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Theorem 7.1. [Duplication Formula] Let \( E \) denote an elliptic curve defined over the rationals, and let \( P \) be a point in \( E\left( \mathbb{Q}\right) \) . Then\n\n\[ h\left( {2P}\right) = {4h}\left( P\right) + \mathrm{O}\left( 1\right) \]\n\nwhere the implied constant in \( \mathrm{O} \) depends on \( E \) but... | This will be proved on p. 137 after some more machinery has been developed. | No |
Consider the curve \( E : {y}^{2} = {x}^{3} - {n}^{2}x \) with \( 1 \leq n \in \mathbb{N} \) . Let \( P \) be a rational point on \( E \) . A calculation gives \[ x\left( {2P}\right) = {\left( \frac{{x}^{2} + {n}^{2}}{2y}\right) }^{2} = \frac{{\left( {x}^{2} + {n}^{2}\right) }^{2}}{4\left( {{x}^{3} - {n}^{2}x}\right) }... | It may be checked that any cancellation in Equation (7.2) is bounded: Explicitly, if \( d \) divides both numerator and denominator, then \( d \mid {16}{n}^{6} \) . Examining the cases \( \left| M\right| \geq \left| N\right| \) and \( \left| M\right| < \left| N\right| \) separately shows that \[ \max \left\{ {{\left| {... | Yes |
Example 7.5. To see that \( e > d \) really occurs in the Nullstellensatz, define\n\n\[ f : {\mathbb{P}}^{1}\left( \mathbb{Q}\right) \rightarrow {\mathbb{P}}^{1}\left( \mathbb{Q}\right) \]\n\nby\n\n\[ f : \left\lbrack {{x}_{0},{x}_{1}}\right\rbrack \mapsto \left\lbrack {{x}_{0}^{2},{\left( {x}_{0} + {x}_{1}\right) }^{2... | However,\n\n\[ {x}_{0}^{3} = {x}_{0} \cdot {f}_{0} \]\n\n\[ {x}_{1}^{3} = \left( {2{x}_{0} + 3{x}_{1}}\right) \cdot {f}_{0} + \left( {-2{x}_{0} + {x}_{1}}\right) \cdot {f}_{1}. \] | Yes |
Lemma 7.8. Define a map \( {\phi }_{1} : E\left( \mathbb{Q}\right) \rightarrow \mathfrak{Q} \) by\n\n\[ \n{\phi }_{1}\left( 0\right) = \overline{1} \n\]\n\n\[ \n{\phi }_{1}\left( T\right) = \overline{-1} \n\]\n\n\[ \n{\phi }_{1}\left( \left( {x, y}\right) \right) = \bar{x}\text{otherwise.} \n\]\n\nThen \( {\phi }_{1} \... | Proof of Lemma 7.8. Let \( {P}_{1} \) and \( {P}_{2} \) denote rational points with\n\n\[ \n{P}_{1} + {P}_{2} = {P}_{3} \n\]\n\nWe wish to deduce that \( {\phi }_{1}\left( {P}_{3}\right) = {\phi }_{1}\left( {P}_{1}\right) {\phi }_{1}\left( {P}_{2}\right) \) . There are a number of special cases to be considered before ... | Yes |
Lemma 7.9. The image of \( E\left( \mathbb{Q}\right) \) under \( {\phi }_{1} \) is a finite subgroup of \( \mathfrak{Q} \) . | Proof. Suppose \( \bar{r} \) lies in the image of \( {\phi }_{1} \) . Without loss of generality, assume \( r \) is a square-free integer. We claim that \( r \mid n \) . To prove this, suppose \( p \) is a prime with \( p \mid r \), then we will show \( p \mid n \) . The statement \( {\phi }_{1}\left( \left( {x, y}\rig... | Yes |
Lemma 7.10. The kernel of \( \phi \) is precisely \( {2E}\left( \mathbb{Q}\right) \) . In other words the rational point \( P = \left( {x, y}\right) \) is the double of a rational point if and only if \( x \) and \( x \pm n \) are all rational squares. Explicitly, write | \[ x = {r}_{1}^{2} \] \[ x + n = {r}_{2}^{2} \] \[ x - n = {r}_{3}^{2}\text{ for }{r}_{i} \in \mathbb{Q}. \] Then \( P = {2Q} \) where \( Q = \left( {X, Y}\right) \) and \( X \) and \( Y \) are given by the formulas \[ X = x + {r}_{1}{r}_{2} + {r}_{1}{r}_{3} + {r}_{2}{r}_{3} \] \[ Y = \left( {{r}_{1} + {r}_{2} + {r}_{3... | Yes |
Let \( n = 6 \) . The point \( Q = \left( {-3,9}\right) \) doubles to the point \( \left( {\frac{25}{4}, - \frac{35}{8}}\right) \) on the curve \( {y}^{2} = {x}^{3} - {36x} \) . | This is verified by taking \( {r}_{1} = \frac{5}{2},{r}_{2} = - \frac{7}{2},{r}_{3} = \frac{1}{2} \) . As expected,\n\n\[ \nX = \frac{25}{4} - \frac{5}{2} \cdot \frac{7}{2} + \frac{5}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \frac{7}{2} = - 3, \n\] \n\nand similarly \( Y = 9 \) . | Yes |
Theorem 7.12. For any rational point \( P \) on an elliptic curve \( E \) defined over the rationals,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{h\left( {{2}^{n}P}\right) }{{4}^{n}} = \widehat{h}\left( P\right) \]\n\nexists. The limit \( \widehat{h}\left( P\right) \) is called the canonical height of \(... | Proof. Let \( {a}_{N} = \frac{1}{{4}^{N}}h\left( {{2}^{N}P}\right) \) . If \( N > M \geq 1 \), then\n\n\[ {a}_{M} - {a}_{N} = \frac{1}{{4}^{M}}h\left( {{2}^{M}P}\right) - \frac{1}{{4}^{N}}h\left( {{2}^{N}P}\right) \]\n\n\[ = \frac{1}{{4}^{M}}h\left( {{2}^{M}P}\right) - \frac{1}{{4}^{M + 1}}h\left( {{2}^{M + 1}P}\right)... | Yes |
Theorem 7.13. Let \( E \) be an elliptic curve defined over the rationals.\n\n(1) For every point \( P \in E\left( \mathbb{Q}\right) \) ,\n\n\[ \widehat{h}\left( P\right) = h\left( P\right) + \mathrm{O}\left( 1\right) \]\n\nuniformly.\n\n(2) For all \( P, Q \in E\left( \mathbb{Q}\right) \) ,\n\n\[ \widehat{h}\left( {P ... | This is proved below. | No |
Lemma 7.14. For all \( P, Q \in E\left( \mathbb{Q}\right) \), \[ h\left( {P + Q}\right) + h\left( {P - Q}\right) = {2h}\left( P\right) + {2h}\left( Q\right) + \mathrm{O}\left( 1\right) \] uniformly. | Proof of Theorem 7.13. (1) By iterating the relation \[ h\left( P\right) = \frac{1}{4}\left( {h\left( {2P}\right) + \mathrm{O}\left( 1\right) }\right) \] we have \[ h\left( P\right) = \frac{1}{4}\left( {\left( {\frac{h\left( {{2}^{2}P}\right) }{4} + \frac{1}{4}\mathrm{O}\left( 1\right) }\right) + \mathrm{O}\left( 1\rig... | Yes |
Lemma 7.17. Assume that \( 4{a}^{3} + {27}{b}^{2} \neq 0 \) . Then the map \( {\mathbb{P}}^{2}\left( \mathbb{Q}\right) \rightarrow {\mathbb{P}}^{2}\left( \mathbb{Q}\right) \) defined by\n\n\[ \left\lbrack {u, v, t}\right\rbrack = \left\lbrack {{2u}\left( {{at} + v}\right) + {4b}{t}^{2},{\left( v - at\right) }^{2} - {4b... | Proof of Lemma 7.17. Suppose the three polynomials vanish:\n\n\[ {2u}\left( {{at} + v}\right) + {4b}{t}^{2} = 0 \]\n\n\( \left( {7.20}\right) \)\n\n\[ {\left( v - at\right) }^{2} - {4btu} = 0\text{, and} \]\n\n(7.21)\n\n\[ {u}^{2} - {4tv} = 0. \]\n\n(7.22)\n\nIf \( t = 0 \), then \( u = 0 \) by Equation (7.22), so \( v... | Yes |
Theorem 8.1. [PRIME NUMBER THEOREM] Asymptotically, the number of primes is given by\n\n\\[ \n\\pi \\left( x\\right) = \\left| {\\{ p \\in \\mathbb{P} \\mid p \\leq x\\} }\\right| \\sim \\frac{x}{\\log x}.\n\\] | The Prime Number Theorem was first proved in 1896 by two mathematicians independently - Hadamard and de la Vallée Poussin. Their proofs used the Riemann zeta function and they were able to give an estimate for the error term in the formula, based upon an estimate for a zero-free region of the zeta function. | No |
