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Lemma 10.3.3. Let \( \gamma \left( s\right) \) be a gamma product. Denote by \( W\left( t\right) \) the inverse Mellin transform of \( \gamma \left( s\right) \), in other words,\n\n\[ W\left( t\right) = \frac{1}{2i\pi }{\int }_{\delta - i\infty }^{\delta + i\infty }{t}^{-z}\gamma \left( z\right) {dz} \]\n\nand let \( {\sigma }_{0} \) be the real part of the rightmost pole of \( \gamma \) .\n\n(1) The integral defining \( W\left( t\right) \) converges absolutely and is independent of \( \delta > {\sigma }_{0} \n\n(2) When \( t \) tends to \( \infty \), the function \( W\left( t\right) \) tends to 0 faster than any power of \( t \) . More precisely, using the notation of Proposition 10.3.2, as \( t \) tends to \( \infty \) we have\n\n\[ W\left( t\right) \leq A{t}^{\left( {2{B}_{r} - n + 1}\right) /N}{e}^{-\left( {{\pi N}/4}\right) {\left( t/P\right) }^{2/N}} \]\n\nfor some explicit constant \( A > 0 \) .\n\n(3) For \( \operatorname{Re}\left( s\right) > {\sigma }_{0} \) we have\n\n\[ \gamma \left( s\right) = {\int }_{0}^{\infty }{t}^{s}W\left( t\right) \frac{dt}{t} \]
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Proof. (1). Since \( \gamma \left( z\right) \) decreases exponentially when \( \left| {\operatorname{Im}\left( z\right) }\right| \) tends to infinity with \( \operatorname{Re}\left( z\right) \) fixed, it is clear that the integral defining \( W\left( t\right) \) converges absolutely. For the same reason, by integrating over the rectangle \( \left\lbrack {{\delta }_{1},{\delta }_{2}}\right\rbrack \times \left\lbrack {-T, T}\right\rbrack \) and letting \( T \) tend to \( \infty \), it is clear that the integral is independent of \( \delta > {\sigma }_{0} \) , proving (1).\n\n(2). For any \( \delta > {\sigma }_{0} \), we clearly have\n\n\[ \left| {W\left( t\right) }\right| \leq \frac{1}{2\pi }{t}^{-\delta }{\int }_{-\infty }^{\infty }\left| {\gamma \left( {\delta + {iT}}\right) }\right| {dT} \]\n\nand this last integral is convergent, so \( W\left( t\right) \) tends to 0 faster than any power of \( t \) since \( \delta \) can be chosen arbitrarily large. More precisely, if for simplicity we set \( {C}_{2} = {B}_{r} - n/2 \) and if \( {C}_{1} \) is suitably large, Proposition 10.3.2 implies that\n\n\[ \left| {\gamma \left( {\delta + {iT}}\right) }\right| \leq {C}_{1}{P}^{\delta }{\left| T\right| }^{{N\delta }/2 + {C}_{2}}{e}^{-\left( {\pi /4}\right) N\left| T\right| } \]\n\nfrom which it follows as above that\n\n\[ W\left( t\right) \leq \frac{{C}_{1}{\left( P/t\right) }^{\delta }}{\pi }\frac{\Gamma \left( {{N\delta }/2 + {C}_{2} + 1}\right) }{{\left( \pi N/4\right) }^{{N\delta }/2 + {C}_{2} + 1}}. \]\n\nWe choose \( \delta \) close to the smallest possible value of the right-hand side, for example, \( \delta = \left( {\pi /2}\right) {\left( t/P\right) }^{2/N} \), which will indeed be larger than \( {\sigma }_{0} \) for \( t \) sufficiently large. A small computation gives (2) with an explicit constant \( A \) (see Exercise 18).\n\n(3). This is simply Mellin's inversion formula, which is applicable here since \( \gamma \left( s\right) \) and \( W\left( t\right) \) are rapidly decreasing functions as \( \left| {\operatorname{Im}\left( s\right) }\right| \rightarrow \infty \) and \( t \rightarrow \infty \), respectively.
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Yes
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Theorem 10.3.4. For \( i = 1 \) and \( i = 2 \), let \( {L}_{i}\left( s\right) = \mathop{\sum }\limits_{{n \geq 1}}{a}_{i}\left( n\right) {n}^{-s} \) be Dirichlet series such that the \( {a}_{i}\left( n\right) \) have at most polynomial growth (or, equivalently, the series \( {L}_{i}\left( s\right) \) converge in some right half-plane \( \operatorname{Re}\left( s\right) \geq {\sigma }_{0} \) ). For \( i = 1 \) and \( i = 2 \), let \( {\gamma }_{i}\left( s\right) \) be functions such that the following conditions hold.\n\n(1) The functions \( {\gamma }_{i}\left( s\right) \) are gamma products with the same degree \( N \) .\n\n(2) The functions \( {\Lambda }_{i}\left( s\right) = {\gamma }_{i}\left( s\right) {L}_{i}\left( s\right) \) extend analytically to the whole complex plane into meromorphic functions of finite type having a finite number of poles.\n\n(3) There exists a functional equation\n\n\[ \n{\Lambda }_{1}\left( {k - s}\right) = w \cdot {\Lambda }_{2}\left( s\right) \n\]\n\nfor some constant \( w \in {\mathbb{C}}^{ * } \), some real number \( k \), valid for all \( s \) different from the poles of \( {\Lambda }_{2}\left( s\right) \) .\n\nDefine the functions \( {F}_{i}\left( {s, x}\right) \) by\n\n\[ \n{F}_{i}\left( {s, x}\right) = \frac{{x}^{s}}{2i\pi }{\int }_{\delta - i\infty }^{\delta + i\infty }\frac{{x}^{-z}{\gamma }_{i}\left( z\right) }{z - s}{dz} = \frac{1}{2i\pi }{\int }_{\delta - i\infty }^{\delta + i\infty }\frac{{x}^{-z}{\gamma }_{i}\left( {z + s}\right) }{z}{dz} \n\]\n\nwhere \( \delta \) is sufficiently large, and the functions \( {p}_{i}\left( {s, x}\right) \) by\n\n\[ \n{p}_{i}\left( {s, x}\right) = \mathop{\sum }\limits_{{a \neq s}}{\operatorname{Res}}_{z = a}\left( \frac{{x}^{z - s}{\Lambda }_{i}\left( z\right) }{s - z}\right) , \n\]\n\nwhere the sum is over all poles a of \( {\Lambda }_{i}\left( z\right) \) different from s.\n\nThen for all \( {t}_{0} > 0 \), we have\n\n\[ \n{\Lambda }_{1}\left( s\right) = \mathop{\sum }\limits_{{n \geq 1}}\frac{{a}_{1}\left( n\right) }{{n}^{s}}{F}_{1}\left( {s, n{t}_{0}}\right) + w\mathop{\sum }\limits_{{n \geq 1}}\frac{{a}_{2}\left( n\right) }{{n}^{k - s}}{F}_{2}\left( {k - s,\frac{n}{{t}_{0}}}\right) + {p}_{1}\left( {s,\frac{1}{{t}_{0}}}\right) \n\]\n\nand symmetrically\n\n\[ \n{\Lambda }_{2}\left( s\right) = \mathop{\sum }\limits_{{n \geq 1}}\frac{{a}_{2}\left( n\right) }{{n}^{s}}{F}_{2}\left( {s,\frac{n}{{t}_{0}}}\right) + {w}^{-1}\mathop{\sum }\limits_{{n \geq 1}}\frac{{a}_{1}\left( n\right) }{{n}^{k - s}}{F}_{1}\left( {k - s, n{t}_{0}}\right) + {p}_{2}\left( {s,{t}_{0}}\right) . \n\]\n\nIf, in addition, \( {t}_{0} = 1 \), we have \( {p}_{i}\left( {s,1}\right) = {\phi }_{i}\left( s\right) \), where \( {\phi }_{i}\left( s\right) \) is the polar part of \( {\Lambda }_{i}\left( s\right) \) : in other words, the unique rational function such that \( {\Lambda }_{i}\left( s\right) - {\phi }_{i}\left( s\right) \) is an entire function and \( {\phi }_{i}\left( s\right) \) tends to 0 as \( \left| s\right| \) tends to \( \infty \) .
