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Theorem 6.5 Let \( r \) be a nonnegative integer. For \( x \geq 1 \) , \[ \mathop{\sum }\limits_{{n \leq x}}\frac{{\log }^{r}n}{n} = \frac{1}{r + 1}{\log }^{r + 1}x + O\left( 1\right) \] where the implied constant depends only on \( r \) .
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Proof. The function \( f\left( t\right) = {\log }^{r}t/t \) is nonnegative and unimodal on \( \lbrack 1,\infty ) \) with maximum value \( {\left( r/e\right) }^{r} \) at \( {t}_{0} = {e}^{r} \) . By Theorem 6.3, \[ \mathop{\sum }\limits_{{n \leq x}}\frac{{\log }^{r}n}{n} = {\int }_{1}^{x}\frac{{\log }^{r}{tdt}}{t} + O\left( 1\right) = \frac{1}{r + 1}{\log }^{r + 1}x + O\left( 1\right) . \] This completes the proof.
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Yes
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Theorem 6.6 Let \( k \) be a nonnegative integer. For \( x \geq 1 \) , \[ \mathop{\sum }\limits_{{n \leq x}}\frac{{\log }^{k}\left( {x/n}\right) }{n} = \frac{1}{k + 1}{\log }^{k + 1}x + O\left( {{\log }^{k}x}\right) \] where the implied constant depends only on \( k \) .
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Proof. The idea is to expand \( {\log }^{k}\left( {x/n}\right) \) by the binomial theorem and apply Theorem 6.5. We have \[ \mathop{\sum }\limits_{{n \leq x}}\frac{{\log }^{k}\left( {x/n}\right) }{n} = \mathop{\sum }\limits_{{n \leq x}}\frac{{\left( \log x - \log n\right) }^{k}}{n} \] \[ = \mathop{\sum }\limits_{{n \leq x}}\frac{1}{n}\mathop{\sum }\limits_{{r = 0}}^{k}\left( \begin{array}{l} k \\ r \end{array}\right) {\left( -1\right) }^{r}{\log }^{k - r}x{\log }^{r}n \] \[ = \mathop{\sum }\limits_{{r = 0}}^{k}\left( \begin{array}{l} k \\ r \end{array}\right) {\left( -1\right) }^{r}{\log }^{k - r}x\mathop{\sum }\limits_{{n \leq x}}\frac{{\log }^{r}n}{n} \] \[ = \mathop{\sum }\limits_{{r = 0}}^{k}\left( \begin{array}{l} k \\ r \end{array}\right) {\left( -1\right) }^{r}{\log }^{k - r}x\left( {\frac{1}{r + 1}{\log }^{r + 1}x + O\left( 1\right) }\right) \] \[ = \mathop{\sum }\limits_{{r = 0}}^{k}\left( \begin{array}{l} k \\ r \end{array}\right) \frac{{\left( -1\right) }^{r}}{r + 1}{\log }^{k + 1}x + O\left( {\mathop{\sum }\limits_{{r = 0}}^{k}\left( \begin{array}{l} k \\ r \end{array}\right) {\log }^{k - r}x}\right) \] \[ = \frac{1}{k + 1}{\log }^{k + 1}x + O\left( {{\log }^{k}x}\right) \] since \[ \mathop{\sum }\limits_{{r = 0}}^{k}\frac{{\left( -1\right) }^{r}}{r + 1}\left( \begin{array}{l} k \\ r \end{array}\right) = \frac{1}{k + 1} \] by Exercise 8.
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Yes
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Theorem 6.7 Let \( k \) be a positive integer. Then\n\n\[ \mathop{\sum }\limits_{{{n}_{1}\cdots {n}_{k} \leq x}}\frac{1}{{n}_{1}\cdots {n}_{k}} = \frac{1}{k!}{\log }^{k}x + O\left( {{\log }^{k - 1}x}\right) ,\]
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Proof. By induction on \( k \) . For \( k = 1 \), we set \( r = 0 \) in Theorem 6.5 and\n\nobtain\n\n\[ \mathop{\sum }\limits_{{{n}_{1} \leq x}}\frac{1}{{n}_{1}} = \log x + O\left( 1\right) \]\n\nAssume that the result holds for the positive integer \( k \) . Then\n\n\[ \mathop{\sum }\limits_{{{n}_{1}\cdots {n}_{k}{n}_{k + 1} \leq x}}\frac{1}{{n}_{1}\cdots {n}_{k}{n}_{k + 1}} \]\n\n\[ = \mathop{\sum }\limits_{{{n}_{k + 1} \leq x}}\frac{1}{{n}_{k + 1}}\mathop{\sum }\limits_{{{n}_{1}\cdots {n}_{k} \leq x/{n}_{k + 1}}}\frac{1}{{n}_{1}\cdots {n}_{k}} \]\n\n\[ = \mathop{\sum }\limits_{{{n}_{k + 1} \leq x}}\frac{1}{{n}_{k + 1}}\left( {\frac{1}{k!}{\log }^{k}\left( {x/{n}_{k + 1}}\right) + O\left( {{\log }^{k - 1}\left( {x/{n}_{k + 1}}\right) }\right) }\right) \]\n\n\[ = \mathop{\sum }\limits_{{{n}_{k + 1} \leq x}}\frac{1}{k!{n}_{k + 1}}{\left( \log x - \log {n}_{k + 1}\right) }^{k} \]\n\n\[ + O\left( {{\log }^{k - 1}x\mathop{\sum }\limits_{{{n}_{k + 1} \leq x}}\frac{1}{{n}_{k + 1}}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\frac{1}{k!n}{\left( \log x - \log n\right) }^{k} + O\left( {{\log }^{k}x}\right) . \]\n\nWe use the binomial theorem and Theorem 6.5 to compute the main term.\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{1}{k!n}{\left( \log x - \log n\right) }^{k} = \mathop{\sum }\limits_{{n \leq x}}\frac{1}{k!n}\mathop{\sum }\limits_{{r = 0}}^{k}{\left( -1\right) }^{r}\left( \begin{array}{l} k \\ r \end{array}\right) {\log }^{k - r}x{\log }^{r}n \]\n\n\[ = \mathop{\sum }\limits_{{r = 0}}^{k}\frac{{\left( -1\right) }^{r}}{k!}\left( \begin{array}{l} k \\ r \end{array}\right) {\log }^{k - r}x\mathop{\sum }\limits_{{n \leq x}}\frac{{\log }^{r}n}{n} \]\n\n\[ = \mathop{\sum }\limits_{{r = 0}}^{k}\frac{{\left( -1\right) }^{r}}{k!}\left( \begin{array}{l} k \\ r \end{array}\right) {\log }^{k - r}x\left( {\frac{1}{r + 1}{\log }^{r + 1}x + O\left( 1\right) }\right) \]\n\n\[ = \frac{1}{k!}{\log }^{k + 1}x\mathop{\sum }\limits_{{r = 0}}^{k}\frac{{\left( -1\right) }^{r}}{r + 1}\left( \begin{array}{l} k \\ r \end{array}\right) + O\left( {{\log }^{k}x}\right) \]\n\n\[ = \frac{1}{\left( {k + 1}\right) !}{\log }^{k + 1}x + O\left( {{\log }^{k}x}\right) \]\n\nby Exercise 8.
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Yes
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Theorem 6.8 (Partial summation) Let \( f\left( n\right) \) and \( g\left( n\right) \) be arithmetic functions. Consider the sum function\n\n\[ F\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}f\left( n\right) \]\n\nLet \( a \) and \( b \) be nonnegative integers with \( a < b \) . Then\n\n\[ \mathop{\sum }\limits_{{n = a + 1}}^{b}f\left( n\right) g\left( n\right) = F\left( b\right) g\left( b\right) - F\left( a\right) g\left( {a + 1}\right) - \mathop{\sum }\limits_{{n = a + 1}}^{{b - 1}}F\left( n\right) \left( {g\left( {n + 1}\right) - g\left( n\right) }\right) . \]
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Proof. Identity (6.6) is a straightforward calculation:\n\n\[ \mathop{\sum }\limits_{{n = a + 1}}^{b}f\left( n\right) g\left( n\right) \]\n\n\[ = \mathop{\sum }\limits_{{n = a + 1}}^{b}\left( {F\left( n\right) - F\left( {n - 1}\right) }\right) g\left( n\right) \]\n\n\[ = \mathop{\sum }\limits_{{n = a + 1}}^{b}F\left( n\right) g\left( n\right) - \mathop{\sum }\limits_{{n = a}}^{{b - 1}}F\left( n\right) g\left( {n + 1}\right) \]\n\n\[ = F\left( b\right) g\left( b\right) - F\left( a\right) g\left( {a + 1}\right) - \mathop{\sum }\limits_{{n = a + 1}}^{{b - 1}}F\left( n\right) \left( {g\left( {n + 1}\right) - g\left( n\right) }\right) . \]
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Yes
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Theorem 6.9 For \( x \geq 1 \) ,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{1}{n} = \log x + \gamma + r\left( x\right) \]\n\nwhere\n\n\[ 0 < \gamma = 1 - {\int }_{1}^{\infty }\frac{\{ t\} }{{t}^{2}}{dt} < 1 \]\n\nand\n\n\[ \left| {r\left( x\right) }\right| < \frac{1}{x} \]\n\nThe number \( \gamma = {0.577}\ldots \) is called Euler’s constant. A famous unsolved problem in number theory is to determine whether \( \gamma \) is rational or irrational.
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Proof. Since \( 0 \leq \{ t\} < 1 \) for all \( t \), we have\n\n\[ 0 < {\int }_{1}^{\infty }\frac{\{ t\} }{{t}^{2}}{dt} < {\int }_{1}^{\infty }\frac{1}{{t}^{2}}{dt} = 1 \]\n\nand so \( \gamma \in \left( {0,1}\right) \) .\n\nWe apply partial summation to the functions \( f\left( n\right) = 1 \) and \( g\left( t\right) = 1/t \) . Then \( F\left( t\right) = \mathop{\sum }\limits_{{n \leq t}}1 = \left\lbrack t\right\rbrack \) and\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{1}{n} = \mathop{\sum }\limits_{{n \leq x}}f\left( n\right) g\left( n\right) \]\n\n\[ = \frac{\left\lbrack x\right\rbrack }{x} + {\int }_{1}^{x}\frac{\left\lbrack t\right\rbrack }{{t}^{2}}{dt} \]\n\n\[ = 1 - \frac{\{ x\} }{x} + {\int }_{1}^{x}\frac{1}{t}{dt} - {\int }_{1}^{x}\frac{\{ t\} }{{t}^{2}}{dt} \]\n\n\[ = \log x + \left( {1 - {\int }_{1}^{\infty }\frac{\{ t\} }{{t}^{2}}{dt}}\right) + {\int }_{x}^{\infty }\frac{\{ t\} }{{t}^{2}}{dt} - \frac{\{ x\} }{x} \]\n\n\[ = \log x + \gamma + r\left( x\right) \]\n\nwhere\n\n\[ r\left( x\right) = {\int }_{x}^{\infty }\frac{\{ t\} }{{t}^{2}}{dt} - \frac{\{ x\} }{x}. \]\n\nMoreover, \( \left| {r\left( x\right) }\right| < 1/x \) since \( 0 \leq \{ x\} /x < 1 \) and\n\n\[ 0 < {\int }_{x}^{\infty }\frac{\{ t\} }{{t}^{2}}{dt} < {\int }_{x}^{\infty }\frac{1}{{t}^{2}}{dt} = \frac{1}{x}. \]
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Yes
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Theorem 6.10 Let \( A = {\left\{ {a}_{i}\right\} }_{i = 1}^{\infty } \) be an infinite set of positive integers with \( {a}_{1} < {a}_{2} < {a}_{3} < \cdots \) . If\n\n\[ A\left( x\right) = \mathop{\sum }\limits_{{{a}_{i} \leq x}}1 = O\left( \frac{x}{{\log }^{2}x}\right) \]\n\nfor \( x \geq 2 \), then the series\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{\infty }\frac{1}{{a}_{i}} \]\n\nconverges.
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Proof. Let \( {\chi }_{A}\left( n\right) \) be the characteristic function of \( A \), that is,\n\n\[ {\chi }_{A}\left( n\right) = \left\{ \begin{array}{ll} 1 & \text{ if }n \in A \\ 0 & \text{ if }n \notin A \end{array}\right. \]\n\nThere exists a number \( c \) such that\n\n\[ A\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}{\chi }_{A}\left( n\right) \leq \frac{cx}{{\log }^{2}x} \]\n\nfor all \( x \geq 2 \), and \( A\left( x\right) \leq 1 \) for \( 1 \leq x < 2 \) . Applying partial summation, we obtain\n\n\[ \mathop{\sum }\limits_{{{a}_{i} \leq x}}\frac{1}{{a}_{i}} = \mathop{\sum }\limits_{{n \leq x}}\frac{{\chi }_{A}\left( n\right) }{n} \]\n\n\[ = \frac{A\left( x\right) }{x} + {\int }_{1}^{x}\frac{A\left( t\right) {dt}}{{t}^{2}} \]\n\n\[ \leq \frac{c}{{\log }^{2}x} + \frac{1}{2} + c{\int }_{2}^{x}\frac{dt}{t{\log }^{2}t} \]\n\n\[ = \frac{c}{{\log }^{2}x} + \frac{1}{2} + c{\int }_{\log 2}^{\log x}\frac{du}{{u}^{2}} \]\n\n\[ < \infty \text{.} \]\n\nThis completes the proof.
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Yes
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Theorem 6.11 For \( x \geq 2 \) , \n\n\[ \mathop{\sum }\limits_{{n \leq x}}{\log }^{2}n = x{\log }^{2}x - {2x}\log x + {2x} + O\left( {{\log }^{2}x}\right) . \]
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Proof. We use partial summation with \( f\left( n\right) = 1 \) and \( g\left( t\right) = {\log }^{2}t \) . Then \n\n\( F\left( t\right) = \left\lbrack t\right\rbrack \) and \( {g}^{\prime }\left( t\right) = 2\log t/t \) . Then \n\n\[ \mathop{\sum }\limits_{{n \leq x}}{\log }^{2}n = \left\lbrack x\right\rbrack {\log }^{2}x - 2{\int }_{1}^{x}\frac{\left\lbrack t\right\rbrack \log t}{t}{dt} \] \n\n\[ = \left( {x-\{ x\} }\right) {\log }^{2}x - 2{\int }_{1}^{x}\frac{\left( {t-\{ t\} }\right) \log t}{t}{dt} \] \n\n\[ = x{\log }^{2}x + O\left( {{\log }^{2}x}\right) - 2{\int }_{1}^{x}\log {tdt} + 2{\int }_{1}^{x}\frac{\{ t\} \log t}{t}{dt} \] \n\n\[ = x{\log }^{2}x - {2x}\log x + {2x} + O\left( {{\log }^{2}x}\right) . \] \n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 6.12 For \( x \geq 2 \) ,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}{\log }^{2}\frac{x}{n} = {2x} + O\left( {{\log }^{2}x}\right) \]
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Proof. From Theorem 6.4 and Theorem 6.11, we obtain\n\n\[ \mathop{\sum }\limits_{{n \leq x}}{\log }^{2}\frac{x}{n} = \mathop{\sum }\limits_{{n \leq x}}{\left( \log x - \log n\right) }^{2} \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\left( {{\log }^{2}x - 2\log x\log n + {\log }^{2}n}\right) \]\n\n\[ = \left\lbrack x\right\rbrack {\log }^{2}x - 2\log x\mathop{\sum }\limits_{{n \leq x}}\log n + \mathop{\sum }\limits_{{n \leq x}}{\log }^{2}n \]\n\n\[ = \;x{\log }^{2}x - 2\log x\left( {x\log x - x}\right) + x{\log }^{2}x - {2x}\log x + {2x} + O\left( {{\log }^{2}x}\right) \]\n\n\[ = {2x} + O\left( {{\log }^{2}x}\right) \text{.} \]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 6.13 The Möbius function \( \mu \left( n\right) \) is multiplicative, and\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = \left\{ \begin{array}{ll} 1 & \text{ if }n = 1 \\ 0 & \text{ if }n > 1 \end{array}\right. \]
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Proof. Multiplicativity follows immediately from the definition of the Möbius function, since if \( m \) and \( n \) are relatively prime square-free integers with \( k \) and \( \ell \) prime factors, respectively, then \( {mn} \) is square-free with \( k + \ell \) factors, and\n\n\[ \mu \left( m\right) \mu \left( n\right) = {\left( -1\right) }^{k}{\left( -1\right) }^{\ell } = {\left( -1\right) }^{k + \ell } = \mu \left( {mn}\right) .\n\nNext we prove the convolution formula (6.9). If \( n = 1 \), then\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = \mu \left( 1\right) = 1 \]\n\nFor \( n \geq 2 \), let\n\n\[ n = {p}_{1}^{{r}_{1}}\cdots {p}_{k}^{{r}_{k}} \]\n\nbe the standard factorization of the integer \( n \) . Then \( r \geq 1 \) . Recall that the radical of \( n \) is the largest square-free divisor of \( n \), that is,\n\n\[ \operatorname{rad}\left( n\right) = {p}_{1}\cdots {p}_{r} \]\n\nis the product of the distinct primes dividing \( n \) . Let \( m = \operatorname{rad}\left( n\right) \) . If \( d \) divides \( n \) and \( \mu \left( d\right) \neq 0 \), then \( d \) is square-free, and so \( d \) divides \( m \) . Since \( m \) is the product of \( k \) primes, it follows that there are exactly \( \left( \begin{array}{l} k \\ i \end{array}\right) \) divisors of \( m \) that can be written as the product of \( i \) distinct primes, that is, the number of divisors \( d \) of \( m \) such that \( \omega \left( d\right) = i \) is \( \left( \begin{matrix} k \\ i \end{matrix}\right) \) . Therefore,\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = \mathop{\sum }\limits_{{d \mid m}}\mu \left( d\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{k}\mathop{\sum }\limits_{\substack{{d \mid m} \\ {\omega \left( d\right) = i} }}\mu \left( d\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{k}\mathop{\sum }\limits_{\substack{{d \mid m} \\ {\omega \left( d\right) = i} }}{\left( -1\right) }^{i} \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{k}\left( \begin{array}{l} k \\ i \end{array}\right) {\left( -1\right) }^{i} \]\n\n\[ = {\left( 1 - 1\right) }^{k} \]\n\n\[ = 0\text{. } \]\n\nThis completes the proof.
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Yes
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Theorem 6.14 (Möbius inversion) If \( f \) is any arithmetic function, and \( g \) is the arithmetic function defined by\n\n\[ g\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) \]\n\nthen\n\n\[ f\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mu \left( \frac{n}{d}\right) g\left( d\right) \]
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Proof. We use Theorem 6.13 and the commutativity and associativity of Dirichlet convolution. The definition\n\n\[ g\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) \]\n\nis equivalent to\n\n\[ g = f * 1\text{.} \]\n\nThen\n\n\[ g * \mu = \left( {f * 1}\right) * \mu = f * \left( {1 * \mu }\right) = f * \delta = f. \]\n\nSimilarly, if\n\n\[ f = g * \mu \]\n\nthen\n\n\[ f * 1 = \left( {g * \mu }\right) * 1 = g * \left( {\mu * 1}\right) = g * \delta = g. \]\n\nThis completes the proof.
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Yes
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Theorem 6.16\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\mu \left( n\right) }{n} = O\left( 1\right) \]
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Proof. Applying Theorem 6.15 with \( f\left( n\right) = \mu \left( n\right) \) and\n\n\[ M\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}\mu \left( n\right) \]\n\nwe obtain\n\n\[ \mathop{\sum }\limits_{{m \leq x}}M\left( \frac{x}{m}\right) = \mathop{\sum }\limits_{{d \leq x}}\mu \left( d\right) \left\lbrack \frac{x}{d}\right\rbrack = \mathop{\sum }\limits_{{n \leq x}}\mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = 1, \]\n\nby Theorem 6.13. Since\n\n\[ \mathop{\sum }\limits_{{d \leq x}}\mu \left( d\right) \left\lbrack \frac{x}{d}\right\rbrack = x\mathop{\sum }\limits_{{d \leq x}}\frac{\mu \left( d\right) }{d} - \mathop{\sum }\limits_{{d \leq x}}\mu \left( d\right) \left\{ \frac{x}{d}\right\} = x\mathop{\sum }\limits_{{d \leq x}}\frac{\mu \left( d\right) }{d} + O\left( x\right) ,\]\n\nit follows that\n\n\[ x\mathop{\sum }\limits_{{d \leq x}}\frac{\mu \left( d\right) }{d} + O\left( x\right) = 1 \]\n\nTherefore,\n\n\[ x\mathop{\sum }\limits_{{d \leq x}}\frac{\mu \left( d\right) }{d} = O\left( x\right) \]\n\nand so\n\n\[ \mathop{\sum }\limits_{{d \leq x}}\frac{\mu \left( d\right) }{d} = O\left( 1\right) \]
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Yes
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Theorem 6.17\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\mu \left( n\right) }{{n}^{2}} = \frac{6}{{\pi }^{2}} + O\left( \frac{1}{x}\right) . \]
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Proof. The Riemann zeta function\n\n\[ \zeta \left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{s}} \]\n\nconverges absolutely for \( s > 1 \) . Similarly, the function\n\n\[ G\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\mu \left( n\right) }{{n}^{s}} \]\n\nconverges absolutely for \( s > 1 \) . Therefore,\n\n\[ \zeta \left( s\right) G\left( s\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{k}^{s}}\mathop{\sum }\limits_{{d = 1}}^{\infty }\frac{\mu \left( d\right) }{{d}^{s}} \]\n\n\[ = \mathop{\sum }\limits_{{k = 1}}^{\infty }\mathop{\sum }\limits_{{d = 1}}^{\infty }\frac{\mu \left( d\right) }{{\left( kd\right) }^{s}} \]\n\n\[ = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{s}}\mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) \]\n\n\[ = 1\text{,} \]\n\nby Theorem 6.13, and so\n\n\[ \frac{1}{\zeta \left( s\right) } = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\mu \left( n\right) }{{n}^{s}} \]\n\nfor \( s > 1 \) . Since\n\n\[ \zeta \left( 2\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{2}} = \frac{{\pi }^{2}}{6} \]\n\nit follows that\n\n\[ \frac{1}{\zeta \left( 2\right) } = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\mu \left( n\right) }{{n}^{2}} = \frac{6}{{\pi }^{2}} \]\n\nand so\n\n\[ \left| {\mathop{\sum }\limits_{{n \leq x}}\frac{\mu \left( n\right) }{{n}^{2}} - \frac{6}{{\pi }^{2}}}\right| = \left| {\mathop{\sum }\limits_{{n > x}}\frac{\mu \left( n\right) }{{n}^{2}}}\right| < \mathop{\sum }\limits_{{n > x}}\frac{1}{{n}^{2}} \ll \frac{1}{x}. \]\n\nThis completes the proof.
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Yes
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Theorem 6.18 If \( f \) is a multiplicative function, then\n\n\[ f\left( \left\lbrack {m, n}\right\rbrack \right) f\left( \left( {m, n}\right) \right) = f\left( m\right) f\left( n\right) \]\n\nfor all positive integers \( m \) and \( n \) .
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Proof. Let \( {p}_{1},\ldots ,{p}_{r} \) be the prime numbers that divide \( m \) or \( n \) . Then\n\n\[ n = \mathop{\prod }\limits_{{i = 1}}^{r}{p}_{i}^{{k}_{i}} \]\n\nand\n\n\[ m = \mathop{\prod }\limits_{{i = 1}}^{r}{p}_{i}^{{\ell }_{i}} \]\n\nwhere \( {k}_{1},\ldots ,{k}_{r},{\ell }_{1},\ldots ,{\ell }_{r} \) are nonnegative integers. Then\n\n\[ \left\lbrack {m, n}\right\rbrack = \mathop{\prod }\limits_{{i = 1}}^{r}{p}_{i}^{\max \left( {{k}_{i},{\ell }_{i}}\right) } \]\n\nand\n\n\[ \left( {m, n}\right) = \mathop{\prod }\limits_{{i = 1}}^{r}{p}_{i}^{\min \left( {{k}_{i},{\ell }_{i}}\right) }.\]\n\nSince\n\n\[ \left\{ {\max \left( {{k}_{i},{\ell }_{i}}\right) ,\min \left( {{k}_{i},{\ell }_{i}}\right) }\right\} = \left\{ {{k}_{i},{\ell }_{i}}\right\} \]\n\nand since \( f \) is multiplicative, it follows that\n\n\[ f\left( \left\lbrack {m, n}\right\rbrack \right) f\left( \left( {m, n}\right) \right) = \mathop{\prod }\limits_{{i = 1}}^{r}f\left( {p}_{i}^{\max \left( {{k}_{i},{\ell }_{i}}\right) }\right) \mathop{\prod }\limits_{{i = 1}}^{r}f\left( {p}_{i}^{\min \left( {{k}_{i},{\ell }_{i}}\right) }\right) \]\n\n\[ = \mathop{\prod }\limits_{{i = 1}}^{r}f\left( {p}_{i}^{{k}_{i}}\right) \mathop{\prod }\limits_{{i = 1}}^{r}f\left( {p}_{i}^{{\ell }_{i}}\right) \]\n\n\[ = f\left( m\right) f\left( n\right) \text{.} \]\n\nThis completes the proof.
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Yes
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Theorem 6.19 Let \( f \) be a multiplicative function with \( f\left( 1\right) = 1 \) . Then\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) f\left( d\right) = \mathop{\prod }\limits_{{p \mid n}}\left( {1 - f\left( p\right) }\right) \]
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Proof. The identity holds for \( n = 1 \) . For \( n \geq 2 \), let \( m = \operatorname{rad}\left( n\right) \) be the product of the distinct primes dividing \( n \) . Since \( \mu \left( d\right) = 0 \) if \( d \) is not square-free, it follows that\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) f\left( d\right) = \mathop{\sum }\limits_{{d \mid m}}\mu \left( d\right) f\left( d\right) = \mathop{\prod }\limits_{{p \mid m}}\left( {1 - f\left( p\right) }\right) = \mathop{\prod }\limits_{{p \mid n}}\left( {1 - f\left( p\right) }\right) . \]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 6.20 Let \( f\left( n\right) \) be a multiplicative function. If\n\n\[ \mathop{\lim }\limits_{{{p}^{k} \rightarrow \infty }}f\left( {p}^{k}\right) = 0 \]\n\nas \( {p}^{k} \) runs through the sequence of all prime powers, then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( n\right) = 0 \]
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Proof. Since \( \mathop{\lim }\limits_{{{p}^{k} \rightarrow \infty }}f\left( {p}^{k}\right) = 0 \), it follows that there exist only finitely many prime powers \( {p}^{k} \) such that \( \left| {f\left( {p}^{k}\right) }\right| \geq 1 \), and so we can define\n\n\[ A = \mathop{\prod }\limits_{{\left| {f\left( {p}^{k}\right) }\right| \geq 1}}\left| {f\left( {p}^{k}\right) }\right| \]\n\nThen \( A \geq 1 \) .\n\nLet \( 0 < \varepsilon < 1 \) . There exist only finitely many prime powers \( {p}^{k} \) such that \( \left| {f\left( {p}^{k}\right) }\right| \geq \varepsilon /A \), and so there are only finitely many integers \( n \) such that\n\n\[ \left| {f\left( {p}^{k}\right) }\right| \geq \frac{\varepsilon }{A} \]\n\nfor every prime power \( {p}^{k} \) that exactly divides \( n \) . Therefore, if \( n \) is sufficiently large, then \( n \) is divisible by at least one prime power \( {p}^{k} \) such that \( \left| {f\left( {p}^{k}\right) }\right| < \) \( \varepsilon /A \), and so \( n \) can be written in the form\n\n\[ n = \mathop{\prod }\limits_{{i = 1}}^{r}{p}_{i}^{{k}_{i}}\mathop{\prod }\limits_{{i = r + 1}}^{{r + s}}{p}_{i}^{{k}_{i}}\mathop{\prod }\limits_{{i = r + s + 1}}^{{r + s + t}}{p}_{i}^{{k}_{i}} \]\n\nwhere \( {p}_{1},\ldots ,{p}_{r + s + t} \) are distinct prime numbers such that\n\n\[ \mid f\left( {{p}_{i}^{{k}_{i}} \mid \geq 1\;}\right. \text{ for }i = 1,\ldots, r \]\n\n\[ \frac{\varepsilon }{A} \leq \left| {f\left( {{p}_{i}^{{k}_{i}} \mid < 1\;}\right. }\right| \;\text{ for }i = r + 1,\ldots, r + s, \]\n\n\[ \left| {f\left( {{p}_{i}^{{k}_{i}}\left| {\; < \frac{\varepsilon }{A}}\right. }\right. }\right| \;\text{ for }i = r + s + 1,\ldots, r + s + t, \]\n\nand\n\n\[ t \geq 1\text{.} \]\n\nSince \( f \) is multiplicative,\n\n\[ \left| {f\left( n\right) }\right| = \mathop{\prod }\limits_{{i = 1}}^{r}\left| {f\left( {p}_{i}^{{k}_{i}}\right) }\right| \mathop{\prod }\limits_{{i = r + 1}}^{{r + s}}\left| {f\left( {p}_{i}^{{k}_{i}}\right) }\right| \mathop{\prod }\limits_{{i = r + s + 1}}^{{r + s + t}}\left| {f\left( {p}_{i}^{{k}_{i}}\right) }\right| < A{\left( \varepsilon /A\right) }^{t} \leq \varepsilon . \]\n\nThis completes the proof.