Theorem 8.2. Let \( a < b \) be real numbers, and suppose that \( \mathrm{f} \) is a complex-valued function defined on \( \left\lbrack {a, b}\right\rbrack \) with a continuous derivative on \( \left( {a, b}\right) \) . Then\n\n\[ \mathop{\sum }\limits_{{a < n \leq b}}\mathrm{f}\left( n\right) = {\int }_{a}^{b}\mathrm{... | Proof. We give the proof in the case \( a, b \in \mathbb{N} \) for simplicity.\n\nSuppose that \( a < n - 1 < n \leq b \) . Now\n\n\[ {\int }_{n - 1}^{n}\lfloor t\rfloor {\mathrm{f}}^{\prime }\left( t\right) {dt} = \left( {n - 1}\right) \left\lbrack {\mathrm{f}\left( n\right) - \mathrm{f}\left( {n - 1}\right) }\right\r... | No |
Theorem 8.5. \[ \mathop{\sum }\limits_{{n = 1}}^{N}\mathrm{\;d}\left( n\right) = N\log N + \left( {{2\gamma } - 1}\right) N + \mathrm{O}\left( \sqrt{N}\right) . \] | Proof. The Euler Summation Formula in the usual form with integer boundaries gives a much larger remainder term of the form \( \mathrm{O}\left( N\right) \), which swamps the \( \left( {{2\gamma } - 1}\right) N \) term. For the sharper result with a \( \mathrm{O}\left( \sqrt{N}\right) \) error term, we apply the Euler S... | Yes |
Theorem 8.6. [Stirling's Formula]\n\n\\[ \n\\log N! = N\\log N - N + \\mathrm{O}\\left( {\\log N}\\right) \n\\] | Proof. Clearly \\( \\log N! = \\mathop{\\sum }\\limits_{{n = 2}}^{N}\\log n \\) . Put \\( \\mathrm{f}\\left( t\\right) = \\log t \\), and then by the Euler Summation Formula\n\n\\[ \n\\log N! = {\\int }_{1}^{N}\\log {tdt} + {\\int }_{1}^{N}\\frac{\\{ t\\} }{t}{dt} = N\\log N - N + \\mathrm{O}\\left( {\\log N}\\right) .... | Yes |
Theorem 8.8. The Möbius function is multiplicative. Moreover, \[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = \left\{ \begin{array}{ll} 1 & \text{ if }n = 1 \\ 0 & \text{ otherwise. } \end{array}\right. \] | Proof. Let \( m \) and \( n \) be integers with \( \gcd \left( {m, n}\right) = 1 \), and factorize \( m \) and \( n \) as products of prime powers. The primes involved must all be distinct. If, in the factorization there is an exponent of 2 or more, then we are done since both sides of the equation \( \mu \left( {mn}\r... | Yes |
Theorem 8.9. For all integers \( n \geq 1,\phi \left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) \frac{n}{d} \) . | Proof. By Corollary 3.6, \( \phi \left( n\right) = n\mathop{\prod }\limits_{{{p}_{i} \mid n}}\left( {1 - \frac{1}{{p}_{i}}}\right) \), so\n\n\[ \frac{\phi \left( n\right) }{n} = 1 - \mathop{\sum }\limits_{{{p}_{i} \mid n}}\frac{1}{{p}_{i}} + \mathop{\sum }\limits_{{i \neq j}}\frac{1}{{p}_{i}{p}_{j}} - \cdots = \mathop{... | Yes |
Theorem 8.11. Convolution is commutative and associative. In other words,\n\n\[ \n\\mathrm{f} * \\mathrm{g} = \\mathrm{g} * \\mathrm{f}\\;\\text{ and }\\;\\left( {\\mathrm{f} * \\mathrm{g}}\\right) * \\mathrm{h} = \\mathrm{f} * \\left( {\\mathrm{g} * \\mathrm{h}}\\right) \n\]\n\nfor any arithmetic functions \( f, g \),... | Proof. The sum in\n\[ \n\\mathop{\\sum }\\limits_{{d \\mid n}}\\mathrm{f}\\left( d\\right) \\mathrm{g}\\left( \\frac{n}{d}\\right) \n\]\n\nruns over all pairs \( d, e \\in \\mathbb{N} \) with \( {de} = n \), so it is equal to\n\n\[ \n\\mathop{\\sum }\\limits_{{{de} = n}}\\mathrm{f}\\left( d\\right) \\mathrm{g}\\left( e... | Yes |
Lemma 8.12. Define the arithmetic function \( \mathrm{I} \) by \( \mathrm{I}\left( 1\right) = 1 \) and \( \mathrm{I}\left( n\right) = 0 \) for all \( n > 1 \) . Then, for any arithmetic function \( \mathrm{f} \) , \[ f * I = I * f = f. \] | Proof. \[ \left( {\mathrm{f} * \mathrm{I}}\right) \left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mathrm{f}\left( d\right) \mathrm{I}\left( \frac{n}{d}\right) = \mathrm{f}\left( n\right) \mathrm{I}\left( 1\right) = \mathrm{f}\left( n\right) \] since all the other summands are zero by the definition of I. | Yes |
Theorem 8.13. If \( \mathrm{f} \) is an arithmetic function with \( \mathrm{f}\left( 1\right) \neq 0 \), then there is a unique arithmetic function \( \mathrm{g} \) such that \( \mathrm{f} * \mathrm{g} = \mathrm{I} \) . This function is denoted \( {\mathrm{f}}^{-1} \) . | Proof. The equation \( \left( {f * g}\right) \left( 1\right) = f\left( 1\right) g\left( 1\right) \) determines \( g\left( 1\right) \) . Then define \( g \) recursively as follows. Assuming that \( \mathrm{g}\left( 1\right) ,\ldots ,\mathrm{g}\left( {n - 1}\right) \) have been defined uniquely, the equation\n\n\[ \left(... | Yes |
Theorem 8.15. [Mößlus INVERSION FORMULA] Given arithmetic functions \( f \) and \( \mathrm{g},\mathrm{f}\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mathrm{\;g}\left( d\right) \) if and only if \( \mathrm{g}\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mathrm{f}\left( d\right) \mu \left( \frac{n}{d}\righ... | Proof. Assume that \( \mathrm{f}\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mathrm{\;g}\left( d\right) \), and let \( \mathrm{u}\left( n\right) = 1 \) for all \( n \) as in Example 8.14. Then \( \mathrm{f} = \mathrm{g} * \mathrm{u} \) . Convolve both sides of \( \mathrm{f} = \mathrm{g} * \mathrm{u} \) with \( ... | Yes |
Example 8.16. If \( \mathrm{f} * \mathrm{\;g} = \mathrm{I} \), then \( \mathrm{F}\left( \sigma \right) \mathrm{G}\left( \sigma \right) = 1 \), so | \[ \frac{1}{\zeta \left( \sigma \right) } = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\mu \left( n\right) }{{n}^{\sigma }} \] | No |
Theorem 8.17. If \( \mathrm{f} \) is a multiplicative arithmetic function, and \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\mathrm{f}\left( n\right) \) converges absolutely, then\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\mathrm{f}\left( n\right) = \mathop{\prod }\limits_{p}\left( {\mathrm{f}\left( 1\right) + \mat... | Proof. Let\n\n\[ P\left( x\right) = \mathop{\prod }\limits_{{p \leq x}}\left( {\mathrm{f}\left( 1\right) + \mathrm{f}\left( p\right) + \mathrm{f}\left( {p}^{2}\right) + \cdots }\right) . \]\n\nEach factor is absolutely convergent by hypothesis, and there are a finite number of factors, so\n\n\[ P\left( x\right) = \math... | Yes |
Let\n\n\[ \n{\mathrm{f}}_{n}\left( x\right) = \frac{{x}^{2}}{{\left( 1 + {x}^{2}\right) }^{n}} \n\]\n\nfor \( n \geq 1 \), and let \( \mathrm{f}\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\mathrm{f}}_{n}\left( x\right) \) . We can sum the \( {\mathrm{f}}_{n} \) because they form a geometric progression, | \n\[ \n\mathop{\sum }\limits_{{n = 0}}^{N}{\mathrm{f}}_{n}\left( x\right) = \frac{{x}^{2}}{1 - \frac{1}{1 + {x}^{2}}} - \frac{{x}^{2}}{\left( {1 - \frac{1}{1 + {x}^{2}}}\right) {\left( 1 + {x}^{2}\right) }^{N + 1}}. \n\]\n\nNow when \( x \neq 0 \) we can let \( N \) tend to infinity, the second term tends to zero, and ... | Yes |
