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Problem: (Bateman 1949) Let Let \(\Phi_q(z)\) denote the \(q^{\text{th}}\) cyclotomic polynomial,\[ \Phi_q(z) \;=\; \prod_{\substack{a=1 \\ (a,q)=1}}^{q} \bigl(z - e(a/q)\bigr), \]where \(e(\theta) = e^{2\pi i \theta}\).

(a) Show that \[ \prod_{d \mid q} \Phi_d(z) \;=\; z^{\,q} - 1. \]

(b) Show that \[ \Phi_q(z) \;=\; \prod_{d \mid q}\bigl(z^{\,d}-1\bigr)^{\mu(q/d)}. \]

(c) If \(P(z)=\sum p_n z^n\) and \(Q(z)=\sum q_n z^n\) are polynomials with real coefficients, then we say that \(P \preccurlyeq Q\) if \(|p_n|\le q_n\) for all non-negative integers \(n\). Show that if \(P_1 \preccurlyeq Q_1\) and \(P_2 \preccurlyeq Q_2\), then \(P_1+P_2 \preccurlyeq Q_1+Q_2\) and \(P_1P_2 \preccurlyeq Q_1Q_2\).

(d) Show that \(\Phi_q(z) \preccurlyeq Q_q(z)\) where \[ Q_q(z) \;=\; \prod_{d \mid q}\Bigl(1 + z^{d} + z^{2d} + \cdots + z^{q-d}\Bigr). \]

(e) Show that \(Q_q(1) = q^{d(q)/2}\).

(f) Show that for any \(\varepsilon>0\) there is a \(q_0(\varepsilon)\) such that if \(q>q_0(\varepsilon)\), then all coefficients of \(\Phi_q\) have absolute value not exceeding \[ \exp\!\Bigl(q^{(\log 2+\varepsilon)/\log\log q}\Bigr). \]

Solution:

Problem: (cf. Bateman & Grosswald 1958) Let \(\mathcal{F}\) be the set of 'power-full' numbers where \(n\) is power-full if \(p|n \Rightarrow p^2 |n\).

(a) Show that \[\sum_{n \in \mathcal{F}} n^{-s} = \frac{\zeta(2s)\zeta(3s)}{\zeta(6s)}\] for \(\sigma \gt 1/2\).

(b) Show that \[\sum_{\substack{a,b,c\\a^2 b^3 c^6 = n}}\mu(c) = \begin{cases}1 & \text{if }n\in\mathcal{F}, \\ 0 & \text{otherwise.}\end{cases}\]

(c) Show that \[\sum_{a^2 b^3 \le y} 1 = \zeta(3/2) y^{1/2} + \zeta(2/3)y^{1/3} + O\left(y^{1/5}\right).\]

(d) Show that \[\sum_{\substack{n \le x\\n \in \mathcal{F}}} 1 = \frac{\zeta(3/2)}{\zeta(3)} x^{1/2} + \frac{\zeta(2/3)}{\zeta(2)}x^{1/3} + O\left(x^{1/5}\right).\]

Solution: (a) \begin{align*} \sum_{n \in \mathcal{F}} n^{-s} &= \prod_{p} \left(1 + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} \cdots \right)\\ &= \prod_{p} \left(1 + \frac{p^{-2s}}{1-p^{-s}}\right)\\ &= \prod_{p} \left(\frac{1 - p^{-6s} }{(1-p^{-2s})(1-p^{-3s})}\right)\\ &= \frac{\zeta(2s)\zeta(3s)}{\zeta(6s)} \end{align*}

(b) \begin{align*}\sum_{n \in \mathcal{F}} n^{-s} = \frac{\zeta(2s)\zeta(3s)}{\zeta(6s)} = \sum_{n=1}^{\infty}\left(\sum_{\substack{a,b,c\\a^2 b^3 c^6 = n}}\mu(c)\right) n^{-s} \end{align*}

