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Problem: Let \(r(n)\) be the number of ordered pairs \((a,b)\) of integers for which \(a^2+b^2=n\).

(a) Show that \[ \sum_{n \le x} r(n) = 1 + 4\left[ \sqrt{x} \right] + 8 \sum_{1 \le n \le \sqrt{x}/2} \left[ \sqrt{x - n^2} \right] - 4 \left[ \sqrt{x/2} \right]^2. \] (b) Show that \[ \sum_{1 \le n \le \sqrt{x}/2} \sqrt{x - n^2} = \left( \frac{\pi}{8} + \frac{1}{2} \right) x - B_1\!\left(\left\{\sqrt{x/2}\right\}\right) - \frac{1}{2}\sqrt{x} + O(1). \] (c) Write \(\sum_{0 \le n \le x} r(n) = \pi x + R(x)\). Show that \[ R(x) = -8 \sum_{1 \le n \le \sqrt{x}/2} B_1\!\left(\left\{\sqrt{x - n^2}\right\}\right) + O(1). \]

Solution: (a) \begin{align*} \sum_{n \le x} r(n) \end{align*}

Problem:Let \(B_1(x) = x - 1/2\), as in Appendix B.

(a) Show that \[ \sum_{n \le x} \frac{1}{n} = \log x + C_0 - {B_1(\{x\})}/{x} + O(1/x^2). \] (b) Write \(\sum_{n \le x} d(n) = x \log x + (2C_0 - 1)x + \Delta(x)\). Show that \[ \Delta(x) = -2 \sum_{n \le \sqrt{x}} B_1(\{x/n\}) + O(1). \] (c) Show that \(\int_0^X \Delta(x)\, dx \ll X\).

(d) Deduce that \begin{align*} \sum_{n \le X} d(n)(X-n) &= \int_0^X \left( \sum_{n \le x} d(n) \right)dx \\ &= \frac{1}{2} X^2 \log X + \left( C_0 - \frac{3}{4} \right) X^2 + O(X). \end{align*}

Solution: (a) \begin{align*} \sum_{n \le x} \frac{1}{n} &= \int_{1-}^{x} \frac{1}{u} d[u] = \int_{1-}^{x} \frac{1}{u} du - \int_{1-}^{x} \frac{1}{u} d\{u\} \\ &=\log x - \int_{1-}^{x} \frac{1}{u} d\left(\{u\}- 1/2 \right)=\log x - \int_{1-}^{x} \frac{1}{u} dB_{1}(\{u\})\\ &=\log x - \frac{1}{u} B_{1}(\{u\})\Bigg|_{1}^{x} - \int_{1-}^{x} \frac{1}{u^2} B_{1}(\{u\}) du\\ &= \log x - {B_1(\{x\})}/{x} + \frac{1}{2} - \int_{1}^{x} \frac{1}{u^2}\left(\{u\}- 1/2 \right) du\\\ &= \log x - {B_1(\{x\})}/{x} + 1 - \int_{1}^{\infty} \frac{\{u\}}{u^2} du + \int_{x}^{\infty}\frac{1}{u^2}\left(\{u\}- 1/2 \right) du \\ &= \log x + C_0 - {B_1(\{x\})}/{x} + O(1/x^2). \end{align*}

(b) Note that \begin{align*} \sum_{n \le \sqrt{x}}[x/n] &= \sum_{n \le \sqrt{x}}\left(x/n- 1/2 - B_1(\{x/n\}) \right) \\ &= x\left(\frac{1}{2}\log x + C_0 - {B_1(\{\sqrt{x}\})}/{\sqrt{x}} + O(1/x) \right) - [\sqrt{x}]/2 -\sum_{n \le \sqrt{x}} B_1(\{x/n\})\\ &= \frac{1}{2}x\log x+ C_0 x - \sqrt{x} \{\sqrt{x}\} + \sqrt{x}/2 - [\sqrt{x}]/2 -\sum_{n \le \sqrt{x}} B_1(\{x/n\})+ O(1) . \end{align*} Hence, \begin{align*} \sum_{n \le x} d(n) &= 2 \sum_{n \le \sqrt{x}}[x/n] - [\sqrt{x}]^2 \\ &= x\log x+2 C_0 x - 2\sqrt{x} \{\sqrt{x}\} + \sqrt{x} - [\sqrt{x}]- [\sqrt{x}]^2 -2\sum_{n \le \sqrt{x}} B_1(\{x/n\})+ O(1)\\ &= x\log x+2 C_0 x - 2\sqrt{x} \{\sqrt{x}\} + \{\sqrt{x}\}- x + 2\sqrt{x}\{\sqrt{x}\} + \{\sqrt{x}\}^2 + \Delta (x)\\ &=x \log x + (2C_0 - 1)x + \Delta(x). \end{align*}