Theorem 8.21. Suppose that the sequence of functions \( \left( {\mathrm{F}}_{n}\right) \) converges to \( \mathrm{F} \) uniformly on \( S \) . If each \( {\mathrm{F}}_{n} \) is continuous on \( S \), then \( \mathrm{F} \) is continuous on \( S \) . | Proof. Fix \( {s}_{0} \in S \) and \( \epsilon > 0 \) . Choose \( N \) such that, for all \( s \in S \) ,\n\n\[ \left| {\mathrm{F}\left( s\right) - {\mathrm{F}}_{N}\left( s\right) }\right| < \frac{\epsilon }{3} \]\n\n(8.20)\n\nThis is possible because the sequence of functions \( \left( {\mathrm{F}}_{n}\right) \) is co... | Yes |
Theorem 8.22. For every \( \delta > 0 \), the partial sums of the Riemann zeta function converge uniformly on \( {S}_{1 + \delta } = \{ s \in \mathbb{C} : \Re \left( s\right) > 1 + \delta \} \) . That is,\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{N}\frac{1}{{n}^{s}} \rightarrow \zeta \left( s\right) \text{ as }N \rightarr... | Proof of Theorem 8.22. Notice that for any \( s \in {S}_{1 + \delta } \) ,\n\n\[ \left| {\zeta \left( s\right) - \mathop{\sum }\limits_{{n = 1}}^{N}\frac{1}{{n}^{s}}}\right| = \left| {\mathop{\sum }\limits_{{n = N + 1}}^{\infty }\frac{1}{{n}^{s}}}\right| \]\n\n\[ < \mathop{\sum }\limits_{{n = N + 1}}^{\infty }\frac{1}{... | Yes |
The function \( {\mathrm{F}}_{N}\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{N}\frac{1}{{n}^{s}} \) is analytic and converges uniformly to \( \zeta \left( s\right) \) on every \( {S}_{1 + \delta },\delta > 0 \) as in Theorem 8.22, so by Theorem 8.23, the Riemann zeta function is analytic on \( {S}_{1 + \delta } \... | Proof of Theorem 8.23. Given a fixed point \( a \in S \), we have to prove that \( \mathrm{F} \) is analytic on a neighborhood of \( a \) . We use complex analysis, in particular Cauchy’s formula. Let \( \gamma \) be a closed simple curve, that is a finite join of smooth curves such that \( a \in \operatorname{Int}\lef... | Yes |
Lemma 8.25. Suppose a sequence of continuous functions \( {\mathrm{G}}_{N} : \gamma \rightarrow \mathbb{C} \) converges uniformly on \( \gamma \) to a function \( \mathrm{G} : \gamma \rightarrow \mathbb{C} \) . Then \( \mathrm{G} \) is continuous and\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty }}{\int }_{\gamma ... | Proof. The continuity of \( \mathrm{G} \) follows from Theorem 8.21, so in particular \( \mathrm{G} \) is integrable. Now\n\n\[ \left| {{\int }_{\gamma }\mathrm{G}\left( s\right) {ds} - {\int }_{\gamma }{\mathrm{G}}_{N}\left( s\right) {ds}}\right| = \left| {{\int }_{\gamma }\left\lbrack {\mathrm{G}\left( s\right) - {\m... | Yes |
Corollary 8.26. For all \( s \) with \( \Re \left( s\right) > 1 \) , | \[ \frac{d}{ds}\zeta \left( s\right) = - \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\log n}{{n}^{s}} \] | Yes |
Consider the function defined by the power series\n\n\[ \mathrm{g}\left( s\right) = 1 + s + {s}^{2} + \cdots ,\]\n\nwhich converges for \( \left| s\right| < 1 \) . Then \( \mathrm{g} \) can be continued to a function that is analytic on the whole of \( \mathbb{C} \) except for a simple pole at \( s = 1 \) . | To see this, notice that for \( \left| s\right| < 1,\mathrm{\;g}\left( s\right) = \frac{-1}{s - 1} \) . The latter expression is defined on \( \mathbb{C} \) apart from a simple pole at \( s = 1 \) with residue -1 . Of course, \( \mathrm{g} \) is not defined by the series for \( \left| s\right| \geq 1 \) . | Yes |
Theorem 8.29. The Riemann zeta function has an analytic continuation to the set \( \{ s \in \mathbb{C} \mid \Re \left( s\right) > 0\} \) with the exception of a simple pole at \( s = 1 \) with residue 1. | FIRST (STANDARD) PROOF OF THEOREM 8.29. This involves a careful use of the Euler Summation Formula - the care is needed because the formula as stated only applies to finite intervals. Assume first that \( \Re \left( s\right) > 1 \) ; then, by the Euler Summation Formula,\n\n\[ \mathop{\sum }\limits_{{n = 2}}^{N}\frac{1... | Yes |
Lemma 8.30. The integral in Equation (8.26) represents an analytic function on the range \( \Re \left( s\right) > 0 \) . | Proof. Write\n\n\[ \mathrm{I}\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\mathrm{f}}_{n}\left( s\right) \]\n\nwhere \( {\mathrm{f}}_{n}\left( s\right) = {\int }_{n}^{n + 1}\frac{\{ t\} }{{t}^{s + 1}}{dt} \) . We will prove that for any \( \delta > 0 \)\n\n(a) the series for \( \mathrm{I}\left( s\right) ... | Yes |
Lemma 9.2. The Gamma function has the following properties.\n\n(1) For all \( s \) with \( \Re \left( s\right) > 0 \) ,\n\n\[ \Gamma \left( {s + 1}\right) = {s\Gamma }\left( s\right) \] | Proof. The first relation is found by integrating,\n\n\[ \Gamma \left( {s + 1}\right) = {\int }_{0}^{\infty }{e}^{-t}{t}^{s}{dt} = {\left\lbrack -{e}^{-t}{t}^{s}\right\rbrack }_{0}^{\infty } + s{\int }_{0}^{\infty }{e}^{-t}{t}^{s - 1}{dt}. \]\n\nThe first term vanishes at \( t = 0 \) because \( \Re \left( s\right) > 0 ... | Yes |
Proposition 9.3. The Gamma function can be analytically continued to all of \( \mathbb{C} \), where it is analytic apart from simple poles at \( 0, - 1, - 2,\ldots \) and so on. | Proof. By Lemma 9.2(1), we may write\n\n\[ \Gamma \left( s\right) = \frac{1}{s}\Gamma \left( {s + 1}\right) \]\n\nThe right-hand side is defined for \( \Re \left( s\right) > - 1 \) apart from \( s = 0 \), where it has a simple pole with residue \( \Gamma \left( 1\right) = 1 \) . Iterating this gives\n\n\[ \Gamma \left(... | Yes |
Theorem 9.4. \( \Gamma \left( s\right) \neq 0 \) for all \( s \in \mathbb{C} \) . | This will be proved in Section 9.6. | No |
The function \( \mathrm{F} \) has an analytic continuation to the whole complex plane apart from poles at 1 and 0 . The Riemann zeta function has an analytic continuation to the complex plane where it is analytic apart from a simple pole at \( s = 1 \) . The zeta function vanishes at negative even integers. | Expand Theorem 9.5 to give\n\n\[ \n{\pi }^{-\left( {1 - s}\right) /2}\Gamma \left( \frac{1 - s}{2}\right) \zeta \left( {1 - s}\right) = {\pi }^{-s/2}\Gamma \left( \frac{s}{2}\right) \zeta \left( s\right) , \n\]\n\n\[ \n\zeta \left( {1 - s}\right) = \frac{{\pi }^{-s + 1/2}\Gamma \left( \frac{s}{2}\right) \zeta \left( s\... | Yes |
The Gaussian function \( \mathrm{f}\left( x\right) = {e}^{-{x}^{2}} \) is in \( \mathcal{S} \). | Notice that \( \mathcal{S} \) is a complex vector space and that any function \( f \in \mathcal{S} \) is integrable,\n\n\[\n\left| {{\int }_{-\infty }^{\infty }\mathrm{f}\left( x\right) {dx}}\right| \leq {\int }_{-\infty }^{\infty }\left| {\mathrm{f}\left( x\right) }\right| {dx} \leq C{\int }_{-\infty }^{\infty }\frac{... | No |
