Abstract: This article delves into the intricacies of a specific mathematical problem: determining the sum of (mu(d)phi(d)) over all divisors (d) of a given integer (n). We will explore two different approaches to solve this problem, one by John Calligy and another by Brian Sittinger. By understanding these methods, we can gain insight into the behavior of multiplicative functions in number theory.

Introduction

The question at hand revolves around an integer (n) with (p_1, p_2, ldots, p_k) denoting its odd prime divisors. This paper aims to provide a detailed explanation of the sum of (mu(d)phi(d)) over all divisors (d) of (n). We will elaborate on the approaches given by John Calligy and Brian Sittinger, providing a comprehensive understanding of the problem and its solutions.

Approach by John Calligy

1. Problem Formulation: Let (p_1, p_2, ldots, p_k) be the odd prime divisors of (n). If the list is empty, (n) is a power of 2 because 2 must divide (n).

2. Contribution Analysis: If (m^2 mid d), all terms contributing to [sum_{d mid n} mu(d)phi(d)] come from products of distinct primes and the empty product, i.e., (d1). Thus, (d) is formed from elements of ({2, p_1, p_2, ldots, p_k}).

3. Pairwise Contributions: Partition the (2^{k 1}) subsets of this collection into pairs (S cup {2}), where (S) is a subset of ({p_1, p_2, ldots, p_k}). For a typical set (S), the contribution to the sum is:

[mu(d)phi(d)mu(2d)phi(2d) 0], since (mu(2d) -mu(d)) and (phi(2d) phi(d)).

Each subset (S) contributes 0 to the sum, leading to the desired sum being 0.

Approach by Brian Sittinger

1. One-to-One Correspondence: There is a one-to-one correspondence between odd squarefree divisors of (n) and even squarefree divisors of (n). For an odd divisor (d), we have:

[mu(2d)phi(2d) -mu(d)phi(d)].

2. Multiplicative Property: Since (mu) and (phi) are multiplicative functions, their product is also multiplicative. Hence, the sum in question is multiplicative. We can write:

[sum_{d mid n} mu(d)phi(d) prod_{p text{ prime}} sum_{d mid p^k, p^k parallel n} mu(d)phi(d) prod_{p text{ prime}} sum_{d mid p} mu(d)phi(d)], where (mu^p 0) for (j geq 2).

3. Simplification: Since (mu(p) -1) and (phi(p) p-1), we get:

[sum_{d mid n} mu(d)phi(d) prod_{p text{ prime}} (1 - frac{p-1}{p}) prod_{p text{ prime}} 1 - 1 frac{1}{p}].

If (n) is even, (2 mid n), the formula reduces to:

[sum_{d mid n} mu(d)phi(d) 0].

Conclusion

The methods presented by John Calligy and Brian Sittinger provide comprehensive insights into the behavior of multiplicative functions over divisors of integers. By leveraging the properties of odd and even squarefree divisors, and the behavior of (mu) and (phi) over these divisors, we can effectively solve the problem and gain a deeper understanding of number theory.