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Yes
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Theorem 6.21 For \( x \geq 1 \) ,\n\n\[ \Phi \left( x\right) = \mathop{\sum }\limits_{{n \leq x}}\varphi \left( n\right) = \frac{3{x}^{2}}{{\pi }^{2}} + O\left( {x\log x}\right) . \]
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Proof. We have\n\n\[ \Phi \left( x\right) = \mathop{\sum }\limits_{{n \leq x}}\varphi \left( n\right) \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\mathop{\sum }\limits_{{{d}^{\prime }d = n}}{d}^{\prime }\mu \left( d\right) \]\n\n\[ = \mathop{\sum }\limits_{{d \leq x}}\mu \left( d\right) \mathop{\sum }\limits_{{{d}^{\prime } \leq x/d}}{d}^{\prime } \]\n\n\[ = \frac{1}{2}\mathop{\sum }\limits_{{d \leq x}}\mu \left( d\right) \left\lbrack \frac{x}{d}\right\rbrack \left( {\left\lbrack \frac{x}{d}\right\rbrack + 1}\right) \]\n\n\[ = \frac{1}{2}\mathop{\sum }\limits_{{d \leq x}}\mu \left( d\right) \left( {{\left( \frac{x}{d}\right) }^{2} + O\left( \frac{x}{d}\right) }\right) \]\n\n\[ = \frac{{x}^{2}}{2}\mathop{\sum }\limits_{{d \leq x}}\frac{\mu \left( d\right) }{{d}^{2}} + O\left( {x\mathop{\sum }\limits_{{d \leq x}}\frac{1}{d}}\right) \]\n\n\[ = \frac{{x}^{2}}{2}\mathop{\sum }\limits_{{d = 1}}^{\infty }\frac{\mu \left( d\right) }{{d}^{2}} - \frac{{x}^{2}}{2}\mathop{\sum }\limits_{{d > x}}\frac{\mu \left( d\right) }{{d}^{2}} + O\left( {x\log x}\right) \]\n\n\[ = \frac{3{x}^{2}}{{\pi }^{2}} + O\left( {x\log x}\right) . \]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 6.22 The probability that two positive integers are relatively prime is \( 6/{\pi }^{2} \) .
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Proof. Let \( N \geq 1 \) . The number of ordered pairs of positive integers \( \left( {m, n}\right) \) such that \( 1 \leq m \leq n \leq N \) is \( N + \left( \begin{matrix} N \\ 2 \end{matrix}\right) = N\left( {N + 1}\right) /2 \) . The number of positive integers \( m \leq n \) that are relatively prime is \( \varphi \left( n\right) \), and so the number of pairs of positive integers \( \left( {m, n}\right) \) such that \( 1 \leq m \leq n \leq N \) and \( m \) and \( n \) are relatively prime is\n\n\[ \n\mathop{\sum }\limits_{{n \leq N}}\varphi \left( n\right) = \frac{3{N}^{2}}{{\pi }^{2}} + O\left( {N\log N}\right) \n\]\n\nTherefore, the frequency of relatively prime pairs of positive integers not exceeding \( N \) is\n\n\[ \n\frac{\frac{3{N}^{2}}{{\pi }^{2}} + O\left( {N\log N}\right) }{N\left( {N + 1}\right) /2} = \frac{6}{{\pi }^{2}} + O\left( \frac{\log N}{N}\right) \rightarrow \frac{6}{{\pi }^{2}} \n\]\n\nas \( N \rightarrow \infty \) . This completes the proof.
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Yes
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Theorem 7.1 The divisor function \( d\left( n\right) \) is multiplicative.
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Proof. Let \( m \) and \( n \) be relatively prime integers,\n\n\[ m = \mathop{\prod }\limits_{{p \mid m}}{p}^{{v}_{p}\left( m\right) }\n\]\n\nand\n\n\[ n = \mathop{\prod }\limits_{{q \mid n}}{q}^{{v}_{q}\left( n\right) }\n\]\n\nSince \( \left( {m, n}\right) = 1 \), the set of primes that divide \( m \) and the set of primes that divide \( n \) are disjoint. Therefore,\n\n\[ {mn} = \mathop{\prod }\limits_{{p \mid m}}{p}^{{v}_{p}\left( m\right) }\mathop{\prod }\limits_{{q \mid n}}{q}^{{v}_{q}\left( n\right) }\n\]\n\nis the standard factorization of \( {mn} \), and\n\n\[ d\left( {mn}\right) = \mathop{\prod }\limits_{{p \mid m}}\left( {{v}_{p}\left( m\right) + 1}\right) \mathop{\prod }\limits_{{q \mid n}}\left( {{v}_{q}\left( n\right) + 1}\right) = d\left( m\right) d\left( n\right) .\n\]\n\nThis completes the proof.
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Yes
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Theorem 7.2 For every \( \varepsilon > 0 \) ,\n\n\[ d\left( n\right) { \ll }_{\varepsilon }{n}^{\varepsilon } \]
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Proof. Let \( \varepsilon > 0 \) . The function \( f\left( n\right) = d\left( n\right) /{n}^{\varepsilon } \) is multiplicative. Therefore, by Theorem 6.20, it suffices to prove that\n\n\[ \mathop{\lim }\limits_{{{p}^{k} \rightarrow \infty }}f\left( {p}^{k}\right) = 0 \]\n\nfor every prime \( p \) . We observe that\n\n\[ \frac{k + 1}{{2}^{{k\varepsilon }/2}} \]\n\nis bounded for \( k \geq 1 \), and so\n\n\[ f\left( {p}^{k}\right) = \frac{d\left( {p}^{k}\right) }{{p}^{k\varepsilon }} \]\n\n\[ = \frac{k + 1}{{p}^{k\varepsilon }} \]\n\n\[ = \left( \frac{k + 1}{{p}^{{k\varepsilon }/2}}\right) \left( \frac{1}{{p}^{{k\varepsilon }/2}}\right) \]\n\n\[ \leq \left( \frac{k + 1}{{2}^{{k\varepsilon }/2}}\right) \left( \frac{1}{{p}^{{k\varepsilon }/2}}\right) \]\n\n\[ \ll \frac{1}{{p}^{{k\varepsilon }/2}} \]
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Yes
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Theorem 7.3 For \( x \geq 1 \) ,\n\n\[ D\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}d\left( n\right) = x\log x + \left( {{2\gamma } - 1}\right) x + O\left( \sqrt{x}\right) .
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Proof. We can interpret the divisor function \( d\left( n\right) \) and the sum function \( D\left( x\right) \) geometrically. A lattice point in the plane is a point whose coordinates are integers. A positive lattice point in the plane is a point whose coordinates are positive integers. In the \( {uv} \) -plane,\n\n\[ d\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}1 = \mathop{\sum }\limits_{{n = {uv}}}1 \]\n\ncounts the number of lattice points \( \left( {u, v}\right) \) on the rectangular hyperbola \( {uv} = n \) that lie in the quadrant \( u > 0, v > 0 \) . The sum function \( D\left( x\right) \) counts the number of lattice points in this quadrant that lie on or under the hyperbola \( {uv} = x \), that is, the number of positive lattice points \( \left( {u, v}\right) \) such that \( 1 \leq u \leq x \) and \( 1 \leq v \leq x/u \) . These lattice points can be divided into three pairwise disjoint classes:\n\n(i)\n\n\[ 1 \leq u \leq \sqrt{x}\;\text{ and }\;1 \leq v \leq \sqrt{x} \]\n\n(ii)\n\n\[ 1 \leq u \leq \sqrt{x}\;\text{ and }\;\sqrt{x} < v \leq x/u \]\n\n(iii)\n\n\[ \sqrt{x} < u \leq x\;\text{ and }\;1 \leq v \leq x/u \]\n\nThe third class consists of the lattice points \( \left( {u, v}\right) \) such that\n\n\[ 1 \leq v \leq \sqrt{x}\;\text{ and }\;\sqrt{x} < u \leq x/v. \]\n\nIt follows from Theorem 6.9 that\n\n\[ D\left( x\right) = {\left\lbrack \sqrt{x}\right\rbrack }^{2} + \mathop{\sum }\limits_{{1 \leq u \leq \sqrt{x}}}\left( {\left\lbrack \frac{x}{u}\right\rbrack - \left\lbrack \sqrt{x}\right\rbrack }\right) + \mathop{\sum }\limits_{{1 \leq v \leq \sqrt{x}}}\left( {\left\lbrack \frac{x}{v}\right\rbrack - \left\lbrack \sqrt{x}\right\rbrack }\right) \]\n\n\[ = {\left\lbrack \sqrt{x}\right\rbrack }^{2} + 2\mathop{\sum }\limits_{{1 \leq u \leq \sqrt{x}}}\left( {\left\lbrack \frac{x}{u}\right\rbrack - \left\lbrack \sqrt{x}\right\rbrack }\right) \]\n\n\[ = 2\mathop{\sum }\limits_{{1 \leq u \leq \sqrt{x}}}\left\lbrack \frac{x}{u}\right\rbrack - {\left\lbrack \sqrt{x}\right\rbrack }^{2} \]\n\n\[ = 2\mathop{\sum }\limits_{{1 \leq u \leq \sqrt{x}}}\left( {\frac{x}{u} - \left\{ \frac{x}{u}\right\} }\right) - {\left( \sqrt{x} - \{ \sqrt{x}\} \right) }^{2} \]\n\n\[ = {2x}\mathop{\sum }\limits_{{1 \leq u \leq \sqrt{x}}}\frac{1}{u} - 2\mathop{\sum }\limits_{{1 \leq u \leq \sqrt{x}}}\left\{ \frac{x}{u}\right\} - x + O\left( \sqrt{x}\right) \]\n\n\[ = {2x}\left( {\log \sqrt{x} + \gamma + O\left( \frac{1}{\sqrt{x}}\right) }\right) - x + O\left( \sqrt{x}\right) \]\n\n\[ = x\log x + \left( {{2\gamma } - 1}\right) x + O\left( \sqrt{x}\right) \text{.} \]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 7.4 For \( x \geq 1 \) ,\n\n\[ \Delta \left( x\right) = \mathop{\sum }\limits_{{n \leq x}}\left( {\log n - d\left( n\right) + {2\gamma }}\right) = O\left( {x}^{1/2}\right) . \]
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Proof. By Theorem 7.3 we have\n\n\[ \mathop{\sum }\limits_{{n \leq x}}d\left( n\right) = x\log x + \left( {{2\gamma } - 1}\right) x + O\left( {x}^{1/2}\right) . \]\n\nBy Theorem 6.4 we have\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\log n = x\log x - x + O\left( {\log x}\right) \]\n\nSubtracting the first equation from the second, we obtain\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\left( {\log n - d\left( n\right) + {2\gamma }}\right) = O\left( {x}^{1/2}\right) - {2\gamma }\{ x\} + O\left( {\log x}\right) = O\left( {x}^{1/2}\right) . \]
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Yes
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Theorem 7.5 For every \( \ell \geq 1 \), the function \( {d}_{\ell }\left( n\right) \) is multiplicative, and\n\n\[ \n{d}_{\ell }\left( {p}^{a}\right) = \left( \begin{matrix} a + \ell - 1 \\ \ell - 1 \end{matrix}\right) \n\] \n\nfor all prime powers \( {p}^{a} \) .
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Proof. Let \( \left( {m, n}\right) = 1 \) . For every ordered factorization of \( {mn} \) into \( \ell \) factors we can construct ordered factorizations of \( m \) and \( n \) into \( \ell \) parts, as follows. If \( {mn} = {d}_{1}\cdots {d}_{\ell } \) is an ordered factorization of \( {mn} \) into \( \ell \) parts, then, by Exercise 20 in Section 1.4, for each \( i = 1,\ldots ,\ell \) there exist unique integers \( {e}_{i} \) and \( {f}_{i} \) such that \( {e}_{i} \) divides \( m,{f}_{i} \) divides \( n \), and \( {d}_{i} = {e}_{i}{f}_{i} \) . Then \( m = {e}_{1}\cdots {e}_{\ell } \) and \( n = {f}_{1}\cdots {f}_{\ell } \) are ordered factorizations of \( m \) and \( n \) , respectively. This construction is reversible, and so establishes a bijection between ordered factorizations of \( {mn} \) and pairs of ordered factorizations of \( m \) and \( n \) . It follows that \( {d}_{\ell }\left( {mn}\right) = {d}_{\ell }\left( m\right) {d}_{\ell }\left( n\right) \), and so the divisor function \( {d}_{\ell } \) is multiplicative.\n\nAn ordered factorization of the prime power \( {p}^{a} \) can be written uniquely in the form \( {p}^{a} = {p}^{{b}_{1}}\cdots {p}^{{b}_{\ell }} \), where \( \left( {{b}_{1},\ldots ,{b}_{\ell }}\right) \) is an ordered \( \ell \) -tuple of nonnegative integers such that \( {b}_{1} + \cdots + {b}_{\ell } = a \) . It follows that \( {d}_{\ell }\left( {p}^{a}\right) \) is exactly the number of ordered partitions of \( a \) into exactly \( \ell \) nonnegative parts. Imagine a sequence of \( a + \ell - 1 \) red squares. If we choose \( \ell - 1 \) of these squares and color them blue, then the remaining \( a \) red squares are divided into exactly \( \ell \) subsequences (possibly empty) of consecutive red squares, separated by blue squares. Every ordered partition of \( a \) into \( \ell \) nonnegative parts can be uniquely constructed in this way, and so \( {d}_{\ell }\left( {p}^{a}\right) \) is the number of ways to choose \( \ell - 1 \) squares from a set of \( a + \ell - 1 \) squares, that is,\n\n\[ \n{d}_{\ell }\left( {p}^{a}\right) = \left( \begin{matrix} a + \ell - 1 \\ \ell - 1 \end{matrix}\right) . \n\] \n\nThis completes the proof.
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Yes
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Theorem 7.6 For \( \ell \geq 2 \) , \[ {D}_{\ell }\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}{d}_{\ell }\left( n\right) = \frac{1}{\left( {\ell - 1}\right) !}x{\log }^{\ell - 1}x + O\left( {x{\log }^{\ell - 2}x}\right) . \]
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Proof. The proof is by induction on \( \ell \) . By Theorem 7.3, \( {D}_{2}\left( x\right) = x\log x + \) \( O\left( x\right) \) . Now assume that the result holds for some integer \( \ell \geq 2 \) . The notation \( \mathop{\sum }\limits_{{{d}_{1}\cdots {d}_{\ell }}} \) means a sum over all ordered \( \ell \) -tuples \( \left( {{d}_{1},\ldots ,{d}_{\ell }}\right) \) of positive integers. Applying Theorem 6.7, we obtain \[ {D}_{\ell + 1}\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}{d}_{\ell + 1}\left( n\right) \] \[ = \mathop{\sum }\limits_{{n \leq x}}\mathop{\sum }\limits_{{{d}_{1}\cdots {d}_{\ell + 1} = n}}1 \] \[ = \mathop{\sum }\limits_{{n \leq x}}\mathop{\sum }\limits_{{{d}_{1}\cdots {d}_{\ell } \mid n}}1 \] \[ = \mathop{\sum }\limits_{{{d}_{1}\cdots {d}_{\ell } \leq x}}\left\lbrack \frac{x}{{d}_{1}\cdots {d}_{\ell }}\right\rbrack \] \[ = x\mathop{\sum }\limits_{{{d}_{1}\cdots {d}_{\ell } \leq x}}\frac{1}{{d}_{1}\cdots {d}_{\ell }} + O\left( {\mathop{\sum }\limits_{{{d}_{1}\cdots {d}_{\ell } \leq x}}1}\right) \] \[ = \frac{x{\log }^{\ell }x}{\ell !} + O\left( {x{\log }^{\ell - 1}x}\right) + O\left( {{D}_{\ell }\left( x\right) }\right) \] \[ = \frac{x{\log }^{\ell }x}{\ell !} + O\left( {x{\log }^{\ell - 1}x}\right) . \] This completes the proof. \( ▱ \)
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Yes
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Theorem 7.7\n\n\[ \n{d}^{2}\left( n\right) = \mathop{\sum }\limits_{{{\delta }^{2} \mid n}}\mu \left( \delta \right) {d}_{4}\left( \frac{n}{{\delta }^{2}}\right) .\n\]
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Proof. Define the arithmetic function \( \widetilde{\mu } \) as follows:\n\n\[ \n\widetilde{\mu }\left( n\right) = \left\{ \begin{array}{ll} \mu \left( \sqrt{n}\right) & \text{ if }n\text{ is a square,} \\ 0 & \text{ otherwise. } \end{array}\right.\n\]\n\nBy Exercise 1, the function \( \widetilde{\mu } \) is multiplicative. Since the Dirichlet convolution of multiplicative functions is multiplicative (Exercise 3 in Section 6.4), the function \( \widetilde{\mu } * {d}_{4} \) is multiplicative, and\n\n\[ \n\widetilde{\mu } * {d}_{4}\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\widetilde{\mu }\left( d\right) {d}_{4}\left( \frac{n}{d}\right)\n\]\n\n\[ \n= \mathop{\sum }\limits_{{{\delta }^{2} \mid n}}\mu \left( \delta \right) {d}_{4}\left( \frac{n}{{\delta }^{2}}\right)\n\]\n\nWe shall prove that \( \widetilde{\mu } * {d}_{4}\left( {p}^{a}\right) = {\left( a + 1\right) }^{2} \) for every prime power \( {p}^{a} \) . By Theorem 7.5,\n\n\[ \n{d}_{4}\left( {p}^{a}\right) = \left( \begin{matrix} a + 3 \\ 3 \end{matrix}\right)\n\]\n\nand so\n\n\[ \n\widetilde{\mu } * {d}_{4}\left( p\right) = \mathop{\sum }\limits_{{{\delta }^{2} \mid p}}\mu \left( \delta \right) {d}_{4}\left( \frac{p}{{\delta }^{2}}\right) = {d}_{4}\left( p\right) = \left( \begin{array}{l} 4 \\ 3 \end{array}\right) = 4.\n\]\n\nIf \( a \geq 2 \), then\n\n\[ \n\widetilde{\mu } * {d}_{4}\left( {p}^{a}\right) = \mathop{\sum }\limits_{{{\delta }^{2} \mid {p}^{a}}}\mu \left( \delta \right) {d}_{4}\left( \frac{{p}^{a}}{{\delta }^{2}}\right)\n\]\n\n\[ \n= {d}_{4}\left( {p}^{a}\right) - {d}_{4}\left( {p}^{a - 2}\right)\n\]\n\n\[ \n= \left( \begin{matrix} a + 3 \\ 3 \end{matrix}\right) - \left( \begin{matrix} a + 1 \\ 3 \end{matrix}\right)\n\]\n\n\[ \n= {\left( a + 1\right) }^{2}\text{. }\n\]\n\nSince \( d\left( {p}^{a}\right) = a + 1 \), it follows that\n\n\[ \n{d}^{2}\left( {p}^{a}\right) = {\left( a + 1\right) }^{2} = \widetilde{\mu } * {d}_{4}\left( {p}^{a}\right)\n\]\n\nfor all prime powers \( {p}^{a} \) . The functions \( {d}^{2} \) and \( \widetilde{\mu } * {d}_{4} \) are both multiplicative. Since multiplicative functions are completely determined by their values on prime powers (Exercise 2 in Section 6.4), it follows that\n\n\[ \n{d}^{2}\left( n\right) = \widetilde{\mu } * {d}_{4}\left( n\right)\n\]\n\nfor all positive integers \( n \) .
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No
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Theorem 7.8 (Ramanujan)\n\n\\[ \n\\mathop{\\sum }\\limits_{{n \\leq x}}{d}^{2}\\left( n\\right) \\sim \\frac{1}{{\\pi }^{2}}x{\\left( \\log x\\right) }^{3} \n\\]\n\nas \\( x \\rightarrow \\infty \\) .
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Proof. Applying Theorem 7.6 with \\( \\ell = 4 \\), we obtain\n\n\\[ \n{D}_{4}\\left( x\\right) = \\frac{x{\\log }^{3}x}{6} + O\\left( {x{\\log }^{2}x}\\right) .\n\\]\n\nBy Theorem 7.7 we have\n\n\\[ \n\\mathop{\\sum }\\limits_{{n \\leq x}}{d}^{2}\\left( n\\right) = \\mathop{\\sum }\\limits_{{n \\leq x}}\\mathop{\\sum }\\limits_{{{\\delta }^{2} \\mid n}}\\mu \\left( \\delta \\right) {d}_{4}\\left( \\frac{n}{{\\delta }^{2}}\\right)\n\\]\n\n\\[ \n= \\mathop{\\sum }\\limits_{{{\\delta }^{2}k \\leq x}}\\mu \\left( \\delta \\right) {d}_{4}\\left( k\\right)\n\\]\n\n\\[ \n= \\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\mu \\left( \\delta \\right) \\mathop{\\sum }\\limits_{{k \\leq x/{\\delta }^{2}}}{d}_{4}\\left( k\\right)\n\\]\n\n\\[ \n= \\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\mu \\left( \\delta \\right) {D}_{4}\\left( \\frac{x}{{\\delta }^{2}}\\right)\n\\]\n\n\\[ \n= \\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\mu \\left( \\delta \\right) \\left( {\\frac{x}{6{\\delta }^{2}}{\\log }^{3}\\frac{x}{{\\delta }^{2}} + O\\left( {\\frac{x}{{\\delta }^{2}}{\\log }^{2}\\frac{x}{{\\delta }^{2}}}\\right) }\\right)\n\\]\n\n\\[ \n= \\frac{x}{6}\\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\frac{\\mu \\left( \\delta \\right) }{{\\delta }^{2}}{\\log }^{3}\\frac{x}{{\\delta }^{2}} + O\\left( {x\\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\frac{1}{{\\delta }^{2}}{\\log }^{2}\\frac{x}{{\\delta }^{2}}}\\right) .\n\\]\n\nWe estimate these sums separately. The first term is\n\n\\[ \n\\frac{x}{6}\\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\frac{\\mu \\left( \\delta \\right) }{{\\delta }^{2}}{\\log }^{3}\\frac{x}{{\\delta }^{2}}\n\\]\n\n\\[ \n= \\frac{x}{6}\\mathop{\\sum }\\limits_{{i = 0}}^{3}\\left( \\begin{array}{l} 3 \\ i \\end{array}\\right) {\\left( -1\\right) }^{i}\\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\frac{\\mu \\left( \\delta \\right) }{{\\delta }^{2}}{\\log }^{3 - i}x{\\log }^{i}{\\delta }^{2}\n\\]\n\n\\[ \n= \\frac{x}{6}{\\log }^{3}x\\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\frac{\\mu \\left( \\delta \\right) }{{\\delta }^{2}} + O\\left( {x{\\log }^{2}x\\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\frac{{\\log }^{3}\\delta }{{\\delta }^{2}}}\\right)\n\\]\n\n\\[ \n= \\frac{x}{6}\\left( {\\frac{6}{{\\pi }^{2}} + O\\left( \\frac{1}{\\sqrt{x}}\\right) }\\right) {\\log }^{3}x + O\\left( {x{\\log }^{2}x\\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\frac{{\\log }^{3}\\delta }{{\\delta }^{2}}}\\right)\n\\]\n\n\\[ \n= \\frac{x{\\log }^{3}x}{{\\pi }^{2}} + + O\\left( {x{\\log }^{2}x}\\right)\n\\]\n\nby Theorem 6.17. Similarly,\n\n\\[ \nx\\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\frac{1}{{\\delta }^{2}}{\\log }^{2}\\frac{x}{{\\delta }^{2}} \\leq x{\\log }^{2}x\\mathop{\\sum }\\limits_{{\\delta \\leq \\sqrt{x}}}\\frac{1}{{\\delta }^{2}} \\ll x{\\log }^{2}x.\n\\]\n\nThis completes the proof of Ramanujan’s theorem. \\( ▱ \\)
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Yes
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Theorem 7.9 The arithmetic function \( \sigma \left( n\right) \) is multiplicative.
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Proof. Let \( m \) and \( n \) be relatively prime positive integers. Since no prime divides both \( m \) and \( n \), we have\n\n\[ \sigma \left( {mn}\right) = \mathop{\prod }\limits_{{p \mid {mn}}}\frac{{p}^{{v}_{p}\left( {mn}\right) + 1} - 1}{p - 1} \]\n\n\[ = \mathop{\prod }\limits_{{p \mid m}}\frac{{p}^{{v}_{p}\left( m\right) + 1} - 1}{p - 1}\mathop{\prod }\limits_{{p \mid n}}\frac{{p}^{{v}_{p}\left( n\right) + 1} - 1}{p - 1} \]\n\n\[ = \sigma \left( m\right) \sigma \left( n\right) \text{.} \] \n\nThis completes the proof.
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Yes
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Theorem 7.10 (Euler) An even integer \( n \) is perfect if and only if there exist prime numbers \( p \) and \( q \) such that\n\n\[ q = {2}^{p} - 1 \]\n\nand\n\n\[ n = {2}^{p - 1}q. \]
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Proof. If \( n \) is of this form, then \( q \) is odd and \( {2n} = {2}^{p}q \) . It follows that\n\n\[ \sigma \left( n\right) = \sigma \left( {2}^{p - 1}\right) \sigma \left( q\right) \]\n\n\[ = \left( {{2}^{p} - 1}\right) \left( {q + 1}\right) \]\n\n\[ = {2}^{p}q + \left( {{2}^{p} - q - 1}\right) \]\n\n\[ = {2n} \]\n\nand so \( n \) is perfect.\n\nConversely, if \( n \) is an even perfect number, then \( \sigma \left( n\right) = {2n} \) . Writing \( n \) in the form\n\n\[ n = {2}^{k - 1}m \]\n\nwhere \( m \) is odd and \( k \geq 2 \) (since \( n \) is even), we have\n\n\[ {2}^{k}m = {2n} = \sigma \left( n\right) = \sigma \left( {{2}^{k - 1}m}\right) = \sigma \left( {2}^{k - 1}\right) \sigma \left( m\right) = \left( {{2}^{k} - 1}\right) \sigma \left( m\right) . \]\n\nSince \( {2}^{k} - 1 \) divides \( {2}^{k}m \) and \( {2}^{k} - 1 \) is relatively prime to \( {2}^{k} \), Euclid’s lemma implies that \( {2}^{k} - 1 \) divides \( m \), and so\n\n\[ m = \left( {{2}^{k} - 1}\right) \ell \]\n\nfor some odd integer \( \ell \) . Then\n\n\[ {2}^{k}\left( {{2}^{k} - 1}\right) \ell = \left( {{2}^{k} - 1}\right) \sigma \left( {\left( {{2}^{k} - 1}\right) \ell }\right) . \]\n\nIf \( \ell > 1 \), then \( 1,\ell \), and \( \left( {{2}^{k} - 1}\right) \ell \) are distinct divisors of \( \left( {{2}^{k} - 1}\right) \ell \), and\n\n\[ {2}^{k}\ell = \sigma \left( {\left( {{2}^{k} - 1}\right) \ell }\right) \geq 1 + \ell + \left( {{2}^{k} - 1}\right) \ell = {2}^{k}\ell + 1, \]\n\nwhich is impossible. Therefore, \( \ell = 1 \) and\n\n\[ {2}^{k} = \sigma \left( {{2}^{k} - 1}\right) = 1 + \left( {{2}^{k} - 1}\right) + \mathop{\sum }\limits_{\substack{{d \mid \left( {{2}^{k} - 1}\right) } \\ {1 < d < {2}^{k} - 1} }}d \]\n\nit follows that \( {2}^{k} - 1 \) has no proper divisors, that is, \( {2}^{k} - 1 \) is a prime number. If the exponent \( k \) were composite, then \( k = {k}_{1}{k}_{2} \) with \( 1 < {k}_{1} \leq {k}_{2} < k \) , and\n\n\[ {2}^{k} - 1 = {\left( {2}^{{k}_{1}}\right) }^{{k}_{2}} - 1 = \left( {{2}^{{k}_{1}} - 1}\right) \left( {1 + {2}^{{k}_{1}} + {2}^{2{k}_{1}} + \cdots + {2}^{{k}_{1}\left( {{k}_{2} - 1}\right) }}\right) \]\n\nwould be composite, which is false. Therefore, \( k = p \) is also prime, and \( m = q = {2}^{p} - 1 \) . This completes the proof. \( ▱ \)
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Yes
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Lemma 7.1 For every \( x \geq 1 \) ,\n\n\[ \mathop{\sum }\limits_{\substack{{{uv} \leq x} \\ {\left( {u, v}\right) = 1} }}\frac{1}{uv} = \frac{3}{{\pi }^{2}}{\log }^{2}x + O\left( {\log x}\right) \]
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Proof. We define\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{\substack{{{uv} \leq x} \\ {\left( {u, v}\right) = 1} }}\frac{1}{uv} \]\n\n(7.3)\n\nand\n\n\[ g\left( x\right) = \mathop{\sum }\limits_{{{st} \leq x}}\frac{1}{st} = \mathop{\sum }\limits_{{n \leq x}}\frac{d\left( n\right) }{n}. \]\n\nIf \( {st} \leq x \) and \( r \) is a common divisor of \( s \) and \( t \), then \( {r}^{2} \leq {st} \leq x \), and so \( r \leq \sqrt{x} \) and\n\n\[ g\left( x\right) = \mathop{\sum }\limits_{{{st} \leq x}}\frac{1}{st} \]\n\n\[ = \mathop{\sum }\limits_{{r \leq {x}^{1/2}}}\mathop{\sum }\limits_{\substack{{{st} \leq x} \\ {\left( {s, t}\right) = r} }}\frac{1}{st} \]\n\n\[ = \mathop{\sum }\limits_{{r \leq {x}^{1/2}}}\frac{1}{{r}^{2}}\mathop{\sum }\limits_{\substack{{{uv} \leq x/{r}^{2}} \\ {\left( {u, v}\right) = 1} }}\frac{1}{uv} \]\n\n\[ = \mathop{\sum }\limits_{{r \leq {x}^{1/2}}}\frac{1}{{r}^{2}}f\left( \frac{x}{{r}^{2}}\right) \]\n\nApplying Möbius inversion (Exercise 7 of Section 6.3 with \( \alpha = 2 \) ), we obtain\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{r \leq {x}^{1/2}}}\frac{\mu \left( r\right) }{{r}^{2}}g\left( \frac{x}{{r}^{2}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{r \leq {x}^{1/2}}}\frac{\mu \left( r\right) }{{r}^{2}}\mathop{\sum }\limits_{{n \leq x/{r}^{2}}}\frac{d\left( n\right) }{n} \]\n\n\[ = \mathop{\sum }\limits_{{n{r}^{2} \leq x}}\frac{\mu \left( r\right) d\left( n\right) }{n{r}^{2}} \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\frac{d\left( n\right) }{n}\mathop{\sum }\limits_{{r \leq {\left( x/n\right) }^{1/2}}}\frac{\mu \left( r\right) }{{r}^{2}} \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\frac{d\left( n\right) }{n}\left( {\frac{6}{{\pi }^{2}} + O\left( {\left( \frac{n}{x}\right) }^{1/2}\right) }\right) \]\n\n\[ = \frac{6}{{\pi }^{2}}\mathop{\sum }\limits_{{n \leq x}}\frac{d\left( n\right) }{n} + O\left( {\frac{1}{{x}^{1/2}}\mathop{\sum }\limits_{{n \leq x}}\frac{d\left( n\right) }{{n}^{1/2}}}\right) \]\n\nby Theorem 6.17. Since\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{d\left( n\right) }{n} = \frac{{\log }^{2}x}{2} + O\left( {\log x}\right) \]\n\nand\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{d\left( n\right) }{{n}^{1/2}} = 2{x}^{1/2}\log x + O\left( {x}^{1/2}\right) \]\n\nby Exercises 13 and 14 of Section 7.1, it follows that\n\n\[ f\left( x\right) = \frac{3}{{\pi }^{2}}{\log }^{2}x + O\left( {\log x}\right) . \]\n\nThis completes the proof.