Lemma 9.11. If \( \mathrm{f}\left( y\right) = {e}^{-\pi {y}^{2}} \), then \( \widehat{\mathrm{f}}\left( y\right) = \mathrm{f}\left( y\right) \) . | Proof.\n\[ \widehat{\mathrm{f}}\left( y\right) = {\int }_{-\infty }^{\infty }{e}^{-\pi {x}^{2}}{e}^{-{2\pi }\mathrm{i}{xy}}{dx}. \]\n\nThe idea is to complete the square,\n\n\[ - \pi \left( {{x}^{2} + 2\mathrm{i}{xy}}\right) = - \pi \left\lbrack {{\left( x + \mathrm{i}y\right) }^{2} + {y}^{2}}\right\rbrack \]\n\nso the... | Yes |
Lemma 9.13. If \( \mathrm{g} \) is periodic and twice differentiable with continuous second derivative, then there exists a constant \( C > 0 \), depending only upon \( \mathrm{g} \), such that\n\n\[ \left| {c}_{k}\right| \leq \frac{C}{{k}^{2}} \]\n\nfor all \( k \neq 0 \) . | Proof. Integrate by parts:\n\n\[ {c}_{k} = {\left\lbrack \frac{-{e}^{-{2\pi }\mathrm{i}{kx}}\mathrm{\;g}\left( x\right) }{{2\pi }\mathrm{i}k}\right\rbrack }_{0}^{1} + {\int }_{0}^{1}\frac{{e}^{-{2\pi }\mathrm{i}{kx}}{\mathrm{\;g}}^{\prime }\left( x\right) }{{2\pi }\mathrm{i}k}{dx}. \]\n\nNow the bracketed term vanishes... | Yes |
Theorem 9.14. Any function \( \mathrm{g} \) that is periodic and differentiable infinitely often has a Fourier series expansion\n\n\[ \mathrm{g}\left( x\right) = \mathop{\sum }\limits_{{k \in \mathbb{Z}}}{c}_{k}{e}^{{2\pi }\mathrm{i}{kx}} \]\n\nthat is uniformly convergent on \( \mathbb{R} \) . | Proof. Let \( \mathrm{G} \) be the Fourier series of \( \mathrm{g} \), and apply Lemma 9.13:\n\n\[ \left| {\mathrm{G}\left( x\right) - \mathop{\sum }\limits_{{k = - n}}^{n}{c}_{k}{e}^{{2\pi }\mathrm{i}{kx}}}\right| \leq C\mathop{\sum }\limits_{{\left| k\right| > n}}\frac{1}{{k}^{2}} \]\n\nwhere the last sum tends to ze... | Yes |
Consider the sequence of functions \( \left( {D}_{K}\right) \) defined by\n\n\[ \n{D}_{K}\left( x\right) = \mathop{\sum }\limits_{{k = - K}}^{K}{e}^{{2\pi }\mathrm{i}{kx}},\;\text{ for }K \in \mathbb{N},\n\]\n\ncalled the Dirichlet kernel. Then\n\n\[ \n{\int }_{0}^{1}{D}_{K}\left( x\right) {dx} = 1\n\]\n\n(9.5)\n\n\[ \... | Proof of Lemma 9.15. Equation (9.5) follows from the fact that\n\n\[ \n{\int }_{0}^{1}{e}^{{2\pi }\mathrm{i}{kx}}{dx} = 0\text{ for all }k \neq 0.\n\]\n\nEquation (9.6) is proved by induction on \( k \) or directly by summation of a geometric progression. Equation (9.7) follows since\n\n![05842f43-827d-4037-8783-9728eb... | Yes |
Lemma 9.16. [Riemann-Lebesgue Lemma] Let \( \mathbf{g} \) be a continuous periodic function, and let \( {c}_{k} \) be the \( k \) th Fourier coefficient of \( \mathrm{g} \) . Then\n\n\[ \mathop{\lim }\limits_{{\left| k\right| \rightarrow \infty }}{c}_{k} = 0 \] | Proof of Lemma 9.16. Define for continuous complex-valued periodic functions \( u, v \) the inner product\n\n\[ \left( {\mathrm{u},\mathrm{v}}\right) = {\int }_{0}^{1}\mathrm{u}\left( x\right) \overline{\mathrm{v}\left( x\right) }{dx} \]\n\nand the norm\n\n\[ \parallel u\parallel = \sqrt{\left( u, u\right) }.\]\n\nLet ... | Yes |
For real \( y > 0 \), define the theta function by\n\n\[ \theta \left( y\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{e}^{-{n}^{2}{\pi y}} \]\n\nThen\n\n\[ \theta \left( \frac{1}{y}\right) = \sqrt{y}\theta \left( y\right) \] | Proof. This relation is far from obvious and looks barely possible. The series defining \( \theta \) converges uniformly in the range \( y > \delta \) for any fixed \( \delta > 0 \) . Fix some real \( b > 0 \) and define, with \( \mathrm{f}\left( y\right) = {e}^{-\pi {y}^{2}} \) as in Lemma 9.11,\n\n\[ {\mathrm{f}}_{b}... | Yes |
Lemma 9.19. For all \( z \in \mathbb{C} \), the function\n\n\[ \n\mathrm{G}\left( z\right) = {\int }_{1}^{\infty }{y}^{z}\mathrm{\;g}\left( y\right) \frac{dy}{y}\n\]\n\nis analytic. | Proof of Lemma 9.19. Write \( \mathrm{G}\left( z\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\mathrm{G}}_{n}\left( z\right) \), where\n\n\[ \n{\mathrm{G}}_{n}\left( z\right) = {\int }_{n}^{n + 1}{y}^{z}\mathrm{\;g}\left( y\right) {dy}.\n\]\n\nWe will prove that the \( {\mathrm{G}}_{n} \) are analytic functions o... | Yes |
Define a function \( f \) by\n\n\[ \mathrm{f}\left( s\right) = s{e}^{\gamma s}\mathop{\prod }\limits_{{n = 1}}^{\infty }\left\lbrack {\left( {1 + \frac{s}{n}}\right) {e}^{-s/n}}\right\rbrack \]\n\nwhere \( \gamma \) is the Euler-Mascheroni constant. Then \( \mathrm{f} \) is an analytic function on the whole of the comp... | Proof of Theorem 9.20. Consider the function\n\n\[ \mathrm{g}\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\mathrm{g}}_{n}\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\left\lbrack {\log \left( {1 + \frac{s}{n}}\right) - \frac{s}{n}}\right\rbrack .\n\n(9.18)\n\nEach \( {\mathrm{g}}_{n} \) is ... | Yes |
Theorem 9.22. [EULER] For all \( s \neq 0, - 1, - 2,\ldots \) , \n\n\[ \n\frac{1}{\mathrm{f}\left( s\right) } = \frac{1}{s}\mathop{\prod }\limits_{{n = 1}}^{\infty }\left\lbrack {{\left( 1 + \frac{1}{n}\right) }^{s}{\left( 1 + \frac{s}{n}\right) }^{-1}}\right\rbrack \n\] | Proof. We use the definition of the Euler-Mascheroni constant \( \gamma \) (see Exercise 1.2 on p. 10 or Theorem 8.3). \n\n\[ \n\mathrm{f}\left( s\right) = s\mathop{\lim }\limits_{{m \rightarrow \infty }}{e}^{s\left( {1 + 1/2 + \cdots + 1/m - \log m}\right) }\mathop{\lim }\limits_{{N \rightarrow \infty }}\mathop{\prod ... | No |
Corollary 9.23. For all \( s \in \mathbb{C} \) ,\n\n\[ \n\frac{1}{\mathrm{f}\left( s\right) } = \mathop{\lim }\limits_{{m \rightarrow \infty }}\frac{1 \cdot 2\cdots \left( {m - 1}\right) {m}^{s}}{s\left( {s + 1}\right) \cdots \left( {s + m - 1}\right) }.\n\] | Proof. By Theorem 9.22, for \( s \neq 0, - 1, - 2,\ldots \)\n\n\[ \n\frac{1}{f\left( s\right) } = \mathop{\lim }\limits_{{m \rightarrow \infty }}\frac{1}{s}\mathop{\prod }\limits_{{n = 1}}^{{m - 1}}{\left( 1 + \frac{1}{n}\right) }^{s}{\left( 1 + \frac{s}{n}\right) }^{-1}\n\]\n\n\[ \n= \mathop{\lim }\limits_{{m \rightar... | Yes |
Theorem 9.24. For all \( s \) such that \( \Re \left( s\right) > 0 \) , \[ \frac{1}{\mathrm{f}\left( s\right) } = \Gamma \left( s\right) = {\int }_{0}^{\infty }{e}^{-t}{t}^{s - 1}{dt}. \] | Proof of Theorem 9.24. For \( n \in \mathbb{N} \), define \[ {\Gamma }_{n}\left( s\right) = {\int }_{0}^{n}{\left( 1 - \frac{t}{n}\right) }^{n}{t}^{s - 1}{dt}. \] Evaluate \( {\Gamma }_{n}\left( s\right) \) using integration by parts. Substitute \( t = {n\tau } \) to give \[ {\Gamma }_{n}\left( s\right) = {n}^{s}{\int ... | Yes |