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Proof. Before beginning the proof itself, we will make a few remarks about convergence. First, it is clear that \( {L}_{i}\left( s\right) \) converges in some right half-plane if and only if \( {a}_{i}\left( n\right) \) has at most polynomial growth. Since the functions \( {\Lambda }_{i}\left( s\right) \) have only a finite number of poles, we may choose a real number \( {\sigma }_{0} \) large enough so that for \( i = 1 \) and \( i = 2 \), the functions \( {L}_{i}\left( s\right) \) converge absolutely for \( \operatorname{Re}\left( s\right) > {\sigma }_{0} \) and all the poles of \( {\Lambda }_{i}\left( s\right) \) are in the strip \( k - {\sigma }_{0} < \operatorname{Re}\left( s\right) < {\sigma }_{0} \) .\n\nFor \( i = 1 \) and \( i = 2 \), denote by \( {W}_{i}\left( t\right) \) the inverse Mellin transform of \( {\gamma }_{i}\left( s\right) \) and by \( {\sigma }_{i} \) the real part of the rightmost pole of \( {\gamma }_{i}\left( s\right) \) . The definition and main properties of this
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Yes
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(1) For all fixed \( r \in \mathbb{R} \), there exists \( e\left( r\right) \) such that as \( \left| T\right| \rightarrow \infty \), we have\n\n\[ \n{L}_{i}\left( {r + {iT}}\right) = O\left( {\left| T\right| }^{e\left( r\right) }\right) \n\]
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Proof. We prove (1) for \( {L}_{1} \), the result following by symmetry. For \( r > \sigma \) , we have \( {L}_{1}\left( {r + {iT}}\right) = O\left( 1\right) \) since the series converges absolutely. For \( r < k - \sigma \) , we apply the functional equation, which gives us\n\n\[ \n{L}_{1}\left( {r + {iT}}\right) = w{L}_{2}\left( {k - r - {iT}}\right) \frac{{\gamma }_{2}\left( {k - r - {iT}}\right) }{{\gamma }_{1}\left( {r + {iT}}\right) }. \n\]\n\nSince \( {\gamma }_{1} \) and \( {\gamma }_{2} \) are gamma products with the same degree \( N \) and \( {L}_{2}(k - r - \) \( {iT}) = O\left( 1\right) \) for \( r < k - \sigma \), it follows that for \( r < k - \sigma \) we have\n\n\[ \n{L}_{1}\left( {r + {iT}}\right) = O\left( {\left| T\right| }^{c\left( r\right) }\right) \n\]\n\nfor some constant \( c\left( r\right) \) depending only on \( r \) . Since \( {\Lambda }_{1}\left( s\right) \) is a function of finite type and \( {\gamma }_{1}\left( s\right) \) is a gamma product, the function \( {L}_{1}\left( s\right) \) is of finite type in any strip \( {\sigma }_{1} \leq \operatorname{Re}\left( s\right) \leq {\sigma }_{2} \) . Thus we may apply the Phragmén-Lindelöf convexity theorem on the strip \( \;\left\lbrack {k - \sigma ,\sigma }\right\rbrack \) (see, for example, \( \;\left\lbrack {Lan3}\right\rbrack ), \) which implies that \( {L}_{1}\left( {r + {iT}}\right) = O\left( {\left| T\right| }^{e\left( r\right) }\right) \) for any \( r \) in the strip \( k - \sigma < r < \sigma \), with an explicit value of the exponent \( e\left( r\right) \) (which we do not need), proving (1).
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Yes
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For any rational function \( \phi \left( z\right) \) that tends to 0 when \( \left| z\right| \) tends to \( \infty \), we have the identity\n\n\[ \mathop{\sum }\limits_{{a \neq s}}{\operatorname{Res}}_{z = a}\left( \frac{\phi \left( z\right) }{s - z}\right) = \phi \left( s\right) \]\n\nwhere the sum is over all the poles a of \( \phi \left( z\right) \) different from \( s \) .
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Proof. Let \( R \) be a real positive number larger than the modulus of the poles of \( \phi \left( z\right) \) and of \( \left| s\right| \), and let \( {C}_{R} \) be the circle of radius \( R \) centered at the origin. By the residue theorem we have\n\n\[ \mathop{\sum }\limits_{{a \neq s}}{\operatorname{Res}}_{z = a}\left( \frac{\phi \left( z\right) }{s - z}\right) - \phi \left( s\right) = \frac{1}{2i\pi }{\int }_{{C}_{R}}\frac{\phi \left( z\right) }{s - z}{dz}. \]\n\nSince \( \phi \left( z\right) \) is a rational function that tends to 0 when \( \left| z\right| \) tends to infinity, there exists \( B \) such that \( \left| {\phi \left( z\right) }\right| \leq B/\left| z\right| \) for \( \left| z\right| \) sufficiently large. It follows\n\nthat\n\[ \left| {{\int }_{{C}_{R}}\frac{\phi \left( z\right) }{s - z}{dz}}\right| \leq {2\pi R}\frac{B}{R}\frac{1}{R - \left| s\right| } = \frac{2\pi B}{R - \left| s\right| } \]\n\nand this tends to zero when \( R \) tends to infinity, proving the lemma and hence the theorem.
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Yes
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Lemma 2. Let \( \mathfrak{D} \) be a bounded \( {}^{ * } \) -derivation on \( \mathfrak{A} = \mathcal{L}\left( H\right) \) . Then there exists a skew-adjoint operator \( A \in \mathcal{L}\left( H\right) \) such that \[ \mathfrak{D}\left( T\right) = {AT} - {TA}\;\text{ for all }T \in \mathcal{L}\left( H\right) . \]
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Proof. For each pair of elements \( x, y \in H \) we define the rank-one operator \( x \otimes y \) by \[ z \mapsto x \otimes y\left( z\right) \mathrel{\text{:=}} \left( {x \mid z}\right) y. \] Take now some fixed \( z \in H,\parallel z\parallel = 1 \), and define \( A \in \mathcal{L}\left( H\right) \) by \[ {Ay} \mathrel{\text{:=}} \mathfrak{D}\left( {z \otimes y}\right) \left( z\right) \;\text{ for all }y \in H. \] For \( T \in \mathcal{L}\left( H\right) \) we obtain \[ \left( {{AT} - {TA}}\right) \left( y\right) = \mathfrak{D}\left( {z \otimes {Ty}}\right) \left( z\right) - T\left( {\mathfrak{D}\left( {z \otimes y}\right) \left( z\right) }\right) \] \[ = \mathfrak{D}\left( {T\left( {z \otimes y}\right) }\right) \left( z\right) - \left( {T \cdot \mathfrak{D}\left( {z \otimes y}\right) }\right) \left( z\right) \] \[ = \left( {\mathfrak{D}T \cdot \left( {z \otimes y}\right) }\right) \left( z\right) + \left( {T \cdot \mathfrak{D}\left( {z \otimes y}\right) }\right) \left( z\right) - \left( {T \cdot \mathfrak{D}\left( {z \otimes y}\right) }\right) \left( z\right) \] \[ = \mathfrak{D}T\left( y\right) \;\text{ for all }y \in H. \] In the final step we show that the operator \( A \in \mathcal{L}\left( H\right) \) for which \[ \mathfrak{D}\left( T\right) = {AT} - {TA}\;\text{ for all }T \in H \] can be chosen as a skew-adjoint operator. In fact, since \( \mathfrak{D} \) is a \( {}^{ * } \) -derivation, we have \[ A{T}^{ * } - {T}^{ * }A = \mathfrak{D}\left( {T}^{ * }\right) = \mathfrak{D}{\left( T\right) }^{ * } = - {A}^{ * }{T}^{ * } + {T}^{ * }{A}^{ * }, \] and hence \[ \left( {A + {A}^{ * }}\right) {T}^{ * } - {T}^{ * }\left( {A + {A}^{ * }}\right) = 0 \] for all \( T \in \mathcal{L}\left( H\right) \) . Therefore, \[ \widetilde{A} \mathrel{\text{:=}} A - \frac{{A}^{ * } + A}{2} = \frac{A - {A}^{ * }}{2} \] is a skew-adjoint operator still satisfying \[ \widetilde{A}T - T\widetilde{A} = {AT} - {TA}\;\text{ for all }T \in H. \]
