For integers $s \geq 0$ and $n \geq 1$ let $\sigma_s(n) := 1^s + \dotsb + n^s$ and let

\begin{equation} \sigma_s^*(n) \ : = \ \sum_{\substack{\scriptstyle 1 \, \leq \, m \, \leq \, n \\ \gcd(m,n) = 1}} m^s. \end{equation}

It is not to difficult to check that $\sigma^*$ can be expressed as the Dirichlet convolution $\bigl( \mu \cdot \Bbb{i}^s \bigr) * \sigma_s$ where $\mu$ is the Möbius funtion, $\Bbb{i}$ is the identity function, namely $\Bbb{i}(n) = n$ for all integers $n \geq 1$, and $\mu \cdot \Bbb{i}^s$ is the point-wise product, namely $\bigl( \mu \cdot \Bbb{i}^s \bigr)(n) = \mu(n) \, n^s$ for all integers $n \geq 1$. A cute little exercise in Vinogradov's text asserts that when $s=0, 1, 2$ formulas for $\sigma_s^*(n)$ with $n > 1$ are given by

\begin{align*} &\phi(n) &\text{when} \quad s = 0 \\ \\ &\Bigl( {\displaystyle 1 \over 2} \,n \Bigr) \, \phi(n) &\text{when} \quad s=1 \\ \\ & \Bigl( {\displaystyle 1 \over 3 } \, n^2 + {\displaystyle 1 \over 6} \, \mu\bigl(\operatorname{rad}(n) \bigr) \, \operatorname{rad}(n)\Bigr) \, \phi(n) &\text{when} \quad s=2 \end{align*}

where $\phi$ is the Euler totient function and $\operatorname{rad}(n)$ denotes the radical of the integer $n$.

**Question**: Is $\sigma_s^*$ always "formally" divisible by $\phi$ for any integer $s \geq 0$ ?