Corollary 9.25. For all \( s \in \mathbb{C}, s \notin \mathbb{N} \) , \[ \Gamma \left( s\right) \Gamma \left( {1 - s}\right) = \frac{\pi }{\sin \left( {\pi s}\right) }.\] | Proof. By Theorem 9.20, \[ \Gamma \left( s\right) \Gamma \left( {-s}\right) = - \frac{1}{{s}^{2}}\mathop{\prod }\limits_{{n = 1}}^{\infty }{\left( 1 + \frac{s}{n}\right) }^{-1}{e}^{s/n}\mathop{\prod }\limits_{{n = 1}}^{\infty }{\left( 1 - \frac{s}{n}\right) }^{-1}{e}^{-s/n} \] \[ = - \frac{1}{{s}^{2}}\mathop{\prod }\li... | Yes |
Theorem 9.26. Define \( N\left( T\right) \) to be the number of zeros of the Riemann zeta function in the critical strip up to height \( T \) , \[ N\left( T\right) = \left| {\{ s \in \mathbb{C} : 0 \leq \Re \left( s\right) \leq 1,\zeta \left( s\right) = 0,0 < \Im \left( s\right) < T\} }\right| . \] Then there is an asy... | The proof makes use of Stirling's Formula extended to the complex plane, \[ \log \Gamma \left( s\right) = - s + \left( {s - \frac{1}{2}}\right) \log s + \mathrm{O}\left( 1\right) \] provided \( \left| {\operatorname{Arg}\left( s\right) }\right| < \pi - \delta \) . | Yes |
Proposition 10.1. There are infinitely many primes congruent to 3 modulo 4. | Proof. This proceeds like Euclid's proof that there are infinitely many prime numbers. Suppose that the proposition is false, and there are only \( r \) such primes \( {p}_{1},\ldots ,{p}_{r} \) . Let\n\n\[ N = {\left( {p}_{1}\cdots {p}_{r}\right) }^{2} + 2. \]\n\nSince\n\n\[ {p}_{1}^{2} \equiv \cdots \equiv {p}_{r}^{2... | Yes |
Proposition 10.2. There are infinitely many primes congruent to \( 1\mathrm{{mod}} \) - ulo 4. | FIRST Proof of Proposition 10.2. This proof is slightly different. Rather than deriving a contradiction, we will show that, for any given \( N > 1 \), there exists a prime congruent to 1 modulo 4 and greater than \( N \) . Given \( N > 1 \) , define\n\n\[ M = {\left( N!\right) }^{2} + 1 \]\n\n(10.1)\n\nClearly, \( M \)... | Yes |
Lemma 10.3. The series \( L\left( {s,\chi }\right) \) converges for \( s = 1 \), and\n\n\[ L\left( {1,\chi }\right) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \frac{\pi }{4}. \] | Proof. Consider the integral\n\n\[ {\int }_{0}^{1}\frac{dt}{1 + {t}^{2}} = {\left\lbrack {\tan }^{-1}\left( t\right) \right\rbrack }_{0}^{1} = \frac{\pi }{4}. \]\n\n(10.5)\n\nSubstitute into this integral the expansion\n\n\[ \frac{1}{1 + {t}^{2}} = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{{\left( -{t}^{2}\righ... | Yes |
Lemma 10.4. The functions \( \chi \) and \( {\chi }_{0} \) are completely multiplicative (see Definition 3.4). | Proof. Check all the possible values of \( m \) and \( n \) modulo 4 . | No |
Example 10.6. Let \( n = 5 \), so \( U\left( {\mathbb{Z}/5\mathbb{Z}}\right) = \{ 1,2,3,4\} \) is a cyclic group, and 2 is a generator. The multiplication table of \( U\left( {\mathbb{Z}/5\mathbb{Z}}\right) \) is | \n\nso \( U\left( {\mathbb{Z}/5\mathbb{Z}}\right) \cong \{ 1, i, - 1, - i\} \) | No |
Lemma 10.8. Let \( G \) be a finite Abelian group, and let \( \chi \) be a character of \( G \) . Then \( \chi \left( {1}_{G}\right) = 1 \) and \( \chi \left( g\right) \) is a root of unity for any \( g \in G \) . In particular, \( \left| {\chi \left( g\right) }\right| = 1 \) . Thus \( \chi \left( g\right) \) lies on t... | Proof. Clearly\n\n\[ \chi \left( {1}_{G}\right) = \chi \left( {{1}_{G} \cdot {1}_{G}}\right) = \chi \left( {1}_{G}\right) \chi \left( {1}_{G}\right) \]\n\nso \( \chi \left( {1}_{G}\right) = 1 \) since \( \chi \left( {1}_{G}\right) \neq 0 \) . As to the second statement, we use the fact that for every \( g \in G \) ther... | Yes |
Theorem 10.10. Let \( G \) be a finite Abelian group. Then the characters of \( G \) form a group with respect to the multiplication\n\n\[ \left( {\chi \cdot \psi }\right) \left( g\right) = \chi \left( g\right) \psi \left( g\right) \]\n\ndenoted \( \widehat{G} \) . The identity in \( \widehat{G} \) is the trivial chara... | Proof of Theorem 10.10. Use the structure theorem for finite Abelian groups, which says that \( G \) is isomorphic to a product of cyclic groups,\n\n\[ G \cong \mathop{\prod }\limits_{{j = 1}}^{k}{C}_{{n}_{j}} \]\n\nChoose a generator \( {g}_{j} \) for each of the factors \( {C}_{{n}_{j}} \) and define characters on \(... | Yes |
Corollary 10.11. Let \( G \) be a finite Abelian group. For any \( 1 \neq g \in G \), there exists \( \chi \in \widehat{G} \) such that \( \chi \left( g\right) \neq 1 \) . | Proof. Looking again at the proof of Theorem 10.10, we may write\n\n\[ g = \left( {*,\ldots ,*,{g}_{j}^{r},*,\ldots , * }\right) \]\n\nwith some entry \( {g}_{j}^{r} \neq 1,0 < r < {n}_{j} \) . Then \( {\chi }^{\left( j\right) }\left( g\right) = {e}^{{2\pi }\mathrm{i}r/{n}_{j}} \neq 1 \) . | Yes |
Theorem 10.12. Let \( G \) be a finite Abelian group. Then, for any element \( h \in \) \( G \) and any character \( \psi \in \widehat{G} \) , \n\n\[ \n\mathop{\sum }\limits_{{g \in G}}\psi \left( g\right) = \left\{ \begin{matrix} \left| G\right| & \text{ if }\psi = {\chi }_{0} \\ 0 & \text{ if }\psi \neq {\chi }_{0} \... | Proof. Consider Equation (10.13) first. The case \( \psi = {\chi }_{0} \) is trivial, so assume \( \psi \neq {\chi }_{0} \) . There is an element \( h \in G \) such that \( \psi \left( h\right) \neq 1 \) . Then \n\n\[ \n\psi \left( h\right) \mathop{\sum }\limits_{{g \in G}}\psi \left( g\right) = \mathop{\sum }\limits_{... | Yes |
Corollary 10.13. For all \( g, h \in G \), we have\n\n\[ \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\chi \left( g\right) \overline{\chi \left( h\right) } = \left\{ \begin{array}{ll} \left| G\right| & \text{ if }g = h \\ 0 & \text{ if }g \neq h \end{array}\right. \] | Proof. Note that\n\n\[ \chi \left( {h}^{-1}\right) = \chi {\left( h\right) }^{-1} = \overline{\chi \left( h\right) } \]\n\nsince \( \chi \left( h\right) \) is on the unit circle in \( \mathbb{C} \) . Then use Theorem 10.12 with \( g{h}^{-1} \) in place of \( h \) . | Yes |
Theorem 10.16. A Dirichlet character is completely multiplicative, and the associated L-function therefore has an Euler product expansion. | Proof. Let \( \chi \) be a Dirichlet character modulo \( q \) . If two integers \( m, n \) are given, and at least one of them is not coprime to \( q \), then neither is the product \( {mn} \) . Thus, \( \chi \left( {mn}\right) = 0 = \chi \left( m\right) \chi \left( n\right) \) . If on the other hand, both \( m \) and ... | Yes |
