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Prove that if $f(n)$ is multiplicative, so is $g(n) = \sum\limits_{d/n}f(d)$.

Any ideas? Thanks

Mateusz Wojtczak
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2 Answers2

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Let $(m,n)\in\mathbb{N}^2$ such that $m\wedge n=1$, then notice that $d\vert nm$ if and only $d=d_1d_2$ with $d_1\vert m$ and $d_2\vert n$. Therefore, one has: \begin{align} g(mn)&=\sum_{d\vert nm}f(d)\\ &=\sum_{d_1\vert m,d_2\vert n}f(d_1d_2)\\ &=\sum_{d_1\vert m,d_2\vert n}f(d_1)f(d_2)\tag{1}\\ &=\left(\sum_{d_1\vert m}f(d_1)\right)\left(\sum_{d_2\vert n}f(d_2)\right)\\ &=g(m)g(n) \end{align} Where $(1)$ follows from the multiplicativity of $f$. Whence the result.

C. Falcon
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  • Why You have assumed that $(m,n)\in\mathbb{N}^2$ ? – Mateusz Wojtczak Feb 05 '17 at 01:03
  • My definition of an arithmetic map is $\mathbb{N}\rightarrow\mathbb{C}$, but if yours is with $\mathbb{Z}$, then just replace $\mathbb{N}^2$ with $\mathbb{Z}^2$, the proof works the same. – C. Falcon Feb 05 '17 at 01:05
  • Ok, and by saying $(m,n)\in\mathbb{N}^2$ such that $m\wedge n=1$ You wanted to state that m and n are relatively prime? There is a small typo in last sum there should be $n$. – Mateusz Wojtczak Feb 05 '17 at 01:28
  • Yes exactly, sorry if this is not a conventional notation. Thank you for the correction! – C. Falcon Feb 05 '17 at 01:38
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Assume that the factorization of $n$ is given by $$ n=p_1^{\alpha_1}\cdots p_{k}^{\alpha_k} \tag{1} $$ There is a bijection between the set of divisors of $n$ and the elements of $[0,\alpha_1]\times\ldots\times[0,\alpha_k]$.
By exploiting the multiplicativity of $f$, we get that: $$ (f*1)(n)\stackrel{\text{def}}{=}\sum_{d\mid n}f(d) = \prod_{j=1}^{k}\sum_{\alpha=0}^{\alpha_k}f(p_j^\alpha) =\prod_{j=1}^{k}(f*1)(p_j^{\alpha_k})\tag{2}$$ If $a$ and $b$ are coprime numbers, the sets of prime divisors of $a$ and $b$ are disjoint and the set of prime divisors of $ab$ is just the union of them. By $(2)$ it follows that $$ (f*1)(ab) = (f*1)(a)\cdot (f*1)(b)\tag{3} $$ as wanted.

Jack D'Aurizio
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