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Yes
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Lemma 7.2 Let \( A \) be a nonempty set of positive integers, and let \( {A}^{ * } \) be the subset of \( A \) consisting of all integers \( a \in A \) not divisible by any other element of \( A \) . Then \( {A}^{ * } \) is a primitive set, and\n\n\[ M\left( A\right) = M\left( {A}^{ * }\right) \]
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Proof. The primitivity of the set \( {A}^{ * } \) follows immediately from the definition.\n\nIf \( b \in M\left( A\right) \), then \( b \) is a multiple of \( a \) for some integer \( a \in A \) . If \( a \notin {A}^{ * } \) , then \( a \) has a proper divisor that belongs to \( A \) . Let \( {a}^{\prime } \) be the smallest element of \( A \) that divides \( a \) . Then \( {a}^{\prime } \in {A}^{ * } \), and \( b \) is a multiple of \( {a}^{\prime } \) . This completes the proof. \( ▱ \)
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Yes
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Lemma 7.3 If \( {A}_{1} \) and \( {A}_{2} \) are nonempty sets of positive integers such that \( M\left( {A}_{1}\right) = M\left( {A}_{2}\right) \), then \( M\left( {{A}_{1} \cap {A}_{2}}\right) = M\left( {A}_{1}\right) \) .
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Proof. By Exercise \( 4, M\left( {{A}_{1} \cap {A}_{2}}\right) \) is a subset of \( M\left( {A}_{1}\right) \) . If \( M\left( {{A}_{1} \cap {A}_{2}}\right) \) is a proper subset of \( M\left( {A}_{1}\right) \), then there exists a smallest integer \( b \in M\left( {A}_{1}\right) \smallsetminus \) \( M\left( {{A}_{1} \cap {A}_{2}}\right) \) . Since \( b \in M\left( {A}_{1}\right) = M\left( {A}_{2}\right) \), we have\n\n\[ b = {m}_{1}{a}_{1} = {m}_{2}{a}_{2} \]\n\nfor positive integers \( {m}_{1},{m}_{2},{a}_{1},{a}_{2} \) with \( {a}_{1} \in {A}_{1},{a}_{2} \in {A}_{2} \) . Moreover, \( {a}_{1} \neq \) \( {a}_{2} \) since \( b \notin M\left( {{A}_{1} \cap {A}_{2}}\right) \) . Suppose \( {a}_{1} < {a}_{2} \) . Since \( {a}_{1} \in M\left( {A}_{1}\right) \) and\n\n\[ {a}_{1} < {a}_{2} \leq {m}_{2}{a}_{2} = b \]\n\nthe minimality of \( b \) implies that \( a \in M\left( {{A}_{1} \cap {A}_{2}}\right) \) . Then \( {a}_{1} = {ma} \) for some \( a \in {A}_{1} \cap {A}_{2} \), and so \( b = {m}_{1}{a}_{1} = {m}_{1}{ma} \in M\left( {{A}_{1} \cap {A}_{2}}\right) \), which is absurd. It follows that \( M\left( {A}_{1}\right) = M\left( {{A}_{1} \cap {A}_{2}}\right) \) .
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No
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Theorem 7.13 Let \( B \) be a set of multiples. There exists a unique primitive set \( {A}^{ * } \) such that \( B = M\left( {A}^{ * }\right) \) .
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Proof. Let \( B = M\left( A\right) \) for some set \( A \), and let \( {A}^{ * } \) be the primitive subset of \( A \) constructed in Lemma 7.2. Then \( B = M\left( {A}^{ * }\right) \) . Let \( {A}^{\prime } \) be any set of positive integers such that \( B = M\left( {A}^{\prime }\right) \) . By Lemma 7.3,\n\n\[ B = M\left( {A}^{\prime }\right) = M\left( {{A}^{\prime } \cap {A}^{ * }}\right) = M\left( {A}^{ * }\right) . \]\n\nSince \( {A}^{\prime } \cap {A}^{ * } \) is a subset of \( {A}^{ * } \), it follows from Exercise 4 that \( {A}^{\prime } \cap {A}^{ * } = {A}^{ * } \) . Thus, \( {A}^{ * } \) is a subset of every set \( {A}^{\prime } \) such that \( M\left( {A}^{\prime }\right) = B \), and so \( {A}^{ * } \) is the primitive set uniquely defined by\n\n\[ {A}^{ * } = \mathop{\bigcap }\limits_{\substack{{{A}^{\prime } \subseteq \mathbf{N}} \\ {M\left( {A}^{\prime }\right) = B} }}{A}^{\prime } \]\n\nThis completes the proof.
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Yes
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Theorem 7.15 If \( A \) is an infinite set of integers with counting function\n\n\[ A\left( x\right) = O\left( \frac{x}{{\log }^{2}x}\right) \]\n\nfor \( x \geq 2 \), then the set of multiples \( M\left( A\right) \) has an asymptotic density.
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Proof. By Theorem 6.10, the infinite series \( \mathop{\sum }\limits_{{a \in A}}{a}^{-1} \) converges. It follows from Theorem 7.14 that the set of multiples \( M\left( A\right) \) has an asymptotic density. \( ▱ \)
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Yes
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Lemma 7.4 The number of positive integers \( n \leq x \) divisible by some prime power \( {p}^{r} \geq {\log }^{4}x \) with \( r \geq 2 \) is \( O\left( {x{\log }^{-2}x}\right) \) .
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Proof. If \( p \) is a prime such that \( p \geq {\log }^{2}x \) and \( {p}^{2} \) divides \( n \), then \( n \) is divisible by a prime power \( {p}^{r} \geq {\log }^{4}x \) with \( r \geq 2 \) . The number of such integers \( n \leq x \) is \( \left\lbrack {x/{p}^{2}}\right\rbrack \) .\n\nIf \( p < {\log }^{2}x \), let \( {u}_{p} \) be the least integer such that \( {p}^{{u}_{p}} \geq {\log }^{4}x \) . The number of integers \( n \leq x \) divisible by a prime power \( {p}^{r} \geq {\log }^{4}x \) is \( \left\lbrack {x/{p}^{{u}_{p}}}\right\rbrack \) .\n\nLet \( {N}_{1}\left( x\right) \) denote the number of integers \( n \leq x \) divisible by a prime power \( {p}^{r} \geq {\log }^{4}x \) . Then\n\n\[ \n{N}_{1}\left( x\right) \leq \mathop{\sum }\limits_{{p \geq {\log }^{2}x}}\left\lbrack \frac{x}{{p}^{2}}\right\rbrack + \mathop{\sum }\limits_{{p < {\log }^{2}x}}\left\lbrack \frac{x}{{p}^{{u}_{p}}}\right\rbrack \n\]\n\n\[ \n\leq x\mathop{\sum }\limits_{{p \geq {\log }^{2}x}}\frac{1}{{p}^{2}} + \left( \frac{x}{{\log }^{4}x}\right) \mathop{\sum }\limits_{{p < {\log }^{2}x}}1 \n\]\n\n\[ \n\leq x\mathop{\sum }\limits_{{n \geq {\log }^{2}x}}\frac{1}{{n}^{2}} + \left( \frac{x}{{\log }^{4}x}\right) {\log }^{2}x \n\]\n\n\[ \n\ll \frac{x}{{\log }^{2}x} \n\]\n\nThis completes the proof.
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Yes
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Lemma 7.5 Let \( x \geq {e}^{e} \) and \( y = \log \log x \) . The number of positive integers \( n \leq x \) such that either \( \omega \left( n\right) \geq {5y} \) or \( P\left( n\right) \leq {x}^{1/\left( {6y}\right) } \) is \( O\left( {x{\log }^{-2}x}\right) \) for all sufficiently large \( x \) .
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Proof. Let \( {N}_{2}\left( x\right) \) denote the number of positive integers \( n \leq x \) with \( \omega \left( n\right) \geq {5y} \) . By Exercise 9 in Section 7.1,\n\n\[ \n{N}_{2}\left( x\right) \ll \frac{x}{{\left( \log x\right) }^{5\log 2 - 1}} \leq \frac{x}{{\log }^{2}x}.\n\]\n\nLet \( p \) be a prime. If \( {p}^{r} \leq x \), then \( 0 \leq r \leq \log x/\log p \leq \log x/\log 2 \), and so the number of prime powers \( {p}^{r} \leq x \) with \( p \leq {x}^{1/\left( {6y}\right) } \) does not exceed\n\n\[ \n\left( {1 + \frac{\log x}{\log 2}}\right) {x}^{1/\left( {6y}\right) } \ll {x}^{1/\left( {6y}\right) }\log x.\n\]\n\nLet \( {N}_{3}\left( x\right) \) denote the number of integers \( n \leq x \) such that \( \omega \left( n\right) < {5y} \) and \( P\left( n\right) \leq {x}^{1/\left( {6y}\right) } \) . Then\n\n\[ \n{N}_{3}\left( x\right) \ll {\left( {x}^{1/\left( {6y}\right) }\log x\right) }^{5y} \ll \frac{x}{{\log }^{2}x}\n\]\n\nfor all sufficiently large \( x \) . \( ▱ \)
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Yes
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Lemma 7.7 Let \( n \leq x \) be a primitive \( k \) -abundant number satisfying conditions (i), (ii), and (iii) of Lemma 7.6. Then \( n \) is divisible by a prime \( p \) such that\n\n\[{\log }^{4}x \leq p \leq {x}^{1/\left( {13y}\right) }.\]
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Proof. If not, then we can write \( n = {ab} \), where \( a \) is a product of primes less than \( {\log }^{4}x \), and \( b \) is a product of primes greater than \( {x}^{1/\left( {13y}\right) } \). Since \( {x}^{1/\left( {13y}\right) } < {x}^{1/\left( {6y}\right) } \), condition (iii) implies that \( b > 1 \).\n\nBy condition (ii), \( \omega \left( b\right) \leq \omega \left( n\right) < {5y} \). Then\n\n\[\\frac{\\sigma \\left( b\\right) }{b} < \\mathop{\\prod }\\limits_{{p \\mid b}}\\left( {1 + \\frac{1}{p} + \\frac{1}{{p}^{2}} + \\cdots }\\right)\]\n\n\[\\leq \\mathop{\\prod }\\limits_{{p \\mid b}}\\left( {1 + \\frac{2}{p}}\\right)\]\n\n\[< {\\left( 1 + \\frac{2}{{x}^{1/\\left( {13y}\\right) }}\\right) }^{\\omega \\left( b\\right) }\]\n\n\[< {\\left( 1 + \\frac{2}{{x}^{1/\\left( {13y}\\right) }}\\right) }^{5y}\]\n\n\[< 1 + \\frac{20y}{{x}^{1/\\left( {13y}\\right) }}\]\n\nif \( x \) is sufficiently large (by Exercise 4 with \( c = 2 \) ). Every prime that divides \( a \) is less than \( {\\log }^{4}x \), and, by condition (i), every prime power that divides \( n \), and hence \( a \), is also less than \( {\\log }^{4}x \). Since \( \\omega \\left( a\\right) \\leq \\omega \\left( n\\right) < {5y} \) by condition (ii), it follows that\n\n\[1 \\leq a < {\\left( {\\log }^{4}x\\right) }^{5y} = {\\left( \\log x\\right) }^{20y}.\]\n\nBy condition (iii), \( b > 1 \), and so \( a < n \). Since \( a \) is a proper divisor of the primitive \( k \) -abundant number \( n \), we have\n\n\[\\sigma \\left( a\\right) < {ka}.\\text{.}\]\n\nSince \( k \) is an integer, we have\n\n\[\\sigma \\left( a\\right) \\leq {ka} - 1\]\n\nand so\n\n\[\\frac{\\sigma \\left( a\\right) }{a} \\leq k - \\frac{1}{a} < k - \\frac{1}{{\\left( \\log x\\right) }^{20y}}\]\n\nSince \( \\sigma \\left( n\\right) \) is multiplicative and \( n = {ab} \) with \( \\left( {a, b}\\right) = 1 \), we have, for \( x \) sufficiently large,\n\n\[\\frac{\\sigma \\left( n\\right) }{n} = \\frac{\\sigma \\left( a\\right) }{a}\\frac{\\sigma \\left( b\\right) }{b}\]\n\n\[< \\left( {k - \\frac{1}{{\\left( \\log x\\right) }^{20y}}}\\right) \\left( {1 + \\frac{20y}{{x}^{1/\\left( {13y}\\right) }}}\\right)\]\n\n\[< k + \\frac{20ky}{{x}^{1/\\left( {13y}\\right) }} - \\frac{1}{{\\left( \\log x\\right) }^{20y}}\]\n\n\[< k,\\text{,}\]\n\nwhich is impossible, since the integer \( n \) is \( k \) -abundant. Therefore, \( n \) must be divisible by a prime \( p \) in the interval (7.10).
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Yes
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Lemma 7.7 Let \( n \leq x \) be a primitive \( k \) -abundant number satisfying conditions (i), (ii), and (iii) of Lemma 7.6. Then \( n \) is divisible by a prime \( p \) such that\n\n\[{\log }^{4}x \leq p \leq {x}^{1/\left( {13y}\right) }.\]\n\n(7.10)
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Proof. If not, then we can write \( n = {ab} \), where \( a \) is a product of primes less than \( {\log }^{4}x \), and \( b \) is a product of primes greater than \( {x}^{1/\left( {13y}\right) } \) . Since \( {x}^{1/\left( {13y}\right) } < {x}^{1/\left( {6y}\right) } \), condition (iii) implies that \( b > 1 \) .\n\nBy condition (ii), \( \omega \left( b\right) \leq \omega \left( n\right) < {5y} \) . Then\n\n\[ \frac{\sigma \left( b\right) }{b} < \mathop{\prod }\limits_{{p \mid b}}\left( {1 + \frac{1}{p} + \frac{1}{{p}^{2}} + \cdots }\right)\]\n\n\[ \leq \mathop{\prod }\limits_{{p \mid b}}\left( {1 + \frac{2}{p}}\right)\]\n\n\[ < {\left( 1 + \frac{2}{{x}^{1/\left( {13y}\right) }}\right) }^{\omega \left( b\right) }\]\n\n\[ < {\left( 1 + \frac{2}{{x}^{1/\left( {13y}\right) }}\right) }^{5y}\]\n\n\[ < 1 + \frac{20y}{{x}^{1/\left( {13y}\right) }}\]\n\nif \( x \) is sufficiently large (by Exercise 4 with \( c = 2 \) ). Every prime that divides \( a \) is less than \( {\log }^{4}x \), and, by condition (i), every prime power that divides \( n \), and hence \( a \), is also less than \( {\log }^{4}x \) . Since \( \omega \left( a\right) \leq \omega \left( n\right) < {5y} \) by condition (ii), it follows that\n\n\[ 1 \leq a < {\left( {\log }^{4}x\right) }^{5y} = {\left( \log x\right) }^{20y}.\]\n\nBy condition (iii), \( b > 1 \), and so \( a < n \) . Since \( a \) is a proper divisor of the primitive \( k \) -abundant number \( n \), we have\n\n\[ \sigma \left( a\right) < {ka}.\text{.}\]\n\nSince \( k \) is an integer, we have\n\n\[ \sigma \left( a\right) \leq {ka} - 1 \]\n\nand so\n\n\[ \frac{\sigma \left( a\right) }{a} \leq k - \frac{1}{a} < k - \frac{1}{{\left( \log x\right) }^{20y}}\]\n\nSince \( \sigma \left( n\right) \) is multiplicative and \( n = {ab} \) with \( \left( {a, b}\right) = 1 \), we have, for \( x \) sufficiently large,\n\n\[ \frac{\sigma \left( n\right) }{n} = \frac{\sigma \left( a\right) }{a}\frac{\sigma \left( b\right) }{b}\]\n\n\[ < \left( {k - \frac{1}{{\left( \log x\right) }^{20y}}}\right) \left( {1 + \frac{20y}{{x}^{1/\left( {13y}\right) }}}\right)\]\n\n\[ < k + \frac{20ky}{{x}^{1/\left( {13y}\right) }} - \frac{1}{{\left( \log x\right) }^{20y}}\]\n\n\[ < k,\]\n\nwhich is impossible, since the integer \( n \) is \( k \) -abundant. Therefore, \( n \) must be divisible by a prime \( p \) in the interval (7.10).
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Yes
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Lemma 7.8 If \( x \) is sufficiently large and \( n \leq x \) is a primitive \( k \) -abundant number satisfying conditions (i), (ii), and (iii) of Lemma 7.6, then\n\n\[ k \leq \frac{\sigma \left( n\right) }{n} < k + \frac{k}{{x}^{1/\left( {6y}\right) }} \]
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Proof. By condition (iii), the integer \( n \) is divisible by a prime \( p \) such that\n\n\[ p \geq P\left( n\right) > {x}^{1/\left( {6y}\right) }.\]\n\nSince \( {p}^{2} > {x}^{1/\left( {3y}\right) } > {\log }^{4}x \) for \( x \) sufficiently large, condition (i) implies that \( {p}^{2} \) does not divide \( n \) . Therefore \( n = {mp} \), where \( \left( {m, p}\right) = 1 \) and \( \sigma \left( m\right) < {km} \) since \( n \) is primitive \( k \) -abundant. It follows that\n\n\[ \frac{\sigma \left( n\right) }{n} = \frac{\sigma \left( m\right) }{m}\frac{\sigma \left( p\right) }{p} < k\left( {1 + \frac{1}{p}}\right) < k + \frac{k}{{x}^{1/\left( {6y}\right) }}.\]\n\nThis completes the proof.
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Yes
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Theorem 7.16 For every integer \( k \geq 2 \), let \( P{A}_{k}\left( x\right) \) denote the number of primitive \( k \) -abundant numbers not exceeding \( x \) . Then\n\n\[ P{A}_{k}\left( x\right) \ll \frac{x}{{\log }^{2}x} \]\n\nand the set \( {A}_{k} \) of \( k \) -abundant numbers possesses an asymptotic density
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Proof. By Lemma 7.6 there are only \( O\left( {x{\log }^{-2}x}\right) \) primitive \( k \) -abundant integers that fail to satisfy conditions (i), (ii), and (iii) of Lemma 7.6.\n\nLet \( t \) be the number of primitive \( k \) -abundant integers \( n \leq x \) that do satisfy these three conditions. We denote these numbers by \( {n}_{1},\ldots ,{n}_{t} \) . By Lemma 7.7, corresponding to each integer \( {n}_{i} \) there is a prime \( {p}_{i} \) such that \( {p}_{i} \) exactly divides \( {n}_{i} \) and\n\n\[ {\log }^{4}x \leq {p}_{i} \leq {x}^{1/\left( {13y}\right) }.\]\n\nLet \( {n}_{i} = {p}_{i}{m}_{i} \) . Then \( \left( {{p}_{i},{m}_{i}}\right) = 1 \) and\n\n\[ 1 \leq {m}_{i} \leq \frac{x}{{\log }^{4}x} \]\n\nIt suffices to prove that the integers \( {m}_{i} \) are distinct.\n\nSuppose that \( {m}_{i} = {m}_{j} \) for some \( i \neq j \) . Then \( {p}_{i} \neq {p}_{j} \) . Since\n\n\[ \frac{\sigma \left( {n}_{i}\right) }{{n}_{i}} = \frac{\left( {p}_{i} + 1\right) }{{p}_{i}}\frac{\sigma \left( {m}_{i}\right) }{{m}_{i}} \]\n\nand\n\n\[ \frac{\sigma \left( {n}_{j}\right) }{{n}_{j}} = \frac{\left( {p}_{j} + 1\right) }{{p}_{j}}\frac{\sigma \left( {m}_{i}\right) }{{m}_{i}} \]\n\nit follows that\n\n\[ \frac{\sigma \left( {n}_{i}\right) {n}_{j}}{{n}_{i}\sigma \left( {n}_{j}\right) } = \frac{\left( {{p}_{i} + 1}\right) {p}_{j}}{{p}_{i}\left( {{p}_{j} + 1}\right) }.\]\n\nSince \( {p}_{i} \) and \( {p}_{j} \) are distinct primes, it follows that \( \left( {{p}_{i} + 1}\right) {p}_{j} \neq {p}_{i}\left( {{p}_{j} + 1}\right) \) . We can assume that \( \left( {{p}_{i} + 1}\right) {p}_{j} > {p}_{i}\left( {{p}_{j} + 1}\right) \), and so\n\n\[ \frac{\sigma \left( {n}_{i}\right) {n}_{j}}{{n}_{i}\sigma \left( {n}_{j}\right) } = \frac{\left( {{p}_{i} + 1}\right) {p}_{j}}{{p}_{i}\left( {{p}_{j} + 1}\right) } \]\n\n\[ \geq 1 + \frac{1}{{p}_{i}\left( {{p}_{j} + 1}\right) } \]\n\n\[ \geq 1 + \frac{1}{{x}^{1/\left( {13y}\right) }\left( {{x}^{1/\left( {13y}\right) } + 1}\right) } \]\n\n\[ \geq 1 + \frac{1}{2{x}^{2/\left( {13y}\right) }} \]\n\nBy Lemma 7.8,\n\n\[ \frac{\sigma \left( {n}_{i}\right) {n}_{j}}{{n}_{i}\sigma \left( {n}_{j}\right) } < \left( {k + \frac{k}{{x}^{1/\left( {6y}\right) }}}\right) \frac{1}{k} < 1 + \frac{1}{{x}^{1/\left( {6y}\right) }}. \]\n\nThis is a contradiction, since\n\n\[ 2{x}^{2/\left( {13y}\right) } < {x}^{1/\left( {6y}\right) } \]\n\nfor all sufficiently large \( x \) . It follows that the numbers \( {m}_{1},\ldots ,{m}_{t} \) are distinct, and so \( t \leq x{\log }^{-4}x \) . This completes the proof. \( ▱ \)
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Yes
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Lemma 8.1 Let \( n \geq 1 \) and \( 1 \leq k \leq n \) . Then\n\n\[ \left( \begin{matrix} n \\ k - 1 \end{matrix}\right) < \left( \begin{array}{l} n \\ k \end{array}\right) \;\text{ if and only if }k < \frac{n + 1}{2}, \]\n\n\[ \left( \begin{matrix} n \\ k - 1 \end{matrix}\right) > \left( \begin{array}{l} n \\ k \end{array}\right) \;\text{ if and only if }k > \frac{n + 1}{2}, \]\n\n\[ \left( \begin{matrix} n \\ k - 1 \end{matrix}\right) = \left( \begin{array}{l} n \\ k \end{array}\right) \;\text{ if and only if }n\text{ is odd and }k = \frac{n + 1}{2}. \]
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Proof. Consider the ratio\n\n\[ r\left( k\right) = \frac{\left( \begin{array}{l} n \\ k \end{array}\right) }{\left( \begin{matrix} n \\ k - 1 \end{matrix}\right) } = \frac{\frac{n!}{k!\left( {n - k}\right) !}}{\frac{n!}{\left( {k - 1}\right) !\left( {n - k + 1}\right) !}} = \frac{\left( {k - 1}\right) !\left( {n - k + 1}\right) !}{k!\left( {n - k}\right) !} = \frac{n - k + 1}{k}. \]\n\nThen \( r\left( k\right) > 1 \) if and only if \( k < \left( {n + 1}\right) /2 \), and \( r\left( k\right) < 1 \) if and only if \( k > \left( {n + 1}\right) /2 \) .
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Yes
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Lemma 8.2 For all positive integers \( n \) , \n\n\[ \n\frac{{2}^{2n}}{2n} \leq \left( \begin{matrix} {2n} \\ n \end{matrix}\right) < {2}^{2n} \n\]
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Proof. By the binomial theorem, \n\n\[ \n{2}^{2n} = {\left( 1 + 1\right) }^{2n} = \mathop{\sum }\limits_{{k = 0}}^{{2n}}\left( \begin{array}{l} n \\ k \end{array}\right) > \left( \begin{matrix} {2n} \\ n \end{matrix}\right) . \n\] \n\nBy Lemma 8.1, the middle binomial coefficient \( \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \) is the largest binomial coefficient in the expansion of \( {\left( 1 + 1\right) }^{2n} \) . Therefore, \n\n\[ \n{2}^{2n} = \mathop{\sum }\limits_{{k = 0}}^{{2n}}\left( \begin{matrix} {2n} \\ k \end{matrix}\right) = 1 + \mathop{\sum }\limits_{{k = 1}}^{{{2n} - 1}}\left( \begin{matrix} {2n} \\ k \end{matrix}\right) + 1 \n\] \n\n\[ \n\leq 2 + \left( {{2n} - 1}\right) \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \n\] \n\n\[ \n\leq {2n}\left( \begin{matrix} {2n} \\ n \end{matrix}\right) \n\] \n\nThis completes the proof.