Theorem 10.17. [ABEL] Let a be an arithmetic function, and define\n\n\[ \n\\mathrm{A}\\left( x\\right) = \\mathop{\\sum }\\limits_{{n \\leq x}}\\mathrm{a}\\left( n\\right) \n\]\n\nLet \( \\mathrm{f} : \\left\\lbrack {x, y}\\right\\rbrack \\rightarrow \\mathbb{C} \) be differentiable with a continuous derivative. Then\n... | Proof. Assume \( x, y \\in \\mathbb{N} \) are integral for simplicity, \( m = y \) and \( k = x \), so the left-hand side of Equation (10.18) is\n\n\[ \n\\mathop{\\sum }\\limits_{{n = k + 1}}^{m}\\mathrm{a}\\left( n\\right) \\mathrm{f}\\left( n\\right) = \\mathop{\\sum }\\limits_{{n = k + 1}}^{m}\\left( {\\mathrm{\\;A}... | Yes |
Lemma 10.18. Define another arithmetic function \( \mathrm{g} \) by\n\n\[ \mathbf{g}\left( n\right) = \left\{ \begin{array}{l} 1\text{ if }n\text{ is a square; } \\ 0\text{ otherwise. } \end{array}\right.\n\]\n\nThen \( \mathrm{f}\left( n\right) \geq \mathrm{g}\left( n\right) \) for all \( n \in \mathbb{N} \) . | Proof. Note that both \( \mathrm{f} \) and \( \mathrm{g} \) are multiplicative arithmetic functions, so it is enough to consider the case \( n = {p}^{k} \), a prime power. We have\n\n\[ \mathrm{f}\left( {p}^{k}\right) = 1 + \psi \left( p\right) + \cdots + \psi {\left( p\right) }^{k} = \begin{cases} 1 & \text{ if }\psi ... | Yes |
Theorem 10.19. [ABEL] Given a real power series\n\n\\[ \n\\mathrm{f}\\left( x\\right) = \\mathop{\\sum }\\limits_{{n = 0}}^{\\infty }{a}_{n}{x}^{n}\n\\] \n\nthat converges for all \\( x \\) with \\( 0 < x < {x}_{0} \\), suppose that the limit\n\n\\[ \nL = \\mathop{\\sum }\\limits_{{n = 0}}^{\\infty }{a}_{n}{x}_{0}^{n} ... | Proof. For any \\( \\epsilon > 0 \\), we will show that, for all \\( x \\) sufficiently close to \\( {x}_{0} \\) and for all \\( N \\) sufficiently large,\n\n\\[ \n\\left| {\\mathop{\\sum }\\limits_{{n = 0}}^{N}{a}_{n}\\left( {{x}_{0}^{n} - {x}^{n}}\\right) }\\right| \\leq \\epsilon\n\\] \n\n(10.24)\n\nTo do this, rewr... | Yes |
Theorem 11.2. Suppose a and \( n \) are positive, nonzero, coprime integers. Then\n\n\[ \left( \frac{a}{n}\right) = {\left( -1\right) }^{\left( {a - 1}\right) /2 \cdot \left( {n - 1}\right) /2}\left( \frac{n}{a}\right) \text{ if }a\text{ and }n\text{ are odd; } \]\n\n\[ \left( \frac{2}{n}\right) = {\left( -1\right) }^{... | Exercise 11.1. Prove Theorem 11.2. (Hint: Consider the primes \( p \equiv 3{\;\operatorname{mod}\; - } \) ulo 4 dividing \( a \) and \( n \) .) | No |
Since 2 is not a quadratic residue modulo 3, it is not a quadratic residue modulo 15. | \[ \left( \frac{2}{15}\right) = \left( \frac{2}{3}\right) \left( \frac{2}{5}\right) \text{by definition} \] \[ = \left( {-1}\right) \left( {-1}\right) \text{by Theorem 11.2} \] \[ = 1\text{.} \] | No |
Theorem 11.7. Let \( {\zeta }_{\mathbb{K}} \) denote the Dedekind zeta function of a quadratic number field with discriminant \( D \) . Let \( h \) denote the class number of \( {O}_{\mathbb{K}} \), let \( w \) denote the number of units in \( {O}_{\mathbb{K}} \) if \( D < 0 \) and let \( u \) denote a fundamental unit... | Outline Proof of THEOREM 11.7. The following proof varies from that in most of the textbooks; references are provided in the notes at the end of the chapter.\n\nThe two cases \( D > 0 \) and \( D < 0 \) vary ultimately but begin in the same way. The idea is to sum over each ideal class, so fix an ideal class \( \mathfr... | Yes |
Lemma 11.8. Let \( N > 1 \) denote a positive integer and suppose \( 1 \leq a < N \) .\n\nThen\n\[ \n- \log \left( {1 - {e}^{{2\pi a}\mathrm{i}/N}}\right) = - \log \left( {2\sin \frac{\pi a}{N}}\right) + \frac{\pi \mathrm{i}}{2}\left( {1 - \frac{2a}{N}}\right) .\n\] | Proof. This is elementary, relying only upon the definition of the principal branch of the complex logarithm and some manipulation with the half-angle formulas. | No |
Lemma 11.9. Let \( \zeta = {e}^{{2\pi }\mathrm{i}/\left| D\right| } \) and define\n\n\[ G = \mathop{\sum }\limits_{{a = 1}}^{{\left| D\right| - 1}}\chi \left( a\right) {\zeta }^{a} \]\n\nThen\n\n\[ {G}^{2} = D\text{.} \] | Proof. The claim about \( {G}^{2} \) is proved in exactly the same way as Equation (3.12) on p. 70 was proved. The product \( {G}^{2} \) may be written\n\n\[ \mathop{\sum }\limits_{\substack{{a = 1;} \\ {\left( {a,\left| D\right| }\right) = 1} }}^{{\left| D\right| - 1}}\chi \left( a\right) {\zeta }^{a}\mathop{\sum }\li... | Yes |
Theorem 11.10. [DIRICHLET] Let \( N \) denote a positive integer and define\n\n\[ H = \mathop{\sum }\limits_{{k = 0}}^{{N - 1}}{e}^{{2\pi }\mathrm{i}{k}^{2}/N} \]\n\nThen\n\n\[ H = \left\{ \begin{matrix} \left( {1 + \mathrm{i}}\right) \sqrt{N} & \text{ if }N \equiv 0 & \left( {\;\operatorname{mod}\;4}\right) , \\ \sqrt... | Proof. The functions \( {\left\{ x \mapsto {e}^{{2\pi }\mathrm{i}{nx}/N}\right\} }_{n \in \mathbb{Z}} \) form an orthonormal family with respect to the inner product\n\n\[ < f, g > = \frac{1}{N}{\int }_{0}^{N}f\left( t\right) \overline{g\left( t\right) }{dt}. \]\n\nThe Fourier expansion \( {}^{3} \) of the map \( x \ma... | Yes |
Example 12.3. The calculation \( {23} + 7 = {30} \) in binary takes five bit operations: |  | Yes |
Lemma 12.6. For \( k \geq \ell \) , \( \mathrm{C} \) (multiply a \( k \) -bit number by an \( \ell \) -bit number) \( = \mathrm{O}\left( {k}^{2}\right) \) . | Proof. Notice that all we need is an upper bound for the number of bit operations required. The long multiplication can be done in the shape shown in Figure 12.2, with arbitrary bits denoted \( * \) . We have the following upper bounds. There are no more than \( \left( {k + 2}\right) \) operations needed to add the low... | Yes |
Lemma 12.7. For \( k \geq \ell \) ,\n\n\[ \mathrm{C}\text{(divide a}k\text{-bit number by an}\ell \text{-bit number)} = \mathrm{O}\left( {k}^{2}\right) \text{.} \] | Proof. This calculation amounts to repeatedly subtracting an \( \ell \) -bit number from a \( k \) -bit number and testing to see if the answer is less than the \( \ell \) -bit number. By Lemma 12.5, this takes no more than \( k\mathrm{O}\left( k\right) = \mathrm{O}\left( {k}^{2}\right) \) bit operations. | Yes |
C(multiply a \( k \) -bit number by a \( k \) -bit number) \( = \mathrm{O}\left( {k}^{1.59}\right) \) | Proof. This relies on chopping up the integers \( m \) and \( n \) to be multiplied in a clever way (as all these methods do). Assume for simplicity that \( k \) is even. If \( m \) has \( k \) bits, then we can write\n\n\[ m = a \cdot {2}^{k/2} + b \]\n\nwhere \( a \) and \( b \) have at most \( k/2 \) bits. Similarly... | Yes |
Example 12.9. [The Knapsack Problem] Given natural numbers \( {n}_{1},\ldots ,{n}_{d} \) and \( N \), decide if there is a subset \( I \) of the index set \( \{ 1,2,\ldots, d\} \) with the property that\n\n\[ \mathop{\sum }\limits_{{i \in I}}{n}_{i} = N \] | The most naïve approach is to simply try all the subsets, and there are \( {2}^{d} \) of them. No algorithm is known to decide this problem in polynomial time. | No |
