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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 \( r \) . | 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\l... | Yes |
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 \( k \) . | 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 \le... | Yes |
Theorem 6.7 Let \( k \) be a positive integer. Then\n\n\[ \mathop{\sum }\limits_{{{n}_{1}\cdots {n}_{k} \leq x}}\frac{1}{{n}_{1}\cdots {n}_{k}} = \frac{1}{k!}{\log }^{k}x + O\left( {{\log }^{k - 1}x}\right) ,\] | 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}_... | Yes |
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_{... | 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... | Yes |
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 \( \gamma = {0.577}\ldots \) ... | 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) ... | Yes |
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\[ ... | 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... | Yes |
Theorem 6.11 For \( x \geq 2 \) , \n\n\[ \mathop{\sum }\limits_{{n \leq x}}{\log }^{2}n = x{\log }^{2}x - {2x}\log x + {2x} + O\left( {{\log }^{2}x}\right) . \] | 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 }^... | Yes |
Theorem 6.12 For \( x \geq 2 \) ,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}{\log }^{2}\frac{x}{n} = {2x} + O\left( {{\log }^{2}x}\right) \] | 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\r... | Yes |
Theorem 6.13 The Möbius function \( \mu \left( n\right) \) is multiplicative, and\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = \left\{ \begin{array}{ll} 1 & \text{ if }n = 1 \\ 0 & \text{ if }n > 1 \end{array}\right. \] | 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\r... | Yes |
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\[ f\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mu \left( \frac{n}{d}\right) g\left( d\right) \] | 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... | Yes |
Theorem 6.16\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\mu \left( n\right) }{n} = O\left( 1\right) \] | 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{... | Yes |
Theorem 6.17\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\mu \left( n\right) }{{n}^{2}} = \frac{6}{{\pi }^{2}} + O\left( \frac{1}{x}\right) . \] | 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... | Yes |
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 \( n \) . | 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 ... | Yes |
Theorem 6.19 Let \( f \) be a multiplicative function with \( f\left( 1\right) = 1 \) . Then\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) f\left( d\right) = \mathop{\prod }\limits_{{p \mid n}}\left( {1 - f\left( p\right) }\right) \] | 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( ... | Yes |
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\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( n\right) = 0 \] | 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}\ri... | Yes |
Theorem 6.21 For \( x \geq 1 \) ,\n\n\[ \Phi \left( x\right) = \mathop{\sum }\limits_{{n \leq x}}\varphi \left( n\right) = \frac{3{x}^{2}}{{\pi }^{2}} + O\left( {x\log x}\right) . \] | 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}^{\p... | Yes |
Theorem 6.22 The probability that two positive integers are relatively prime is \( 6/{\pi }^{2} \) . | 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 \( \varph... | Yes |
Theorem 7.1 The divisor function \( d\left( n\right) \) is multiplicative. | 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 ... | Yes |
Theorem 7.2 For every \( \varepsilon > 0 \) ,\n\n\[ d\left( n\right) { \ll }_{\varepsilon }{n}^{\varepsilon } \] | 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... | Yes |
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) . | 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\[... | Yes |
Theorem 7.4 For \( x \geq 1 \) ,\n\n\[ \Delta \left( x\right) = \mathop{\sum }\limits_{{n \leq x}}\left( {\log n - d\left( n\right) + {2\gamma }}\right) = O\left( {x}^{1/2}\right) . \] | 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 t... | Yes |
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 \( {p}^{a} \) . | 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, t... | Yes |
Theorem 7.6 For \( \ell \geq 2 \) , \[ {D}_{\ell }\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}{d}_{\ell }\left( n\right) = \frac{1}{\left( {\ell - 1}\right) !}x{\log }^{\ell - 1}x + O\left( {x{\log }^{\ell - 2}x}\right) . \] | 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 ... | Yes |
Theorem 7.7\n\n\[ \n{d}^{2}\left( n\right) = \mathop{\sum }\limits_{{{\delta }^{2} \mid n}}\mu \left( \delta \right) {d}_{4}\left( \frac{n}{{\delta }^{2}}\right) .\n\] | 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 multiplica... | No |
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 \\) . | 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 ... | Yes |
Theorem 7.9 The arithmetic function \( \sigma \left( n\right) \) is multiplicative. | 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}\le... | Yes |
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\[ n = {2}^{p - 1}q. \] | 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\[ = {2n} \]\... | Yes |
Lemma 7.1 For every \( x \geq 1 \) ,\n\n\[ \mathop{\sum }\limits_{\substack{{{uv} \leq x} \\ {\left( {u, v}\right) = 1} }}\frac{1}{uv} = \frac{3}{{\pi }^{2}}{\log }^{2}x + O\left( {\log x}\right) \] | 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 \( {st} \leq x \) ... | Yes |
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\[ M\left( A\right) = M\left( {A}^{ * }\right) \] | 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 s... | Yes |
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 \( M\left( {{A}_{1} \cap {A}_{2}}\right) = M\left( {A}_{1}\right) \) . | 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} \c... | No |