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Yes
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Theorem 8.1 For every positive integer \( n \) , \n\n\[ \mathop{\prod }\limits_{{p \leq n}}p < {4}^{n} \]
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Proof. Let \( m \geq 1 \) . We consider the binomial coefficients \n\n\[ M = \left( \begin{matrix} {2m} + 1 \\ m \end{matrix}\right) = \left( \begin{matrix} {2m} + 1 \\ m + 1 \end{matrix}\right) \] \n\n\[ = \frac{\left( {{2m} + 1}\right) {2m}\left( {{2m} - 1}\right) \left( {{2m} - 2}\right) \cdots \left( {m + 2}\right) }{m!}. \] \n\nThis is an integer, since \( M \) is a binomial coefficient. Moreover, \n\n\[ {2M} = \left( \begin{matrix} {2m} + 1 \\ m \end{matrix}\right) + \left( \begin{matrix} {2m} + 1 \\ m + 1 \end{matrix}\right) \] \n\n\[ < \mathop{\sum }\limits_{{k = 0}}^{{{2m} + 1}}\left( \begin{matrix} {2m} + 1 \\ k \end{matrix}\right) \] \n\n\[ = {2}^{{2m} + 1}\text{,} \] \nand so \n\n\[ M < {4}^{m}\text{.} \] \n\nIf \( p \) is a prime number such that \( m + 2 \leq p \leq {2m} + 1 \), then \( p \) divides the product \n\n\[ \left( {{2m} + 1}\right) {2m}\left( {{2m} - 1}\right) \left( {{2m} - 2}\right) \cdots \left( {m + 2}\right) , \] \n\nbut \( p \) does not divide \( m \) !. It follows that \( p \) divides \( M \), and so \n\n\[ \mathop{\prod }\limits_{{m + 2 \leq p \leq {2m} + 1}}p \] \n\ndivides \( M \) . Therefore, \n\n\[ \mathop{\prod }\limits_{{m + 2 \leq p \leq {2m} + 1}}p \leq M < {4}^{m} \] \n\n(8.3) \n\nfor all positive integers \( m \) . \n\nWe shall prove inequality (8.1) by induction on \( n \) . This inequality holds for \( n = 1 \) and \( n = 2 \), since \( 1 < {4}^{1} \) and \( 2 < {4}^{2} \), respectively. Let \( n \geq 3 \), and assume that (8.1) holds for all positive integers \( m < n \) . If \( n \) is even, then \n\n\[ \mathop{\prod }\limits_{{p \leq n}}p = \mathop{\prod }\limits_{{p \leq n - 1}}p < {4}^{n - 1} < {4}^{n} \] \n\nIf \( n \) is odd, then \( n = {2m} + 1 \) for some \( m \geq 1 \), and \n\n\[ \mathop{\prod }\limits_{{p \leq n}}p = \mathop{\prod }\limits_{{p \leq m + 1}}p\mathop{\prod }\limits_{{m + 2 \leq p \leq {2m} + 1}}p. \] \n\nBy the induction hypothesis we have \n\n\[ \mathop{\prod }\limits_{{p \leq m + 1}}p < {4}^{m + 1} \] \n\n(8.4) \n\nIt follows from (8.3) and (8.4) that \n\n\[ \mathop{\prod }\limits_{{p \leq n}}p = \mathop{\prod }\limits_{{p \leq m + 1}}p\mathop{\prod }\limits_{{m + 2 \leq p \leq {2m} + 1}}p < {4}^{m + 1}{4}^{m} = {4}^{{2m} + 1} = {4}^{n}. \] \n\nThis proves (8.1). \n\nInequality (8.2) follows from (8.1) as follows. If \( x \geq 1 \), then \( n = \left\lbrack x\right\rbrack \geq 1 \) \n\nand \n\n\[ \vartheta \left( x\right) = \vartheta \left( n\right) = \log \mathop{\prod }\limits_{{p \leq n}}p < n\log 4 \leq x\log 4. \] \n\nThe proof that (8.2) implies (8.1) is similar. \( ▱ \)
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Yes
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Theorem 8.3 Let \( {p}_{n} \) denote the nth prime number. There exist positive constants \( a \) and \( b \) such that\n\n\[ \n\text{an}\log n \leq {p}_{n} \leq {bn}\log n \n\]\n\nfor all \( n \geq 2 \) .
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Proof. By Chebyshev’s inequality (8.5), there exist positive constants \( A \) and \( B \) such that\n\n\[ \nA{p}_{n} \leq \pi \left( {p}_{n}\right) \log {p}_{n} = n\log {p}_{n} \leq B{p}_{n}. \n\]\n\nLet \( a = {B}^{-1} > 0 \) . Since \( {p}_{n} \geq n \), we have\n\n\[ \n{p}_{n} \geq {B}^{-1}n\log {p}_{n} \geq {an}\log n. \n\]\n\nSimilarly,\n\n\[ \n{p}_{n} \leq {A}^{-1}n\log {p}_{n} \n\]\n\nFor \( n \) sufficiently large,\n\n\[ \n\log {p}_{n} \leq \log n + \log \log {p}_{n} - \log A \n\]\n\n\[ \n\leq \log n + 2\log \log {p}_{n} \n\]\n\n\[ \n\leq \log n + \left( {1/2}\right) \log {p}_{n} \n\]\n\nand so\n\n\[ \n\log {p}_{n} \leq 2\log n \n\]\n\nTherefore, there exists an integer \( {n}_{0} \geq 2 \) such that\n\n\[ \n{p}_{n} \leq {A}^{-1}n\log {p}_{n} \leq 2{A}^{-1}n\log n \n\]\n\nfor all \( n \geq {n}_{0} \) . Since \( {p}_{n}/n\log n \) is bounded for \( 2 \leq n \leq {n}_{0} \), there exists a constant \( b \) such that \( {p}_{n} \leq {bn}\log n \) for all \( n \geq 2 \) . This completes the proof.
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Yes
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Theorem 8.4 For \( x \geq 2 \) ,\n\n\[ \mathop{\sum }\limits_{{m \leq x}}\psi \left( \frac{x}{m}\right) = \mathop{\sum }\limits_{{d \leq x}}\Lambda \left( d\right) \left\lbrack \frac{x}{d}\right\rbrack = x\log x - x + O\left( {\log x}\right) . \]
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Proof. With \( f\left( n\right) = \Lambda \left( n\right) \) in Theorem 6.15, we have\n\n\[ F\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}\Lambda \left( n\right) = \psi \left( x\right) \]\n\nand so\n\n\[ \mathop{\sum }\limits_{{m \leq x}}\psi \left( \frac{x}{m}\right) = \mathop{\sum }\limits_{{d \leq x}}\Lambda \left( d\right) \left\lbrack \frac{x}{d}\right\rbrack \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\mathop{\sum }\limits_{{d \mid n}}\Lambda \left( d\right) \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\log n \]\n\n\[ = x\log x - x + O\left( {\log x}\right) \text{.} \]\n\nThe last identity comes from Theorem 6.4.
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Yes
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Theorem 8.5 (Mertens) For \( x \geq 1 \) ,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\Lambda \left( n\right) }{n} = \log x + O\left( 1\right) \]\n\n(8.10)\n\nand\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p} = \log x + O\left( 1\right) \]\n\n(8.11)
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Proof. Since \( \psi \left( x\right) = O\left( x\right) \) by Chebyshev’s theorem, we have\n\n\[ x\log x - x + O\left( {\log x}\right) = \mathop{\sum }\limits_{{d \leq x}}\Lambda \left( d\right) \left\lbrack \frac{x}{d}\right\rbrack \]\n\n\[ = \mathop{\sum }\limits_{{d \leq x}}\Lambda \left( d\right) \left( {\frac{x}{d} - \left\{ \frac{x}{d}\right\} }\right) \]\n\n\[ = x\mathop{\sum }\limits_{{d \leq x}}\frac{\Lambda \left( d\right) }{d} - \mathop{\sum }\limits_{{d \leq x}}\Lambda \left( d\right) \left\{ \frac{x}{d}\right\} \]\n\n\[ = x\mathop{\sum }\limits_{{d \leq x}}\frac{\Lambda \left( d\right) }{d} + O\left( {\psi \left( x\right) }\right) \]\n\n\[ = x\mathop{\sum }\limits_{{d \leq x}}\frac{\Lambda \left( d\right) }{d} + O\left( x\right) \]\n\nWe obtain equation (8.10) by dividing by \( x \) .\n\nNext, we observe that\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\Lambda \left( n\right) }{n} - \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p} = \mathop{\sum }\limits_{\substack{{{p}^{k} \leq x} \\ {k \geq 2} }}\frac{\log p}{{p}^{k}} \]\n\n\[ \leq \mathop{\sum }\limits_{{p \leq x}}\log p\mathop{\sum }\limits_{{k = 2}}^{\infty }\frac{1}{{p}^{k}} \]\n\n\[ \leq \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p\left( {p - 1}\right) } \]\n\n\[ \ll \text{1 .} \]\n\nThis proves (8.11). \( ▱ \)
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Yes
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Theorem 8.6\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\vartheta \left( n\right) }{{n}^{2}} = \log x + O\left( 1\right) \]
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Proof. We begin with the convergent series\n\n\[ \mathop{\sum }\limits_{{k \leq x}}\frac{\ell \left( k\right) }{{k}^{2}} \leq \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{\ell \left( k\right) }{{k}^{2}} < \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{\log k}{{k}^{2}} < \infty . \]\n\nBy Theorem 6.3 applied to the function \( f\left( t\right) = 1/{t}^{2} \), we have\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\vartheta \left( n\right) }{{n}^{2}} = \mathop{\sum }\limits_{{n \leq x}}\mathop{\sum }\limits_{{k \leq n}}\frac{\ell \left( k\right) }{{n}^{2}} \]\n\n\[ = \mathop{\sum }\limits_{{k \leq x}}\ell \left( k\right) \mathop{\sum }\limits_{{k \leq n \leq x}}\frac{1}{{n}^{2}} \]\n\n\[ = \mathop{\sum }\limits_{{k \leq x}}\ell \left( k\right) \left( {\frac{1}{k} - \frac{1}{x} + O\left( \frac{1}{{k}^{2}}\right) }\right) \]\n\n\[ = \mathop{\sum }\limits_{{k \leq x}}\frac{\ell \left( k\right) }{k} - \frac{\vartheta \left( x\right) }{x} + O\left( {\mathop{\sum }\limits_{{k \leq x}}\frac{\ell \left( k\right) }{{k}^{2}}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p} + O\left( 1\right) \]\n\n\[ = \log x + O\left( 1\right) \]\n\nby Theorem 8.5. \( ▱ \)
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Yes
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Theorem 8.7 (Mertens) There exists a constant \( {b}_{1} \) such that\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p} = \log \log x + {b}_{1} + O\left( \frac{1}{\log x}\right) \]\n\nfor \( x \geq 2 \) .
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Proof. We can write\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p} = \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p}\frac{1}{\log p} = \mathop{\sum }\limits_{{2 \leq n \leq x}}f\left( n\right) g\left( n\right) ,\]\n\nwhere\n\n\[ f\left( n\right) = \left\{ \begin{array}{ll} \frac{\log p}{p} & \text{ if }n = p \\ 0 & \text{ otherwise,} \end{array}\right. \]\n\nand\n\n\[ g\left( t\right) = \frac{1}{\log t}\;\text{ for }t > 1. \]\n\nLet\n\n\[ F\left( t\right) = \mathop{\sum }\limits_{{n \leq t}}f\left( n\right) = \mathop{\sum }\limits_{{p \leq t}}\frac{\log p}{p}. \]\n\nThen \( F\left( t\right) = 0 \) for \( t < 2 \) . By Theorem 8.5,\n\n\[ F\left( t\right) = \log t + r\left( t\right) ,\;\text{ where }r\left( t\right) = O\left( 1\right) . \]\n\nTherefore, the integral\n\n\[ {\int }_{2}^{\infty }\frac{r\left( t\right) }{t{\left( \log t\right) }^{2}}{dt} \]\n\nconverges absolutely, and\n\n\[ {\int }_{x}^{\infty }\frac{r\left( t\right) {dt}}{t{\left( \log t\right) }^{2}} = O\left( \frac{1}{\log x}\right) . \]\n\nBy partial summation, we obtain\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p} = \mathop{\sum }\limits_{{n \leq x}}f\left( n\right) g\left( n\right) \]\n\n\[ = F\left( x\right) g\left( x\right) - {\int }_{2}^{x}F\left( t\right) {g}^{\prime }\left( t\right) {dt} \]\n\n\[ = \frac{\log x + r\left( x\right) }{\log x} + {\int }_{2}^{x}\frac{\log t + r\left( t\right) }{t{\left( \log t\right) }^{2}}{dt} \]\n\n\[ = 1 + O\left( \frac{1}{\log x}\right) + {\int }_{2}^{x}\frac{1}{t\log t}{dt} + {\int }_{2}^{x}\frac{r\left( t\right) }{t{\left( \log t\right) }^{2}}{dt} \]\n\n\[ = \log \log x + 1 - \log \log 2 + {\int }_{2}^{\infty }\frac{r\left( t\right) }{t{\left( \log t\right) }^{2}}{dt} \]\n\n\[ - {\int }_{x}^{\infty }\frac{r\left( t\right) }{t{\left( \log t\right) }^{2}}{dt} + O\left( \frac{1}{\log x}\right) \]\n\n\[ = \log \log x + {b}_{1} + O\left( \frac{1}{\log x}\right) \]\n\nwhere\n\n\[ {b}_{1} = 1 - \log \log 2 + {\int }_{2}^{\infty }\frac{r\left( t\right) }{t{\left( \log t\right) }^{2}}{dt}. \]\n\n(8.12)\n\nThis completes the proof.
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Yes
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Theorem 8.8 (Mertens’s formula) There exists a constant \( \gamma \) such that for \( x \geq 2 \) ,\n\n\[ \mathop{\prod }\limits_{{p \leq x}}{\left( 1 - \frac{1}{p}\right) }^{-1} = {e}^{\gamma }\log x + O\left( 1\right) \]
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Proof. We begin with two observations. First, the series \( \mathop{\sum }\limits_{p}\mathop{\sum }\limits_{{k = 2}}^{\infty }{p}^{-k}/k \) converges, since\n\n\[ \mathop{\sum }\limits_{p}\mathop{\sum }\limits_{{k = 2}}^{\infty }\frac{1}{k{p}^{k}} < \mathop{\sum }\limits_{p}\mathop{\sum }\limits_{{k = 2}}^{\infty }\frac{1}{{p}^{k}} = \mathop{\sum }\limits_{p}\frac{1}{p\left( {p - 1}\right) } < \mathop{\sum }\limits_{{n = 2}}^{\infty }\frac{1}{n\left( {n - 1}\right) } < \infty . \]\n\nLet\n\n\[ {b}_{2} = \mathop{\sum }\limits_{p}\mathop{\sum }\limits_{{k = 2}}^{\infty }\frac{1}{k{p}^{k}} > 0 \]\n\nSecond, for \( x \geq 2 \) ,\n\n\[ 0 < \mathop{\sum }\limits_{{p > x}}\mathop{\sum }\limits_{{k = 2}}^{\infty }\frac{1}{k{p}^{k}} < \mathop{\sum }\limits_{{p > x}}\frac{1}{p\left( {p - 1}\right) } < \mathop{\sum }\limits_{{n > x}}\frac{1}{n\left( {n - 1}\right) } \]\n\n\[ = \mathop{\sum }\limits_{{n = \left\lbrack x\right\rbrack + 1}}^{\infty }\left( {\frac{1}{n - 1} - \frac{1}{n}}\right) = \frac{1}{\left\lbrack x\right\rbrack } \]\n\n\[ \leq \frac{2}{x} \]\n\nFrom the Taylor series\n\n\[ - \log \left( {1 - t}\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{t}^{k}}{k}\;\text{ for }\left| t\right| < 1 \]\n\nand Theorem 8.7 we obtain\n\n\[ \log \mathop{\prod }\limits_{{p \leq x}}{\left( 1 - \frac{1}{p}\right) }^{-1} = \mathop{\sum }\limits_{{p \leq x}}\log {\left( 1 - \frac{1}{p}\right) }^{-1} \]\n\n\[ = \mathop{\sum }\limits_{{p \leq x}}\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{k{p}^{k}} \]\n\n\[ = \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p} + \mathop{\sum }\limits_{{p \leq x}}\mathop{\sum }\limits_{{k = 2}}^{\infty }\frac{1}{k{p}^{k}} \]\n\n\[ = \log \log x + {b}_{1} + O\left( \frac{1}{\log x}\right) + {b}_{2} - \mathop{\sum }\limits_{{p > x}}\mathop{\sum }\limits_{{k = 2}}^{\infty }\frac{1}{k{p}^{k}} \]\n\n\[ = \log \log x + {b}_{1} + {b}_{2} + O\left( \frac{1}{\log x}\right) + O\left( \frac{1}{x}\right) \]\n\n\[ = \log \log x + {b}_{1} + {b}_{2} + O\left( \frac{1}{\log x}\right) . \]\n\nLet \( \gamma = {b}_{1} + {b}_{2} \) . Then\n\n\[ \mathop{\prod }\limits_{{p \leq x}}{\left( 1 - \frac{1}{p}\right) }^{-1} = {e}^{\gamma }\log x\exp \left( {O\left( \frac{1}{\log x}\right) }\right) . \]\n\nSince \( \exp \left( t\right) = 1 + O\left( t\right) \) for \( t \) in any bounded interval \( \left\lbrack {0,{t}_{0}}\right\rbrack \), and since \( O\left( {1/\log x}\right) \) is bounded for \( x \geq 2 \), we have\n\n\[ \exp \left( {O\left( \frac{1}{\log x}\right) }\right) = 1 + O\left( \frac{1}{\log x}\right) . \]\n\nTherefore,\n\n\[ \mathop{\prod }\limits_{{p \leq x}}{\left( 1 - \frac{1}{p}\right) }^{-1} = {e}^{\gamma }\log x\exp \left( {O\left( \frac{1}{\log x}\right) }\right) \]\n\n\[ = {e}^{\gamma }\log x\left( {1 + O\left( \frac{1}{\log x}\right) }\right) \]\n\n\[ = {e}^{\gamma }\log x + O\left( 1\right) . \]\n\nThis is Mertens’s formula. \( ▱ \)
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Yes
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Theorem 8.9 For \( x \geq 2 \) ,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\omega \left( n\right) = x\log \log x + {b}_{1}x + O\left( \frac{x}{\log x}\right) ,\]\n\nwhere \( {b}_{1} \) is the positive real number defined by (8.12).
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Proof. Applying Chebyshev's theorem (Theorem 8.2) and Mertens's theorem (Theorem 8.7), we obtain\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\omega \left( n\right) = \mathop{\sum }\limits_{{n \leq x}}\mathop{\sum }\limits_{{p \mid n}}1 = \mathop{\sum }\limits_{{p \leq x}}\mathop{\sum }\limits_{\substack{{n \leq x} \\ {p \mid n} }}1 \]\n\n\[ = \mathop{\sum }\limits_{{p \leq x}}\left\lbrack \frac{x}{p}\right\rbrack = \mathop{\sum }\limits_{{p \leq x}}\frac{x}{p} + O\left( {\pi \left( x\right) }\right) \]\n\n\[ = x\mathop{\sum }\limits_{{p \leq x}}\frac{1}{p} + O\left( \frac{x}{\log x}\right) \]\n\n\[ = x\left( {\log \log x + {b}_{1} + O\left( \frac{1}{\log x}\right) }\right) + O\left( \frac{x}{\log x}\right) \]\n\n\[ = x\log \log x + {b}_{1}x + O\left( \frac{x}{\log x}\right) . \]
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Yes
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Theorem 8.10 For \( x \geq 2 \) ,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\omega {\left( n\right) }^{2} = x{\left( \log \log x\right) }^{2} + O\left( {x\log \log x}\right) . \]
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Proof. We have\n\n\[ \omega {\left( n\right) }^{2} = {\left( \mathop{\sum }\limits_{{p \mid n}}1\right) }^{2} = \left( {\mathop{\sum }\limits_{{{p}_{1} \mid n}}1}\right) \left( {\mathop{\sum }\limits_{{{p}_{2} \mid n}}1}\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{p}_{1}{p}_{2} \mid n} \\ {{p}_{1} \neq {p}_{2}} }}1 + \mathop{\sum }\limits_{{p \mid n}}1 = \mathop{\sum }\limits_{\substack{{{p}_{1}{p}_{2} \mid n} \\ {{p}_{1} \neq {p}_{2}} }}1 + \omega \left( n\right) . \]\n\nBy Theorem 8.9,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\omega {\left( n\right) }^{2} = \mathop{\sum }\limits_{{n \leq x}}\mathop{\sum }\limits_{\substack{{{p}_{1}{p}_{2} \mid n} \\ {{p}_{1} \neq {p}_{2}} }}1 + \mathop{\sum }\limits_{{n \leq x}}\omega \left( n\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{p}_{1}{p}_{2} \leq x} \\ {{p}_{1} \neq {p}_{2}} }}\mathop{\sum }\limits_{\substack{{n \leq x} \\ {{p}_{1}{p}_{2} \mid n} }}1 + x\log \log x + O\left( x\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{p}_{1}{p}_{2} \leq x} \\ {{p}_{1} \neq {p}_{2}} }}\left\lbrack \frac{x}{{p}_{1}{p}_{2}}\right\rbrack + O\left( {x\log \log x}\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{{p}_{1}{p}_{2} \leq x} \\ {{p}_{1} \neq {p}_{2}} }}\frac{x}{{p}_{1}{p}_{2}} + O\left( {\mathop{\sum }\limits_{\substack{{{p}_{1}{p}_{2} \leq x} \\ {{p}_{1} \neq {p}_{2}} }}1}\right) + O\left( {x\log \log x}\right) \]\n\n\[ = x\mathop{\sum }\limits_{\substack{{{p}_{1}{p}_{2} \leq x} \\ {{p}_{1} \neq {p}_{2}} }}\frac{1}{{p}_{1}{p}_{2}} + O\left( {x\log \log x}\right) \]\n\nsince, by the Fundamental Theorem of Arithmetic, there are at most \( {2x} \) ordered pairs \( \left( {{p}_{1},{p}_{2}}\right) \) of distinct primes such that \( {p}_{1}{p}_{2} \leq x \). From Theorem 8.7, we obtain\n\n\[ \mathop{\sum }\limits_{\substack{{{p}_{1}{p}_{2} \leq x} \\ {{p}_{1} \neq {p}_{2}} }}\frac{1}{{p}_{1}{p}_{2}} \leq {\left( \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p}\right) }^{2} \]\n\n\[ = {\left( \log \log x + O\left( 1\right) \right) }^{2} \]\n\n\[ = {\left( \log \log x\right) }^{2} + O\left( {\log \log x}\right) \]\n\nand\n\n\[ \mathop{\sum }\limits_{\substack{{{p}_{1}{p}_{2} \leq x} \\ {{p}_{1} \neq {p}_{2}} }}\frac{1}{{p}_{1}{p}_{2}} \geq {\left( \mathop{\sum }\limits_{{p \leq \sqrt{x}}}\frac{1}{p}\right) }^{2} - \mathop{\sum }\limits_{{p \leq \sqrt{x}}}\frac{1}{{p}^{2}} \]\n\n\[ = {\left( \log \log \sqrt{x} + O\left( 1\right) \right) }^{2} + O\left( 1\right) \]\n\n\[ = {\left( \log \log x\right) }^{2} + O\left( {\log \log x}\right) . \]\n\nTherefore,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\omega {\left( n\right) }^{2} = x{\left( \log \log x\right) }^{2} + O\left( {x\log \log x}\right) . \]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 8.11 (Chebyshev’s inequality) Let \( S \) be a finite set of integers, and let \( f \) be a real-valued function defined on \( S \) . Let \( \mu \) and \( t \) be real numbers with \( t > 0 \) . Then the number of integers \( n \in S \) such that\n\n\[ \left| {f\left( n\right) - \mu }\right| \geq t \]\n\ndoes not exceed\n\n\[ \frac{1}{{t}^{2}}\mathop{\sum }\limits_{{n \in S}}{\left( f\left( n\right) - \mu \right) }^{2} \]
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Proof. If \( \left| {f\left( n\right) - \mu }\right| \geq t \), then\n\n\[ 1 \leq \frac{{\left( f\left( n\right) - \mu \right) }^{2}}{{t}^{2}} \]\n\nand\n\n\[ \operatorname{card}\{ n \in S : \left| {f\left( n\right) - \mu }\right| \geq t\} = \mathop{\sum }\limits_{\substack{{n \in S} \\ {\left| {f\left( n\right) - \mu }\right| \geq t} }}1 \]\n\n\[ \leq \mathop{\sum }\limits_{\substack{{n \in S} \\ {\left| {f\left( n\right) - \mu }\right| \geq t} }}\frac{{\left( f\left( n\right) - \mu \right) }^{2}}{{t}^{2}} \]\n\n\[ \leq \frac{1}{{t}^{2}}\mathop{\sum }\limits_{{n \in S}}{\left( f\left( n\right) - \mu \right) }^{2} \]
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Yes
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Theorem 8.12 (Hardy-Ramanujan) For every \( \delta > 0 \), the number of integers \( n \leq x \) such that\n\n\[ \left| {\omega \left( n\right) - \log \log n}\right| \geq {\left( \log \log x\right) }^{\frac{1}{2} + \delta } \]\n\nis \( o\left( x\right) \) .
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Proof. (Turán [143]) Let \( S \) be the set of positive integers \( n \) not exceeding \( x, f\left( n\right) = \omega \left( n\right) \), and \( \mu = \log \log x \) . Applying Chebyshev’s inequality, we see that for any \( t > 0 \), the number of integers \( n \leq x \) such that \( \mid \omega \left( n\right) - \) \( \log \log x \mid \geq t \) is at most\n\n\[ \frac{1}{{t}^{2}}\mathop{\sum }\limits_{{n \leq x}}{\left( \omega \left( n\right) - \log \log x\right) }^{2} \]\n\nWe use Theorem8.9 and Theorem8.10 to evaluate this sum as follows:\n\n\[ \mathop{\sum }\limits_{{n \leq x}}{\left( \omega \left( n\right) - \log \log x\right) }^{2} \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\omega {\left( n\right) }^{2} - 2\log \log x\mathop{\sum }\limits_{{n \leq x}}\omega \left( n\right) + \mathop{\sum }\limits_{{n \leq x}}{\left( \log \log x\right) }^{2} \]\n\n\[ = x{\left( \log \log x\right) }^{2} + O\left( {x\log \log x}\right) - 2\log \log x\left( {x\log \log x + O\left( x\right) }\right) \]\n\n\[ + x{\left( \log \log x\right) }^{2} + O\left( {\left( \log \log x\right) }^{2}\right) \]\n\n\[ = O\left( {x\log \log x}\right) \text{.} \]\n\nLet \( \delta > 0 \) and \( t = {\left( \log \log x\right) }^{\frac{1}{2} + \delta } - 1 \) . Then\n\n\[ {t}^{2} > {\left( \log \log x\right) }^{1 + {2\delta }} - 2{\left( \log \log x\right) }^{\frac{1}{2} + \delta } \]\n\n\[ = {\left( \log \log x\right) }^{1 + \delta }\left( {{\left( \log \log x\right) }^{\delta } - 2{\left( \log \log x\right) }^{-1/2}}\right) \]\n\n\[ \geq {\left( \log \log x\right) }^{1 + \delta } \]\n\nfor \( x \) sufficiently large. Therefore, if\n\n\[ T = \left\{ {n \in S : \left| {\omega \left( n\right) - \log \log x}\right| \geq {\left( \log \log x\right) }^{\frac{1}{2} + \delta } - 1}\right\} \]\n\nthen\n\n\[ \left| T\right| \ll \frac{x\log \log x}{{\left( {\left( \log \log x\right) }^{\frac{1}{2} + \delta } - 1\right) }^{2}} \]\n\n\[ < \frac{x\log \log x}{{\left( \log \log x\right) }^{1 + \delta }} \]\n\n\[ = \;\frac{x}{{\left( \log \log x\right) }^{\delta }} \]\n\n\[ = o\left( x\right) \text{.} \]\n\nLet \( x > {e}^{e} \) . If\n\n\[ {x}^{1/e} \leq n \leq x \]\n\nthen\n\n\[ 0 < \log \log x - 1 \leq \log \log n \leq \log \log x. \]\n\nIf\n\n\[ \left| {\omega \left( n\right) - \log \log n}\right| \geq {\left( \log \log x\right) }^{\frac{1}{2} + \delta } \]\n\nthen\n\n\[\left| {\omega \left( n\right) - \log \log x}\right| \geq \left| {\omega \left( n\right) - \log \log n}\right| - \left| {\log \log x - \log \log n}\right| \]\n\n\[ \geq {\left( \log \log x\right) }^{\frac{1}{2} + \delta } - 1\]\n\n\[ = t\text{.}\]\n\nTherefore, if\n\n\[ U = \left\{ {n \in S : \left| {\omega \left( n\right) - \log \log n}\right| \geq {\left( \log \log x\right) }^{\frac{1}{2} + \delta }}\right\} ,\]\n\nthen \( U \subseteq T \) and so\n\n\[ \left| U\right| \leq {x}^{1/e} + \left| T\right| = o\left( x\right) \]\n\nThis completes the proof.
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Yes
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Theorem 9.1 For every positive integer \( n \) , \n\n\[ \n{\Lambda }_{2}\left( n\right) = \Lambda \left( n\right) \log n + \Lambda * \Lambda \left( n\right) . \n\]
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Proof. Recall that pointwise multiplication by the logarithm function \( L\left( n\right) \) is a derivation on the ring of arithmetic functions (Theorem 6.2). Multiplying the identity \( L = 1 * \Lambda \) by \( L \), we obtain \n\n\[ \n{L}^{2} = L \cdot L \n\] \n\n\[ \n= L \cdot \left( {1 * \Lambda }\right) \n\] \n\n\[ \n= \;1 * \left( {L \cdot \Lambda }\right) + \left( {L \cdot 1}\right) * \Lambda \n\] \n\n\[ \n= 1 * \left( {\Lambda \cdot L}\right) + L * \Lambda \text{.} \n\] \n\nTherefore, \n\n\[ \n{\Lambda }_{2} = \mu * {L}^{2} = \mu * 1 * \left( {\Lambda \cdot L}\right) + \mu * L * \Lambda = \Lambda \cdot L + \Lambda * \Lambda , \n\] \n\nwhich is the formula we want.