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Yes
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Proposition 2. The resolvent \( R\left( {\lambda, A}\right) \) for \( \operatorname{Re}\lambda > 0 \) of the differentiation operator \( A \) with maximal domain \( D\left( A\right) \) (i.e., of the generator of the left translation semigroup) on any of the above spaces \( X \) is given by
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\[ \left( {R\left( {\lambda, A}\right) f}\right) \left( s\right) = {\int }_{s}^{\infty }{\mathrm{e}}^{-\lambda \left( {\tau - s}\right) }f\left( \tau \right) {d\tau }\;\text{ for }f \in X, s \in \mathbb{R}. \]
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Yes
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For every \( f \in \mathrm{C}\left\lbrack {0,1}\right\rbrack \), we have\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{B}_{n}f = f \]
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This lemma, which also re-proves the Weierstrass Approximation Theorem 4.11, is one of the fundamental results of classical approximation theory and can be proved in many different ways. A very elegant proof uses Korovkin's theorem, which assures that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{B}_{n}f = f\;\text{ for all }f \in \mathrm{C}\left\lbrack {0,1}\right\rbrack \]\n\nif this convergence holds for the three functions\n\n\[ \mathbb{1}\left( s\right) \mathrel{\text{:=}} 1,\;{id}\left( s\right) \mathrel{\text{:=}} s,\;\text{ and }\;i{d}^{2}\left( s\right) \mathrel{\text{:=}} {s}^{2}\;\text{ for }s \in \left\lbrack {0,1}\right\rbrack .\n\nThis, however, is straightforward, since for \( s \in \left\lbrack {0,1}\right\rbrack \) and \( n \in \mathbb{N} \) ,\n\n\[ {B}_{n}\mathbb{1}\left( s\right) = 1 \]\n\n\[ {B}_{n}{id}\left( s\right) = s \]\n\n\[ {B}_{n}i{d}^{2}\left( s\right) = {s}^{2} + \frac{s\left( {1 - s}\right) }{n}. \]
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Yes
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Lemma 1. The semigroup \( {\left( T\left( t\right) \right) }_{t \geq 0} \) satisfies\n\n\[ T\left( t\right) f\left( s\right) = \left\{ \begin{array}{ll} {\mathrm{e}}^{{\int }_{s}^{0}m\left( \sigma \right) {d\sigma }}\left\lbrack {{\mathrm{e}}^{\left( {s + t}\right) m\left( 0\right) }f\left( 0\right) }\right. & \\ \left. {\; + {\int }_{0}^{s + t}{\mathrm{e}}^{{\tau m}\left( 0\right) }{LT}\left( {s + t - \tau }\right) {fd\tau }}\right\rbrack & \text{ for }s + t > 0, \\ {\mathrm{e}}^{{\int }_{s}^{s + t}m\left( \sigma \right) {d\sigma }}f\left( {s + t}\right) & \text{ for }s + t \leq 0. \end{array}\right. \]
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Proof. For \( f \in D\left( A\right) \) we have\n\n\[ \frac{d}{dr}\left( {{\mathrm{e}}^{{rm}\left( 0\right) }\left( {T\left( {t - r}\right) f}\right) \left( 0\right) + {\int }_{0}^{r}{\mathrm{e}}^{{\tau m}\left( 0\right) }{LT}\left( {t - \tau }\right) {fd\tau }}\right) = 0. \]\n\nThis implies\n\n\[ \left( {T\left( t\right) f}\right) \left( 0\right) = {\mathrm{e}}^{{tm}\left( 0\right) }f\left( 0\right) + {\int }_{0}^{t}{\mathrm{e}}^{{\tau m}\left( 0\right) }{LT}\left( {t - \tau }\right) {fd\tau }. \]\n\nOn the other hand, we have\n\n\[ \frac{d}{dr}\left( {{\mathrm{e}}^{{\int }_{s}^{s + r}m\left( \sigma \right) {d\sigma }}\left( {T\left( {t - r}\right) f}\right) \left( {s + r}\right) }\right) = 0. \]\n\nTherefore, we obtain\n\n\[ \left( {T\left( t\right) f}\right) \left( s\right) = \left\{ \begin{array}{ll} {\mathrm{e}}^{{\int }_{s}^{0}m\left( \sigma \right) {d\sigma }}\left( {T\left( {s + t}\right) f}\right) \left( 0\right) & \text{ for }s + t > 0, \\ {\mathrm{e}}^{{\int }_{s}^{s + t}m\left( \sigma \right) {d\sigma }}f\left( {s + t}\right) & \text{ for }s + t \leq 0. \end{array}\right. \]
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Yes
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Lemma 2. If \( m\left( {-\infty }\right) < 0 \), then the semigroup \( {\left( T\left( t\right) \right) }_{t \geq 0} \) is quasi-compact.
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Proof. We define operators \( K\left( t\right) \in \mathcal{L}\left( X\right) \) by\n\n\[ K\left( t\right) f\left( s\right) \mathrel{\text{:=}} \left\{ \begin{array}{ll} {\mathrm{e}}^{{\int }_{s}^{0}m\left( \sigma \right) {d\sigma }}\left\lbrack {{\mathrm{e}}^{\left( {s + t}\right) m\left( 0\right) }f\left( 0\right) }\right. & \\ \left. {\; + {\int }_{0}^{s + t}{\mathrm{e}}^{\left( {s + t - \tau }\right) m\left( 0\right) }{LT}\left( \tau \right) {fd\tau }}\right\rbrack & \text{ for }0 < s + t, \\ \left( {t + s + 1}\right) \cdot {\mathrm{e}}^{{\int }_{s}^{0}m\left( \sigma \right) {d\sigma }}f\left( 0\right) & \text{ for } - 1 < s + t \leq 0, \\ 0 & \text{ for }s + t \leq - 1. \end{array}\right. \]\n\nThese operators are compact by the Arzelà-Ascoli theorem. On the other hand, since \( m\left( {-\infty }\right) < 0 \), we have\n\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}\parallel T\left( t\right) - K\left( t\right) \parallel = 0. \]\n\nTherefore, the semigroup \( {\left( T\left( t\right) \right) }_{t \geq 0} \) is quasi-compact.
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Yes
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Lemma 3. Suppose that \( m\left( {-\infty }\right) < 0 \) . If \( \xi \left( 0\right) \leq 0 \), i.e., \( L{g}_{0} \geq - m\left( 0\right) \), then the characteristic function \( \xi \) has a unique zero \( {\lambda }_{0} \geq 0 \) that is a dominant eigenvalue of the operator \( A \) .
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Proof. The function \( \xi : {\mathbb{R}}_{ + } \ni \lambda \mapsto \lambda - {L}_{0}{g}_{\lambda } - a - m\left( 0\right) \) is strictly increasing from \( \xi \left( 0\right) \) to \( \infty \) . Consequently, if \( \xi \left( 0\right) \leq 0 \), it has a unique zero \( {\lambda }_{0} \) that is an eigenvalue of \( A \) . Now take an arbitrary eigenvalue \( \lambda \) of \( A \) with \( \operatorname{Re}\lambda \geq {\lambda }_{0} \) . Then, we have\n\n\[ \left| {\lambda - a - m\left( 0\right) }\right| = \left| {{L}_{0}{g}_{\lambda }}\right| \leq {L}_{0}{g}_{{\lambda }_{0}} = {\lambda }_{0} - a - m\left( 0\right) .\n\]\n\nThis implies \( \lambda = {\lambda }_{0} \), and therefore \( {\lambda }_{0} \) is a dominant eigenvalue of \( A \) .
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Yes
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Theorem 1.1 (Division algorithm) Let \( a \) and \( d \) be integers with \( d \geq 1 \) . There exist unique integers \( q \) and \( r \) such that\n\n\[ a = {dq} + r \]\n\n(1.1)\n\nand\n\n\[ 0 \leq r \leq d - 1 \]\n\n(1.2)\n\n\n\nThe integer \( q \) is called the quotient and the integer \( r \) is called the remainder in the division of \( a \) by \( d \) .
