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Problem: (Jarník 1926; cf. Bombieri & Pila 1989) Let \(\mathcal{C}\) be a simple closed curve in the plane, of arc length \(L\). Show that the number of 'lattice points' \((m,n)\), \(m,n\in\mathbb{Z}\), lying on \(\mathcal{C}\) is at most \(L+1\). Show that if \(\mathcal{C}\) is strictly convex then the number of lattice points on \(\mathcal{C}\) is \(\ll 1 + L^{2/3}\), and that this estimate is best possible.

Solution: Please refer to this document.

Problem: (cf. Hartman & Wintner 1947) Suppose that \(\sum |f(n)|d(n) \lt \infty\), and that \(\sum |F(n)|d(n) \lt \infty\). Show that \[F(n) = \underset{n|m}{\sum_{m}} f(m)\] for all \(n\) if and only if \[f(n) = \underset{n|m}{\sum_{m}} \mu (m/n) F(m)\]

Solution: (\( \Rightarrow\)) \begin{align*} \underset{n|m}{\sum_{m}} \mu (m/n) F(m) &= \mu (1) F(n) + \mu(2) F(2n) + \cdots \\ &= \mu(1) (f(n)+ f(2n) + \cdots) + \mu(2) (f(2n) + f(4n)+ \cdots) + \cdots \\ &= \sum_{k} \left( \sum_{d|k} \mu(d) \right) f(kn) = f(n) \end{align*} (\(\Leftarrow\)) \begin{align*} \underset{n|m}{\sum_{m}} f(m) & = f(n) + f(2n) + \cdots \\ &= \underset{n|m}{\sum_{m}} \mu (m/n) F(m) + \underset{2n|m}{\sum_{m}} \mu (m/2n) F(m) + \cdots \\ &= (\mu (1) F(n) + \mu(2) F(2n) + \cdots) + (\mu (1) F(2n) + \mu(2) F(4n) + \cdots) + \cdots \\ &= \mu(1) (F(n)+ F(2n) + \cdots) + \mu(2) (F(2n) + F(4n)+ \cdots) + \cdots \\ &= \sum_{k} \left( \sum_{d|k} \mu(d) \right) F(kn) = F(n) \end{align*}

Problem: (cf. Hille 1937) Suppose that \(f(x)\) and \(F(x)\) are complex-valued functions defined on \([1, \infty)\). Show that \[F(x) = \sum_{n \le x} f(x/n)\] for all \(x\) if and only if \[f(x) = \sum_{n \le x} \mu (n) F(x/n)\] for all \(x\).

Solution: (\(\Rightarrow\)) \begin{align*} \sum_{n \le x } \mu (n) F(x/n) &= \sum_{mn \le x} \mu (n) f(x/mn)\\ &= \sum_{k \le x} \left( {\sum_{d|k}}\mu (d) \right) f(x/k) = f(x) \end{align*} (\(\Leftarrow\)) \begin{align*} \sum_{n \le x } f(x/n) &= \sum_{nm \le x } \mu (m) F(x/mn) \\& = \sum_{k \le x} \left( {\sum_{d|k}}\mu (d) \right) F(x/k)=F(x) \end{align*}

Problem: (cf. Evelyn & Linfoot 1930) Let \(N\) be a positive integer, and suppose that \(P\) is square-free.

(a) Show that the number of residue classes \(n \pmod{P^2}\) for which \((n,\ P^2)\) is square-free and \((N-n,\ P^2)\) is square-free is \[P^2 \underset{p^2 | N}{\prod_{p|P}} \left(1- \frac{1}{p^2} \right)\underset{p^2 \nmid N}{\prod_{p|P}} \left(1- \frac{2}{p^2} \right).\]

(b) Show that the number of integers \(n\), \(0 \lt n \lt N\), for which \((n,\ P^2)\) is square-free and \((N-n,\ P^2)\) is square-free is \[N \underset{p^2 | N}{\prod_{p|P}} \left(1- \frac{1}{p^2} \right)\underset{p^2 \nmid N}{\prod_{p|P}} \left(1- \frac{2}{p^2} \right) + O(P^2).\]

(c) Show that the number of \(n\), \(0 \lt n \lt N\), such that \(n\) is divisible by the square of a prime \(\gt y\) is \(\ll N/y\).

(d) Take \(P\) to be the product of all primes not exceeding \(y\). By letting \(y\) tend to infinity slowly, show that the number of ways of writing \(N\) as a sum of two square-free integers is \(\sim c(N)N\) where \[c(N) = a \prod_{p^2 | N}\left( 1+ \frac{1}{p^2-2} \right),\quad a = \prod_{p} \left(1-\frac{2}{p^2} \right).\]

Solution:

(a) (i) \((N-n,\ P^2)\) is not square-free, i.e., there exists a prime \(p |P\) such that \(p^2|(N-n)\).