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Yes
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Lemma 9.1 For \( x > e \) ,\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p\left( {1 + \log \frac{x}{p}}\right) } \ll \log \log x \]
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Proof. By Mertens's theorem (Theorem 8.5), for every positive integer \( j \) we have\n\n\[ \mathop{\sum }\limits_{{\frac{x}{{e}^{j}} < p \leq \frac{x}{{e}^{j - 1}}}}\frac{\log p}{p} = \left( {\log \frac{x}{{e}^{j - 1}} + O\left( 1\right) }\right) - \left( {\log \frac{x}{{e}^{j}} + O\left( 1\right) }\right) = O\left( 1\right) .\n\nMoreover, if\n\n\[ \frac{x}{{e}^{j}} < p \leq \frac{x}{{e}^{j - 1}} \]\n\nthen\n\n\[ j \leq 1 + \log \frac{x}{p} < j + 1 \]\n\nand so\n\n\[ \mathop{\sum }\limits_{{\frac{x}{{e}^{j}} < p \leq \frac{x}{{e}^{j - 1}}}}\frac{\log p}{p\left( {1 + \log \frac{x}{p}}\right) } \leq \frac{1}{j}\mathop{\sum }\limits_{{\frac{x}{{e}^{j}} < p \leq \frac{x}{{e}^{j - 1}}}}\frac{\log p}{p} \ll \frac{1}{j}. \]\n\nTherefore,\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p\left( {1 + \log \frac{x}{p}}\right) } = \mathop{\sum }\limits_{{j = 1}}^{{\left\lbrack {\log x}\right\rbrack + 1}}\mathop{\sum }\limits_{{\frac{x}{{e}^{j}} < p \leq \frac{x}{{e}^{j - 1}}}}^{{\left\lbrack {\log x}\right\rbrack + 1}}\frac{\log p}{p\left( {1 + \log \frac{x}{p}}\right) } \]\n\n\[ \ll \mathop{\sum }\limits_{{j = 1}}^{{\left\lbrack {\log x}\right\rbrack + 1}}\frac{1}{j} \]\n\n\[ \ll \log \log x\text{.} \]\n\nThis completes the proof.
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Yes
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Theorem 9.6 For \( k \geq 1 \), let\n\n\[ \n{\Pi }_{k}^{ * }\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}{r}_{k}\left( n\right) = \mathop{\sum }\limits_{\substack{{\left( {{p}_{1},\ldots ,{p}_{k}}\right) \in {\mathbf{P}}^{k}} \\ {{p}_{1}\cdots {p}_{k} \leq x} }}1. \n\]\n\nThen\n\n\[ \nk!{\pi }_{k}\left( x\right) \leq {\Pi }_{k}^{ * }\left( x\right) \leq k!{\pi }_{k}^{ * }\left( x\right) \ll x. \n\]\n\nFor \( k \geq 2 \) ,\n\n\[ \n0 \leq {\pi }_{k}^{ * }\left( x\right) - {\pi }_{k}\left( x\right) \leq {\Pi }_{k - 1}^{ * }\left( x\right) \n\]
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Proof. We have\n\n\[ \n{\Pi }_{k}^{ * }\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}{r}_{k}\left( n\right) \leq k!\mathop{\sum }\limits_{\substack{{n \leq x} \\ {{r}_{k}\left( n\right) > 0} }}1 = k!{\pi }_{k}^{ * }\left( x\right) \leq k!x \ll x \n\]\n\nand\n\n\[ \n{\Pi }_{k}^{ * }\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}{r}_{k}\left( n\right) \geq k!\mathop{\sum }\limits_{\substack{{n \leq x} \\ {{r}_{k}\left( n\right) = k!} }}1 = k!{\pi }_{k}\left( x\right) . \n\]\n\nLet \( k \geq 2 \) . The function \( {\pi }_{k}^{ * }\left( x\right) - {\pi }_{k}\left( x\right) \) counts the number of positive integers \( n \leq x \) that can be written as a product of \( k \) primes but not as a product of \( k \) distinct primes. Every such integer is of the form \( n = \) \( {p}_{1}\cdots {p}_{k - 2}{p}_{k - 1}^{2} \) . Therefore,\n\n\[ \n{\pi }_{k}^{ * }\left( x\right) - {\pi }_{k}\left( x\right) \leq \mathop{\sum }\limits_{\substack{{\left( {{p}_{1},\ldots ,{p}_{k - 1}}\right) \in {\mathbf{P}}^{k - 1}} \\ {{p}_{1}\cdots {p}_{k - 1}^{2} \leq x} }}1 \n\]\n\n\[ \n\leq \mathop{\sum }\limits_{\substack{{\left( {{p}_{1},\ldots ,{p}_{k - 1}}\right) \in {\mathbf{P}}^{k - 1}} \\ {{p}_{1}\cdots {p}_{k - 1} \leq x} }}1 \n\]\n\n\[ \n= {\Pi }_{k - 1}^{ * }\left( x\right) \text{.} \n\]\n\nThis completes the proof.
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Yes
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Theorem 9.7 Let \( {S}_{0}\left( x\right) = 1 \) . For \( k \geq 1 \), let\n\n\[ \n{S}_{k}\left( x\right) = \mathop{\sum }\limits_{\substack{{\left( {{p}_{1},\ldots ,{p}_{k}}\right) \in {\mathbf{P}}^{k}} \\ {{p}_{1}\cdots {p}_{k} \leq x} }}\frac{1}{{p}_{1}\cdots {p}_{k}} = \mathop{\sum }\limits_{{n \leq x}}\frac{{r}_{k}\left( n\right) }{n}.\n\]\n\nThen\n\n\[ \n{S}_{k}\left( x\right) \sim {\left( \log \log x\right) }^{k} \n\]\n\nand\n\n\[ \n{S}_{k}\left( x\right) = \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p}{S}_{k - 1}\left( \frac{x}{p}\right) .\n\]
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Proof. By Theorem 8.7,\n\n\[ \n{S}_{1}\left( x\right) = \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p} \sim \log \log x \n\]\n\nand so\n\n\[ \n{S}_{1}\left( {x}^{1/k}\right) \sim \log \log {x}^{1/k} \sim \log \log x \n\]\n\nfor all \( k \geq 1 \) . Since\n\n\[ \n{\left( {S}_{1}\left( {x}^{1/k}\right) \right) }^{k} = {\left( \mathop{\sum }\limits_{{p \leq {x}^{1/k}}}\frac{1}{p}\right) }^{k} = \mathop{\sum }\limits_{\substack{{\left( {{p}_{1},\ldots ,{p}_{k}}\right) \in {\mathbf{P}}^{k}} \\ {{p}_{i} \leq {x}^{1/k}} }}\frac{1}{{p}_{1}\cdots {p}_{k}} \n\]\n\n\[ \n\leq \mathop{\sum }\limits_{\substack{{\left( {{p}_{1},\ldots ,{p}_{k}}\right) \in {\mathbf{P}}^{k}} \\ {{p}_{1}\cdots {p}_{k} \leq x} }}\frac{1}{{p}_{1}\cdots {p}_{k}} = {S}_{k}\left( x\right) \n\]\n\n\[ \n\leq {\left( \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p}\right) }^{k} = {S}_{1}{\left( x\right) }^{k} \n\]\n\n\nit follows that\n\n\[ \n{S}_{k}\left( x\right) \sim {\left( \log \log x\right) }^{k}. \n\]\n\nAlso,\n\n\[ \n{S}_{k}\left( x\right) = \mathop{\sum }\limits_{\substack{{\left( {{p}_{1},\ldots ,{p}_{k - 1},{p}_{k}}\right) \in {\mathbf{P}}^{k}} \\ {{p}_{1}\cdots {p}_{k - 1}{p}_{k} \leq x} }}\frac{1}{{p}_{1}\cdots {p}_{k - 1}{p}_{k}} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{{p}_{k} \leq x}}\frac{1}{{p}_{k}}\mathop{\sum }\limits_{\substack{{\left( {{p}_{1},\ldots ,{p}_{k - 1}}\right) \in {\mathbf{P}}^{k - 1}} \\ {{p}_{1}\cdots {p}_{k - 1} \leq x/{p}_{k}} }}\frac{1}{{p}_{1}\cdots {p}_{k - 1}} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{{p}_{k} \leq x}}\frac{1}{{p}_{k}}{S}_{k - 1}\left( \frac{x}{{p}_{k}}\right) \n\]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 9.9 For \( k \geq 1 \) , \[ {\pi }_{k}\left( x\right) \sim {\pi }_{k}^{ * }\left( x\right) \sim \frac{x{\left( \log \log x\right) }^{k - 1}}{k\log x}. \]
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Proof. This follows from Theorem 9.8 by partial summation. We have \[ {\vartheta }_{k}\left( x\right) = \mathop{\sum }\limits_{\substack{{\left( {{p}_{1},\ldots ,{p}_{k}}\right) \in {\mathbf{P}}^{k}} \\ {{p}_{1}\cdots {p}_{k} \leq x} }}\log {p}_{1}\cdots {p}_{k} = \mathop{\sum }\limits_{{n \leq x}}{r}_{k}\left( n\right) \log n \] and, by Theorem 9.6, the arithmetic function \( {r}_{k}\left( n\right) \) has mean value \[ {\Pi }_{k}^{ * }\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}{r}_{k}\left( n\right) = O\left( x\right) \] Then \[ {\vartheta }_{k}\left( x\right) = {\Pi }_{k}^{ * }\left( x\right) \log x - {\int }_{1}^{x}\frac{{\Pi }_{k}^{ * }\left( t\right) {dt}}{t} \] \[ = {\Pi }_{k}^{ * }\left( x\right) \log x + O\left( x\right) \text{.} \] It follows that \[ {\Pi }_{k}^{ * }\left( x\right) = \frac{{\vartheta }_{k}\left( x\right) }{\log x} + O\left( \frac{x}{\log x}\right) \sim \frac{{kx}{\left( \log \log x\right) }^{k - 1}}{\log x}. \] For \( k \geq 2 \) , \[ {\Pi }_{k - 1}^{ * }\left( x\right) = o\left( {{\Pi }_{k}^{ * }\left( x\right) }\right) . \] By Theorem 9.6, \[ {\Pi }_{k}^{ * }\left( x\right) \leq k!{\pi }_{k}^{ * }\left( x\right) \leq k!{\pi }_{k}\left( x\right) + k!{\Pi }_{k - 1}^{ * }\left( x\right) \leq {\Pi }_{k}^{ * }\left( x\right) + k!{\Pi }_{k - 1}^{ * }\left( x\right) , \] and so \[ {\pi }_{k}^{ * }\left( x\right) \sim {\pi }_{k}\left( x\right) \sim \frac{{\Pi }_{k}^{ * }\left( x\right) }{k!} \sim \frac{x{\left( \log \log x\right) }^{k - 1}}{\left( {k - 1}\right) !\log x}. \] This completes the proof.
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Yes
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Theorem 10.1 (Orthogonality relations) Let \( \mathop{\sum }\limits_{{a{\;(\operatorname{mod}\;m)}}} \) denote the sum over a complete set of residue classes modulo \( m \), and let \( \mathop{\sum }\limits_{{\chi \;\left( {\;\operatorname{mod}\;m}\right) }} \) denote the sum over the \( \varphi \left( m\right) \) Dirichlet characters modulo \( m \) . If \( \chi \) is a Dirichlet character modulo \( m \), then\n\n\[ \mathop{\sum }\limits_{{a{\;(\operatorname{mod}\;m)}}}\chi \left( a\right) = \left\{ \begin{array}{ll} \varphi \left( m\right) & \text{ if }\chi = {\chi }_{0}, \\ 0 & \text{ if }\chi \neq {\chi }_{0} \end{array}\right. \]\n\nIf \( a \) is an integer, then\n\n\[ \mathop{\sum }\limits_{{\chi {\;(\operatorname{mod}\;m)}}}\chi \left( a\right) = \left\{ \begin{array}{lll} \varphi \left( m\right) & \text{ if }a \equiv 1 & \left( {\;\operatorname{mod}\;m}\right) , \\ 0 & \text{ if }a ≢ 1 & \left( {\;\operatorname{mod}\;m}\right) . \end{array}\right. \]
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Proof. This is simply Theorem 4.6 applied to the multiplicative group \( {\left( \mathbf{Z}/m\mathbf{Z}\right) }^{ \times } \) .
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Yes
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Theorem 10.2 (Orthogonality relations) Let \( \mathop{\sum }\limits_{{a\;\left( {\;\operatorname{mod}\;m}\right) }} \) denote the sum over a complete set of residue classes modulo \( m \), and let \( \mathop{\sum }\limits_{\chi }\left( {\;\operatorname{mod}\;m}\right) \) denote the sum over the \( \varphi \left( m\right) \) Dirichlet characters modulo \( m \) . If \( {\chi }_{1} \) and \( {\chi }_{2} \) are Dirichlet characters modulo \( m \), then\n\n\[ \mathop{\sum }\limits_{{a{\;(\operatorname{mod}\;m)}}}{\chi }_{1}\left( a\right) \overline{{\chi }_{2}}\left( a\right) = \left\{ \begin{array}{ll} \varphi \left( m\right) & \text{ if }{\chi }_{1} = {\chi }_{2}, \\ 0 & \text{ if }{\chi }_{1} \neq {\chi }_{2}. \end{array}\right. \]\n\nIf \( a \) and \( b \) are integers, then\n\n\[ \mathop{\sum }\limits_{{\chi {\;(\operatorname{mod}\;m)}}}\chi \left( a\right) \overline{\chi }\left( b\right) = \left\{ \begin{array}{ll} \varphi \left( m\right) & \text{ if }\left( {a, m}\right) = \left( {b, m}\right) = 1\;\text{ and }\;a \equiv b{\;(\operatorname{mod}\;m)}, \\ 0 & \text{ otherwise. } \end{array}\right. \]
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Proof. This is Theorem 4.7 applied to the multiplicative group \( {\left( \mathbf{Z}/m\mathbf{Z}\right) }^{ \times } \) . 口
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Yes
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Theorem 10.4 Let \( \\chi \) be a nonprincipal character modulo \( m \) . The Dirichlet \( L \) -function \( L\\left( {s,\\chi }\\right) \) is analytic in the half-plane \( \\sigma > 0 \) . Let \( K \) be a compact set in the half-plane \( \\sigma > 0 \) . For \( s \\in K \) and \( x \\geq 1 \) ,
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Proof. To prove that \( L\\left( {s,\\chi }\\right) \) is analytic in \( \\sigma > 0 \), it suffices to prove that the series \( \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }\\chi \\left( n\\right) {n}^{-s} \) converges uniformly on every compact subset of the right half-plane \( \\sigma > 0 \) .\n\nLet \( K \) be a compact subset of the right half-plane. There exist positive constants \( \\delta \) and \( \\Delta \) such that \( \\sigma \\geq \\delta \) and \( \\left| s\\right| \\leq \\Delta \) for every \( s = \\sigma + {it} \\in K \) . We use partial summation (Theorem 6.8) with\n\n\[ f\\left( n\\right) = \\chi \\left( n\\right) \]\n\nand\n\n\[ g\\left( t\\right) = \\frac{1}{{t}^{s}} \]\n\nBy Exercise 6 in Section 10.1, \( F\\left( t\\right) = \\mathop{\\sum }\\limits_{{n \\leq t}}\\chi \\left( n\\right) \\ll 1 \) and\n\n\[ \\mathop{\\sum }\\limits_{{N < n \\leq M}}\\frac{\\chi \\left( n\\right) }{{n}^{s}} = \\mathop{\\sum }\\limits_{{N < n \\leq M}}f\\left( n\\right) g\\left( n\\right) \]\n\n\[ = F\\left( M\\right) g\\left( M\\right) - F\\left( N\\right) g\\left( N\\right) - {\\int }_{N}^{M}F\\left( t\\right) {g}^{\\prime }\\left( t\\right) {dt} \]\n\n\[ = \\frac{F\\left( M\\right) }{{M}^{s}} - \\frac{F\\left( N\\right) }{{N}^{s}} + s{\\int }_{N}^{M}\\frac{F\\left( t\\right) }{{t}^{s + 1}}{dt} \]\n\n\[ \\ll \\frac{1}{{M}^{\\sigma }} + \\frac{1}{{N}^{\\sigma }} + \\left| s\\right| {\\int }_{N}^{M}\\frac{1}{{t}^{\\sigma + 1}}{dt} \]\n\n\[ \\ll \\frac{1}{{N}^{\\sigma }} + \\frac{\\left| s\\right| }{\\sigma {N}^{\\sigma }} \]\n\n\[ \\ll \\left( {1 + \\frac{\\Delta }{\\delta }}\\right) \\frac{1}{{N}^{\\sigma }} \]\n\n\[ \\ll \\frac{1}{{N}^{\\delta }} \]\n\nwhere the implied constants depend on the modulus \( m \) and the compact set \( K \) . It follows that the partial sums of the series \( L\\left( {s,\\chi }\\right) \) are uniformly\n\nCauchy on \( K \), and so \( L\\left( {s,\\chi }\\right) \) converges uniformly on \( K \) and is analytic in the right half-plane.\n\nSince\n\n\[ \\mathop{\\sum }\\limits_{{N < n \\leq M}}\\frac{\\chi \\left( n\\right) }{{n}^{s}} \\ll \\frac{1}{{N}^{\\sigma }} \]\n\nfor all \( M > N \), it follows that\n\n\[ L\\left( {s,\\chi }\\right) - \\mathop{\\sum }\\limits_{{n = 1}}^{N}\\frac{\\chi \\left( n\\right) }{{n}^{s}} = \\mathop{\\sum }\\limits_{{n = N}}^{\\infty }\\frac{\\chi \\left( n\\right) }{{n}^{s}} \\ll \\frac{1}{{N}^{\\sigma }}. \]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 10.5 Let \( {\chi }_{0} \) be the principal character modulo \( m \) . For \( \sigma > 1 \) ,\n\n\[ L\left( {s,{\chi }_{0}}\right) = \zeta \left( s\right) \mathop{\prod }\limits_{{p \mid m}}\left( {1 - \frac{1}{{p}^{s}}}\right) \]\n\nand\n\n\[ \mathop{\lim }\limits_{{\sigma \rightarrow {1}^{ + }}}\left( {\sigma - 1}\right) L\left( {\sigma ,{\chi }_{0}}\right) = \mathop{\prod }\limits_{{p \mid m}}\left( {1 - \frac{1}{p}}\right) . \]\n\nFor \( 1 < \sigma < 2 \) ,\n\n\[ \log L\left( {\sigma ,{\chi }_{0}}\right) = \log \left( \frac{1}{\sigma - 1}\right) + O\left( 1\right) \]
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Proof. The Riemann zeta function is not analytic at \( s = 1 \), since for \( \sigma > 1 \) and \( n \geq 1 \) we have\n\n\[ {\int }_{n}^{n + 1}\frac{dx}{{x}^{\sigma }} < \frac{1}{{n}^{\sigma }} < {\int }_{n - 1}^{n}\frac{dx}{{x}^{\sigma }} \]\n\nand so\n\n\[ 0 < \frac{1}{\sigma - 1} = {\int }_{1}^{\infty }\frac{dx}{{x}^{\sigma }} < \zeta \left( \sigma \right) < 1 + {\int }_{1}^{\infty }\frac{dx}{{x}^{\sigma }} = \frac{\sigma }{\sigma - 1}. \]\n\nTherefore,\n\n\[ 1 < \left( {\sigma - 1}\right) \zeta \left( \sigma \right) < \sigma \]\n\nand\n\n\[ \mathop{\lim }\limits_{{\sigma \rightarrow {1}^{ + }}}\left( {\sigma - 1}\right) \zeta \left( \sigma \right) = 1 \]\n\n(10.3)\n\nIf \( 1 < \sigma < 2 \), then\n\n\[ \log \left( \frac{1}{\sigma - 1}\right) < \log \zeta \left( \sigma \right) < \log \left( \frac{1}{\sigma - 1}\right) + \log \sigma < \log \left( \frac{1}{\sigma - 1}\right) + \log 2 \]\n\nand so\n\n\[ \log \zeta \left( \sigma \right) = \log \left( \frac{1}{\sigma - 1}\right) + O\left( 1\right) \]\n\n(10.4)\n\nIf \( {\chi }_{0} \) is the principal character modulo \( m \), then\n\n\[ L\left( {s,{\chi }_{0}}\right) = \mathop{\prod }\limits_{p}{\left( 1 - \frac{{\chi }_{0}\left( p\right) }{{p}^{s}}\right) }^{-1} \]\n\n\[ = \mathop{\prod }\limits_{{\left( {p, m}\right) = 1}}{\left( 1 - \frac{1}{{p}^{s}}\right) }^{-1} \]\n\n\[ = \mathop{\prod }\limits_{p}{\left( 1 - \frac{1}{{p}^{s}}\right) }^{-1}\mathop{\prod }\limits_{{p \mid m}}\left( {1 - \frac{1}{{p}^{s}}}\right) \]\n\n\[ = \zeta \left( s\right) \mathop{\prod }\limits_{{p \mid m}}\left( {1 - \frac{1}{{p}^{s}}}\right) \]\n\nLet \( 1 < \sigma < 2 \) . Then\n\n\[ \left( {\sigma - 1}\right) L\left( {\sigma ,{\chi }_{0}}\right) = \left( {\sigma - 1}\right) \zeta \left( \sigma \right) \mathop{\prod }\limits_{{p \mid m}}\left( {1 - \frac{1}{{p}^{\sigma }}}\right) ,\]\n\nand (10.3) implies that\n\n\[ \mathop{\lim }\limits_{{\sigma \rightarrow {1}^{ + }}}\left( {\sigma - 1}\right) L\left( {\sigma ,{\chi }_{0}}\right) = \mathop{\prod }\limits_{{p \mid m}}\left( {1 - \frac{1}{p}}\right) . \]\n\nMoreover,\n\n\[ \log L\left( {\sigma ,{\chi }_{0}}\right) = \log \zeta \left( \sigma \right) + \log \mathop{\prod }\limits_{{p \mid m}}\left( {1 - \frac{1}{{p}^{\sigma }}}\right) \]\n\n\[ = \log \left( \frac{1}{\sigma - 1}\right) + O\left( 1\right) \]\n\nby (10.4). \( ▱ \)
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Yes
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Theorem 10.6 For \( 1 < \sigma < 2 \) , \n\n\[ \n\mathop{\sum }\limits_{{p \equiv 1\left( {\;\operatorname{mod}\;4}\right) }}\frac{1}{{p}^{\sigma }} = \frac{1}{2}\log \frac{1}{\sigma - 1} + O\left( 1\right) \n\] \n\nand \n\n\[ \n\mathop{\sum }\limits_{{p \equiv 3\left( {\;\operatorname{mod}\;4}\right) }}\frac{1}{{p}^{\sigma }} = \frac{1}{2}\log \frac{1}{\sigma - 1} + O\left( 1\right) \n\] \n\nIn particular, there exist infinitely many primes \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) and infinitely many primes \( p \equiv 3\;\left( {\;\operatorname{mod}\;4}\right) \) .
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Proof. Since \( L\left( {s,{\chi }_{4}}\right) \) is continuous for \( \sigma > 0 \), it follows that \n\n\[ \n\log L\left( {\sigma ,{\chi }_{4}}\right) = O\left( 1\right) \;\text{ for }1 \leq \sigma \leq 2. \n\] \n\nLet \( 1 < \sigma < 2 \) . By (10.1) of Theorem 10.3, we have \n\n\[ \n\log L\left( {\sigma ,{\chi }_{0}}\right) = \mathop{\sum }\limits_{{p \geq 3}}\frac{1}{{p}^{\sigma }} + O\left( 1\right) \n\] \n\nand \n\n\[ \n\log L\left( {\sigma ,{\chi }_{4}}\right) = \mathop{\sum }\limits_{{p \geq 3}}\frac{{\left( -1\right) }^{\left( {p - 1}\right) /2}}{{p}^{\sigma }} + O\left( 1\right) . \n\] \n\nTherefore, \n\n\[ \n\mathop{\sum }\limits_{{p \equiv 1\left( {\;\operatorname{mod}\;4}\right) }}\frac{1}{{p}^{\sigma }} = \frac{1}{2}\left( {\log L\left( {\sigma ,{\chi }_{0}}\right) + \log L\left( {\sigma ,{\chi }_{4}}\right) }\right) + O\left( 1\right) \n\] \n\n\[ \n= \frac{1}{2}\log L\left( {\sigma ,{\chi }_{0}}\right) + O\left( 1\right) \n\] \n\n\[ \n= \frac{1}{2}\log \frac{1}{\sigma - 1} + O\left( 1\right) \n\] \n\nby Theorem 10.5. Since \n\n\[ \n\mathop{\lim }\limits_{{\sigma \rightarrow {1}^{ + }}}\log \frac{1}{\sigma - 1} = \infty \n\] \n\nit follows that there exist infinitely many primes congruent to 1 modulo 4. Similarly, \n\n\[ \n\mathop{\sum }\limits_{{p \equiv 3\left( {\;\operatorname{mod}\;4}\right) }}\frac{1}{{p}^{\sigma }} = \frac{1}{2}\log \frac{1}{\sigma - 1} + O\left( 1\right) \n\] \n\nand there exist infinitely many primes congruent to 3 modulo 4.
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Yes
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Lemma 10.1 Let \( {\chi }_{0} \) be the principal character modulo \( m \) . Then\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{{\chi }_{0}\left( n\right) \Lambda \left( n\right) }{n} = \log x + O\left( 1\right) \]
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Proof. Observe that\n\n\[ \mathop{\sum }\limits_{\substack{{n \leq x} \\ {\left( {n, m}\right) > 1} }}\frac{\Lambda \left( n\right) }{n} = \mathop{\sum }\limits_{{p \mid m}}\mathop{\sum }\limits_{\substack{{{p}^{k} \leq x} \\ {k \geq 1} }}\frac{\Lambda \left( {p}^{k}\right) }{{p}^{k}} \]\n\n\[ < \mathop{\sum }\limits_{{p \mid m}}\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{\log p}{{p}^{k}} \]\n\n\[ = \mathop{\sum }\limits_{{p \mid m}}\frac{\log p}{p - 1} \]\n\n\[ = O\left( 1\right) \text{.} \]\n\nBy Mertens's theorem (Theorem 8.5), we have\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{{\chi }_{0}\left( n\right) \Lambda \left( n\right) }{n} = \mathop{\sum }\limits_{\substack{{n \leq x} \\ {\left( {n, m}\right) = 1} }}\frac{\Lambda \left( n\right) }{n} \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\frac{\Lambda \left( n\right) }{n} - \mathop{\sum }\limits_{\substack{{n \leq x} \\ {\left( {n, m}\right) > 1} }}\frac{\Lambda \left( n\right) }{n} \]\n\n\[ = \log x + O\left( 1\right) \text{.} \]\n\nThis completes the proof. \( ▱ \)
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Yes
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Lemma 10.2 Let \( \chi \) be a nonprincipal character modulo \( m \) . If \( L\left( {1,\chi }\right) \neq 0 \) , then\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = O\left( 1\right) \]
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Proof. Recall that \( F\left( t\right) = \mathop{\sum }\limits_{{k \leq t}}\chi \left( k\right) \ll 1 \) (Exercise 6 in Section 10.1). By partial summation, we have\n\n\[ \mathop{\sum }\limits_{{k \leq x}}\frac{\chi \left( k\right) \log k}{k} = \frac{F\left( x\right) \log x}{x} - {\int }_{1}^{x}\frac{F\left( t\right) \left( {1 - \log t}\right) }{{t}^{2}}{dt} \]\n\n\[ \ll \frac{\log x}{x} + {\int }_{1}^{\infty }\frac{1 + \log t}{{t}^{2}}{dt} \]\n\n\[ \ll 1\text{.} \]\n\nBy Theorem 10.4, we have\n\n\[ L\left( {1,\chi }\right) = \mathop{\sum }\limits_{{d \leq x/n}}\frac{\chi \left( d\right) }{d} + O\left( \frac{n}{x}\right) . \]\n\nUsing the identity \( \log k = \mathop{\sum }\limits_{{n \mid k}}\Lambda \left( n\right) \), we obtain\n\n\[ \mathop{\sum }\limits_{{k \leq x}}\frac{\chi \left( k\right) \log k}{k} = \mathop{\sum }\limits_{{k \leq x}}\frac{\chi \left( k\right) }{k}\mathop{\sum }\limits_{{n \mid k}}\Lambda \left( n\right) \]\n\n\[ = \mathop{\sum }\limits_{{{nd} \leq x}}\frac{\chi \left( {nd}\right) \Lambda \left( n\right) }{nd} \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n}\mathop{\sum }\limits_{{d \leq x/n}}\frac{\chi \left( d\right) }{d} \]\n\n\[ = \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n}\left( {L\left( {1,\chi }\right) + O\left( \frac{n}{x}\right) }\right) \]\n\n\[ = L\left( {1,\chi }\right) \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} + \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n}O\left( \frac{n}{x}\right) \]\n\n\[ = L\left( {1,\chi }\right) \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} + O\left( 1\right) \]\n\nsince\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n}O\left( \frac{n}{x}\right) \ll \frac{1}{x}\mathop{\sum }\limits_{{n \leq x}}\Lambda \left( n\right) = \frac{\psi \left( x\right) }{x} \ll 1 \]\n\nby Chebyshev's theorem (Theorem 8.2). Therefore,\n\n\[ L\left( {1,\chi }\right) \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = O\left( 1\right) \]\n\nIf \( L\left( {1,\chi }\right) \neq 0 \), then\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = O\left( 1\right) \]\n\nThis completes the proof.