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Proof. Consider the set \( S \) of nonnegative integers of the form\n\n\[ a - {dx} \]\n\nwith \( x \in \mathbf{Z} \) . If \( a \geq 0 \), then \( a = a - d \cdot 0 \in S \) . If \( a < 0 \), let \( x = - y \), where \( y \) is a positive integer. Since \( d \) is positive, we have \( a - {dx} = a + {dy} \in S \) if \( y \) is sufficiently large. Therefore, \( S \) is a nonempty set of nonnegative integers. By the minimum principle, \( S \) contains a smallest element \( r \), and \( r = a - {dq} \geq 0 \) for some \( q \in \mathbf{Z} \) . If \( r \geq d \), then\n\n\[ 0 \leq r - d = a - d\left( {q + 1}\right) < r \]\n\nand \( r - d \in S \), which contradicts the minimality of \( r \) . Therefore, \( q \) and \( r \) satisfy conditions (1.1) and (1.2).\n\nLet \( {q}_{1},{r}_{1},{q}_{2},{r}_{2} \) be integers such that\n\n\[ a = d{q}_{1} + {r}_{1} = d{q}_{2} + {r}_{2}\;\text{ and }\;0 \leq {r}_{1},{r}_{2} \leq d - 1. \]\n\nThen\n\n\[ \left| {{r}_{1} - {r}_{2}}\right| \leq d - 1 \]\n\nand\n\n\[ d\left( {{q}_{1} - {q}_{2}}\right) = {r}_{2} - {r}_{1} \]\n\nIf \( {q}_{1} \neq {q}_{2} \), then\n\n\[ \left| {{q}_{1} - {q}_{2}}\right| \geq 1 \]\n\nand\n\n\[ d \leq d\left| {{q}_{1} - {q}_{2}}\right| = \left| {{r}_{2} - {r}_{1}}\right| \leq d - 1 \]\n\nwhich is impossible. Therefore, \( {q}_{1} = {q}_{2} \) and \( {r}_{1} = {r}_{2} \) . This proves that the quotient and remainder are unique.
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Yes
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Theorem 1.3 Let \( H \) be a subgroup of the integers under addition. There exists a unique nonnegative integer \( d \) such that \( H \) is the set of all multiples of \( d \), that is,\n\n\[ H = \{ 0, \pm d, \pm {2d},\ldots \} = d\mathbf{Z}. \]
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Proof. We have \( 0 \in H \) for every subgroup \( H \) . If \( H = \{ 0\} \) is the zero subgroup, then we choose \( d = 0 \) and \( H = 0\mathbf{Z} \) . Moreover, \( d = 0 \) is the unique generator of this subgroup.\n\nIf \( H \neq \{ 0\} \), then there exists \( a \in H, a \neq 0 \) . Since \( - a \) also belongs to \( H \) , it follows that \( H \) contains positive integers. By the minimum principle, \( H \) contains a least positive integer \( d \) . By Exercise \( {21},{dq} \in H \) for every integer \( q \), and so \( d\mathbf{Z} \subseteq H \) .\n\nLet \( a \in H \) . By the division algorithm, we can write \( a = {dq} + r \), where \( q \) and \( r \) are integers and \( 0 \leq r \leq d - 1 \) . Since \( {dq} \in H \) and \( H \) is closed under subtraction, it follows that\n\n\[ r = a - {dq} \in H. \]\n\nSince \( 0 \leq r < d \) and \( d \) is the smallest positive integer in \( H \), we must have \( r = 0 \), that is, \( a = {dq} \in d\mathbf{Z} \) and \( H \subseteq d\mathbf{Z} \) . It follows that \( H = d\mathbf{Z} \).\n\nIf \( H = d\mathbf{Z} = {d}^{\prime }\mathbf{Z} \), where \( d \) and \( {d}^{\prime } \) are positive integers, then \( {d}^{\prime } \in d\mathbf{Z} \) implies that \( {d}^{\prime } = {dq} \) for some integer \( q \), and \( d \in {d}^{\prime }\mathbf{Z} \) implies that \( d = {d}^{\prime }{q}^{\prime } \) for some integer \( {q}^{\prime } \) . Therefore,\n\n\[ d = {d}^{\prime }{q}^{\prime } = {dq}{q}^{\prime } \]\n\nand so \( q{q}^{\prime } = 1 \), hence \( q = {q}^{\prime } = \pm 1 \) and \( d = \pm {d}^{\prime } \) . Since \( d \) and \( {d}^{\prime } \) are positive, we have \( d = {d}^{\prime } \), and \( d \) is the unique positive integer that generates the subgroup \( H \) . \( ▱ \)
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No
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Theorem 1.4 Let \( A \) be a nonempty set of integers, not all zero. Then \( A \) has a unique greatest common divisor, and there exist integers \( {a}_{1},\ldots ,{a}_{k} \in \) A and \( {x}_{1},\ldots ,{x}_{k} \) such that\n\n\[ \gcd \left( A\right) = {a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k} \]
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Proof. Let \( H \) be the subset of \( \mathbf{Z} \) consisting of all integers of the form\n\n\[ {a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k}\;\text{ with }{a}_{1},\ldots ,{a}_{k} \in A\text{ and }{x}_{1},\ldots ,{x}_{k} \in \mathbf{Z}. \]\n\nThen \( H \) is a subgroup of \( \mathbf{Z} \) and \( A \subseteq H \) . By Theorem 1.3, there exists a unique positive integer \( d \) such that \( H = d\mathbf{Z} \), that is, \( H \) consists of all multiples of \( d \) . In particular, every integer \( a \in A \) is a multiple of \( d \), and so \( d \) is a common divisor of \( A \) . Since \( d \in H \), there exist integers \( {a}_{1},\ldots ,{a}_{k} \in A \) and \( {x}_{1},\ldots ,{x}_{k} \) such that\n\n\[ d = {a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k} \]\n\nIt follows that every common divisor of \( A \) must divide \( d \), hence \( d \) is a greatest common divisor of \( A \) .\n\nIf the positive integers \( d \) and \( {d}^{\prime } \) are both greatest common divisors, then \( d \mid {d}^{\prime } \) and \( {d}^{\prime } \mid d \), and so \( d = {d}^{\prime } \) . It follows that \( \gcd \left( A\right) \) is unique.
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Yes
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Theorem 1.5 Let \( {a}_{1},\ldots ,{a}_{k} \) be integers, not all zero. Then \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) = \) 1 if and only if there exist integers \( {x}_{1},\ldots ,{x}_{k} \) such that\n\n\[ \n{a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k} = 1.\n\]
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Proof. This follows immediately from Theorem 1.4. \( ▱ \)
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No
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Theorem 1.6 Let \( a \) and \( b \) be integers with \( b \geq 1 \) . If the Euclidean algorithm for \( a \) and \( b \) has length \( n \) with sequence of partial quotients \( {q}_{0},{q}_{1},\ldots ,{q}_{n - 1} \) , then \[ \frac{a}{b} = \left\langle {{q}_{0},{q}_{1},\ldots ,{q}_{n - 1}}\right\rangle \]
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Proof. Let \( {r}_{0} = a \) and \( {r}_{1} = b \) . The proof is by induction on \( n \) . If \( n = 1 \) , then \[ {r}_{0} = {r}_{1}{q}_{0} \] and \[ \frac{a}{b} = \frac{{r}_{0}}{{r}_{1}} = {q}_{0} = \left\langle {q}_{0}\right\rangle \] If \( n = 2 \), then \[ {r}_{0} = {r}_{1}{q}_{0} + {r}_{2} \] \[ {r}_{1} = {r}_{2}{q}_{1} \] and \[ \frac{a}{b} = \frac{{r}_{0}}{{r}_{1}} = {q}_{0} + \frac{{r}_{2}}{{r}_{1}} = {q}_{0} + \frac{1}{\frac{{r}_{1}}{{r}_{2}}} = {q}_{0} + \frac{1}{{q}_{1}} = \left\langle {{q}_{0},{q}_{1}}\right\rangle . \] Let \( n \geq 2 \), and assume that the theorem is true for integers \( a \) and \( b \geq 1 \) whose Euclidean algorithm has length \( n \) . Let \( a \) and \( b \geq 1 \) be integers whose Euclidean algorithm has length \( n + 1 \) and whose sequence of partial quotients is \( \left\langle {{q}_{0},{q}_{1},\ldots ,{q}_{n}}\right\rangle \) . Let \[ {r}_{0} = {r}_{1}{q}_{0} + {r}_{2} \] \[ {r}_{1} = {r}_{2}{q}_{1} + {r}_{3} \] \[ \text{:} \] \[ {r}_{n - 1} = {r}_{n}{q}_{n - 1} + {r}_{n + 1} \] \[ {r}_{n} = {r}_{n + 1}{q}_{n} \] be the \( n + 1 \) equations in the Euclidean algorithm for \( a = {r}_{0} \) and \( b = {r}_{1} \) . The Euclidean algorithm for the positive integers \( {r}_{1} \) and \( {r}_{2} \) has length \( n \) with sequence of partial quotients \( {q}_{1},\ldots ,{q}_{n} \) . It follows from the induction hypothesis that \[ \frac{{r}_{1}}{{r}_{2}} = \left\langle {{q}_{1},\ldots ,{q}_{n}}\right\rangle \] and so \[ \frac{a}{b} = \frac{{r}_{0}}{{r}_{1}} = {q}_{0} + \frac{1}{\frac{{r}_{1}}{{r}_{2}}} = {q}_{0} + \frac{1}{\left\langle {q}_{1},\ldots ,{q}_{n}\right\rangle } = \left\langle {{q}_{0},{q}_{1},\ldots ,{q}_{n}}\right\rangle . \] This completes the proof.