(c) Note that \[\left|\int_{a}^{b} B_{1}\left( \{x\} \right)dx \right| \le 1/4\] for every \(a,b\). Hence, \begin{align*} \left| \int_{0}^{X} \sum_{n \le \sqrt{x}} B_{1} (\{x/n\}) dx\right| &= \left|\sum_{0 \le k \le \sqrt{X}-1} \int_{k^2}^{(k+1)^2} \sum_{n \le k} B_{1} (\{x/n\})dx \right|\\ &\le \sum_{0 \le k \le \sqrt{X}-1} \sum_{n \le k}\left| \int_{k^2}^{(k+1)^2} B_{1} (\{x/n\})dx \right|\\ &\le \sum_{0 \le k \le \sqrt{X}-1} \frac{1}{4}k \ll X. \end{align*}

(d) \begin{align*} \int_0^X \left( \sum_{n \le x} d(n) \right)dx &= \int_0^X ( x \log x + (2C_0 - 1)x + \Delta(x)) dx\\ &= \frac{1}{2} X^2 \log X + \left( C_0 - \frac{3}{4} \right) X^2 + O(X). \end{align*}

Problem:(Feller & Tornier 1932) Let \(f(n)\) denote the multiplicative function such that \(f(p) = 1\) for all primes \(p\), and \(f(p^k) = -1\) whenever \(k > 1\).

(a) Show that \[ \sum_{n=1}^\infty \frac{f(n)}{n^s} = \zeta(s)\prod_p \left(1 - \frac{2}{p^{2s}}\right) \] for \(\sigma > 1\).

(b) Deduce that \[ f(n) = \sum_{d^2 \mid n} \mu(d) 2^{\omega(d)}. \]

(c) Explain why \(2^{\omega(n)} \leq d(n)\) for all \(n\).

(d) Show that \[ \sum_{n \leq x} f(n) = ax + O\!\left(x^{1/2}\log x\right) \] where \(a\) is the constant of Exercise 4.

(e) Let \(g(n)\) denote the number of primes \(p\) such that \(p^2 \mid n\). Show that the set of \(n\) for which \(g(n)\) is even has asymptotic density \((1+a)/2\).

(f) Put \[ e_k = \frac{1}{k} \sum_{d \mid k} \mu(d)\, 2^{k/d}. \] Show that if \(|z| \lt 1\), then \[ \log(1 - 2z) = \sum_{k=1}^\infty e_k \log(1 - z^k). \]

(g) Deduce that \[ a = \prod_{k=1}^\infty \zeta(2k)^{e_k}. \]

Solution: (a) \begin{align*} \sum_{n=1}^\infty \frac{f(n)}{n^s} &= \prod_{p} \left( 1 + f(p)p^{-s} + f(p^2)p^{-2s} + \cdots \right)\\ &= \prod_{p} \left( 1 + p^{-s} - \frac{p^{-2s}}{1-p^{-s}}\right) = \prod_{p} \left( \frac{1 -2 p^{-2s}}{1-p^{-s}}\right)\\ &=\zeta(s)\prod_p \left(1 - \frac{2}{p^{2s}}\right) \end{align*}

(b) Let \(h(d)\) be \(h(d) = \mu(\sqrt{d})2^{\omega(\sqrt{d})}\) if \(d\) is a perfect square, and \(h(d) = 0\) otherwise. Then, \[\sum_{n=1}^\infty \frac{h(n)}{n^s} = \prod_p \left(1 - \frac{2}{p^{2s}}\right).\] Hence, \[ f(n) =\sum_{d | n} h(d) = \sum_{d^2 \mid n} \mu(d) 2^{\omega(d)}. \]

(c) \[2^{\omega(p^{k})} = 2 \le k+1 = d(p^{k})\]

