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I would like someone to verify my proof of the following claim, which I have been using to solve some problems about proving series identities in Ch. 11 of Apostol's analytic number theory text. Let $f$ and $g$ be multiplicative arithmetic. Then the arithmetic function $$h(n) = \sum_{d^2 \mid n}f(d)g(n/d^2)$$ is multiplicative.

Suppose $(m,n) = 1$. We have \begin{equation*} \begin{aligned} h(m)h(n) &= \mathrel{\phantom{=}} \sum_{q^2 \mid m}f(q)g(m/q^2)\sum_{d^2 \mid n}f(d)g(n/d^2) \\ &= \sum_{q^2 \mid m, d^2 \mid n}f(q)f(d)g(m/q^2)g(n/d^2) \\ &= \sum_{q^2d^2 \mid mn}f(qd)g(mn/q^2d^2) \\ &= \sum_{d^2 \mid mn}f(d)g(mn/d^2) \\ &= h(mn), \end{aligned} \end{equation*} where we use the multiplicativity of $f$ and $g$ in the third equality and in the fourth equality we use the fact that $(m,n) = 1$ implies that specifying a pair $(q^2,d^2)$ with $q^2 \mid m$ and $d^2 \mid n$ is equivalent to specifying a square divisor of $mn$.

Thanks!

Ethan Alwaise
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1 Answers1

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Your proof is correct. An alternative way to obtain the result is to consider the function $\tilde{f}$ defined by $\tilde{f}(n) = 0$ if $n$ is not a square, and $\tilde{f}(k^2) = f(k)$. This function is multiplicative if and only if $f$ is multiplicative, and we have $h = \tilde{f} \ast g$ with the ordinary Dirichlet convolution. By multiplicativity preservation of Dirichlet convolution, the multiplicativity of $h$ follows. Another perspective on the same argument looks at the (formal) Dirichlet series

\begin{gather} F(s) = \sum_{n = 1}^{\infty} \frac{f(n)}{n^s} = \prod_p \biggl(1 + \sum_{k = 1}^{\infty} \frac{f(p^k)}{p^{ks}}\biggr)\,, \\ G(s) = \sum_{n = 1}^{\infty} \frac{g(n)}{n^s} = \prod_p \biggl(1 + \sum_{k = 1}^{\infty} \frac{g(p^k)}{p^{ks}}\biggr)\,. \end{gather}

Then

$$H(s) = \sum_{n = 1}^{\infty} \frac{h(n)}{n^s} = F(2s)G(s) = \prod_p \Biggl[\biggl(1 + \sum_{k = 1}^{\infty} \frac{f(p^k)}{p^{2ks}}\biggr)\biggl(1 + \sum_{k = 1}^{\infty} \frac{g(p^k)}{p^{ks}}\biggr)\Biggr]$$

has an Euler product, so $h$ is multiplicative.

Daniel Fischer
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