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Yes
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Theorem 1.7 (Euclid’s lemma) Let \( a, b, c \) be integers. If \( a \) divides \( {bc} \) and \( \left( {a, b}\right) = 1 \), then a divides \( c \) .
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Proof. Since \( a \) divides \( {bc} \), we have \( {bc} = {aq} \) for some integer \( q \) . Since \( a \) and \( b \) are relatively prime, Theorem 1.5 implies that there exist integers \( x \) and \( y \) such that\n\n\[ 1 = {ax} + {by}. \]\n\nMultiplying by \( c \), we obtain\n\n\[ c = {acx} + {bcy} = {acx} + {aqy} = a\left( {{cx} + {qy}}\right) ,\]\n\nand so \( a \) divides \( c \) . This completes the proof.
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Yes
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Theorem 1.8 Let \( k \geq 2 \), and let \( a,{b}_{1},{b}_{2},\ldots ,{b}_{k} \) be integers. If \( \left( {a,{b}_{i}}\right) = 1 \) for all \( i = 1,\ldots, k \), then \( \left( {a,{b}_{1}{b}_{2}\cdots {b}_{k}}\right) = 1 \) .
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Proof. The proof is by induction on \( k \) . Let \( k = 2 \) and \( d = \left( {a,{b}_{1}{b}_{2}}\right) \) . We must show that \( d = 1 \) . Since \( d \) divides \( a \) and \( \left( {a,{b}_{1}}\right) = 1 \), it follows that \( \left( {d,{b}_{1}}\right) = 1 \) . Since \( d \) divides \( {b}_{1}{b}_{2} \), Euclid’s lemma implies that \( d \) divides \( {b}_{2} \) . Therefore, \( d \) is a common divisor of \( a \) and \( {b}_{2} \), but \( \left( {a,{b}_{2}}\right) = 1 \) and so \( d = 1 \) .\n\nLet \( k \geq 3 \), and assume that the result holds for \( k - 1 \) . Let \( a,{b}_{1},\ldots ,{b}_{k} \) be integers such that \( \left( {a,{b}_{i}}\right) = 1 \) for \( i = 1,\ldots, k \) . The induction assumption implies that \( \left( {a,{b}_{1}\cdots {b}_{k - 1}}\right) = 1 \) . Since we also have \( \left( {a,{b}_{k}}\right) = 1 \), it follows from the case \( k = 2 \) that \( \left( {a,{b}_{1}\cdots {b}_{k - 1}{b}_{k}}\right) = 1 \) . This completes the proof.
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Yes
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Theorem 1.9 If a prime number \( p \) divides a product of integers, then \( p \) divides one of the factors.
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Proof. Let \( {b}_{1},{b}_{2},\ldots ,{b}_{k} \) be integers such that \( p \) divides \( {b}_{1}\cdots {b}_{k} \) . By Theorem 1.8, we have \( \left( {p,{b}_{i}}\right) > 1 \) for some \( i \) . Since \( p \) is prime, it follows that \( p \) divides \( {b}_{i} \) .
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Yes
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Theorem 1.10 (Fundamental theorem of arithmetic) Every positive integer can be written uniquely (up to order) as the product of prime numbers.
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Proof. First we prove that every positive integer can be written as a product of primes. Since an empty product is equal to 1 , we can write 1 as the empty product of primes. Let \( n \geq 2 \) . Suppose that every positive integer less than \( n \) is a product of primes. If \( n \) is prime, we are done. If \( n \) is composite, then \( n = d{d}^{\prime } \), where \( 1 < d \leq {d}^{\prime } < n \) . By the induction hypothesis, \( d \) and \( {d}^{\prime } \) are both products of primes, and so \( n = d{d}^{\prime } \) is a product of primes.\n\nNext we use induction to prove that this representation is unique. The representation of 1 as the product of the empty set of primes is unique. Let \( n \geq 2 \) and assume that the statement is true for all positive integers less than \( n \) . We must show that if \( n = {p}_{1}\cdots {p}_{k} = {p}_{1}^{\prime }\cdots {p}_{\ell }^{\prime } \), where \( {p}_{1},\ldots ,{p}_{k},{p}_{1}^{\prime },\ldots ,{p}_{\ell }^{\prime } \) are primes, then \( k = \ell \) and there is a permutation \( \sigma \) of \( 1,\ldots, k \) such that \( {p}_{i} = {p}_{\sigma \left( i\right) }^{\prime } \) for \( i = 1,\ldots, k \) . By Theorem 1.9, since \( {p}_{k} \) divides \( {p}_{1}^{\prime }\cdots {p}_{\ell }^{\prime } \), there exists an integer \( {j}_{0} \in \{ 1,\ldots ,\ell \} \) such that \( {p}_{k} \) divides \( {p}_{{j}_{0}}^{\prime } \), and so \( {p}_{k} = {p}_{{j}_{0}}^{\prime } \) since \( {p}_{{j}_{0}}^{\prime } \) is prime. Therefore,\n\n\[ \frac{n}{{p}_{k}} = {p}_{1}\cdots {p}_{k - 1} = \mathop{\prod }\limits_{\substack{{j = 1} \\ {j \neq {j}_{0}} }}^{\ell }{p}_{j}^{\prime } < n. \]\n\nIt follows from the induction hypothesis that \( k - 1 = \ell - 1 \), and there is a one-to-one map \( \sigma \) from \( \{ 1,\ldots, k - 1\} \) into \( \{ 1,\ldots, k\} \smallsetminus \left\{ {j}_{0}\right\} \) such that \( {p}_{i} = {p}_{\sigma \left( i\right) }^{\prime } \) for \( i = 1,\ldots, k - 1 \) . Let \( \sigma \left( k\right) = {j}_{0} \) . This defines the permutation \( \sigma \), and the proof is complete.
