Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 1G / Solution

An arithmetic function $f$ is multiplicative if $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime integers. If $g(n)={\sum }_{d\mid n}f(d)$ and $(m,n)=1$, every positive integer divisor of $mn$ is uniquely $d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Henceso $g$ is multiplicative.

The Möbius function is defined by $\mu (1)=1$, $\mu (n)=0$ if the square of a prime number divides $n$, and $\mu (n)=(-1{)}^r$ if $n$ is a product of $r$ distinct prime numbers. Coprime integers have disjoint sets of prime factors, so this definition immediately gives $\mu (mn)=\mu (m)\mu (n)$ when $(m,n)=1$.

For $n=p_1^{a_1}\cdots p_r^{a_r}>1$, only the square-free divisors contribute, and thereforeFor $n=1$ the sum is $1$. Thus $1\ast \mu =\epsilon$ in the language of Dirichlet convolution. Convolving $g=1\ast f$ with $\mu$ gives the Möbius inversion formula

If $f(n)=n$, its divisor sum is the sum-of-divisors function. Conversely, if $g(n)=n$, Möbius inversion givesusing the standard divisor identity for the Euler totient function. Therefore the two requested answers are