Problem 1. For $r \geq 2,$ show that there are positive constants $c_{1}$ and $c_{2}$ such that $$c_{1} n^{r} \leq J_{r}(n) \leq c_{2} n^{r}$$
Proof . Since each factor of $$\prod_{p \mid n}\left(1-\frac{1}{p^{r}}\right)$$ is less than 1 , we can take $c_{2}=1$. For the lower bound, we have $$\prod_{p \mid d}\left(1-\frac{1}{p^{r}}\right) \geq \prod_{i \leq \nu(n)}\left(1-\frac{1}{p_{i}^{r}}\right)$$ which converges to a nonzero limit as $\nu(n) \rightarrow \infty$. Thus, there is a constant $c_{1}$ such that $$J_{r}(n) \geq c_{1} n^{r}$$
Problem 2. Show that $$\sum_{n \leq x \atop(n, k)=1} n=\frac{\phi(k)}{2 k} x^{2}+O(d(k) x)$$ where $d(k)$ denotes the number of divisors of $k$.
Proof . We have \begin{aligned} \sum_{n \leq x \atop(n, k)=1} n &=\sum_{n \leq x} n \sum_{d|n \atop d| k} \mu(d) \\ &=\sum_{d \mid k} \mu(d) \sum_{n \leq x \atop d \mid n} n \\ &=\sum_{d \mid k} \mu(d) d \sum_{t \leq x / d} t \\ &=\sum_{d \mid k} \mu(d) d\left\{\frac{1}{2}\left[\frac{x}{d}\right]\left(\left[\frac{x}{d}\right]+1\right)\right\} \\ &=\sum_{d \mid k} \mu(d) d\left\{\frac{x^{2}}{2 d^{2}}+O\left(\frac{x}{d}\right)\right\} \end{aligned} which is equal to $$\frac{x^{2} \phi(k)}{2 k}+O(x d(k))$$ as required.
Problem 3. (Dual Möbius inversion formula) Suppose $f(d)=$ $\sum_{d \mid n} g(n),$ where the summation is over all multiples of d. Show that $$g(d)=\sum_{d \mid n} \mu\left(\frac{n}{d}\right) f(n)$$ and conversely (assuming that all the series are absolutely convergent).
Proof . We have \begin{aligned} \sum_{d \mid n} \mu\left(\frac{n}{d}\right) f(n) &=\sum_{t} \mu(t) f(d t) \\ &=\sum_{t} \mu(t) \sum_{r} g(d t r) \\ &=\sum_{m} g(d m)\left(\sum_{t r=m} \mu(t)\right)=g(m), \end{aligned} since the inner sum is 1 if $m=1,$ and zero otherwise. Similarly, for the converse, \begin{aligned} \sum_{d \mid n} g(n) &=\sum_{t} g(d t) \\ &=\sum_{t} \sum_{r} \mu(r) f(d t r) \\ &=\sum_{m} f(d m)\left(\sum_{t r=m} \mu(r)\right)=f(d), \end{aligned} since the inner sum is again 1 if $m=1,$ and zero otherwise.
Problem 4. Let $p$ be a prime not dividing $m .$ Show that $p \mid \phi_{m}(a)$ if and only if the order of $a(\bmod p)$ is $m .$ (Here $\phi_{m}(x)$ is the mth cyclotomic polynomial.)
Proof . Since $$x^{m}-1=\prod_{d \mid m} \phi_{d}(x)$$ we deduce $a^{m} \equiv 1(\bmod p)$. If $k$ is the order of $a(\bmod p)$, then $$a^{k}-1=\prod_{d \mid k} \phi_{d}(a) \equiv 0 \quad(\bmod p)$$ so that $\phi_{d}(a) \equiv 0(\bmod p)$ for some $d \mid k$.