Theorem 7.13 Let \( B \) be a set of multiples. There exists a unique primitive set \( {A}^{ * } \) such that \( B = M\left( {A}^{ * }\right) \) . | 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\... | Yes |
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. | 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. \( ▱ \) | Yes |
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 \( O\left( {x{\log }^{-2}x}\right) \) . | 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 \( p < {\log }^{2}x \), l... | Yes |
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 \( x \) . | 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 \( {... | Yes |
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\[{\log }^{4}x \leq p \leq {x}^{1/\left( {13y}\right) }.\] | 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 ... | Yes |
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) | 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\nB... | Yes |
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\[ k \leq \frac{\sigma \left( n\right) }{n} < k + \frac{k}{{x}^{1/\left( {6y}\right) }} \] | 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 \( n \) . T... | Yes |
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 | 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 num... | Yes |
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{a... | 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 + ... | Yes |
Lemma 8.2 For all positive integers \( n \) , \n\n\[ \n\frac{{2}^{2n}}{2n} \leq \left( \begin{matrix} {2n} \\ n \end{matrix}\right) < {2}^{2n} \n\] | 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... | Yes |
Theorem 8.1 For every positive integer \( n \) , \n\n\[ \mathop{\prod }\limits_{{p \leq n}}p < {4}^{n} \] | 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)... | Yes |
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 \( n \geq 2 \) . | 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}\l... | Yes |
Theorem 8.4 For \( x \geq 2 \) ,\n\n\[ \mathop{\sum }\limits_{{m \leq x}}\psi \left( \frac{x}{m}\right) = \mathop{\sum }\limits_{{d \leq x}}\Lambda \left( d\right) \left\lbrack \frac{x}{d}\right\rbrack = x\log x - x + O\left( {\log x}\right) . \] | 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 ... | Yes |
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) | 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... | Yes |
Theorem 8.6\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\vartheta \left( n\right) }{{n}^{2}} = \log x + O\left( 1\right) \] | 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 fu... | Yes |
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 \( x \geq 2 \) . | 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 \\... | Yes |
Theorem 8.8 (Mertens’s formula) There exists a constant \( \gamma \) such that for \( x \geq 2 \) ,\n\n\[ \mathop{\prod }\limits_{{p \leq x}}{\left( 1 - \frac{1}{p}\right) }^{-1} = {e}^{\gamma }\log x + O\left( 1\right) \] | 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 }\fra... | Yes |
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). | 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 \m... | Yes |
Theorem 8.10 For \( x \geq 2 \) ,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\omega {\left( n\right) }^{2} = x{\left( \log \log x\right) }^{2} + O\left( {x\log \log x}\right) . \] | 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}... | Yes |
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 n... | 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 ... | Yes |
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 \( o\left( x\right) \) . | 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 \( \mid \omega \left( n\right) - \) \... | Yes |
Theorem 9.1 For every positive integer \( n \) , \n\n\[ \n{\Lambda }_{2}\left( n\right) = \Lambda \left( n\right) \log n + \Lambda * \Lambda \left( n\right) . \n\] | 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... | Yes |
Lemma 9.1 For \( x > e \) ,\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p\left( {1 + \log \frac{x}{p}}\right) } \ll \log \log x \] | 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... | Yes |
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)... | 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 }... | Yes |
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... | 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} = {\l... | Yes |
Theorem 9.9 For \( k \geq 1 \) , \[ {\pi }_{k}\left( x\right) \sim {\pi }_{k}^{ * }\left( x\right) \sim \frac{x{\left( \log \log x\right) }^{k - 1}}{k\log x}. \] | 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... | Yes |
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 \( \varphi \left( m\right) \) Diri... | Proof. This is simply Theorem 4.6 applied to the multiplicative group \( {\left( \mathbf{Z}/m\mathbf{Z}\right) }^{ \times } \) . | Yes |
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\righ... | Proof. This is Theorem 4.7 applied to the multiplicative group \( {\left( \mathbf{Z}/m\mathbf{Z}\right) }^{ \times } \) . 口 | Yes |
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 \( x \\geq 1 \) , | 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 co... | Yes |
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... | 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... | Yes |
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}^{\... | 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... | Yes |
Lemma 10.1 Let \( {\chi }_{0} \) be the principal character modulo \( m \) . Then\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{{\chi }_{0}\left( n\right) \Lambda \left( n\right) }{n} = \log x + O\left( 1\right) \] | 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_{... | Yes |
Lemma 10.2 Let \( \chi \) be a nonprincipal character modulo \( m \) . If \( L\left( {1,\chi }\right) \neq 0 \) , then\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = O\left( 1\right) \] | 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) \... | Yes |