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Yes
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Theorem 1.11 Let \( {a}_{1},\ldots ,{a}_{k} \) be positive integers. Then\n\n\[\left( {{a}_{1},\ldots ,{a}_{k}}\right) = \mathop{\prod }\limits_{p}{p}^{\min \left\{ {{v}_{p}\left( {a}_{1}\right) ,\ldots ,{v}_{p}\left( {a}_{k}\right) }\right\} }\]\n\nand\n\n\[\left\lbrack {{a}_{1},\ldots ,{a}_{k}}\right\rbrack = \mathop{\prod }\limits_{p}{p}^{\max \left\{ {{v}_{p}\left( {a}_{1}\right) ,\ldots ,{v}_{p}\left( {a}_{k}\right) }\right\} }.\]
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Proof. This follows immediately from the fundamental theorem of arithmetic. \( ▱ \)
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No
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Theorem 1.12 For every positive integer \( n \) and prime \( p \) , \[ {v}_{p}\left( {n!}\right) = \mathop{\sum }\limits_{{r = 1}}^{\left\lbrack \frac{\log n}{\log p}\right\rbrack }\left\lbrack \frac{n}{{p}^{r}}\right\rbrack . \]
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Proof. Let \( 1 \leq m \leq n \) . If \( {p}^{r} \) divides \( m \), then \( {p}^{r} \leq m \leq n \) and \( r \leq \) \( \log n/\log p \) . Since \( r \) is an integer, we have \( r \leq \left\lbrack {\log n/\log p}\right\rbrack \) and \[ {v}_{p}\left( m\right) = \mathop{\sum }\limits_{\substack{{r = 1} \\ {{p}^{r} \mid m} }}^{\left\lbrack \frac{\log n}{\log p}\right\rbrack }1 \] The number of positive integers not exceeding \( n \) that are divisible by \( {p}^{r} \) is exactly \( \left\lbrack {n/{p}^{r}}\right\rbrack \), and so \[ {v}_{p}\left( {n!}\right) = \mathop{\sum }\limits_{{m = 1}}^{n}{v}_{p}\left( m\right) = \mathop{\sum }\limits_{{m = 1}}^{n}\mathop{\sum }\limits_{\substack{{r = 1} \\ {{p}^{r} \mid m} }}^{\left\lbrack \frac{\log n}{\log p}\right\rbrack }1 \] \[ = \mathop{\sum }\limits_{{r = 1}}^{\left\lbrack \frac{\log n}{\log p}\right\rbrack }\mathop{\sum }\limits_{\substack{{m = 1} \\ {{p}^{r} \mid m} }}^{n}1 = \mathop{\sum }\limits_{{r = 1}}^{\left\lbrack \frac{\log n}{\log p}\right\rbrack }\left\lbrack \frac{n}{{p}^{r}}\right\rbrack . \] This completes the proof.
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Yes
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Theorem 1.13 Let \( m \) and a be nonzero integers. There exists a positive integer \( k \) such that \( m \) divides \( {a}^{k} \) if and only if \( \operatorname{rad}\left( m\right) \) divides \( \operatorname{rad}\left( a\right) \) .
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Proof. We know that \( m \) divides \( {a}^{k} \) if and only if \( {v}_{p}\left( m\right) \leq {v}_{p}\left( {a}^{k}\right) = \) \( k{v}_{p}\left( a\right) \) for every prime \( p \) (Exercise 14). If there exists an integer \( k \) such that \( m \) divides \( {a}^{k} \), then \( {v}_{p}\left( a\right) > 0 \) whenever \( {v}_{p}\left( m\right) > 0 \), and so every prime that divides \( m \) also divides \( a \) . This implies that \( \operatorname{rad}\left( m\right) \) divides \( \operatorname{rad}\left( a\right) \) .\n\nConversely, if \( \operatorname{rad}\left( m\right) \) divides \( \operatorname{rad}\left( a\right) \), then \( {v}_{p}\left( a\right) > 0 \) for every prime \( p \) such that \( {v}_{p}\left( m\right) > 0 \) . Since only finitely many primes divide \( m \), it follows that there exists a positive integer \( k \) such that \( {v}_{p}\left( {a}^{k}\right) = k{v}_{p}\left( a\right) \geq {v}_{p}\left( m\right) \) for all primes \( p \), and so \( m \) divides \( {a}^{k} \) . \( ▱ \)
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No
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Theorem 1.14 (Euclid’s theorem) There are infinitely many primes.
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Proof. Let \( {p}_{1},\ldots ,{p}_{n} \) be any finite set of prime numbers. Consider the integer\n\n\[ N = {p}_{1}\cdots {p}_{n} + 1 \]\n\nSince \( N > 1 \), it follows from the fundamental theorem of arithmetic that \( N \) is divisible by some prime \( p \) . If \( p = {p}_{i} \) for some \( i = 1,\ldots, n \), then \( p \) divides \( N - {p}_{1}\cdots {p}_{n} = 1 \), which is absurd. Therefore, \( p \neq {p}_{i} \) for all \( i = 1,\ldots, n \) . This means that, for any finite set of primes, there always exists a prime that does not belong to the set, and so the number of primes is infinite. \( ▱ \)
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Yes
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Theorem 1.15 Let \( {a}_{1},\ldots ,{a}_{k} \) be integers, not all zero. For any integer \( b \) , there exist integers \( {x}_{1},\ldots ,{x}_{k} \) such that\n\n\[ \n{a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k} = b \n\]\n\nif and only if \( b \) is a multiple of \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) . In particular, the linear equation (1.4) has a solution for every integer \( b \) if and only if the numbers \( {a}_{1},\ldots ,{a}_{k} \) are relatively prime.
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Proof. Let \( d = \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) . If equation (1.4) is solvable in integers \( {x}_{i} \) , then \( d \) divides \( b \) since \( d \) divides each integer \( {a}_{i} \) . Conversely, if \( d \) divides \( b \), then \( b = {dq} \) for some integer \( q \) . By Theorem 1.4, there exist integers \( {y}_{1},\ldots ,{y}_{k} \) such that\n\n\[ \n{a}_{1}{y}_{1} + \cdots + {a}_{k}{y}_{k} = d.\n\]\n\nLet \( {x}_{i} = {y}_{i}q \) for \( i = 1,\ldots, k \) . Then\n\n\[ \n{a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k} = {a}_{1}\left( {{y}_{1}q}\right) + \cdots + {a}_{k}\left( {{y}_{k}q}\right) = {dq} = b \n\]\n\nis a solution of (1.4). It follows that (1.4) is solvable in integers for every \( b \) if and only if \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) = 1 \) . \( ▱ \)
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Yes
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Theorem 1.16 Let \( {a}_{1},\ldots ,{a}_{k} \) be positive integers such that\n\n\[ \left( {{a}_{1},\ldots ,{a}_{k}}\right) = 1 \]\n\nIf\n\n\[ b \geq \left( {{a}_{k} - 1}\right) \mathop{\sum }\limits_{{i = 1}}^{{k - 1}}{a}_{i} \]\n\nthen there exist nonnegative integers \( {x}_{1},\ldots ,{x}_{k} \) such that\n\n\[ {a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k} = b. \]
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Proof. By Theorem 1.15, there exist integers \( {z}_{1},\ldots ,{z}_{k} \) such that\n\n\[ {a}_{1}{z}_{1} + \cdots + {a}_{k}{z}_{k} = b. \]\n\nUsing the division algorithm, we can divide each of the integers \( {z}_{1},\ldots ,{z}_{k - 1} \) by \( {a}_{k} \) so that\n\n\[ {z}_{i} = {a}_{k}{q}_{i} + {x}_{i} \]\n\nand\n\n\[ 0 \leq {x}_{i} \leq {a}_{k} - 1 \]\n\nfor \( i = 1,\ldots, k - 1 \) . Let\n\n\[ {x}_{k} = {z}_{k} + \mathop{\sum }\limits_{{i = 1}}^{{k - 1}}{a}_{i}{q}_{i} \]\n\nThen\n\n\[ b = {a}_{1}{z}_{1} + \cdots + {a}_{k - 1}{z}_{k - 1} + {a}_{k}{z}_{k} \]\n\n\[ = {a}_{1}\left( {{a}_{k}{q}_{1} + {x}_{1}}\right) + \cdots + {a}_{k - 1}\left( {{a}_{k}{q}_{k - 1} + {x}_{k - 1}}\right) + {a}_{k}{z}_{k} \]\n\n\[ = {a}_{1}{x}_{1} + \cdots + {a}_{k - 1}{x}_{k - 1} + {a}_{k}\left( {{z}_{k} + \mathop{\sum }\limits_{{i = 1}}^{{k - 1}}{a}_{i}{q}_{i}}\right) \]\n\n\[ = {a}_{1}{x}_{1} + \cdots + {a}_{k - 1}{x}_{k - 1} + {a}_{k}{x}_{k} \]\n\n\[ \leq \left( {{a}_{k} - 1}\right) \mathop{\sum }\limits_{{i = 1}}^{{k - 1}}{a}_{i} + {a}_{k}{x}_{k} \]\n\nwhere \( {x}_{k} \) is an integer, possibly negative. However, if\n\n\[ b \geq \left( {{a}_{k} - 1}\right) \mathop{\sum }\limits_{{i = 1}}^{{k - 1}}{a}_{i} \]\n\nthen \( {a}_{k}{x}_{k} \geq 0 \) and so \( {x}_{k} \geq 0 \) . This completes the proof.