(d) \begin{align*} \sum_{n \leq x} f(n) &= \sum_{n \leq x}\sum_{d^2 \mid n} \mu(d) 2^{\omega(d)}\\ &=x\sum_{d^2 \le x}\frac{\mu(d) 2^{\omega(d)}}{d^2} + O\left( \sum_{d^2 \le x} 2^{\omega(d)}\right)\\ &=ax + O\left( \sum_{d^2 \le x} d(d)\right) =ax + O\!\left(x^{1/2}\log x\right) \end{align*}

(e) Note that \[f(n) = (-1)^{g(n)}.\] Hence, the asymptotic density of the set of \(n\) for which \(g(n)\) is even is equal to \[\frac{1}{x} \sum_{n \le x} \frac{1 + f(n)}{2} \to \frac{1+a}{2}\] as \(x \to \infty\).

(f) Note that \begin{align*} \sum_{k|n} k{e_k} = \sum_{k|n}\sum_{d \mid k} \mu(d)\, 2^{k/d} = 2^{n}. \end{align*} Hence, \begin{align*} \sum_{k=1}^\infty e_k \log(1 - z^k) &= - \sum_{k=1}^{\infty} e_k \sum_{n=1}^{\infty} \frac{1}{n}z^{nk} =- \sum_{n=1}^{\infty}\sum_{k=1}^{\infty} e_k \frac{1}{n}z^{nk}\\ &=- \sum_{n=1}^{\infty} \sum_{k|n} e_k \frac{n}{k} z^{n} =- \sum_{n=1}^{\infty} \frac{1}{n} (2z)^{n} = \log (1-2z). \end{align*}

(g) \begin{align*} \log a &= \sum_{p} \log(1-2/p^2) = \sum_{p}\sum_{k=1}^\infty e_k \log(1 - p^{-2k})\\ &=\sum_{k=1}^\infty e_k \sum_{p}\log(1 - p^{-2k}) =\sum_{k=1}^\infty e_k \log \zeta(2k) \end{align*}

Problem: (a) Show that if \(a^2\) is the largest perfect square \(\le x\) then \(x-a^{2} \le 2 \sqrt{x}\).

(b) Let \(a^2\) be as above, and let \(b^2\) be the least perfect square such that \(a^2 +b^2 \gt x\). Show that \(a^2+b^2 \lt x + 6 x^{1/4}\). Thus for any \(x \ge 1\), there is a sum of two squares in the interval \((x, x + 6 x^{1/4})\).

Solution: (a) Since \(a \le \sqrt{x} \lt a+1\), \[x-a^2 \lt x- (\sqrt{x}-1)^2 = 2\sqrt{x}-1 \lt 2\sqrt{x} .\]

(b) Since \(a^2 + (b-1)^2 \le x \lt a^2 + b^2\), \[(b-1)^2 \le x-a^2 \le 2\sqrt{x} \Rightarrow b \le 1+ \sqrt{2}x^{1/4} .\] Hence, \[a^2 + b^2 \le x+ 2b-1 \le 1 +2\sqrt{2}x^{1/4} \le x+ 6x^{1/4} .\]

Problem: (Wintner 1944, p. 46) Suppose that \(\sum_d |g(d)| / d \lt \infty\). Show that \(\sum_{d \le x} |g(d)| = o(x)\). Suppose also that \(\sum_{n \le x} f(n) = cx + o(x)\), and put \(h(n) = \sum_{d|n} f(d) g(n/d)\). Show that \[\sum_{n \le x} h(n) = cgx + o(x)\] where \(g = \sum_{d} g(d)/d\).

Solution: Let \(G(x) =\sum_{d \le x} |g(d)|\), then \begin{align*} x\sum_{m \le x} |g(m)|/m = x\int_{1-}^{x} \frac{dG(u)}{u}= x\frac{G(u)}{u} \Bigg|_{1-}^{x} + x\int_{1-}^{x} \frac{G(u)}{u^2}du = o(x). \end{align*} Hence, \begin{align*} \sum_{n \le x} h(n) &= \sum_{n \le x}\sum_{km = n} f(k) g(m)\\ &= \sum_{m \le x} g(m) \left(\sum_{k \le x/m} f(k) \right)\\ &= \sum_{m \le x} g(m) \left(c(x/m) + o(x/m)\right)\\ &= cx\sum_{m \le x} g(m)/m + o\left(x \sum_{m \le x} |g(m)|/m\right)\\ &= cgx + o(x). \end{align*}