Lemma 10.3 Let \( \chi \) be a nonprincipal character modulo \( m \) . If \( L\left( {1,\chi }\right) = 0 \), then \[ \mathop{\sum }\limits_{{n \leq x}}\frac{\chi \left( n\right) \Lambda \left( n\right) }{n} = - \log x + O\left( 1\right) \] | 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 \lef... | Yes |
Theorem 10.7 Let \( \\chi \) be a complex character modulo \( m \) . Then \( L\\left( {1,\\chi }\\right) \\neq 0 \) . | 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) \\m... | Yes |
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 particu... | 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 ... | Yes |
Theorem 10.10 Let \( m \geq 1 \) and a be relatively prime integers. Then\n\n\[ \mathop{\sum }\limits_{\substack{{n \leq x} \\ {n \equiv a\;\left( {\;\operatorname{mod}\;m}\right) } }}\frac{\Lambda \left( n\right) }{n} = \frac{\log x}{\varphi \left( m\right) } + O\left( 1\right) \] | 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 \( L\left( {1,\chi }\right) \neq 0 \) b... | Yes |
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 \( n \in A + B \) . | 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\lbr... | Yes |
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 \( {\mathbf{N}}_{0} \subseteq A + B \) . | 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 \( n \in A + B \) . | Yes |
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. | Proof. This follows immediately from Lemma 11.2 with \( A = B \) . | No |
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\[ \sigma \left( {A + B}\right) \geq \alpha + \beta - {\alpha \beta } \] | 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,... | Yes |
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\[ 1 - \sigma \left( {{A}_{1} + \cdots + {A}_{h}}\right) \leq \mathop{\prod }\limits_{{i = 1}}^{h}\left( {1 - \sigma \left( {A}_{i}\right) }\right) \] | 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... | Yes |
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} ... | 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}\r... | Yes |
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. | Proof. Let \( \alpha = \sigma \left( A\right) \) . The result follows from Theorem 11.3 with \( {A}_{i} = A \) for \( i = 1,\ldots, h\left( \alpha \right) \) . | No |
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. | 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}}\rig... | Yes |
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. | 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... | No |
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 \( B \) . | 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 ... | Yes |
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}\r... | 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| = \le... | Yes |
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 (... | 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\... | No |
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\[... | 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 | No |
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 s... | 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 \( ... | Yes |
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. | Proof. This is a consequence of Theorem 11.9. | No |
Theorem 11.11 (Waring-Hilbert) For every \( k \geq 2 \), the set of nonnegative \( k \) th powers is a basis of finite order. | Proof. This is the special case of Theorem 11.10 applied to the polynomial \( f\left( x\right) = {x}^{k} \) . \( ▱ \) | Yes |
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. | 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... | Yes |
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. | Proof. This follows from Theorem 11.12. | No |
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. | Proof. This follows from Theorem 11.12. \( ▱ \) | No |
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}_{... | 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 }\r... | Yes |
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 ... | 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}}... | Yes |
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 ... | 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 \(... | Yes |
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{\\... | 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 }\\li... | Yes |
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\lef... | 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\... | Yes |
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,\l... | 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... | Yes |
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\l... | 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( {... | Yes |
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 in... | 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,\ld... | Yes |
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 lead... | 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... | Yes |
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. | 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\... | Yes |
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. | 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 d... | Yes |
Lemma 13.3 For every positive integer \( n \) , \[ {\sigma }^{ * }\left( n\right) = 2\mathop{\sum }\limits_{{1 \leq u < \sqrt{n}}}{\left( -1\right) }^{u - 1}{\sigma }^{ * }\left( {n - {u}^{2}}\right) + {\left\{ {\left( -1\right) }^{n - 1}n\right\} }_{n = {\ell }^{2}}. \] | 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 re... | Yes |
Theorem 13.3 (Fermat) An odd prime number \( p \) can be represented by the quadratic form \( {x}^{2} + {y}^{2} \) if and only if \( p \equiv 1\;\left( {\;\operatorname{mod}\;4}\right) \) . | 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,... | Yes |
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 \( p = {a}^{2} + {b}^{2} \) . | 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 \( ... | Yes |
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