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Yes
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Theorem 1.17 Let \( {a}_{1} \) and \( {a}_{2} \) be relatively prime positive integers. Then\n\n\[ G\left( {{a}_{1},{a}_{2}}\right) = \left( {{a}_{1} - 1}\right) \left( {{a}_{2} - 1}\right) . \]
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Proof. We saw in the proof of Theorem 1.15 that for every integer \( b \) there exist integers \( {x}_{1} \) and \( {x}_{2} \) such that\n\n\[ b = {a}_{1}{x}_{1} + {a}_{2}{x}_{2}\;\text{ and }\;0 \leq {x}_{1} \leq {a}_{2} - 1. \]\n\n(1.5)\n\nIf we have another representation\n\n\[ b = {a}_{1}{x}_{1}^{\prime } + {a}_{2}{x}_{2}^{\prime },\;\text{ and }\;0 \leq {x}_{1}^{\prime } \leq {a}_{2} - 1, \]\n\nthen\n\n\[ {a}_{1}\left( {{x}_{1} - {x}_{1}^{\prime }}\right) = {a}_{2}\left( {{x}_{2}^{\prime } - {x}_{2}}\right) \]\n\nSince \( {a}_{2} \) divides \( {a}_{1}\left( {{x}_{1} - {x}_{1}^{\prime }}\right) \) and \( \left( {{a}_{1},{a}_{2}}\right) = 1 \), Euclid’s lemma (Theorem 1.7) implies that \( {a}_{2} \) divides \( {x}_{1} - {x}_{1}^{\prime } \) . Then \( {x}_{1} = {x}_{1}^{\prime } \), since \( \left| {{x}_{1} - {x}_{1}^{\prime }}\right| \leq {a}_{2} - 1 \) . It follows that \( {x}_{2} = {x}_{2}^{\prime } \), and so the representation (1.5) is unique.\n\nIf the integer \( b \) cannot be represented as a nonnegative integral combination of \( {a}_{1} \) and \( {a}_{2} \), then we must have \( {x}_{1} \leq - 1 \) in the representation (1.5). This implies that\n\n\[ b = {a}_{1}{x}_{1} + {a}_{2}{x}_{2} \leq {a}_{1}\left( {{a}_{2} - 1}\right) + {a}_{2}\left( {-1}\right) = \left( {{a}_{1} - 1}\right) \left( {{a}_{2} - 1}\right) - 1, \]\n\nand so \( G\left( {{a}_{1},{a}_{2}}\right) \leq \left( {{a}_{1} - 1}\right) \left( {{a}_{2} - 1}\right) \) . On the other hand, since\n\n\[ {a}_{1}\left( {{a}_{2} - 1}\right) + {a}_{2}\left( {-1}\right) = {a}_{1}{a}_{2} - {a}_{1} - {a}_{2} < {a}_{1}{a}_{2}, \]\n\nit follows that if\n\n\[ {a}_{1}{a}_{2} - {a}_{1} - {a}_{2} = {a}_{1}{x}_{1} + {a}_{2}{x}_{2} \]\n\nfor any nonnegative integers \( {x}_{1} \) and \( {x}_{2} \), then \( 0 \leq {x}_{1} \leq {a}_{2} - 1 \) . By the uniqueness of the representation (1.5), we must have \( {x}_{1} = {a}_{2} - 1 \) and \( {x}_{2} = - 1 \) . Therefore, the integer \( {a}_{1}{a}_{2} - {a}_{1} - {a}_{2} \) cannot be represented as a nonnegative integral linear combination of \( {a}_{1} \) and \( {a}_{2} \), and so \( G\left( {{a}_{1},{a}_{2}}\right) = \) \( \left( {{a}_{1} - 1}\right) \left( {{a}_{2} - 1}\right) \) . \( ▱ \)
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Yes
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Theorem 2.2 Let \( m, a, b \) be integers with \( m \geq 1 \) . Let \( d = \left( {a, m}\right) \) be the greatest common divisor of \( a \) and \( m \) . The congruence\n\n\[ \n{ax} \equiv b\;\left( {\;\operatorname{mod}\;m}\right) \n\]\n\n(2.1)\n\nhas a solution if and only if\n\n\[ \nb \equiv 0\;\left( {\;\operatorname{mod}\;d}\right) \n\]\n\nIf \( b \equiv 0\;\left( {\;\operatorname{mod}\;d}\right) \), then the congruence (2.1) has exactly \( d \) solutions in integers that are pairwise incongruent modulo \( m \) . In particular, if \( \left( {a, m}\right) = 1 \) , then for every integer \( b \) the congruence (2.1) has a unique solution modulo \( m \) .
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Proof. Let \( d = \left( {a, m}\right) \) . Congruence (2.1) has a solution if and only if there exist integers \( x \) and \( y \) such that\n\n\[ \n{ax} - b = {my} \n\]\n\nor, equivalently,\n\n\[ \nb = {ax} - {my}. \n\]\n\nBy Theorem 1.15, this is possible if and only if \( b \equiv 0\left( {\;\operatorname{mod}\;d}\right) \) .\n\nIf \( x \) and \( {x}_{1} \) are solutions of (2.1), then\n\n\[ \na\left( {{x}_{1} - x}\right) \equiv a{x}_{1} - {ax} \equiv b - b \equiv 0\;\left( {\;\operatorname{mod}\;m}\right) , \n\]\n\nand so\n\n\[ \na\left( {{x}_{1} - x}\right) = {mz} \n\]\n\nfor some integer \( z \) . If \( d \) is the greatest common divisor of \( a \) and \( m \), then \( \left( {a/d, m/d}\right) = 1 \) and\n\n\[ \n\left( \frac{a}{d}\right) \left( {x - {x}_{1}}\right) = \left( \frac{m}{d}\right) z \n\]\n\nBy Euclid’s lemma (Theorem 1.7), \( m/d \) divides \( {x}_{1} - x \), and so\n\n\[ \n{x}_{1} = x + \frac{im}{d} \n\]\n\nfor some integer \( i \), that is,\n\n\[ \n{x}_{1} \equiv x\;\left( {\;\operatorname{mod}\;\frac{m}{d}}\right) \n\]\n\nMoreover, every integer \( {x}_{1} \) of this form is a solution of (2.1). An integer \( {x}_{1} \) congruent to \( x \) modulo \( m/d \) is congruent to \( x + {im}/d \) modulo \( m \) for some integer \( i = 0,1,\ldots d - 1 \), and the \( d \) integers \( x + {im}/d \) with \( i = 0,1,\ldots, d - 1 \) are pairwise incongruent modulo \( m \) . Thus, the congruence (2.1) has exactly \( d \) pairwise incongruent solutions. This completes the proof.
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Yes
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Theorem 2.3 If \( p \) is a prime, then \( \mathbf{Z}/p\mathbf{Z} \) is a field.
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Proof. If \( a + p\mathbf{Z} \in \mathbf{Z}/p\mathbf{Z} \) and \( a + p\mathbf{Z} \neq p\mathbf{Z} \), then \( a \) is an integer not divisible by \( p \) . By Theorem 2.2, there exists an integer \( x \) such that \( {ax} \equiv 1 \) \( \left( {\;\operatorname{mod}\;p}\right) \) . This implies that\n\n\[ \left( {a + p\mathbf{Z}}\right) \left( {x + p\mathbf{Z}}\right) = 1 + p\mathbf{Z} \]\n\nand so \( a + p\mathbf{Z} \) is invertible. Thus, every nonzero congruence class in \( \mathbf{Z}/p\mathbf{Z} \) is a unit and \( \mathbf{Z}/p\mathbf{Z} \) is a field.
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Yes
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Lemma 2.1 Let \( p \) be a prime number. Then \( {x}^{2} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \) if and only if \( x \equiv \pm 1\;\left( {\;\operatorname{mod}\;p}\right) \) .
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Proof. If \( x \equiv \pm 1\;\left( {\;\operatorname{mod}\;p}\right) \), then \( {x}^{2} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \) . Conversely, if \( {x}^{2} \equiv 1 \) \( \left( {\;\operatorname{mod}\;p}\right) \), then \( p \) divides \( {x}^{2} - 1 = \left( {x - 1}\right) \left( {x + 1}\right) \), and so \( p \) must divide \( x - 1 \) or \( x + 1 \) .
