplicative functions more effectively and solve equations like in the original exercise. By leveraging this property, we can deduce that if two multiplicative functions agree on the powers of a prime, they will agree on all whole numbers.
Equality of Functions
Equality of functions means that two functions produce the same output for every possible input. With multiplicative functions, if you can show they agree for all powers of any prime number, and they are both multiplicative, then they must be equal everywhere. Here's why:
  • If for every prime \( p \) and an integer \( k \), \( f(p^k) = g(p^k) \), then by multiplicative property, for any integer with prime factorization \( n = p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m} \), the two functions agree.
That's the whole idea behind proving functions are equal in exercises like this. The functions follow clear rules that, once satisfied for prime powers, are satisfied for all integers. By analyzing the functions at these elemental levels (prime factors), you ensure that they are equal across the board.

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Most popular questions from this chapter

The Mangoldt function \(\Lambda\) is defined by $$ \Lambda(n)=\left\\{\begin{array}{ll} \log p & \text { if } n=p^{k}, \text { where } p \text { is a prime and } k \geq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ Prove that \(\Lambda(n)=\sum_{d \mid n} \mu(n / d) \log d=-\sum_{d \mid n} \mu(d) \log d\). [Hint: First show that \(\sum_{d \mid n} \Lambda(d)=\log n\) and then apply the M枚bius inversion formula.]

Find the highest power of 5 dividing \(1000 !\) and the highest power of 7 dividing \(2000 !\)

For any integer \(n \geq 1\), establish the inequality \(\tau(n) \leq 2 \sqrt{n}\). [Hint: If \(d \mid n\), then one of \(d\) or \(n / d\) is less than or equal to \(\sqrt{n}\).]

For any positive integer \(n\), show the following: (a) \(\sum_{d \mid n} \sigma(d)=\sum_{d \mid n}(n / d) \tau(d)\) (b) \(\sum_{d \mid n}(n / d) \sigma(d)=\sum_{d \mid n} d \tau(d)\). [Hint: Because the functions $$ F(n)=\sum_{d \mid n} \sigma(d) \quad \text { and } \quad G(n)=\sum_{d \mid n} \frac{n}{d} \tau(d) $$ are both multiplicative, it suffices to prove that \(F\left(p^{k}\right)=G\left(p^{k}\right)\) for any prime \(p\).]

For any positive integer \(n\), prove that \(\sum_{d \mid n} \tau(d)^{3}=\left(\sum_{d \mid n} \tau(d)\right)^{2} .\) [Hint: Both sides of the equation in question are multiplicative functions of \(n\), so that it suffices to consider the case \(n=p^{k}\), where \(p\) is a prime.]
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