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Yes
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Lemma 10.3 Let \( \chi \) be a nonprincipal character modulo \( m \) . If \( L\left( {1,\chi }\right) = 0 \), then \[ \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = - \log x + O\left( 1\right) \]
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Proof. Since \[ \Lambda \left( n\right) = - \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) \log d \] we have \[ \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = - \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) }{n}\mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) \log d. \] From the identity \[ \log x = \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) }{n}\mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) \log x \] we have \[ \log x + \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} \] \[ = \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) }{n}\mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) \log \frac{x}{d} \] \[ = \mathop{\sum }\limits_{{{dk} \leq x}}\frac{\chi \left( {dk}\right) \mu \left( d\right) }{dk}\log \frac{x}{d} \] \[ = \mathop{\sum }\limits_{{d \leq x}}\frac{\chi \left( d\right) \mu \left( d\right) }{d}\log \frac{x}{d}\mathop{\sum }\limits_{{k \leq x/d}}\frac{\chi \left( k\right) }{k} \] \[ = \mathop{\sum }\limits_{{d \leq x}}\frac{\chi \left( d\right) \mu \left( d\right) }{d}\log \frac{x}{d}\left( {L\left( {1,\chi }\right) + O\left( \frac{d}{x}\right) }\right) \] \[ = L\left( {1,\chi }\right) \mathop{\sum }\limits_{{d \leq x}}\frac{\chi \left( d\right) \mu \left( d\right) }{d}\log \frac{x}{d} + \mathop{\sum }\limits_{{d \leq x}}O\left( \frac{d}{x}\right) \frac{\chi \left( d\right) \mu \left( d\right) }{d}\log \frac{x}{d} \] \[ = L\left( {1,\chi }\right) \mathop{\sum }\limits_{{d \leq x}}\frac{\chi \left( d\right) \mu \left( d\right) }{d}\log \frac{x}{d} + O\left( 1\right) , \] since \[ \mathop{\sum }\limits_{{d \leq x}}O\left( \frac{d}{x}\right) \frac{\chi \left( d\right) \mu \left( d\right) }{d}\log \frac{x}{d} \ll \frac{1}{x}\mathop{\sum }\limits_{{d \leq x}}\log \frac{x}{d} \ll 1 \] by Theorem 6.4. If \( L\left( {1,\chi }\right) = 0 \), then \[ \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = - \log x + O\left( 1\right) \] This completes the proof.
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Yes
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Theorem 10.7 Let \( \\chi \) be a complex character modulo \( m \) . Then \( L\\left( {1,\\chi }\\right) \\neq 0 \) .
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Proof. Let \( N \) denote the number of nonprincipal characters modulo \( m \) such that \( L\\left( {1,\\chi }\\right) = 0 \) . We shall prove that \( N = 0 \) or 1 . By Lemmas 10.1, 10.2, and 10.3, and the orthogonality relations for Dirichlet characters (Theorem 10.1), we have\n\n\\[ \n\\varphi \\left( m\\right) \\mathop{\\sum }\\limits_{\\substack{{n \\leq x} \\\\ {n \\equiv 1\\left( {\\;\\operatorname{mod}\\;m}\\right) } }} \\frac{\\Lambda \\left( n\\right) }{n} = \\mathop{\\sum }\\limits_{{n \\leq x}} \\frac{\\Lambda \\left( n\\right) }{n} \\mathop{\\sum }\\limits_{{\\chi \\left( {\\;\\operatorname{mod}\\;m}\\right) }} \\chi \\left( n\\right) \n\\]\n\n\\[ \n= \\mathop{\\sum }\\limits_{{\\chi \\left( {\\;\\operatorname{mod}\\;m}\\right) }} \\mathop{\\sum }\\limits_{{n \\leq x}} \\frac{\\chi \\left( n\\right) \\Lambda \\left( n\\right) }{n} \n\\]\n\n\\[ \n= \\mathop{\\sum }\\limits_{{n \\leq x}} \\frac{{\\chi }_{0}\\left( n\\right) \\Lambda \\left( n\\right) }{n} + \\mathop{\\sum }\\limits_{{\\chi \\neq {\\chi }_{0}}} \\mathop{\\sum }\\limits_{{n \\leq x}} \\frac{\\chi \\left( n\\right) \\Lambda \\left( n\\right) }{n} \n\\]\n\n\\[ \n= \\log x - N\\log x + O\\left( 1\\right) \n\\]\n\n\\[ \n= \\left( {1 - N}\\right) \\log x + O\\left( 1\\right) \\text{.} \n\\]\n\nSince \( \\Lambda \\left( n\\right) /n \\geq 0 \) for all \( n \\geq 1 \), it follows that both sides of this equation are nonnegative for large \( x \), and so \( N \\leq 1 \) . Therefore, \( L\\left( {1,\\chi }\\right) = 0 \) for at most one nonprincipal character \( \\chi \) .\n\nIf \( \\chi \) is a complex character modulo \( m \), then \( \\bar{\\chi } \) is also a complex character and \( \\chi \\neq \\bar{\\chi } \) . We have\n\n\\[ \n\\overline{L\\left( {1,\\chi }\\right) } = \\overline{\\mathop{\\sum }\\limits_{{n = 1}}^{\\infty } \\frac{\\chi \\left( n\\right) }{n}} = \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty } \\frac{\\bar{\\chi }\\left( n\\right) }{n} = L\\left( {1,\\bar{\\chi }}\\right) , \n\\]\n\nand so \( L\\left( {1,\\chi }\\right) = 0 \) if and only if \( L\\left( {1,\\bar{\\chi }}\\right) = 0 \) . Since \( N \\leq 1 \), we must have \( L\\left( {1,\\chi }\\right) \\neq 0 \) for every complex character \( \\chi \) . This completes the proof.
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Yes
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Theorem 10.9 (Dirichlet) Let \( m \) and a be relatively prime positive integers. For \( 1 < \sigma < 2 \) , \[ \mathop{\sum }\limits_{{p \equiv a\left( {\;\operatorname{mod}\;m}\right) }}\frac{1}{{p}^{\sigma }} = \frac{1}{\varphi \left( m\right) }\log \left( \frac{1}{\sigma - 1}\right) + O\left( 1\right) \] In particular, there exist infinitely primes \( p \) such that \( p \equiv a\;\left( {\;\operatorname{mod}\;m}\right) \) .
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Proof. Let \( 1 < \sigma < 2 \) . Using the orthogonality relations for Dirichlet characters (Theorem 10.2) and the estimate (10.1) for \( \log L\left( {s,\chi }\right) \) from Theorem 10.3, we obtain \[ \mathop{\sum }\limits_{\left( \;\operatorname{mod}\;m\right) }\bar{\chi }\left( a\right) \log L\left( {\sigma ,\chi }\right) = \mathop{\sum }\limits_{{\chi \left( {\;\operatorname{mod}\;m}\right) }}\mathop{\sum }\limits_{p}\frac{\bar{\chi }\left( a\right) \chi \left( p\right) }{{p}^{\sigma }} + O\left( 1\right) \] \[ = \mathop{\sum }\limits_{p}\frac{1}{{p}^{\sigma }}\mathop{\sum }\limits_{\left( \;\operatorname{mod}\;m\right) }\bar{\chi }\left( a\right) \chi \left( p\right) + O\left( 1\right) \] \[ = \varphi \left( m\right) \mathop{\sum }\limits_{{p \equiv a\left( {\;\operatorname{mod}\;m}\right) }}\frac{1}{{p}^{\sigma }} + O\left( 1\right) . \] By Theorem 10.5, the term on the left corresponding to the principal character \( {\chi }_{0} \) is \[ \overline{{\chi }_{0}}\left( a\right) \log L\left( {\sigma ,{\chi }_{0}}\right) = \log \left( \frac{1}{\sigma - 1}\right) + O\left( 1\right) , \] and so \[ \varphi \left( m\right) \mathop{\sum }\limits_{{p \equiv a\left( {\;\operatorname{mod}\;m}\right) }}\frac{1}{{p}^{\sigma }} = \log \left( \frac{1}{\sigma - 1}\right) + \mathop{\sum }\limits_{{\chi \neq {\chi }_{0}}}\bar{\chi }\left( a\right) \log L\left( {\sigma ,\chi }\right) + O\left( 1\right) . \] If \( \chi \) is a nonprincipal character modulo \( m \), then \( L\left( {1,\chi }\right) \neq 0 \) by Theorem 10.7 and Theorem 10.8, and so \( \log L\left( {\sigma ,\chi }\right) = O\left( 1\right) \) for \( 1 \leq \sigma \leq 2 \) . This proves that \[ \mathop{\sum }\limits_{{p \equiv a\left( {\;\operatorname{mod}\;m}\right) }}\frac{1}{{p}^{\sigma }} = \frac{1}{\varphi \left( m\right) }\log \left( \frac{1}{\sigma - 1}\right) + O\left( 1\right) . \] Therefore, the series \( \mathop{\sum }\limits_{{p \equiv a{\;(\operatorname{mod}\;m)}}}{p}^{-\sigma } \) diverges as \( \sigma \rightarrow {1}^{ + } \), and so it must have infinitely many terms, that is, there must exist infinitely primes \( p \) such that \( p \equiv a\;\left( {\;\operatorname{mod}\;m}\right) \) . This completes the proof of Dirichlet’s theorem.
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Yes
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Theorem 10.10 Let \( m \geq 1 \) and a be relatively prime integers. Then\n\n\[ \mathop{\sum }\limits_{\substack{{n \leq x} \\ {n \equiv a\;\left( {\;\operatorname{mod}\;m}\right) } }}\frac{\Lambda \left( n\right) }{n} = \frac{\log x}{\varphi \left( m\right) } + O\left( 1\right) \]
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Proof. For the principal character \( {\chi }_{0} \) we have\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{{\chi }_{0}\left( n\right) \Lambda \left( n\right) }{n} = \log x + O\left( 1\right) \] \n\nby Lemma 10.1. For every nonprincipal character \( \chi \) modulo \( m \), we have \( L\left( {1,\chi }\right) \neq 0 \) by Theorems 10.7 and 10.8, and so\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = O\left( 1\right) \] \n\nby Lemma 10.2. Since \( {\chi }_{0}\left( a\right) = 1 \), it follows that\n\n\[ \mathop{\sum }\limits_{\left( \;\operatorname{mod}\;m\right) }\bar{\chi }\left( a\right) \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = \bar{\chi }\left( a\right) \log x + O\left( 1\right) = \log x + O\left( 1\right) . \] \n\nOn the other hand, by Theorem 10.2,\n\n\[ \mathop{\sum }\limits_{\left( \;\operatorname{mod}\;m\right) }\overline{\chi \left( a\right) }\mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = \mathop{\sum }\limits_{{n \leq x}}\frac{\Lambda \left( n\right) }{n}\mathop{\sum }\limits_{{\chi \left( {\;\operatorname{mod}\;m}\right) }}\overline{\chi \left( a\right) }\chi \left( n\right) \] \n\n\[ = \varphi \left( m\right) \mathop{\sum }\limits_{\substack{{n \leq x} \\ {n \equiv a\left( {\;\operatorname{mod}\;m}\right) } }}\frac{\Lambda \left( n\right) }{n}. \] \n\nThis completes the proof.
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Yes
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Lemma 11.1 Let \( A \) and \( B \) be sets of integers such that \( 0 \in A \) and \( 0 \in B \) . If \( A\left( n\right) + B\left( n\right) \geq n \), then \( n \in A + B \) .
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Proof. If \( n \in A \), then \( n = n + 0 \in A + B \) . Similarly, if \( n \in B \), then \( n = 0 + n \in A + B \) . Suppose that \( n \notin A \cup B \) . Define sets \( {A}^{\prime } \) and \( {B}^{\prime } \) by \[ {A}^{\prime } = \{ n - a : a \in A,1 \leq a \leq n - 1\} \] and \[ {B}^{\prime } = B \cap \left\lbrack {1, n - 1}\right\rbrack \] Then \( \left| {A}^{\prime }\right| = A\left( n\right) \), since \( n \notin A \), and \( \left| {B}^{\prime }\right| = B\left( n\right) \), since \( n \notin B \) . Moreover, \[ {A}^{\prime } \cup {B}^{\prime } \subseteq \left\lbrack {1, n - 1}\right\rbrack \] Since \[ \left| {A}^{\prime }\right| + \left| {B}^{\prime }\right| = A\left( n\right) + B\left( n\right) \geq n \] it follows that \[ {A}^{\prime } \cap {B}^{\prime } \neq \varnothing \text{.} \] Therefore, \( n - a = b \) for some \( a \in A \) and \( b \in B \), and so \( n = a + b \in A + B \) .
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Yes
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Lemma 11.2 Let \( A \) and \( B \) be sets of integers such that \( 0 \in A \) and \( 0 \in B \) . If \( \sigma \left( A\right) + \sigma \left( B\right) \geq 1 \), then \( {\mathbf{N}}_{0} \subseteq A + B \) .
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Proof. We have \( 0 = 0 + 0 \in A + B \) . If \( n \geq 1 \), then\n\n\[ A\left( n\right) + B\left( n\right) \geq \left( {\sigma \left( A\right) + \sigma \left( B\right) }\right) n \geq n, \]\n\nand Lemma 11.1 implies that \( n \in A + B \) .
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Yes
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Lemma 11.3 Let \( A \) be a set of integers such that \( 0 \in A \) and \( \sigma \left( A\right) \geq 1/2 \) . Then \( A \) is a basis of order 2.
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Proof. This follows immediately from Lemma 11.2 with \( A = B \) .
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No
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Theorem 11.1 (Shnirel’man) Let \( A \) and \( B \) be sets of integers such that \( 0 \in A \) and \( 0 \in B \) . Let \( \sigma \left( A\right) = \alpha \) and \( \sigma \left( B\right) = \beta \) . Then\n\n\[ \sigma \left( {A + B}\right) \geq \alpha + \beta - {\alpha \beta } \]
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Proof. Let \( n \geq 1 \) . Let \( {a}_{0} = 0 \) and let\n\n\[ 1 \leq {a}_{1} < \cdots < {a}_{k} \leq n \]\n\nbe the \( k = A\left( n\right) \) positive elements of \( A \) that do not exceed \( n \) . Since \( 0 \in B \) , it follows that \( {a}_{i} = {a}_{i} + 0 \in A + B \) for \( i = 1,\ldots, k \) . For \( i = 0,\ldots, k - 1 \), let\n\n\[ 1 \leq {b}_{1} < \cdots < {b}_{{r}_{i}} \leq {a}_{i + 1} - {a}_{i} - 1 \]\n\nbe the \( {r}_{i} = B\left( {{a}_{i + 1} - {a}_{i} - 1}\right) \) positive integers in \( B \) that are less than \( {a}_{i + 1} - {a}_{i} \) . Then\n\n\[ {a}_{i} < {a}_{i} + {b}_{1} < \cdots < {a}_{i} + {b}_{{r}_{i}} < {a}_{i + 1} \]\n\nand\n\n\[ {a}_{i} + {b}_{j} \in A + B \]\n\nfor \( j = 1,\ldots ,{r}_{i} \) . Let\n\n\[ 1 \leq {b}_{1} < \cdots < {b}_{{r}_{k}} \leq n - {a}_{k} \]\n\nbe the \( {r}_{k} = B\left( {n - {a}_{k}}\right) \) positive integers in \( B \) that do not exceed \( n - {a}_{k} \) .\n\nThen\n\n\[ {a}_{k} < {a}_{k} + {b}_{1} < \cdots < {a}_{k} + {b}_{{r}_{k}} \leq n \]\n\nand\n\n\[ {a}_{k} + {b}_{j} \in A + B \]\n\nfor \( j = 1,\ldots ,{r}_{k} \) . It follows that\n\n\[ \left( {A + B}\right) \left( n\right) \geq A\left( n\right) + \mathop{\sum }\limits_{{i = 0}}^{k}{r}_{i} \]\n\n\[ = A\left( n\right) + \mathop{\sum }\limits_{{i = 0}}^{{k - 1}}B\left( {{a}_{i + 1} - {a}_{i} - 1}\right) + B\left( {n - {a}_{k}}\right) \]\n\n\[ \geq A\left( n\right) + \beta \mathop{\sum }\limits_{{i = 0}}^{{k - 1}}\left( {{a}_{i + 1} - {a}_{i} - 1}\right) + \beta \left( {n - {a}_{k}}\right) \]\n\n\[ = A\left( n\right) + {\beta n} - {\beta k} \]\n\n\[ = \left( {1 - \beta }\right) A\left( n\right) + {\beta n} \]\n\n\[ \geq \left( {1 - \beta }\right) {\alpha n} + {\beta n} \]\n\n\[ = \left( {\alpha + \beta - {\alpha \beta }}\right) n \]\n\nand so\n\n\[ \frac{\left( {A + B}\right) \left( n\right) }{n} \geq \alpha + \beta - {\alpha \beta } \]\n\nfor all positive integers \( n \) . Therefore,\n\n\[ \sigma \left( {A + B}\right) = \inf \left\{ {\frac{\left( {A + B}\right) \left( n\right) }{n} : n = 1,2,\ldots }\right\} \geq \alpha + \beta - {\alpha \beta }. \]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 11.2 Let \( h \geq 1 \), and let \( {A}_{1},\ldots ,{A}_{h} \) be sets of integers with \( 0 \in {A}_{i} \) for \( i = 1,\ldots, h \) . Then\n\n\[ 1 - \sigma \left( {{A}_{1} + \cdots + {A}_{h}}\right) \leq \mathop{\prod }\limits_{{i = 1}}^{h}\left( {1 - \sigma \left( {A}_{i}\right) }\right) \]
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Proof. This is by induction on \( h \) . Let \( \sigma \left( {A}_{i}\right) = {\alpha }_{i} \) for \( i = 1,\ldots, h \) . For \( h = 1 \), there is nothing to prove, and for \( h = 2 \) the inequality is equivalent to (11.3).\n\nLet \( h \geq 3 \), and assume that the theorem holds for \( h - 1 \) sets. Let \( {A}_{1},\ldots ,{A}_{h} \) be \( h \) sets of integers such that \( 0 \in {A}_{i} \) for all \( i \) . Let \( B = \) \( {A}_{1} + \cdots + {A}_{h - 1} \) . We have the induction hypothesis\n\n\[ 1 - \sigma \left( B\right) = 1 - \sigma \left( {{A}_{1} + \cdots + {A}_{h - 1}}\right) \leq \mathop{\prod }\limits_{{i = 1}}^{{h - 1}}\left( {1 - \sigma \left( {A}_{i}\right) }\right) ,\]\n\nand so\n\n\[ 1 - \sigma \left( {{A}_{1} + \cdots + {A}_{h}}\right) = 1 - \sigma \left( {B + {A}_{h}}\right) \]\n\n\[ \leq (1 - \sigma \left( B\right) \left( {1 - \sigma \left( {A}_{h}\right) }\right) \]\n\n\[ \leq \mathop{\prod }\limits_{{i = 1}}^{{h - 1}}\left( {1 - \sigma \left( {A}_{i}\right) }\right) \left( {1 - \sigma \left( {A}_{h}\right) }\right. \]\n\n\[ = \mathop{\prod }\limits_{{i = 1}}^{h}\left( {1 - \sigma \left( {A}_{i}\right) }\right) \]
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Yes
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Theorem 11.3 Let \( 0 < \alpha \leq 1 \) . There exists an integer \( h = h\left( \alpha \right) \) such that if \( {A}_{1},\ldots ,{A}_{h} \) are sets of nonnegative integers with \( 0 \in {A}_{i} \) and \( \sigma \left( {A}_{i}\right) \geq \alpha \) for all \( i = 1,\ldots, h \), then\n\n\[{A}_{1} + \cdots + {A}_{h} = {\mathbf{N}}_{0}\]
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Proof. Since \( 0 \leq 1 - \alpha < 1 \), there exists a positive integer \( {h}_{1} \) such that\n\n\[0 \leq {\left( 1 - \alpha \right) }^{{h}_{1}} \leq \frac{1}{2}\]\n\nLet \( h = 2{h}_{1} \), and let \( {A}_{1},\ldots ,{A}_{h} \) be sets of nonnegative integers with \( 0 \in {A}_{i} \) and \( \sigma \left( {A}_{i}\right) \geq \alpha \) for \( i = 1,\ldots, h \) . We define \( A = {A}_{1} + \cdots + {A}_{{h}_{1}} \) and \( B = \) \( {A}_{{h}_{1} + 1} + \cdots + {A}_{2{h}_{1}} \) . By Theorem 11.2,\n\n\[\sigma \left( A\right) = \sigma \left( {{A}_{1} + \cdots + {A}_{{h}_{1}}}\right) \geq 1 - \mathop{\prod }\limits_{{i = 1}}^{{h}_{1}}\left( {1 - \sigma \left( {A}_{i}\right) }\right) \geq 1 - {\left( 1 - \alpha \right) }^{{h}_{1}} \geq \frac{1}{2}.\n\nSimilarly,\n\n\[\sigma \left( B\right) = \sigma \left( {{A}_{{h}_{1} + 1} + \cdots + {A}_{2{h}_{1}}}\right) \geq \frac{1}{2}.\n\nApplying Lemma 11.3, we obtain\n\n\[{A}_{1} + \cdots + {A}_{h} = A + B = {\mathbf{N}}_{0}.\n\nThis completes the proof.
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Yes
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Theorem 11.4 (Shnirel’man) Let \( A \) be a set of nonnegative integers such that \( 0 \in A \) and \( \sigma \left( A\right) > 0 \) . Then \( A \) is a basis of finite order.
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Proof. Let \( \alpha = \sigma \left( A\right) \) . The result follows from Theorem 11.3 with \( {A}_{i} = A \) for \( i = 1,\ldots, h\left( \alpha \right) \) .
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No
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Theorem 11.5 Let \( A \) be a set of nonnegative integers with \( 0 \in A \) such that \( \sigma \left( {{h}_{1}A}\right) > 0 \) for some positive integer \( {h}_{1} \). Then \( A \) is a basis of finite order.
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Proof. If \( \sigma \left( {{h}_{1}A}\right) > 0 \), then there exists an integer \( {h}_{2} \) such that \( {h}_{1}A \) is a basis of order \( {h}_{2} \), that is, every nonnegative integer is a sum of \( {h}_{2} \) elements of \( {h}_{1}A \). Since\n\n\[ \n{h}_{2}\left( {{h}_{1}A}\right) = \left( {{h}_{1}{h}_{2}}\right) A \n\]\n\nthe set \( A \) is a basis of order \( h = {h}_{1}{h}_{2} \).
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Yes
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Theorem 11.6 Let \( B \) be a set of nonnegative integers with \( 0 \in B \) and \( \gcd \left( B\right) = 1 \) . If \( {d}_{L}\left( B\right) > 0 \), then \( B \) is an asymptotic basis of finite order.
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Proof. The set \( A = B \cup \{ 1\} \) has positive Shnirel’man density (by Exercise 1), and so \( A \) is a basis of order \( {h}_{1} \) for some positive integer \( {h}_{1} \) . It follows that every nonnegative integer can be written in the form \( u + j \) , where \( 0 \leq j \leq {h}_{1} \) and \( u \) is a sum of \( {h}_{1} - j \) elements of \( B \) . Since \( 0 \in B \) ,\n\n\[ u \in \left( {{h}_{1} - j}\right) B \subseteq {h}_{1}B. \]\n\nIf \( B \) is any set of relatively prime positive integers, then, by Theorem 1.16, there exists an integer \( {n}_{0} = {n}_{0}\left( B\right) \) such that every integer \( n \geq {n}_{0} \) can be represented as a sum of elements of \( B \) . Since \( 0 \in B \) and \( \gcd \left( B\right) = 1 \), there exists a positive integer \( {h}_{2} \) such that\n\n\[ {n}_{0} + j \in {h}_{2}B \]\n\nfor \( j = 0,1,\ldots ,{h}_{1} \) . Let \( h = {h}_{1} + {h}_{2} \) . If \( n \geq {n}_{0} \), then \( n - {n}_{0} \geq 0 \) and we can write \( n - {n}_{0} \) in the form \( u + j \), where \( u \in {h}_{1}B \) and \( 0 \leq j \leq {h}_{1} \) . Then\n\n\[ n = u + \left( {{n}_{0} + j}\right) \in {h}_{1}B + {h}_{2}B = {hB}, \]\n\nand so \( B \) is an asymptotic basis of finite order. \( ▱ \)
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No
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Theorem 11.7 Let \( B \) be a set of nonnegative integers with \( \gcd \left( B\right) = d \) . If \( {d}_{L}\left( B\right) > 0 \), then every sufficiently large multiple of \( d \) is the sum of a bounded number of elements of \( B \) .
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Proof. The set \( {d}^{-1} * B = \{ b/d : b \in B\} \) consists of nonnegative integers, and\n\n\[ A = \{ 0\} \cup {d}^{-1} * B \]\n\n is a set of nonnegative integers with \( 0 \in A \) and \( \gcd \left( A\right) = 1 \) . By Theorem 11.6, every sufficiently large integer can be represented as the sum of exactly \( h \) elements of \( A \), and so every sufficiently large multiple of \( d \) can be represented as the sum of at most \( h \) elements of \( B \) .