(c) \begin{align*} \sum_{a^2 b^3 \le y} 1 &= \sum_{a^2 \le y^{2/5}} \sum_{b^{3}\le y a^{-2}}1 + \sum_{b^3 \le y^{3/5}} \sum_{a^2 \le y b^{-3}}1 - \sum_{a^2 \le y^{2/5}}1\sum_{b^3 \le y^{3/5}}1\\ &= \sum_{a \le y^{1/5}} \sum_{b\le y^{1/3} a^{-2/3}}1 + \sum_{b \le y^{1/5}} \sum_{a \le y^{1/2} b^{-3/2}}1 - \sum_{a\le y^{1/5}}1\sum_{b\le y^{1/5}}1\\ &= \sum_{a \le y^{1/5}} y^{1/3} a^{-2/3} + \sum_{b \le y^{1/5}} y^{1/2} b^{-3/2} - y^{2/5} + O(y^{1/5})\\ &= y^{1/3} \left(3y^{1/15} + \zeta(2/3) + O(y^{-2/15}) \right) \\ &\qquad\qquad+ y^{1/2} \left(\zeta(3/2)-2y^{-1/10}+ O(y^{-3/10}) \right)\\ &\qquad\qquad\qquad - y^{2/5} + O(y^{1/5}) \\ &=\zeta(3/2) y^{1/2} + \zeta(2/3)y^{1/3} + O\left(y^{1/5}\right) \end{align*}

(d)\begin{align*} \sum_{\substack{n \le x\\n \in \mathcal{F}}} 1 & = \sum_{\substack{a,b,c\\a^2 b^3 c^6 \le x}}\mu(c)\\ & =\sum_{c^{6} \le x}\mu(c) \sum_{a^2 b^3 \le x/c^{6}} 1\\ & =\sum_{c^{6} \le x}\mu(c) \left(\zeta(3/2) x^{1/2}/c^{3} + \zeta(2/3) x^{1/3}/c^2 + O\left(x^{1/5}/c^{6/5}\right) \right)\\ &=\frac{\zeta(3/2)}{\zeta(3)} x^{1/2} + \frac{\zeta(2/3)}{\zeta(2)}x^{1/3} + O\left(x^{1/5}\right) \end{align*}

Problem: (Davenport 1932) Let \[f(n) = -\sum_{d|n} \frac{\mu(d)\log d}{d}.\]

(a) By recalling Exercise 2.1.16(a), or otherwise, show that \(f(n) \ge 0\) for all \(n\).

(b) Show that \(f(n) \ll \log\log n\) for \(n \ge 3\).

(c) Show that \(f(n) \sim \frac{1}{4}\log\log n\) if \(n = \prod_{y \lt p \le y^2}p\).

(d) Show that \(f(n) \le\left( \frac{1}{4} + o(1)\right)\log\log n\) as \(n\to \infty\).

Solution: (a) By Exercise 2.1.16(a), \[f(n) = -\sum_{d|n} \frac{\mu(d)\log d}{d} = \frac{\varphi(n)}{n} \sum_{p \mid n} \frac{\log p}{p - 1} \ge 0.\]

(b) Note that \begin{align*} f(n) \le \sum_{p \mid n} \frac{\log p}{p - 1}. \end{align*} Since \({\log x}/{(x - 1)}\) is decreasing function for all \(x\gt 0\), it is enough to show only when \(n = \prod_{p \le y}p\). Now, \begin{align*} \sum_{p \mid n} \frac{\log p}{p - 1} &= \sum_{p \le y} \frac{\log p}{p-1}\\ &\ll \log y = \log \log n. \end{align*}

(c) \begin{align*} f(n) &= \frac{\varphi(n)}{n} \sum_{p \mid n} \frac{\log p}{p - 1}\\ &= \prod_{y \lt p \le y^2} \left(1-\frac{1}{p} \right) \sum_{y \lt p \le y^2} \frac{\log p}{p-1}\\ &= \left(\frac{1}{2} + o(1)\right) \left(\log y + O(1) \right)\\ &= \left(\frac{1}{4} + o(1)\right)\log\log n \end{align*}