Lemma 12.10. C(multiply \( a \) modulo \( m \) by \( b \) modulo \( m \) ) \( = \mathrm{O}\left( {\left( \log m\right) }^{2}\right) \) . | Proof. If \( m \) has \( k \) bits, then the residues of \( a \) and \( b \) modulo \( m \) are numbers with no more than \( k \) bits. Computing \( a \cdot b \) takes \( \mathrm{O}\left( {k}^{2}\right) \) bit operations by Lemma 12.6, and \( a \cdot b \) has no more than \( {2k} \) bits. Finding the residue of \( a \c... | Yes |
Theorem 12.11. \( \mathrm{C}\left( {\text{find}\gcd \left( {a, b}\right) }\right) = \mathrm{O}\left( {\left( \log a\right) }^{3}\right) \) | Proof. Each step involves a simple division of integers that are less than \( a \), so each step has complexity \( \mathrm{O}\left( {\left( \log a\right) }^{2}\right) \) . We need to estimate how many steps are taken before the algorithm terminates. Since the remainders \( {r}_{i} \) are nonnegative integers, the numbe... | Yes |
Let \( m = {31} \) and \( a = {12} \). Since \( \gcd \left( {{31},{12}}\right) = 1 \), there is an \( x \) with \( {12x} \equiv 1 \) modulo 31. | Apply the Euclidean Algorithm:\n\n\[ \n{31} = {12} \cdot 2 + 7 \n\]\n\n\[ \n{12} = 7 \cdot 1 + 5 \n\]\n\n\[ \n7 = 5 \cdot 1 + 2 \n\]\n\n\[ \n5 = 2 \cdot 2 + 1\text{.} \n\]\n\nIt follows that\n\n\[ \n1 = 5 - 2 \cdot 2 \n\]\n\n\[ \n= 5 - 2\left( {7 - 5}\right) = 3 \cdot 5 - 2 \cdot 7 \n\]\n\n\[ \n= 3\left( {{12} - 7}\rig... | Yes |
Theorem 12.15. Given \( m \geq 2 \) and \( b, n \in \mathbb{N},0 \leq b < m \) , \[ \left. {\mathrm{C}\left( {\text{ compute }{b}^{n}\text{ modulo }m}\right) = \mathrm{O}\left( {{\left( \log m\right) }^{2}\log n}\right) }\right) . | Proof. Assume that \( n \) is presented as a binary number \[ n = {n}_{0} + 2{n}_{1} + \cdots + {2}^{k}{n}_{k} \] with bits \( {n}_{i} \in \{ 0,1\} \) . Now carry out the following calculations (all modulo \( m \) ). \[ \text{Step 1: compute}{b}_{0} = {b}^{{n}_{0}}\text{;} \] Step 2: compute \( {b}^{2} \) ; Step 3: com... | Yes |
Theorem 12.16. Suppose that \( {m}_{1},\ldots ,{m}_{r} \) are integers with \( \gcd \left( {{m}_{i},{m}_{j}}\right) = 1 \) for all \( i \neq j \) . Then the simultaneous congruences\n\n\[ x \equiv {a}_{1}\;\left( {\;\operatorname{mod}\;{m}_{1}}\right) \]\n\n\[ x \equiv {a}_{2}\;\left( {\;\operatorname{mod}\;{m}_{2}}\ri... | The uniqueness means that if \( x \) and \( y \) are integers solving all the congruences, then \( x \equiv y \) modulo \( M \) .\n\nProof. If \( x \) and \( y \) both satisfy the congruences, then\n\n\[ x - y \equiv 0\;\left( {\;\operatorname{mod}\;{m}_{i}}\right) \text{ for }i = 1,\ldots, r. \]\n\nSince the \( {m}_{i... | Yes |
Consider the simultaneous congruences\n\n\\[ \nx \\equiv 2\\;\\left( {\\;\\operatorname{mod}\\;3}\\right) \n\\]\n\n\\[ \nx \\equiv 3\\;\\left( {\\;\\operatorname{mod}\\;4}\\right) \n\\]\n\n\\[ \nx \\equiv 4\\;\\left( {\\;\\operatorname{mod}\\;5}\\right) . \n\\]\n\nThe Chinese Remainder Theorem predicts a solution that ... | Working through the proof gives\n\n\\[ \n{M}_{1} = {20}\\text{so}{20}{N}_{1} \\equiv 1\\;\\left( {\\;\\operatorname{mod}\\;3}\\right) \\text{and}{N}_{1} = 2\\text{,} \n\\]\n\n\\[ \n{M}_{2} = {15}\\text{so}{15}{N}_{2} \\equiv 1\\;\\left( {\\;\\operatorname{mod}\\;4}\\right) \\text{and}{N}_{2} = 3\\text{,} \n\\]\n\n\\[ \... | Yes |
The fact that 91 is not a prime was readily detected by this method. To see how recalcitrant some numbers can be, consider \( n = {561} \) . The first few numbers \( b \) with \( \gcd \left( {b,{561}}\right) = 1 \) are \( 2,4,5,7,8,{10} \), and we can easily find that | \[ {2}^{560} \equiv 1\;\left( {\;\operatorname{mod}\;{561}}\right) \] \[ {4}^{560} \equiv 1\;\left( {\;\operatorname{mod}\;{561}}\right) \] \[ {5}^{560} \equiv 1\;\left( {\;\operatorname{mod}\;{561}}\right) \] \[ {7}^{560} \equiv 1\;\left( {\;\operatorname{mod}\;{561}}\right) \] \[ {8}^{560} \equiv 1\;\left( {\;\operat... | Yes |
Theorem 12.21. Suppose that \( n \) is odd.\n\n(1) \( n \) is a pseudoprime to base \( b \) if and only if the multiplicative order of \( \bar{b} \) in \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) divides \( \left( {n - 1}\right) \) . | Proof. (1) The congruence \( {b}^{n - 1} \equiv 1 \) modulo \( n \) means that \( {\bar{b}}^{n - 1} \) is the identity in the group \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \), which holds if and only if the order of \( \bar{b} \) divides \( \left( {n - 1}\right) \) . | Yes |
Let \( n = {561} \). We saw in Example 12.20 that 561 is a pseudoprime with respect to the bases \( 2,4,5,7,8 \), and 10 . Knowing that \( {561} = 3 \cdot {11} \cdot {17} \), we can use the Chinese Remainder Theorem to argue as follows. Let \( b \) be any integer with \( \gcd \left( {b,{561}}\right) = 1 \) . | Then\n\n\[ \gcd \left( {b,3}\right) = 1 \Rightarrow {b}^{2} \equiv 1{\;(\operatorname{mod}\;3)} \Rightarrow {b}^{560} = {\left( {b}^{2}\right) }^{280} \equiv 1{\;(\operatorname{mod}\;3)}, \]\n\n\[ \gcd \left( {b,{11}}\right) = 1 \Rightarrow {b}^{10} \equiv 1\;\left( {\;\operatorname{mod}\;{11}}\right) \Rightarrow {b}^{... | Yes |
Lemma 12.24. If \( p \) is a prime, then \( G = {\left( \mathbb{Z}/{p}^{2}\mathbb{Z}\right) }^{ * } \) is a cyclic group. | Proof. For \( p = 2 \), the invertible elements of \( \mathbb{Z}/4\mathbb{Z} \) are the residues 1 and 3, so in this case \( G \) is a cyclic group with two elements. We may thus assume that \( p \) is an odd prime.\n\nWe first claim that if \( \gcd \left( {a, p}\right) = 1 \), then \( 1 + {ap} \) has order \( p \) in ... | Yes |
Theorem 12.25. If \( n \) is a Carmichael number, then \( n \) is square-free. | Proof. Let \( n \) be a Carmichael number with \( {p}^{2} \mid n \) for some prime \( p \) . Let \( {n}^{\prime } \) denote the \( p \) -primary part of \( n \) - that is, \( {n}^{\prime } \) is \( n{p}^{-r} \) if \( n \) is divisible by \( p \) exactly \( r \) times. Let \( g,1 < g < {p}^{2} \), be a generator of \( {... | Yes |
Lemma 12.27. If \( n > 1 \) is an odd composite number, then at least half of all the integers \( b,1 \leq b \leq n \), with \( \gcd \left( {b, n}\right) = 1 \) will satisfy | Proof. This is essentially the same as the proof of Theorem 12.21(3). If \( {b}_{1} \) and \( {b}_{2} \) satisfy \( \gcd \left( {{b}_{1}, n}\right) = \gcd \left( {{b}_{2}, n}\right) = 1 \), but\n\n\[ \left( \frac{{b}_{1}}{n}\right) = {b}_{1}^{\left( {n - 1}\right) /2}\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nand\... | Yes |