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Yes
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Lemma 11.4 Let \( f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{k}{a}_{i}{x}^{i} \) be an integer-valued polynomial of degree \( k \) with leading coefficient \( {a}_{k} > 0 \) . Let\n\n\[ \n{x}^{ * }\left( f\right) = \frac{2\left( {\left| {a}_{k - 1}\right| + \left| {a}_{k - 2}\right| + \cdots + \left| {a}_{0}\right| }\right) }{{a}_{k}}.\n\]\n\n(11.4)\n\nIf \( x > {x}^{ * }\left( f\right) \) is an integer, then\n\n\[ \n\frac{{a}_{k}{x}^{k}}{2} < f\left( x\right) < \frac{3{a}_{k}{x}^{k}}{2}\n\]\n\n(11.5)
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Proof. Since\n\n\[ \nf\left( x\right) = {a}_{k}{x}^{k}\left( {1 + \frac{{a}_{k - 1}}{{a}_{k}x} + \frac{{a}_{k - 2}}{{a}_{k}{x}^{2}} + \cdots + \frac{{a}_{0}}{{a}_{k}{x}^{k}}}\right) ,\n\]\n\nit follows for \( x > {x}^{ * }\left( f\right) \) that\n\n\[ \n\left| {\frac{f\left( x\right) }{{a}_{k}{x}^{k}} - 1}\right| = \left| {\frac{{a}_{k - 1}}{{a}_{k}x} + \frac{{a}_{k - 2}}{{a}_{k}{x}^{2}} + \cdots + \frac{{a}_{0}}{{a}_{k}{x}^{k}}}\right| \n\]\n\n\[ \n\leq \frac{\left| {a}_{k - 1}\right| }{{a}_{k}x} + \frac{\left| {a}_{k - 2}\right| }{{a}_{k}{x}^{2}} + \cdots + \frac{\left| {a}_{0}\right| }{{a}_{k}{x}^{k}} \n\]\n\n\[ \n\leq \frac{\left| {a}_{k - 1}\right| + \left| {a}_{k - 2}\right| + \cdots + \left| {a}_{0}\right| }{{a}_{k}x} \n\]\n\n\[ \n= \frac{{x}^{ * }\left( f\right) }{2x} \n\]\n\n\[ \n< \frac{1}{2}\text{. } \n\]\n\nThis proves (11.5).\n\nIf \( {x}_{1},\ldots ,{x}_{s} \) are integers such that\n\n\[ \n{x}^{ * }\left( f\right) < {x}_{j} \leq {\left( \frac{2N}{3{a}_{k}s}\right) }^{1/k}\n\]\n\nfor \( j = 1,\ldots, s \), then\n\n\[ \n0 < \frac{{a}_{k}{x}_{j}^{k}}{2} < f\left( {x}_{j}\right) < \frac{3{a}_{k}{x}_{j}^{k}}{2} \leq \frac{N}{s}\n\]\n\nand\n\n\[ \n0 < f\left( {x}_{1}\right) + \cdots + f\left( {x}_{s}\right) < N.\n\]\n\nThe number of integers in the interval\n\n\[ \n\left( {{x}^{ * }\left( f\right) ,{\left( \frac{2N}{3{a}_{k}s}\right) }^{1/k}}\right\rbrack\n\]\n\nis greater than\n\n\[ \n{\left( \frac{2N}{3{a}_{k}s}\right) }^{1/k} - {x}^{ * }\left( f\right) - 1\n\]\n\nand so\n\n\[ \n{R}_{f, s}\left( N\right) > {\left( {\left( \frac{2N}{3{a}_{k}s}\right) }^{1/k} - {x}^{ * }\left( f\right) - 1\right) }^{s} \geq \frac{1}{2}{\left( \frac{2N}{3{a}_{k}s}\right) }^{s/k}\n\]\n\nfor \( N \) sufficiently large. This proves (11.6). \( ▱ \)
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Yes
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Lemma 11.5 Let \( f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{k}{a}_{i}{x}^{i} \) be an integer-valued polynomial of degree \( k \) such that \[ A\left( f\right) = \{ f\left( i\right) {\} }_{i = 0}^{\infty } \] is a strictly increasing sequence of nonnegative integers. Define \( {x}^{ * }\left( f\right) \) by (11.4) and let \[ N\left( f\right) = \frac{{x}^{ * }{\left( f\right) }^{k}}{{2k}!}. \] For \( N \geq N\left( f\right) \), if \( {x}_{1},\ldots ,{x}_{s} \) are nonnegative integers with \[ \mathop{\sum }\limits_{{j = 1}}^{s}f\left( {x}_{j}\right) \leq N \] then \[ 0 \leq {x}_{j} \leq {\left( 2k!N\right) }^{1/k}\;\text{ for }j = 1,\ldots, s. \]
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Proof. Recall that \( k!{a}_{k} \geq 1 \) by Exercise 6 in Section 11.1. If \( N \geq N\left( f\right) \) and \( {x}_{j} > {\left( 2k!N\right) }^{1/k} \geq {x}^{ * }\left( f\right) \), then \[ f\left( {x}_{j}\right) > \frac{{a}_{k}{x}_{j}^{k}}{2} \geq k!{a}_{k}N \geq N \] and so \[ \mathop{\sum }\limits_{{i = 1}}^{s}f\left( {x}_{i}\right) \geq f\left( {x}_{j}\right) > N \] This completes the proof. \( ▱ \)
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No
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Theorem 11.8 Let \( \{ s\left( k\right) {\} }_{k = 1}^{\infty } \) be the sequence of integers defined recursively by \( s\left( 1\right) = 1 \) and\n\n\[ s\left( k\right) = {8k}{2}^{\left\lbrack {\log }_{2}s\left( k - 1\right) \right\rbrack }\;\text{ for }k \geq 2.\]\n\nLet \( c \geq 1 \) and \( P \geq 1 \) . If\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{k}{a}_{i}{x}^{i} \]\n\nis an integer-valued polynomial of degree \( k \) such that\n\n\[ \left| {a}_{i}\right| \leq c{P}^{k - i}\;\text{ for }i = 0,1,\ldots, k,\]\n\nthen for every integer \( n \) ,\n\n\[ \text{ NSE }\left\{ \begin{array}{l} \mathop{\sum }\limits_{{j = 1}}^{{s\left( k\right) }}f\left( {x}_{j}\right) = n\;\text{ with }{x}_{j} \in \mathbf{Z} \\ \text{ and }\left| {x}_{j}\right| \leq {cP}\text{ for }j = 1,\ldots, s\left( k\right) \end{array}\right\} { \ll }_{k, c}{P}^{s\left( k\right) - k}. \]
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Proof. Let \( c = {c}_{1} \) and \( {f}_{j}\left( x\right) = f\left( x\right) \) for \( j = 1,\ldots, s\left( k\right) \) in Theorem 12.3. 口
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No
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Theorem 11.9 Let \( f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{k}{a}_{i}{x}^{i} \) be an integer-valued polynomial of degree \( k \) with \( {a}_{k} > 0 \) and \( \gcd \left( {\bar{A}\left( f\right) }\right) = 1 \) . Then \( A\left( f\right) \cup \{ 0\} \) is an asymptotic basis of finite order, that is, for some \( h \) and every sufficiently large integer \( n \) there exists a positive integer \( {h}_{n} \leq h \) and nonnegative integers \( {x}_{1},\ldots ,{x}_{{h}_{n}} \) such that\n\n\[ f\left( {x}_{1}\right) + \cdots + f\left( {x}_{{h}_{n}}\right) = n. \]
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Proof. Define \( N\left( f\right) \) by (11.7), and let \( s = s\left( k\right) \) be the integer constructed in Theorem 11.8. Let \( W = {sA}\left( f\right) \) be the set consisting of all sums of \( s \) integers of the form \( f\left( x\right) \) with \( x \in {\mathbf{N}}_{0} \) . We shall prove that the sumset \( W \) has lower asymptotic density \( {d}_{L}\left( W\right) > 0 \) .\n\nLet \( W\left( N\right) \) be the counting function of \( W \) . Choose \( c \geq {\left( 2k!\right) }^{1/k} \) and choose \( N \geq N\left( f\right) \) sufficiently large that for \( P = {N}^{1/k} \),\n\n\[ \left| {a}_{i}\right| \leq c{P}^{k - i}\;\text{ for }i = 0,1,\ldots, k. \]\n\nThen \( 0 < {a}_{k} \leq c \) . By Lemma 11.5, if \( {x}_{1},\ldots ,{x}_{s} \) are nonnegative integers such that \( \mathop{\sum }\limits_{{j = 1}}^{s}f\left( {x}_{j}\right) \leq N \), then\n\n\[ 0 \leq {x}_{j} \leq {\left( 2k!N\right) }^{1/k} \leq {cP}\;\text{ for }j = 1,\ldots, s. \]\n\nWe get upper bounds for \( {r}_{f, s}\left( n\right) \) and \( {R}_{f, s}\left( N\right) \) as follows: If \( 0 \leq n \leq N \) , then\n\n\[ {r}_{f, s}\left( n\right) \; = \;\operatorname{NSE}\left\{ {f\left( {x}_{1}\right) + \cdots + f\left( {x}_{s}\right) = n : {x}_{i} \in {\mathbf{N}}_{0}}\right\} \]\n\n\[ \leq \;\text{ NSE }\left\{ {f\left( {x}_{1}\right) + \cdots + f\left( {x}_{s}\right) = n : \left| {x}_{j}\right| \leq {cP}}\right\} \]\n\n\[ { \ll }_{k, c}\;{P}^{s - k} \]\nby Theorem 11.8, and so\n\n\[ {R}_{f, s}\left( N\right) = \mathop{\sum }\limits_{{0 \leq n \leq N}}{r}_{f, s}\left( n\right) \]\n\n\[ = \mathop{\sum }\limits_{\substack{{0 \leq n \leq N} \\ {{r}_{f, s}\left( n\right) \geq 1} }}{r}_{f, s}\left( n\right) \]\n\n\[ { \ll }_{k, c}W\left( N\right) {P}^{s - k} \]\n\n\[ { \ll }_{k, c}\left( \frac{W\left( N\right) }{N}\right) {P}^{s} \]\n\nWe can apply Lemma 11.4 to obtain a lower bound for \( {R}_{k, s}\left( N\right) \) . For \( N \) sufficiently large,\n\n\[ {R}_{f, s}\left( N\right) > \frac{1}{2}{\left( \frac{2N}{3{a}_{k}s}\right) }^{s/k} \geq \frac{1}{2}{\left( \frac{2N}{3cs}\right) }^{s/k}{ \gg }_{k, c}{P}^{s}. \]\n\nTherefore,\n\n\[ {P}^{s}{ \ll }_{k, c}{R}_{f, s}\left( N\right) { \ll }_{k, c}\left( \frac{W\left( N\right) }{N}\right) {P}^{s}, \]\n\nand so \( W\left( N\right) /N{ \gg }_{k, c}1 \) . It follows that\n\n\[ {d}_{L}\left( {{sA}\left( f\right) }\right) = {d}_{L}\left( W\right) > 0, \]\n\nand the result follows immediately from Theorem 11.7. \( ▱ \)
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Yes
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Theorem 11.10 Let \( f\left( x\right) \) be an integer-valued polynomial of degree \( k \) with leading coefficient \( {a}_{k} > 0 \) . If \( 0,1 \in A\left( f\right) = \left\{ {f\left( x\right) : x \in {\mathbf{N}}_{0}}\right\} \), then \( A\left( f\right) \) is a basis of finite order.
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Proof. This is a consequence of Theorem 11.9.
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No
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Theorem 11.11 (Waring-Hilbert) For every \( k \geq 2 \), the set of nonnegative \( k \) th powers is a basis of finite order.
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Proof. This is the special case of Theorem 11.10 applied to the polynomial \( f\left( x\right) = {x}^{k} \) . \( ▱ \)
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Yes
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Theorem 11.12 Let \( f\left( x\right) \) be an integer-valued polynomial of degree \( k \) with leading coefficient \( {a}_{k} > 0 \) and \( \gcd \left( {A\left( f\right) }\right) = 1 \) . Then \( A\left( f\right) \cup \{ 0\} \) is an asymptotically stable asymptotic basis of finite order.
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Proof. This requires only minor modifications of the proof of Theorem 11.9. Let \( A\left( f\right) = \{ f\left( i\right) {\} }_{i = 0}^{\infty } \), and let \( B \) be a set of nonnegative integers of lower asymptotic density \( {d}_{L}\left( B\right) = \beta > 0 \) . Then\n\n\[ \n{A}_{B} = \{ f\left( b\right) : b \in B\} .\n\]\n\nLet \( s = s\left( k\right) \) be the integer constructed in Theorem 11.8. The sumset \( {W}_{s} = s{A}_{B} \) consists of all sums of \( s \) integers of the form \( f\left( b\right) \) with \( b \in B \) . Let \( {W}_{s}\left( N\right) \) be the counting function of the sumset \( {W}_{s} \) . Let \( {r}_{f, s}^{\left( B\right) }\left( n\right) \) denote the number of solutions of the equation\n\n\[ \nf\left( {b}_{1}\right) + \cdots + f\left( {b}_{s}\right) = n \n\]\n\nwith \( {b}_{1},\ldots ,{b}_{s} \in B \), and let\n\n\[ \n{R}_{f, s}^{\left( B\right) }\left( N\right) = \mathop{\sum }\limits_{{n = 0}}^{N}{r}_{f, s}^{\left( B\right) }\left( n\right) .\n\]\n\nWe shall again compute upper and lower bounds for \( {R}_{f, s}^{\left( B\right) }\left( n\right) \) .\n\nChoose real numbers \( c \geq {\left( 2k!\right) }^{1/k} \) and \( N \geq N\left( f\right) \) such that for \( P = {N}^{1/k} \),\n\n\[ \n\left| {a}_{i}\right| \leq c{P}^{k - i}\;\text{ for }i = 1,\ldots, k.\n\]\n\nBy Theorem 11.8, we have the upper bound\n\n\[ \n{R}_{k, s}^{\left( B\right) }\left( N\right) \; = \;\mathop{\sum }\limits_{\substack{{n = 0} \\ {{r}_{k, s}^{\left( B\right) }\left( n\right) \geq 1} }}^{N}{r}_{k, s}^{\left( B\right) }\left( n\right) \leq \mathop{\sum }\limits_{\substack{{n = 0} \\ {{r}_{k, s}^{\left( B\right) }\left( n\right) \geq 1} }}^{N}{r}_{k, s}\left( n\right) \n\]\n\n\[ \n{ \ll }_{k, c}{W}_{B}\left( N\right) {P}^{s - k} \n\]\n\n\[ \n{ \ll }_{k, c}\left( \frac{{W}_{B}\left( N\right) }{N}\right) {P}^{s} \n\]\n\nfor all sufficiently large \( N \) .\n\nTo obtain a lower bound, we observe that the number of integers \( b \in B \)\n\nsuch that\n\[ \n{x}^{ * }\left( f\right) < b \leq {\left( \frac{2N}{3{a}_{k}s}\right) }^{1/k} \n\]\n\n(11.8)\n\nis\n\n\[ \nB\left( {\left( \frac{2N}{3{a}_{k}s}\right) }^{1/k}\right) - B\left( {{x}^{ * }\left( f\right) }\right) \geq \left( \frac{\beta }{2}\right) {\left( \frac{2N}{3{a}_{k}s}\right) }^{1/k} - B\left( {{x}^{ * }\left( f\right) }\right) { \gg }_{k, c}P \n\]\n\nfor sufficiently large \( N \) . By Lemma 11.4, if \( b \in B \) satisfies inequality (11.8),\n\nthen\n\[ \n0 \leq f\left( b\right) \leq \frac{N}{s} \n\]\nand so\n\n\[ \n{R}_{f, s}^{\left( B\right) }\left( N\right) { \gg }_{k, c}{P}^{s} \n\]\n\nIt follows that \( {W}_{B}\left( N\right) /N{ \gg }_{k, c}1 \), and so \( {W}_{B} = s{A}_{B} \) has positive lower asymptotic density. The result now follows from Theorem 11.7. \( ▱ \)
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Yes
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Theorem 11.13 Let \( f\left( x\right) \) be an integer-valued polynomial of degree \( k \) with leading coefficient \( {a}_{k} > 0 \) . If \( 0,1 \in A\left( f\right) = \left\{ {f\left( x\right) : x \in {\mathbf{N}}_{0}}\right\} \), then \( A\left( f\right) \) is a stable basis of finite order.
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Proof. This follows from Theorem 11.12.
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No
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Theorem 11.14 (Waring-Shnirel’man) For every \( k \geq 2 \), the set of nonnegative \( k \) th powers is a stable basis of finite order and an asymptotically stable asymptotic basis of finite order.
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Proof. This follows from Theorem 11.12. \( ▱ \)
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No
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Lemma 12.1 Let \( {B}_{1} \) and \( {B}_{2} \) be weighted sets of integers. Define the addition map \( \sigma : {B}_{1} \times {B}_{2} \rightarrow {B}_{1} + {B}_{2} \) by \( \sigma \left( {{b}_{1},{b}_{2}}\right) = {b}_{1} + {b}_{2} \) and the difference maps \( {\delta }_{i} : {B}_{i} \times {B}_{i} \rightarrow {B}_{i} - {B}_{i} \) by \( {\delta }_{i}\left( {{b}_{i},{b}_{i}^{\prime }}\right) = {b}_{i} - {b}_{i}^{\prime } \) for \( i = 1,2 \) . Consider the weighted sumset \( S = {B}_{1} + {B}_{2} \) and the weighted difference sets \( {D}_{1} = {B}_{1} - {B}_{1} \) and \( {D}_{2} = {B}_{2} - {B}_{2} \) . Then for all integers \( n \) ,\n\n\[ \n{w}_{S}^{\left( \sigma \right) }\left( n\right) \leq \frac{1}{2}\left( {{w}_{{D}_{1}}^{\left( {\delta }_{1}\right) }\left( 0\right) + {w}_{{D}_{2}}^{\left( {\delta }_{2}\right) }\left( 0\right) }\right) .\n\]
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Proof. For \( i = 1,2 \) we have\n\n\[ \n{w}_{{D}_{i}}^{\left( {\delta }_{i}\right) }\left( 0\right) = \mathop{\sum }\limits_{\substack{{\left( {{b}_{i},{b}_{i}^{\prime }}\right) \in {B}_{i} \times {B}_{i}} \\ {{b}_{i} - {b}_{i}^{\prime } = 0} }}{w}_{{B}_{i}}\left( {b}_{i}\right) {w}_{{B}_{i}}\left( {b}_{i}^{\prime }\right) = \mathop{\sum }\limits_{{{b}_{i} \in {B}_{i}}}{w}_{{B}_{i}}{\left( {b}_{i}\right) }^{2}.\n\]\n\nTo each \( {b}_{1} \in {B}_{1} \) there exists at most one \( {b}_{2} \in {B}_{2} \) such that \( {b}_{1} + {b}_{2} = n \) . Applying the elementary inequality\n\n\[ \n{xy} \leq \frac{1}{2}\left( {{x}^{2} + {y}^{2}}\right) \;\text{ for }x, y \in \mathbf{R},\n\]\n\nwe obtain\n\n\[ \n{w}_{S}^{\left( \sigma \right) }\left( n\right) = \mathop{\sum }\limits_{\substack{{\left( {{b}_{1},{b}_{2}}\right) \in {B}_{1} \times {B}_{2}} \\ {{b}_{1} + {b}_{2} = n} }}{w}_{{B}_{1}}\left( {b}_{1}\right) {w}_{{B}_{2}}\left( {b}_{2}\right)\n\]\n\n\[ \n\leq \mathop{\sum }\limits_{\substack{{\left( {{b}_{1},{b}_{2}}\right) \in {B}_{1} \times {B}_{2}} \\ {{b}_{1} + {b}_{2} = n} }}\frac{1}{2}\left( {{w}_{{B}_{1}}{\left( {b}_{1}\right) }^{2} + {w}_{{B}_{2}}{\left( {b}_{2}\right) }^{2}}\right)\n\]\n\n\[ \n\leq \frac{1}{2}\left( {\mathop{\sum }\limits_{{{b}_{1} \in {B}_{1}}}{w}_{{B}_{1}}{\left( {b}_{1}\right) }^{2} + \mathop{\sum }\limits_{{{b}_{2} \in {B}_{2}}}{w}_{{B}_{2}}{\left( {b}_{2}\right) }^{2}}\right)\n\]\n\n\[ \n= \frac{1}{2}\left( {{w}_{{D}_{1}}^{\left( {\delta }_{1}\right) }\left( 0\right) + {w}_{{D}_{2}}^{\left( {\delta }_{2}\right) }\left( 0\right) }\right) .\n\]\n\nThis completes the proof.
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Yes
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For \( t \geq 1 \), let \( {B}_{1},\ldots ,{B}_{{2}^{t}} \) be weighted sets of integers, and let \( S \) be the weighted sumset\n\n\[ S = {B}_{1} + \cdots + {B}_{{2}^{t}} \]\n\nwith weight function determined by the addition map \( \sigma : {B}_{1} \times \cdots \times {B}_{{2}^{t}} \rightarrow \) \( {B}_{1} + \cdots + {B}_{{2}^{t}} \) . For \( i = 1,\ldots ,{2}^{t} \), consider the weighted difference sets\n\n\[ {D}_{i} = {2}^{t - 1}{B}_{i} - {2}^{t - 1}{B}_{i} = {2}^{t - 1}\left( {{B}_{i} - {B}_{i}}\right) \]\n\nwith weight functions defined by the maps\n\n\[ {\delta }_{i} : {B}_{i} \times \cdots \times {B}_{i} \rightarrow {D}_{i} \]\n\n\[ {\delta }_{i}\left( {{b}_{i,1},\ldots ,{b}_{i,{2}^{t}}}\right) = \left( {{b}_{i,1} + \cdots + {b}_{i,{2}^{t - 1}}}\right) - \left( {{b}_{i,{2}^{t - 1} + 1} + \cdots + {b}_{i,{2}^{t}}}\right) . \]\n\nThen for all integers \( n \) ,\n\n\[ {w}_{S}^{\left( \sigma \right) }\left( n\right) \leq \frac{1}{{2}^{t}}\mathop{\sum }\limits_{{i = 1}}^{{2}^{t}}{w}_{{D}_{i}}^{\left( {\delta }_{i}\right) }\left( 0\right) . \]
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Proof. The proof of (12.3) is by induction on \( t \) . The case \( t = 1 \) is Lemma 12.1.\n\nLet \( t \geq 2 \), and assume that the lemma holds for \( t - 1 \) . Consider the weighted sumsets\n\n\[ {S}_{1} = {B}_{1} + \cdots + {B}_{{2}^{t - 1}} \]\n\nand\n\n\[ {S}_{2} = {B}_{{2}^{t - 1} + 1} + \cdots + {B}_{{2}^{t}} \]\n\nwith weights \( {w}_{{S}_{1}}^{\left( {\sigma }_{1}\right) } \) and \( {w}_{{S}_{2}}^{\left( {\sigma }_{2}\right) } \), respectively, and the weighted difference sets\n\n\[ {T}_{1} = {S}_{1} - {S}_{1} \]\n\nand\n\n\[ {T}_{2} = {S}_{2} - {S}_{2} \]\n\nwith weights \( {w}_{{T}_{1}}^{\left( {\Delta }_{1}\right) } \) and \( {w}_{{T}_{2}}^{\left( {\Delta }_{2}\right) } \), respectively. Since\n\n\[ S = {S}_{1} + {S}_{2} \]\n\nwe can define an addition map \( {\sigma }^{\prime } : {S}_{1} \times {S}_{2} \rightarrow S \) . By Theorem 12.1,\n\n\[ {w}_{S}^{\left( \sigma \right) }\left( s\right) = {w}_{S}^{\left( {\sigma }^{\prime }\right) }\left( s\right) \]\n\nfor all \( s \in S \) . (Indeed, Theorem 12.1 implies that all of the weight functions constructed in this proof are well-defined.)\n\nBy Lemma 12.1,\n\n\[ {w}_{S}^{\left( \sigma \right) }\left( s\right) \leq \frac{1}{2}\left( {{w}_{{T}_{1}}^{\left( {\Delta }_{1}\right) }\left( 0\right) + {w}_{{T}_{2}}^{\left( {\Delta }_{2}\right) }\left( 0\right) }\right) \]\n\nfor all \( s \in S \) . For \( i = 1,\ldots ,{2}^{t} \), we define the weighted difference sets\n\n\[ {B}_{i}^{\prime } = {B}_{i} - {B}_{i} \]\n\nThen\n\n\[ {T}_{1} = {S}_{1} - {S}_{1} \]\n\n\[ = \left( {{B}_{1} + \cdots + {B}_{{2}^{t - 1}}}\right) - \left( {{B}_{1} + \cdots + {B}_{{2}^{t - 1}}}\right) \]\n\n\[ = \left( {{B}_{1} - {B}_{1}}\right) + \cdots + \left( {{B}_{{2}^{t - 1}} - {B}_{{2}^{t - 1}}}\right) \]\n\n\[ = {B}_{1}^{\prime } + \cdots + {B}_{{2}^{t - 1}}^{\prime }\text{.} \]\n\nSimilarly,\n\n\[ {T}_{2} = {S}_{2} - {S}_{2} = {B}_{{2}^{t - 1} + 1}^{\prime } + \cdots + {B}_{{2}^{t}}^{\prime }. \]\n\nFor \( i = 1,\ldots ,{2}^{t} \), we define the weighted difference sets\n\n\[ {D}_{i}^{\prime } = {2}^{t - 2}{B}_{i}^{\prime } - {2}^{t - 2}{B}_{i}^{\prime } \]\n\nwith weight functions \( {w}_{{D}_{i}^{i}}^{\left( {\delta }_{i}^{\prime }\right) } \) . By induction, the lemma holds for sums of \( {2}^{t - 1} \) weighted sets. Therefore, we have\n\n\[ {w}_{{T}_{1}}^{\left( {\Delta }_{1}\right) }\left( 0\right) \leq \frac{1}{{2}^{t - 1}}\mathop{\sum }\limits_{{i = 1}}^{{2}^{t - 1}}{w}_{{D}_{i}^{\prime }}^{\left( {\delta }_{i}^{\prime }\right) }\left( 0\right) \]\n\nand\n\n\[ {w}_{{T}_{2}}^{\left( {\Delta }_{2}\right) }\left( 0\right) \leq \frac{1}{{2}^{t - 1}}\mathop{\sum }\limits_{{i = {2}^{t - 1} + 1}}^{{2}^{t}}{w}_
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Yes
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Lemma 12.3 Let \( Q \geq 1 \) . Let \( {u}_{1},\ldots ,{u}_{k} \) be relatively prime integers such that\n\n\[ U = \max \left\{ {\left| {u}_{1}\right| ,\ldots ,\left| {u}_{k}\right| }\right\} \leq Q.\]\n\nFor every integer \( m \),\n\n\[ \text{ NSE }\left\{ \begin{array}{l} {u}_{1}{v}_{1} + \cdots + {u}_{k}{v}_{k} = m \\ \text{ with }\left| {v}_{1}\right| ,\ldots ,\left| {v}_{k}\right| \leq Q \end{array}\right\} \leq \frac{\left( {k - 1}\right) !{\left( 3Q\right) }^{k - 1}}{U}. \]
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Proof. The proof is by induction on \( k \) . If \( k = 1 \), then \( \gcd \left( {u}_{1}\right) = 1 \) and \( U = \left| {u}_{1}\right| = 1 \) . The number of solutions of the equation \( {u}_{1}{v}_{1} = m \) with \( \left| {v}_{1}\right| \leq Q \) is at most\n\n\[ 1 = \frac{0!{\left( 3Q\right) }^{0}}{U} \]\n\nLet \( k = 2 \) and \( U = \max \left\{ {\left| {u}_{1}\right| ,\left| {u}_{2}\right| }\right\} = \left| {u}_{2}\right| \) . If\n\n\[ {u}_{1}{v}_{1} + {u}_{2}{v}_{2} = m \]\n\n(12.6)\n\nthen\n\n\[ {u}_{1}{v}_{1} \equiv m\;\left( {\;\operatorname{mod}\;U}\right) \]\n\nSince \( \left( {{u}_{1},{u}_{2}}\right) = \left( {{u}_{1}, U}\right) = 1 \), we have\n\n\[ {v}_{1} \equiv {u}_{1}^{-1}m\;\left( {\;\operatorname{mod}\;U}\right) \]\n\nThe number of integers \( {v}_{1} \) in the congruence class \( {u}_{1}^{-1}m\;\left( {\;\operatorname{mod}\;U}\right) \) with \( \left| {v}_{1}\right| \leq Q \) is at most\n\n\[ \frac{2Q}{U} + 1 \leq \frac{3Q}{U}\;\left( {\text{ since }U \leq Q}\right) \]\n\nFor each such integer \( {v}_{1} \) there is at most one integer \( {v}_{2} \) that satisfies the linear equation (12.6). Therefore,\n\n\[ \text{NSE}\left\{ {{u}_{1}{v}_{1} + {u}_{2}{v}_{2} = m\;\text{with}\left| {v}_{1}\right| ,\left| {v}_{2}\right| \leq Q}\right\} \leq \frac{3Q}{U}\text{.} \]
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Yes
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Lemma 12.4 Let\n\n\\[ \nf\\left( x\\right) = \\mathop{\\sum }\\limits_{{i = 0}}^{k}{a}_{i}{x}^{i} \n\\]\n\nbe a polynomial of degree \\( k \\) with complex coefficients. Then\n\n\\[ \nf\\left( {x + u}\\right) - f\\left( x\\right) = u{g}_{u}\\left( x\\right) \n\\]\n\nwhere\n\n\\[ \n{g}_{u}\\left( x\\right) = \\mathop{\\sum }\\limits_{{i = 0}}^{{k - 1}}{a}_{i}^{\\prime }\\left( u\\right) {x}^{i} \n\\]\n\nis a polynomial of degree \\( k - 1 \\) with coefficients\n\n\\[ \n{a}_{i}^{\\prime }\\left( u\\right) = \\mathop{\\sum }\\limits_{{j = i + 1}}^{k}\\left( \\begin{array}{l} j \\\\ i \\end{array}\\right) {a}_{j}{u}^{j - i - 1}. \n\\]\n\nFor any positive number \\( P \\), if\n\n\\[ \n\\left| x\\right| \\leq {c}_{1}P \n\\]\n\n\\[ \n\\left| u\\right| \\leq 2{c}_{1}P \n\\]\n\nand\n\n\\[ \n\\left| {a}_{i}\\right| \\leq c{P}^{k - i}\\;\\text{ for }i = 0,1,\\ldots, k, \n\\]\n\nthen\n\n\\[ \n\\left| {{a}_{i}^{\\prime }\\left( u\\right) }\\right| \\leq c{\\left( 4{c}_{1}\\right) }^{k}k{P}^{k - 1 - i}\\;\\text{ for }i = 0,1,\\ldots, k - 1 \n\\]\n\nand\n\n\\[ \n\\left| {{g}_{u}\\left( x\\right) }\\right| \\leq c{\\left( 2{c}_{1}\\right) }^{2k}{k}^{2}{P}^{k - 1} \n\\]
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Proof. This is a purely formal calculation. We have\n\n\\[ \nf\\left( {x + u}\\right) - f\\left( x\\right) = \\mathop{\\sum }\\limits_{{j = 0}}^{k}{a}_{j}{\\left( x + u\\right) }^{j} - \\mathop{\\sum }\\limits_{{j = 0}}^{k}{a}_{j}{x}^{j} \n\\]\n\n\\[ \n= \\mathop{\\sum }\\limits_{{j = 1}}^{k}{a}_{j}\\mathop{\\sum }\\limits_{{i = 0}}^{{j - 1}}\\left( \\begin{array}{l} j \\\\ i \\end{array}\\right) {x}^{i}{u}^{j - i} \n\\]\n\n\\[ \n= u\\mathop{\\sum }\\limits_{{i = 0}}^{{k - 1}}\\left( {\\mathop{\\sum }\\limits_{{j = i + 1}}^{k}\\left( \\begin{array}{l} j \\\\ i \\end{array}\\right) {a}_{j}{u}^{j - i - 1}}\\right) {x}^{i} \n\\]\n\n\\[ \n= u{g}_{u}\\left( x\\right) \\text{.} \n\\]\n\nIf \\( \\left| {a}_{i}\\right| \\leq c{P}^{k - i} \\) and \\( \\left| u\\right| \\leq 2{c}_{1}P \\), then\n\n\\[ \n\\left| {{a}_{i}^{\\prime }\\left( u\\right) }\\right| \\leq \\mathop{\\sum }\\limits_{{j = i + 1}}^{k}\\left( \\begin{array}{l} j \\\\ i \\end{array}\\right) \\left| {a}_{j}\\right| {\\left| u\\right| }^{j - i - 1} \\leq \\mathop{\\sum }\\limits_{{j = i + 1}}^{k}{2}^{j}c{P}^{k - j}{\\left( 2{c}_{1}P\\right) }^{j - i - 1} \n\\]\n\n\\[ \n\\leq c{\\left( 4{c}_{1}\\right) }^{k}k{P}^{k - 1 - i}. \n\\]\n\nIf also \\( \\left| x\\right| \\leq {c}_{1}P \\), then\n\n\\[ \n\\left| {{g}_{u}\\left( x\\right) }\\right| \\leq \\mathop{\\sum }\\limits_{{i = 0}}^{{k - 1}}\\left| {{a}_{i}^{\\prime }\\left( u\\right) }\\right| {\\left| x\\right| }^{i} \n\\]\n\n\\[ \n\\leq \\mathop{\\sum }\\limits_{{i = 0}}^{{k - 1}}c{\\left( 4{c}_{1}\\right) }^{k}k{P}^{k - 1 - i}{\\left( {c}_{1}P\\right) }^{i} \n\\]\n\n\\[ \n\\leq c{\\left( 2{c}_{1}\\right) }^{2k}{k}^{2}{P}^{k - 1} \n\\]\n\nThis completes the proof.
