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 \(d \mid k\). 

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