Theorem 12.28. Let \( b \) and \( n \) be integers with \( 1 \leq b \leq n \) . Then\n\n\[ \mathrm{C}\left( {\text{ calculate the Jacobi symbol }\left( \frac{b}{n}\right) }\right) = \mathrm{O}\left( {\left( \log n\right) }^{3}\right) . | Proof. We may assume that \( \gcd \left( {b, n}\right) = 1 \) . Choose \( {b}_{1} \) with \( {b}_{1} \equiv b \) modulo \( n \) and \( - \frac{n}{2} \leq {b}_{1} \leq \frac{n}{2} \) . Then\n\n\[ \left( \frac{b}{n}\right) = \left( {\pm \frac{{b}_{2}}{n}}\right) \text{ with }0 < {b}_{2} \leq \frac{n}{2} \]\n\n\[ = \left(... | Yes |
Lemma 12.29. Let \( n \) denote a positive integer and suppose a is an integer with \( \gcd \left( {a, n}\right) = 1 \) . Then \( n \) is prime if and only if the following congruence holds for polynomials\n\n\[{\left( x - a\right) }^{n} \equiv {x}^{n} - a\;\left( {\;\operatorname{mod}\;n}\right)\]\n\nIn other words, i... | Exercise 12.7. Prove Lemma 12.29. (Hint: It is not much harder than the proof of Fermat's Little Theorem. However, on this occasion, we suggest the use of congruences and the Binomial Theorem rather than Lagrange.) | No |
Theorem 12.30. Suppose \( n \in \mathbb{N} \) and \( s \leq n \) . Suppose primes \( q \) and \( r \) are chosen with the properties that \( q \mid \left( {r - 1}\right) ,{n}^{\left( {r - 1}\right) /q} ≢ 0,1 \) modulo \( r \), and\n\n\[ \left( \begin{matrix} q + s - 1 \\ s \end{matrix}\right) \geq {n}^{2\lfloor \sqrt{r... | This gives a version of the Agrawal-Kayal-Saxena algorithm.\n\n(1) Decide if \( n \) is a power of a natural number. If it is, go to (5).\n\n(2) Choose integers \( q, r \), and \( s \) satisfying the hypotheses of Theorem 12.30.\n\n(3) For \( a = 1,\ldots, s - 1 \), do two checks. If \( a \mid n \), go to (5). If \( {\... | No |
Theorem 12.33. Suppose \( n > 1 \) denotes any odd integer and the Extended Riemann Hypothesis holds for \( L\left( {\cdot ,\chi }\right) \) where \( \chi \) denotes the Jacobi symbol \( \left( \frac{ \cdot }{n}\right) \) . Then either an integer a exists with \( a < 2{\left( \log n\right) }^{2} \) such that \( \gcd \l... | \[ \left( \frac{a}{n}\right) \neq {a}^{\left( {n - 1}\right) /2}\;\left( {\;\operatorname{mod}\;n}\right) \] | No |
Let \( n = {91} \) and \( f\left( x\right) = {x}^{2} + 1 \) . Then take \( {x}_{0} = 1 \) and compute\n\n\[ \n{x}_{1} = 2,{x}_{2} = 5,{x}_{3} = {26},{x}_{4} = {40},\n\]\n\nand so on. (Remember that these are residues modulo \( n \) .) We find\n\n\[ \n\gcd \left( {{x}_{3} - {x}_{2}, n}\right) = \gcd \left( {{26} - 5,{91... | \[ \n{x}_{1} = 2,{x}_{2} = 5,{x}_{3} = {26},{x}_{4} = {40},\n\]\n\nand so on. (Remember that these are residues modulo \( n \) .) We find\n\n\[ \n\gcd \left( {{x}_{3} - {x}_{2}, n}\right) = \gcd \left( {{26} - 5,{91}}\right) = \gcd \left( {{21},{91}}\right) = 7.\n\] | Yes |
Let \( n = {323} \) and \( f\left( x\right) = {x}^{2} + 1 \) . Take \( {x}_{0} = 1 \), then | \[ {x}_{1} = 2,{x}_{2} = 5,{x}_{3} = {26},{x}_{4} = {31}\text{.} \] We find that \[ \gcd \left( {{x}_{5} - {x}_{4}, n}\right) = \gcd \left( {{316} - {31},{323}}\right) = \gcd \left( {{285},{323}}\right) = {19}, \] so 323 is divisible by 19 . | Yes |
Let \( n = {4087} \) and \( f\left( x\right) = {x}^{2} + x + 1 \) . Pick \( {x}_{0} = 2 \) and compute | \[
\gcd \left( {{x}_{1} - {x}_{0}, n}\right) = \gcd \left( {7 - 2,{4087}}\right) = 1,
\]
\[
\gcd \left( {{x}_{2} - {x}_{1}, n}\right) = \gcd \left( {{57} - 7,{4087}}\right) = 1,
\]
\[
\gcd \left( {{x}_{3} - {x}_{1}, n}\right) = \gcd \left( {{3307} - 7,{4087}}\right) = 1,
\]
\[
\gcd \left( {{x}_{4} - {x}_{3}, n}\righ... | Yes |
Solve \( {s}^{2} \equiv {t}^{2}\left( {\;\operatorname{mod}\;n}\right) \): | \[ {\left( -{77}\right) }^{2} \equiv {\left( {67} \cdot {68}\right) }^{2} \equiv {2}^{4}{3}^{4} \equiv {\left( {2}^{2}{3}^{2}\right) }^{2} \equiv {36}^{2}\;\left( {\;\operatorname{mod}\;{4633}}\right) . \] We check \( \gcd \left( {-{77} + {36}, n}\right) = {41} \) and \( \gcd \left( {-{77} - {36}, n}\right) = {113} \),... | Yes |
Let \( n = {4633} \) and \( B = \{ - 1,2,3\} \) . Then the calculation in Example 12.38 gives\n\n\[ \left\lbrack \begin{array}{lll} 1 & 4 & 2 \\ 1 & 0 & 2 \\ 0 & 7 & 0 \end{array}\right\rbrack \equiv \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right\rbrack \;\left( {\;\operatorname{m... | and the first two rows give the desired relation. | No |
Let \( n = {1829} \) and \( B = \{ - 1,2,3,5,7,{11},{13}\} \) . Then\n\n\[{42},{43},{61},{74},{85},{86}\text{are}B\text{-numbers.}\] | Factorizing them gives the vectors\n\n\[{42}^{2} \equiv - 5 \cdot {13} \rightarrow \left( {1,0,0,1,0,0,1}\right)\]\n\n\[{43}^{2} \equiv {2}^{2} \cdot 5 \rightarrow \left( {0,2,0,1,0,0,0}\right)\]\n\n\[{61}^{2} \equiv {3}^{2} \cdot 7 \rightarrow \left( {0,2,0,0,1,0,0}\right)\]\n\n\[{74}^{2} \equiv - {11} \rightarrow \le... | Yes |
Let \( n = {2201} \), and compute a few candidates for \( B \) -numbers modulo \( n \) : | \[ {47}^{2} \equiv 8 = {2}^{3} \] \[ {48}^{2} \equiv {103}\text{has no small factors,} \] \[ {49}^{3} \equiv {200} = {2}^{3} \cdot {5}^{2} \] \[ {50}^{2} \equiv {299} = {13} \cdot {23} \] \[ {51}^{2} \equiv {400} = {2}^{4} \cdot {5}^{2} \] \[ {52}^{2} \equiv {503}\text{has no small factors.} \] So choose \( B = \{ - 1,... | Yes |
Let \( n = {21} \), and choose the elliptic curve\n\n\[ E : {y}^{2} + y = {x}^{3} - x \]\n\n together with the point \( P = \left( {0,0}\right) \). The first few multiples of \( P \) modulo 21 can be computed without any problem:\n\n\[ {2P} = \left( {1,0}\right) ,{3P} = \left( {{20},{20}}\right) ,{4P} = \left( {2,{18}}... | However, the computer will refuse to calculate \( {7P} \) modulo 21 . In order to find this value, it would need to invert 6 modulo 21; this is done using the Euclidean Algorithm, which rapidly detects the factor 3 of 21. Notice that if the initial point is \( P = \left( {0,0}\right) \), then it is \( \gcd \left( {x\le... | Yes |
Example 12.43. For a slightly more impressive example, let \( n = {39701558597} \) . Working with the same elliptic curve and the same point, this time the computer will find\n\n\[ \n{526P} = \left( {{3341173047},{12476794460}}\right) \;\left( {\;\operatorname{mod}\;n}\right) ,\n\]\n\nbut refuses to go any further. The... | Lenstra’s idea was that the flexibility of choosing curves \( E \) and points \( P \) , together with a suitable multiplier \( k \), might make it possible to detect when computing \( {kP} \) modulo \( n \) becomes impossible - in other words, when you stumble onto a factor of \( n \) . There are much fuller accounts o... | No |
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