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Yes
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Theorem 12.3 Let \( \{ s\left( k\right) {\} }_{k = 1}^{\infty } \) be the sequence of integers defined recursively by \( s\left( 1\right) = 1 \) and\n\n\[ s\left( k\right) = {8k}{2}^{\left\lbrack {\log }_{2}s\left( k - 1\right) \right\rbrack }\;\text{ for }k \geq 2. \]\n\nLet \( c \geq 1 \) . For \( j = 1,\ldots, s\left( k\right) \), let\n\n\[ {f}_{j}\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{k}{a}_{ij}{x}^{i} \]\n\nbe a sequence of polynomials with complex coefficients such that\n\n\[ \left| {a}_{kj}\right| \leq c\;\text{ for }j = 1,\ldots, s\left( k\right) . \]\n\nChoose \( P \geq 1 \) such that\n\n\[ \left| {a}_{ij}\right| \leq c{P}^{k - i}\;\text{ for }i = 0,1,\ldots, k - 1\text{ and }j = 1,\ldots, s\left( k\right) . \]\n\nLet \( {c}_{1} \geq 1 \) . For every complex number \( z \) ,\n\n\[ \text{ NSE }\left\{ \begin{array}{l} \mathop{\sum }\limits_{{j = 1}}^{{s\left( k\right) }}{f}_{j}\left( {x}_{j}\right) = z\;\text{ with }{x}_{j} \in \mathbf{Z} \\ \text{ and }\left| {x}_{j}\right| \leq {c}_{1}P\text{ for }j = 1,\ldots, s\left( k\right) \end{array}\right\} { \ll }_{k, c,{c}_{1}}{P}^{s\left( k\right) - k}. \]
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Proof. The proof is by induction on the degree \( k \) of the polynomials.\n\nFor \( k = 1 \) we have \( s\left( 1\right) = 1 \) and \( {f}_{1}\left( x\right) = {a}_{11}x + {a}_{01} \) . For any number \( z \) , there exists at most one integer \( {x}_{1} \) such that \( {f}_{1}\left( {x}_{1}\right) = z \), and so\n\n\[ \text{ NSE }\left\{ \begin{array}{l} {f}_{1}\left( {x}_{1}\right) = z\;\text{ with }{x}_{1} \in \mathbf{Z} \\ \text{ and }\left| {x}_{1}\right| \leq {c}_{1}P \end{array}\right\} \leq 1 = {P}^{s\left( 1\right) - 1}. \]\n\nLet \( k \geq 2 \), and assume that the theorem holds for \( {s}^{\prime } = s\left( {k - 1}\right) \) polynomials of degree \( k - 1 \) . Define\n\n\[ t = t\left( k\right) = \left\lbrack {{\log }_{2}{s}^{\prime }}\right\rbrack + 2 \]\n\nand\n\n\[ s = s\left( k\right) = {2k}{2}^{t} = {8k}{2}^{\left\lbrack {\log }_{2}s\left( k - 1\right) \right\rbrack }. \]\n\nSince \( \left\lbrack x\right\rbrack \leq x < \left\lbrack x\right\rbrack + 1 \) for every real number \( x \), we have\n\n\[ {s}^{\prime } = {2}^{{\log }_{2}{s}^{\prime }} < {2}^{\left\lbrack {{\log }_{2}{s}^{\prime }}\right\rbrack + 1} = {2}^{t - 1}. \]\n\nConsider the weighted set \( \left. \left( {X,{w}_{X}}\right) \right) \), where\n\n\[ X = \left\{ {x \in \mathbf{Z} : \left| x\right| \leq {c}_{1}P}\right\} \]\n\nand \( {w}_{X}\left( x\right) = 1 \) for all \( x \in X \) . For \( j = 1,\ldots, s \) we have the weighted sets\n\n\[ {F}_{j} = \left\{ {{f}_{j}\left( x\right) : x \in X}\right\} = \left\{ {{f}_{j}\left( x\right) : \left| x\right| \leq {c}_{1}P}\right\} \]\n\nwith weights\n\n\[ {w}_{{F}_{j}}^{\left( {f}_{j}\right) }\left( z\right) = \operatorname{NSE}\left\{ {{f}_{j}\left( x\right) = z : \left| x\right| \leq {c}_{1}P}\right\} . \]\n\nLet \( S \) be the weighted sumset\n\n\[ S = {F}_{1} + \cdots + {F}_{s} \]\n\nThen\n\n\[ {w}_{S}\left( z\right) = \operatorname{NSE}\left\{ {\mathop{\sum }\limits_{{j = 1}}^{s}{f}_{j}\left( {x}_{j}\right) = z\;\text{ with }\left| {x}_{j}\right| \leq {c}_{1}P}\right\} . \]\n\nFor\n\n\[ m = \frac{s}{2} = k{2}^{t} \]\n\nwe consider the weighted sumsets\n\n\[ {B}_{1} = {F}_{1} + \cdots + {F}_{m} \]\n\nand\n\n\[ {B}_{2} = {F}_{m + 1} + \cdots + {F}_{2m} \]\n\nand the weighted difference sets\n\n\[ {D}_{1} = {B}_{1} - {B}_{1} = \left\{ {\mathop{\sum }\limits_{{j = 1}}^{m}\left( {{f}_{j}\left( {y}_{j}\right) - {f}_{j}\left( {x}_{j}\right) }\right) : \left| {x}_{j}\right| ,\left| {y}_{j}\right| \leq {c}_{1}P}\right\} \]\n\nand\n\n\[ {D}_{2} = {B}_{2} - {B}_{2} = \left\{ {\mathop{\sum }\limits_{{j = m + 1}}^{{2m}}\left( {{f}_{j}\left( {y}_{j}\right) - {f}_{j}\left( {x}_{j}\right) }\right) : \left| {x}_{j}\right| ,\left| {y}_{j}\right| \leq {c}_{1}P}\right\} . \]\n\nApplying Lemma 12.1 to \( S = {B}_{1} + {B}_{2} \), we obtain\n\n\[ {w}_{S}\left( z\right) \leq \frac{1}{2}\left( {{w}_{{D}_{1}}\left( 0\right) + {w}_{{D}_{2}}\left( 0\right) }\right) . \]
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Yes
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Lemma 12.5 Let \( c \geq 1 \) . Let \( \mathcal{F} = {\left\{ {f}_{j}\left( x\right) \right\} }_{j = 1}^{s} \) be a sequence of integer-valued polynomials of degree \( k \), and let \( {a}_{kj} \) be the leading coefficient of \( {f}_{j}\left( x\right) \) . We assume that\n\n\[ 0 < {a}_{kj} \leq c \]\n\nfor \( j = 1,\ldots, s \) . If \( N \) is sufficiently large, then\n\n\[ {R}_{\mathcal{F}}\left( N\right) > \frac{1}{2}{\left( \frac{2N}{3cs}\right) }^{s/k}. \]\n\n(12.15)
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Proof. Define \( {x}^{ * }\left( {f}_{j}\right) \) by (11.4) for \( j = 1,\ldots, s \) . If the integers \( {x}_{j} \) satisfy the inequalities\n\n\[ {x}^{ * }\left( {f}_{j}\right) \leq {x}_{j} \leq {\left( \frac{2N}{3cs}\right) }^{1/k} \]\n\nthen, by Lemma 11.4,\n\n\[ 0 \leq {f}_{j}\left( {x}_{j}\right) \leq \frac{3{a}_{kj}{x}_{j}^{k}}{2} \leq \frac{3c}{2}\left( \frac{2N}{3cs}\right) = \frac{N}{s} \]\n\nand\n\n\[ 0 \leq {f}_{1}\left( {x}_{1}\right) + \cdots + {f}_{s}\left( {x}_{s}\right) \leq N. \]\n\nTherefore,\n\n\[ {R}_{\mathcal{F}}\left( N\right) > {\left( {\left( \frac{2N}{3cs}\right) }^{1/k} - {x}^{ * }\left( f\right) - 1\right) }^{s} \geq \frac{1}{2}{\left( \frac{2N}{3cs}\right) }^{s/k} \]\n\nfor \( N \) sufficiently large. This proves (12.15). \( ▱ \)
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Yes
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Lemma 12.6 Let \( \mathcal{F} = {\left\{ {f}_{j}\left( x\right) \right\} }_{j = 1}^{s} \) be a sequence of integer-valued polynomials of degree \( k \), and let \( {a}_{kj} \) be the leading coefficient of \( {f}_{j}\left( x\right) \) . Let \( c \geq 1 \) . We assume that\n\n\[ 0 < {a}_{kj} \leq c \]\n\nand that \( A\left( {f}_{j}\right) = \left\{ {{f}_{j}\left( x\right) : x \in {\mathbf{N}}_{0}}\right\} \) is a strictly increasing sequence of nonnegative integers for \( j = 1,\ldots, s \) . There exists a number \( {N}_{1}\left( \mathcal{F}\right) \) such that if \( N \geq {N}_{1}\left( \mathcal{F}\right) \) and \( {x}_{1},\ldots ,{x}_{s} \) are nonnegative integers with\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{s}f\left( {x}_{j}\right) \leq N \]\n\nthen\n\n\[ {x}_{j} \leq {\left( 4k!N\right) }^{1/k}\;\text{ for }j = 1,\ldots, s. \]
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Proof. The proof is the same as the proof of Lemma 11.5. Recall that \( k!{a}_{kj} \geq 1 \) by Exercise 6 in Section 11.1. Define \( {x}^{ * }\left( {f}_{j}\right) \) by (11.4) for \( j = 1,\ldots, s \), and \( {x}^{ * }\left( \mathcal{F}\right) = \max \left\{ {{x}^{ * }\left( {f}_{1}\right) ,\ldots ,{x}^{ * }\left( {f}_{s}\right) }\right\} \) . Let\n\n\[ {N}_{1}\left( \mathcal{F}\right) = \frac{{x}^{ * }{\left( \mathcal{F}\right) }^{k}}{{2k}!}. \]\n\n(12.16)\n\nIf \( N \geq {N}_{1}\left( \mathcal{F}\right) ,1 \leq \ell \leq s \), and \( {x}_{\ell } > {\left( 2k!N\right) }^{1/k} \geq {x}^{ * }\left( \mathcal{F}\right) \geq {x}^{ * }\left( {f}_{\ell }\right) \), then\n\n\[ {f}_{\ell }\left( {x}_{\ell }\right) \geq \frac{{a}_{k\ell }{x}_{\ell }^{k}}{2} > k!{a}_{k}N \geq N \]\n\nand so\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{s}f\left( {x}_{j}\right) \geq f\left( {x}_{\ell }\right) \geq f\left( {x}_{\ell }\right) > N. \]\n\nIt follows that if \( {x}_{1},\ldots ,{x}_{s} \) are nonnegative integers such that\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{s}f\left( {x}_{j}\right) \leq N \]\n\nthen\n\n\[ {x}_{j} \leq {\left( 2k!N\right) }^{1/k}\;\text{ for }j = 1,\ldots, s. \]\n\nThis completes the proof.
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Yes
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For any positive integer \( k \) and real number \( c \geq 1 \), there exists a number \( \delta \left( {k, c}\right) > 0 \) with the following property: If \( s = s\left( k\right) \) is the integer defined by (12.11), and if \( \mathcal{F} = {\left\{ {f}_{j}\left( x\right) \right\} }_{j = 1}^{s} \) is a sequence of integer-valued polynomials of degree \( k \) whose leading coefficients \( {a}_{kj} \) satisfy\n\n\[ 0 < {a}_{kj} \leq c \]\n\nthen the sumset\n\n\[ B = \left\{ {{f}_{1}\left( {x}_{1}\right) + \cdots + {f}_{s}\left( {x}_{s}\right) : {x}_{1},\ldots ,{x}_{s} \in {\mathbf{N}}_{0}}\right\} \]\n\nhas lower asymptotic density\n\n\[ {d}_{L}\left( B\right) \geq \delta \left( {k, c}\right) > 0. \]
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Proof. Replacing the polynomial \( {f}_{j}\left( x\right) \) with \( {f}_{j}\left( {x + {x}_{0}}\right) \) for a sufficiently large integer \( {x}_{0} \), we can assume that \( \left\{ {{f}_{j}\left( x\right) : x \in {\mathbf{N}}_{0}}\right\} \) is a strictly increasing sequence of nonnegative integers for \( j = 1,\ldots, s \) .\n\nDefine \( {N}_{1}\left( \mathcal{F}\right) \) by (12.16). Choose \( {N}_{2}\left( \mathcal{F}\right) \) sufficiently large that for \( N \geq \) \( {N}_{2}\left( \mathcal{F}\right) \) and \( P = {N}^{1/k} \), we have\n\n\[ \left| {a}_{ij}\right| \leq c{P}^{k - i}\;\text{ for }i = 0,1,\ldots, k - 1, \]\n\nand so Theorem 12.3 applies to the polynomials in the sequence \( \mathcal{F} \).\n\nLet \( N\left( \mathcal{F}\right) = \max \left\{ {{N}_{1}\left( \mathcal{F}\right) ,{N}_{2}\left( \mathcal{F}\right) }\right) \) and \( {c}_{1} = {\left( 2k!\right) }^{1/k} \) . By Lemma 12.6, if \( N \geq N\left( \mathcal{F}\right) \) and \( {x}_{1},\ldots ,{x}_{s} \) are nonnegative integers such that\n\n\[ {f}_{1}\left( {x}_{1}\right) + \cdots + {f}_{s}\left( {x}_{s}\right) \leq N \]\n\nthen \( {x}_{j} \leq {c}_{1}P \) for \( j = 1,\ldots, s \) . Therefore, if \( 0 \leq n \leq N \), then\n\n\[ {r}_{\mathcal{F}}\left( n\right) \; = \;\text{ NSE }\left\{ \begin{array}{l} {f}_{1}\left( {x}_{1}\right) + \cdots + {f}_{s}\left( {x}_{s}\right) = n \\ \text{ with }{x}_{j} \in {\mathbf{N}}_{0}\text{ for }j = 1,\ldots, s\left( k\right) \end{array}\right\} \]\n\n\[ = \;\text{ NSE }\left\{ \begin{array}{l} {f}_{1}\left( {x}_{1}\right) + \cdots + {f}_{s}\left( {x}_{s}\right) = n \\ \text{ with }0 \leq {x}_{j} \leq {c}_{1}P\text{ for }j = 1,\ldots, s\left( k\right) \end{array}\right\} \]\n\n\[ { \ll }_{k, c}\;{P}^{s - k} \]\n\nby Theorem 12.3. Let \( B\left( n\right) \) be the counting function of the set \( B \) . We have\n\n\[ {R}_{\mathcal{F}}\left( N\right) \; = \;\mathop{\sum }\limits_{{n = 0}}^{N}{r}_{\mathcal{F}}\left( n\right) = \mathop{\sum }\limits_{\substack{{n = 0} \\ {n \in B} }}^{N}{r}_{\mathcal{F}}\left( n\right) \]\n\n\[ { \ll }_{k, c}\;B\left( N\right) {P}^{s - k} = \frac{B\left( N\right) {P}^{s}}{N}. \]\n\nBy Lemma 12.5,\n\n\[ {R}_{\mathcal{F}}\left( N\right) > \frac{1}{2}{\left( \frac{2}{3cs}\right) }^{s/k}{P}^{s} \]\n\nIt follows that \( B\left( N\right) /N{ \gg }_{k, c}1 \) . This completes the proof. \( ▱ \)
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Yes
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Theorem 12.5 Let \( k \) be a positive integer and \( c \geq 1 \) . There exists a positive integer \( h = h\left( {k, c}\right) \) with the following property: Let \( \mathcal{F} = {\left\{ {f}_{j}\left( x\right) \right\} }_{j = 1}^{h} \) be a sequence of integer-valued polynomials of degree \( k \) such that the leading coefficient \( {a}_{kj} \) of \( {f}_{j}\left( x\right) \) satisfies the inequality \( 0 < {a}_{kj} \leq c \) for \( j = 1,\ldots, h \) . There exists a positive integer \( m \) such that the sumset \[ S = \left\{ {{f}_{1}\left( {x}_{1}\right) + \cdots + {f}_{h}\left( {x}_{h}\right) : {x}_{j} \in {\mathbf{N}}_{0}\;\text{ for }j = 1,\ldots, h}\right\} \] eventually coincides with a union of congruence classes modulo \( m \) .
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Proof. Let \( s = s\left( k\right) \) be the positive integer constructed in Theorem 12.3 and let \( \delta = \delta \left( {k, c}\right) \) be the positive number constructed in Theorem 12.4. We define \[ d = \left\lbrack \frac{1}{\delta }\right\rbrack + 1 \] and \[ h = h\left( {k, c}\right) = {ds}. \] Let \( \mathcal{F} = {\left\{ {f}_{j}\left( x\right) \right\} }_{j = 1}^{h} \) be a sequence of integer-valued polynomials of degree \( k \) whose leading coefficients are positive and not greater than \( c \) . For \( i = 1,\ldots, d \), let \( {\mathcal{F}}_{i} = {\left\{ {f}_{\left( {i - 1}\right) s + j}\left( x\right) \right\} }_{j = 1}^{s} \) . By Theorem 12.4, the sumset \[ {B}_{i} = \left\{ {\mathop{\sum }\limits_{{j = 1}}^{s}{f}_{\left( {i - 1}\right) s + j}\left( {x}_{j}\right) : {x}_{j} \in {\mathbf{N}}_{0}}\right\} \] has lower asymptotic density \( {d}_{L}\left( {B}_{i}\right) \geq \delta > 0 \) . Since \[ S = {B}_{1} + \cdots + {B}_{d} = \left\{ {\mathop{\sum }\limits_{{j = 1}}^{h}{f}_{j}\left( {x}_{j}\right) : {x}_{j} \in {\mathbf{N}}_{0}}\right\} \] and \[ \mathop{\sum }\limits_{{i = 1}}^{d}{d}_{L}\left( {B}_{i}\right) \geq {\delta d} = \delta \left( {\left\lbrack \frac{1}{\delta }\right\rbrack + 1}\right) > 1 \geq {d}_{L}\left( S\right) , \] Kneser’s theorem implies that \( S \) eventually coincides with a union of congruence classes modulo \( m \) for some positive integer \( m \) . \( ▱ \)
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Yes
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Lemma 13.1 Let \( n \) be an odd positive integer. Then \( \sigma \left( n\right) \) is odd if and only if \( n \) is a square.
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Proof. Let\n\n\[ n = \mathop{\prod }\limits_{{p \mid n}}{p}^{{v}_{p}} \]\n\nbe the unique factorization of \( n \) as a product of odd prime numbers. The positive integer \( d \) divides \( n \) if and only if \( d \) can be written in the form\n\n\[ d = \mathop{\prod }\limits_{{p \mid n}}{p}^{{u}_{p}} \]\n\nwhere\n\n\[ 0 \leq {u}_{p} \leq {v}_{p} \]\n\nand so\n\n\[ \sigma \left( n\right) = \mathop{\prod }\limits_{{p \mid n}}\mathop{\sum }\limits_{{{u}_{p} = 0}}^{{v}_{p}}{p}^{{u}_{p}} \]\n\n\[ \equiv \mathop{\prod }\limits_{{p \mid n}}\left( {{u}_{p} + 1}\right) \;\left( {\;\operatorname{mod}\;2}\right) \]\n\n\[ \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) \]\n\nif and only if \( {u}_{p} \) is even for all \( p \), that is, \( {u}_{p} = 2{w}_{p} \) and\n\n\[ n = \mathop{\prod }\limits_{{p \mid n}}{p}^{{v}_{p}} = {\left( \mathop{\prod }\limits_{{p \mid n}}{p}^{{w}_{p}}\right) }^{2} \]\n\nis a square. This completes the proof.
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Yes
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Lemma 13.2 If \( n = {2}^{k}m \), where \( k \geq 0 \) and \( m \) is odd, then \( {\sigma }^{ * }\left( n\right) = \) \( {2}^{k}\sigma \left( m\right) \) . If \( {\sigma }^{ * }\left( n\right) \) is odd, then \( n \) is the square of an odd integer.
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Proof. Let \( d \) be a divisor of \( n \) . If the conjugate divisor \( \delta = n/d \) is odd, then \( {2}^{k} \) must divide \( d \), and so \( d = {2}^{k}{d}^{\prime } \) for some integer \( {d}^{\prime } \) . Then\n\n\[ \n{2}^{k}m = n = {d\delta } = {2}^{k}{d}^{\prime }\delta \n\]\n\nand \( {d}^{\prime } \) is a divisor of \( m \) . Conversely, if \( {d}^{\prime } \) is any divisor of \( m \), then \( {2}^{k}{d}^{\prime } \) is a divisor of \( n \) whose conjugate divisor \( m/{d}^{\prime } \) is odd. Therefore,\n\n\[ \n{\sigma }^{ * }\left( n\right) = {2}^{k}\mathop{\sum }\limits_{{{d}^{\prime } \mid m}}{d}^{\prime } = {2}^{k}\sigma \left( m\right) . \n\]\n\nIf \( {\sigma }^{ * }\left( n\right) \) is odd, then \( k = 0 \) and \( n = m \) is odd. It follows that \( {\sigma }^{ * }\left( n\right) = \) \( \sigma \left( m\right) = \sigma \left( n\right) \) is odd, and so \( n \) is a square by Lemma 13.1. This completes the proof.
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Yes
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Lemma 13.3 For every positive integer \( n \) , \[ {\sigma }^{ * }\left( n\right) = 2\mathop{\sum }\limits_{{1 \leq u < \sqrt{n}}}{\left( -1\right) }^{u - 1}{\sigma }^{ * }\left( {n - {u}^{2}}\right) + {\left\{ {\left( -1\right) }^{n - 1}n\right\} }_{n = {\ell }^{2}}. \]
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Proof. We apply Theorem 13.2 to the odd function \( f\left( y\right) = y \) . If \( n = {\ell }^{2} \) , the right side of the identity is \[ {\left( -1\right) }^{\ell - 1}\ell f\left( \ell \right) = {\left( -1\right) }^{n - 1}{\ell }^{2} = {\left( -1\right) }^{n - 1}n. \] To obtain the left side of the identity, we recall the involution (13.1) on triples \( \left( {u, d,\delta }\right) \) such that \( {u}^{2} + {d\delta } = n \) and \( \delta \) is odd, and obtain \[ \mathop{\sum }\limits_{\substack{{u + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{u}u = 0 \] Then \[ \mathop{\sum }\limits_{\substack{{u + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{u}f\left( {u + d}\right) = \mathop{\sum }\limits_{\substack{{u + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{u}\left( {u + d}\right) \] \[ = \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{u}u + \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{u}d \] \[ = \mathop{\sum }\limits_{\substack{{{u}^{2} + {d\delta } = n} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}{\left( -1\right) }^{u}d \] \[ = \mathop{\sum }\limits_{{{u}^{2} < n}}{\left( -1\right) }^{u}\mathop{\sum }\limits_{\substack{{n - {u}^{2} = {d\delta }} \\ {\delta \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) } }}d \] \[ = \mathop{\sum }\limits_{{\left| u\right| < \sqrt{n}}}{\left( -1\right) }^{u}{\sigma }^{ * }\left( {n - {u}^{2}}\right) \] Therefore, \[ \mathop{\sum }\limits_{{\left| u\right| < \sqrt{n}}}{\left( -1\right) }^{u}{\sigma }^{ * }\left( {n - {u}^{2}}\right) = {\left\{ {\left( -1\right) }^{n - 1}n\right\} }_{n = {\ell }^{2}}. \] This completes the proof.
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Yes
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Theorem 13.3 (Fermat) An odd prime number \( p \) can be represented by the quadratic form \( {x}^{2} + {y}^{2} \) if and only if \( p \equiv 1\;\left( {\;\operatorname{mod}\;4}\right) \) .
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Proof. Since every square is congruent to 0 or 1 modulo 4, it follows that a sum of two squares must be congruent to 0,1 , or 2 modulo 4 , and so no integer congruent to 3 modulo 4 can be represented as the sum of two squares.\n\nLet \( p \) be an odd prime number. Then \( p \) is certainly not a square. By Lemma 13.3,\n\n\[ \n{\sigma }^{ * }\left( p\right) = 2{\sigma }^{ * }\left( {p - 1}\right) - 2{\sigma }^{ * }\left( {p - 4}\right) + 2{\sigma }^{ * }\left( {p - 9}\right) - \cdots .\n\]\n\nSince \( {\sigma }^{ * }\left( p\right) = p + 1 \), we have\n\n\[ \n\frac{p + 1}{2} = {\sigma }^{ * }\left( {p - {1}^{2}}\right) - {\sigma }^{ * }\left( {p - {2}^{2}}\right) + {\sigma }^{ * }\left( {p - {3}^{2}}\right) - \cdots .\n\]\n\nIf \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), then \( \left( {p + 1}\right) /2 \) is an odd integer, and so at least one of the terms on right side of this equation must be odd. Thus, there exists a positive integer \( b < \sqrt{n} \) such that \( {\sigma }^{ * }\left( {p - {b}^{2}}\right) \) is odd. By Lemma 13.2, \( p - {b}^{2} = {a}^{2} \) for some odd integer \( a \) . This completes the proof.
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Yes
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Theorem 13.4 If \( p \) is a prime number such that \( p \equiv 1\\left( {\\;\\operatorname{mod}\\;4}\\right) \), then there exist unique positive integers \( a \) and \( b \) such that \( a \) is odd, \( b \) is even, and \( p = {a}^{2} + {b}^{2} \) .
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Proof. Let\n\n\[ p = {a}_{1}^{2} + {b}_{1}^{2} = {a}_{2}^{2} + {b}_{2}^{2} \]\n\nwhere \( {a}_{1} \) and \( {a}_{2} \) are positive odd integers and \( {b}_{1} \) and \( {b}_{2} \) are positive even integers. We must prove that \( {a}_{1} = {a}_{2} \) and \( {b}_{1} = {b}_{2} \).\n\nIf \( {a}_{1} < {a}_{2} \), then \( {b}_{1} > {b}_{2} \) and there exist positive integers \( x \) and \( y \) such that\n\n\[ {a}_{2} = {a}_{1} + {2x} \]\n\nand\n\n\[ {b}_{2} = {b}_{1} - {2y} \]\n\nThen\n\n\[ p = {a}_{2}^{2} + {b}_{2}^{2} \]\n\n\[ = {\\left( {a}_{1} + 2x\\right) }^{2} + {\\left( {b}_{1} - 2y\\right) }^{2} \]\n\n\[ = {a}_{1}^{2} + 4{a}_{1}x + 4{x}^{2} + {b}_{1}^{2} - 4{b}_{1}y + 4{y}^{2} \]\n\n\[ = p + 4{a}_{1}x + 4{x}^{2} - 4{b}_{1}y + 4{y}^{2}, \]\nand so\n\n\[ x\\left( {{a}_{1} + x}\\right) = y\\left( {{b}_{1} - y}\\right) .\n\nLet \( \\left( {x, y}\\right) = d \). Define the positive integers \( X \) and \( Y \) by \( x = {dX} \) and \( y = {dY} \). Then\n\n\[ X\\left( {{a}_{1} + x}\\right) = Y\\left( {{b}_{1} - y}\\right) .\n\nSince \( \\left( {X, Y}\\right) = 1 \), it follows that there exists a positive integer \( r \) such that\n\n\[ {rY} = {a}_{1} + x = {a}_{1} + {dX} \]\n\nand\n\n\[ {rX} = {b}_{1} - y = {b}_{1} - {dY} \]\n\nThen \( {r}^{2} + {d}^{2} \\geq 2 \) and \( {x}^{2} + {y}^{2} \\geq 2 \), and\n\n\[ p = {a}_{1}^{2} + {b}_{1}^{2} = {\\left( rY - dX\\right) }^{2} + {\\left( rX + dY\\right) }^{2} = \\left( {{r}^{2} + {d}^{2}}\\right) \\left( {{X}^{2} + {Y}^{2}}\\right) ,\n\nwhich is impossible, since \( p \) is prime and not composite. Therefore, \( {a}_{1} = {a}_{2} \) and \( {b}_{1} = {b}_{2} \), and the representation of a prime \( p \\equiv 1\\;\\left( {\\;\\operatorname{mod}\\;4}\\right) \) as a sum of two squares is essentially unique.
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Yes
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