Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 11H (Number Theory)

An arithmetic function $f$ is multiplicative when $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime. If $m$ and $n$ are coprime, every divisor $d$ of $mn$ has a unique factorization $d=d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Hence the Dirichlet convolution of multiplicative functions satisfiesand $(f\star g)(1)=1$, so it is multiplicative.

Let $\epsilon$ be the identity for Dirichlet convolution, with $\epsilon (1)=1$ and $\epsilon (n)=0$ for $n>1$. The Möbius function satisfies $\mu \star 1=\epsilon$. Therefore, using commutativity and associativity,This is the Möbius inversion formula.

Take both functions to be the constant arithmetic function $1$. Thenso the divisor function is multiplicative by part (a). For $s$ with $\mathrm{Re}s>1$, the relevant Dirichlet series are absolutely convergent, so their Cauchy product may be rearranged:Equivalently, Dirichlet convolution becomes multiplication of absolutely convergent Dirichlet series.

Let $P=p_1{p}_2\cdots p_{k+1}$ be the product of the first $k+1$ primes, and take $n=P^a$ for positive integers $a$. The prime factorization givesTheir ratio isThus every sufficiently large $a$ gives $\tau (P^a)\ge (\log P^a{)}^k$, producing infinitely many such $n$.

If $n={\prod }_{p^{\alpha }\parallel n}{p}^{\alpha }$, multiplicativity and the prime-power formula for the divisor function giveChoose $P_0=2^{1/ϵ}$. For $p\ge P_0$ and $\alpha \ge 1$, the elementary bound $\alpha +1\le 2^{\alpha }$ givesFor each of the finitely many primes $p<P_0$, the sequence $(\alpha +1)p^{-\alpha ϵ}$ tends to zero and therefore has a finite maximum $M_p$ over $\alpha \ge 0$. Consequentlyuniformly in $n$, which proves $\tau (n)\le C(ϵ)n^{ϵ}$.