If \(p^2 | n\), then \(p^2 | N\). Hence, \[P^2 \underset{p^2 | N}{\prod_{p|P}}\left(1- \frac{1}{p^2} \right).\]

If \(p^2 \nmid n\), then \(p^2 \nmid N\). Put \(N-n = p^2 M\) for some \((p^2,\ M) = 1\), then \[n = N - p^2 M \pmod{P^2}.\] Hence, \[P^2 \underset{p^2 \nmid N}{\prod_{p|P}}\left(1- \frac{1}{p^2} \right).\]

(ii) \((N-n,\ P^2)\) is square-free and \((n,\ P^2\) is not square-free, i.e., there exists a prime \(p | P\) such that \(p^2 | n\) and \(p^2 \nmid N\). Hence, \[P^2 \underset{p^2 \nmid N}{\prod_{p|P}}\left(1- \frac{1}{p^2} \right).\]

By (i) and (ii), \[P^2 \underset{p^2 | N}{\prod_{p|P}} \left(1- \frac{1}{p^2} \right)\underset{p^2 \nmid N}{\prod_{p|P}} \left(1- \frac{2}{p^2} \right).\]

(b) Let \(f\) be \[f(k) = \# \{ 0 \lt i \lt k:\ (i,\ P^2)\text{ and }(N-i,\ P^2)\text{ is square-free.} \} \] Put \(N = AP^2 + B\) where \(0 \le B \lt P^2\), then \[f(N) = f(AP^2 + B) = Af(P^2) + f(B)\] \[=(AP^2 + B) \frac{f(P^2)}{P^2} + f(B) - \frac{B}{P^2}f(P^2).\] \[\left| f(B) - \frac{B}{P^2}f(P^2) \right| \le P^2 + P^2 = 2P^2 = O(P^2)\] Therefore, \[N \underset{p^2 | N}{\prod_{p|P}} \left(1- \frac{1}{p^2} \right)\underset{p^2 \nmid N}{\prod_{p|P}} \left(1- \frac{2}{p^2} \right) + O(P^2).\]

(c) \[\#\{0\lt n \lt N : n\text{ is divisible by the square of a prime }\gt y \} \le \sum_{y \lt p \lt N}\left[\frac{N}{p^2} \right] \] \[\le N \sum_{y \lt p \lt N}\frac{1}{p^2}\ll N/y\]

(d) \[\# = N \underset{p^2 | N}{\prod_{p|P}} \left(1- \frac{1}{p^2} \right)\underset{p^2 \nmid N}{\prod_{p|P}} \left(1- \frac{2}{p^2} \right) + O(P^2) + O(N/y)\] Let \(y\) be \(\log N\), then \(O(P^2) = O(\log^4 N )\) and \(O(N/y) = O(N/\log N)\). Hence, \[\# / c(N) N \sim \prod_{p^2|N}\left(1- \frac{1}{p^2} \right)\prod_{p^2\nmid N}\left(1- \frac{2}{p^2} \right) /c(N)=1\]

Problem: (Linfoot & Evelyn 1929) Let \(\mathcal{Q}_k\) denote the set of positive \(k^{\text{th}}\) power free integers (i.e., \(q \in \mathcal{Q}_k\) if and only if \(m^{k} | q \Rightarrow m=1\)).

(a) Show that \[\sum_{n \in \mathcal{Q}_k} n^{-s} = \frac{\zeta(s)}{\zeta(ks)}\] for \(\sigma \gt 1\).

(b) Show that for any fixed integer \(k \gt 1\) \[\underset{n \in \mathcal{Q}_k}{\sum_{n\le x}}1 =\frac{x}{\zeta(k)}+ O\left(x^{1/k} \right)\] for \(x \ge 1\).

Solution:

(a) Let a function \(f\) be \(f(n) = 1\) if \(n \in \mathcal{Q} _k\), and \(f(n) = 0\) otherwise. Then, \(f\) is multiplicative. Hence \[\sum_{n \in \mathcal{Q}_k} n^{-s} = \sum_{n=1}^{\infty}f(n) n^{-s}\] \[=\prod_{p} (1 + f(p)p^{-s} + f(p^2)p^{-2s} + f(p^3)p^{-3s}+\cdots)\] \[=\prod_{p} (1 + p^{-s} + \cdots + p^{-(k-1)s}) = \prod_{p} \frac{1-p^{-ks}}{1-p^{-s}} = \frac{\zeta(s)}{\zeta(ks)}\] for \(\sigma \gt 1\).

(b) By (a), \[f(n) = \sum_{d^{k}|n} \mu(d).\] Hence, \[\underset{n \in \mathcal{Q}_k}{\sum_{n\le x}}1 = \sum_{n \le x} f(n) = x \sum_{d^{k} \le x} \frac{\mu(d)}{d^{k}} + O\left(\sum_{d^k \le x}1\right)\] \[=\frac{x}{\zeta(k)}+ O\left(x^{1/k} \right)\] for \(x \ge 1\).