(d) Suppose \(n = \prod_{y^{a} \lt p \le y^2}p\), then \begin{align*} f(n) &= \frac{\varphi(n)}{n} \sum_{p \mid n} \frac{\log p}{p - 1}\\ &= \prod_{y^{a} \lt p \le y^2} \left(1-\frac{1}{p} \right) \sum_{y^{a} \lt p \le y^2} \frac{\log p}{p-1}\\ &= \left(\frac{\log y^{a}}{2\log y} + o(1)\right) \left(\log y^2 - \log y^{a} + O(1) \right)\\ &= \left(\frac{a}{2} + o(1)\right) \left((2-a)\log y+ O(1) \right)\\ &= \left(\frac{a(2-a)}{4} + o(1)\right)\log\log n\le\left( \frac{1}{4} + o(1)\right)\log\log n. \end{align*}

Problem: Let \(d_k (n)\) be as in Exercise 2.1.18. Show that if \(k\) and \(\kappa\) are fixed, then \[\sum_{n\le x} d_k (n) ^{\kappa} \ll x (\log x) ^{k ^{\kappa}-1}.\] for \(x \ge 2\).

Solution: Since \[\sum_{p \le x}d_{k}(p)^{\kappa}\log p \ll x, \qquad \sum_{\substack{p^{m}\\m\ge 2}}\frac{d_{k}(p^{m})^{\kappa}m\log p}{p^{m}} \ll 1,\] by Corollary 2.15, \begin{align*} \sum_{n\le x} d_k (n) ^{\kappa} &\ll \frac{x}{\log x} \prod_{p \le x} \left(1 + \frac{d_k (p) ^{\kappa}}{p}+\frac{d_k (p^2) ^{\kappa}}{p^2}+\cdots \right)\\ &\ll x (\log x) ^{k ^{\kappa}-1}. \end{align*}

Problem: Let \(f(n) = \prod_{p|n}(1+p^{-1/2})\).

(a) Show that there is a constant \(a\) such that if \(n\ge 3\), then \[f(n) \lt \exp (a(\log n)^{1/2}(\log \log n)^{-1}). \]

(b) Show that \(\sum_{n\le x}f(n) = cx + O(x^{1/2})\) where \(c = \prod_{p}(1+p^{-3/2})\).

Solution: (a) It is enough to show only when \(n = \prod_{p\le y} p\). Note that \(\log \log n = \log y + O(1)\) and \[\sum_{p\le y}p^{-1/2} = \int_{2-}^{y} u^{-1/2} d\pi(u) \ll y^{1/2}(\log y)^{-1}.\] Now, \begin{align*} \log f(n) &= \sum_{p\le y}\log(1+p^{-1/2})\\ &=\sum_{p\le y}p^{-1/2} + O \left(\sum_{p\le y}p^{-1}\right)\\ &\ll y^{1/2}(\log y)^{-1} = (\log n)^{1/2} (\log \log n)^{-1}. \end{align*}

(b) Note that \[f(n) = \sum_{d|n} \mu(d)^2 d^{-1/2}.\] Hence, \begin{align*} \sum_{n\le x}f(n) &= \sum_{n\le x}\sum_{d|n} \mu(d)^2 d^{-1/2}\\ &=x\sum_{d\le x}\mu(d)^2 d^{-3/2} + O\left(\sum_{d\le x} \mu(d)^2 d^{-1/2}\right)\\ &= x\sum_{d=1}^{\infty}\mu(d)^2 d^{-3/2} + \sum_{d \gt x}\mu(d)^2 d^{-3/2} + O(x^{1/2})\\ &= cx + O(x^{1/2}). \end{align*}