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Yes
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Theorem 2.4 (Wilson) If \( p \) is prime, then\n\n\[ \left( {p - 1}\right) ! \equiv - 1\;\left( {\;\operatorname{mod}\;p}\right) \]
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Proof. This is true for \( p = 2 \) and \( p = 3 \), since \( 1! \equiv - 1\left( {\;\operatorname{mod}\;2}\right) \) and \( 2! \equiv - 1\;\left( {\;\operatorname{mod}\;3}\right) \) . Let \( p \geq 5 \) . By Theorem 2.2, to each integer \( a \in \) \( \{ 1,2,\ldots, p - 1\} \) there is a unique integer \( {a}^{-1} \in \{ 1,2,\ldots, p - 1\} \) such that \( a{a}^{-1} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \) . By Lemma 2.1, \( a = {a}^{-1} \) if and only if \( a = 1 \) or \( a = \) \( p - 1 \) . Therefore, we can partition the \( p - 3 \) numbers in the set \( \{ 2,3,\ldots, p - 2\} \) into \( \left( {p - 3}\right) /2 \) pairs of integers \( \left\{ {{a}_{i},{a}_{i}^{-1}}\right\} \) such that \( {a}_{i}{a}_{i}^{-1} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \) for \( i = 1,\ldots ,\left( {p - 3}\right) /2 \) . Then\n\n\[ \left( {p - 1}\right) ! \equiv 1 \cdot 2 \cdot 3\cdots \left( {p - 2}\right) \left( {p - 1}\right) \]\n\n\[ \equiv \left( {p - 1}\right) \mathop{\prod }\limits_{{i = 1}}^{{\left( {p - 3}\right) /2}}{a}_{i}{a}_{i}^{-1} \]\n\n\[ \equiv \;p - 1 \]\n\n\[ \equiv - 1\;\left( {\;\operatorname{mod}\;p}\right) \]\n\nThis completes the proof.
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Yes
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Theorem 2.5 Let \( m \) and \( d \) be positive integers such that \( d \) divides \( m \) . If a is an integer relatively prime to \( d \), then there exists an integer \( {a}^{\prime } \) such that \( {a}^{\prime } \equiv a\;\left( {\;\operatorname{mod}\;d}\right) \) and \( {a}^{\prime } \) is relatively prime to \( m \) .
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Proof. Let \( m = \mathop{\prod }\limits_{{i = 1}}^{k}{p}_{i}^{{r}_{i}} \) and \( d = \mathop{\prod }\limits_{{i = 1}}^{k}{p}_{i}^{{s}_{i}} \), where \( {r}_{i} \geq 1 \) and \( 0 \leq {s}_{i} \leq {r}_{i} \) for \( i = 1,\ldots, k \) . Let \( {m}^{\prime } \) be the product of the prime powers that divide \( m \) but not \( d \) . Then\n\n\[ \n{m}^{\prime } = \mathop{\prod }\limits_{\substack{{i = 1} \\ {{s}_{i} = 0} }}^{k}{p}_{i}^{{r}_{i}} \n\]\n\nand\n\n\[ \n\left( {{m}^{\prime }, d}\right) = 1\text{.} \n\]\n\nBy Theorem 2.2, there exists an integer \( x \) such that\n\n\[ \n{dx} \equiv 1 - a\;\left( {\;\operatorname{mod}\;{m}^{\prime }}\right) . \n\]\n\nThen\n\n\[ \n{a}^{\prime } = a + {dx} \equiv 1\;\left( {\;\operatorname{mod}\;{m}^{\prime }}\right) \n\]\n\nand so\n\n\[ \n\left( {{a}^{\prime },{m}^{\prime }}\right) = 1\text{.} \n\]\n\nAlso,\n\n\[ \n{a}^{\prime } \equiv a\;\left( {\;\operatorname{mod}\;d}\right) \n\]\n\nIf \( \left( {{a}^{\prime }, m}\right) \neq 1 \), there exists a prime \( p \) that divides both \( {a}^{\prime } \) and \( m \) . However, \( p \) does not divide \( {m}^{\prime } \) since \( \left( {{a}^{\prime },{m}^{\prime }}\right) = 1 \) . It follows that \( p \) divides \( d \), and so \( p \) divides \( {a}^{\prime } - {dx} = a \), which is impossible since \( \left( {a, d}\right) = 1 \) . Therefore, \( \left( {{a}^{\prime }, m}\right) = 1 \) .
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Yes
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Theorem 2.6 Let \( m \) and \( n \) be relatively prime positive integers. For every integer \( c \) there exist unique integers \( a \) and \( b \) such that\n\n\[ 0 \leq a \leq n - 1 \]\n\n\[ 0 \leq b \leq m - 1 \]\n\nand\n\n\[ c \equiv {ma} + {nb}\;\left( {\;\operatorname{mod}\;{mn}}\right) .\n\]\n\nMoreover, \( \left( {c,{mn}}\right) = 1 \) if and only if \( \left( {a, n}\right) = \left( {b, m}\right) = 1 \) in the representation (2.4).
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Proof. If \( {a}_{1},{a}_{2},{b}_{1},{b}_{2} \) are integers such that\n\n\[ m{a}_{1} + n{b}_{1} \equiv m{a}_{2} + n{b}_{2}\;\left( {\;\operatorname{mod}\;{mn}}\right) ,\]\n\nthen\n\n\[ m{a}_{1} \equiv m{a}_{1} + n{b}_{1} \equiv m{a}_{2} + n{b}_{2} \equiv m{a}_{2}\;\left( {\;\operatorname{mod}\;n}\right) .\n\]\n\nSince \( \left( {m, n}\right) = 1 \), it follows that\n\n\[ {a}_{1} \equiv {a}_{2}\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nand so \( {a}_{1} = {a}_{2} \). Similarly, \( {b}_{1} = {b}_{2} \). It follows that the \( {mn} \) integers \( {ma} + {nb} \) are pairwise incongruent modulo \( {mn} \). Since there are exactly \( {mn} \) distinct congruence classes modulo \( {mn} \), the congruence (2.4) has a unique solution for every integer \( c \).\n\nLet \( c \equiv {ma} + {nb}\;\left( {{\;\operatorname{mod}\;m}n}\right) \). Since \( \left( {m, n}\right) = 1 \), we have\n\n\[ \left( {c, m}\right) = \left( {{ma} + {nb}, m}\right) = \left( {{nb}, m}\right) = \left( {b, m}\right) \]\n\nand\n\n\[ \left( {c, n}\right) = \left( {{ma} + {nb}, n}\right) = \left( {{ma}, n}\right) = \left( {a, n}\right) .\n\nIt follows that \( \left( {c,{mn}}\right) = 1 \) if and only if \( \left( {c, m}\right) = \left( {c, n}\right) = 1 \) if and only if \( \left( {b, m}\right) = \left( {a, n}\right) = 1 \). This completes the proof. \( ▱ \)
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Yes
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Theorem 2.7 The Euler phi function is multiplicative. Moreover,
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Proof. Let \( \\left( {m, n}\\right) = 1 \) . There are \( \\varphi \\left( {mn}\\right) \) congruence classes in the ring \( \\mathbf{Z}/{mn}\\mathbf{Z} \) that are relatively prime to \( {mn} \) . By Theorem 2.6, every congruence class modulo \( {mn} \) can be written uniquely in the form \( {ma} + {nb} + {mn}\\mathbf{Z} \) , where \( a \) and \( b \) are integers such that \( 0 \\leq a \\leq n - 1 \) and \( 0 \\leq b \\leq m - 1 \) . Moreover, the congruence class \( {ma} + {nb} + {mn}\\mathbf{Z} \) is prime to \( {mn} \) if and only if \( \\left( {b, m}\\right) = \\left( {a, n}\\right) = 1 \) . Since there are \( \\varphi \\left( n\\right) \) integers \( a \\in \\left\\lbrack {0, n - 1}\\right\\rbrack \) that are relatively prime to \( n \), and \( \\varphi \\left( m\\right) \) integers \( b \\in \\left\\lbrack {0, m - 1}\\right\\rbrack \) relatively prime to \( m \), it follows that \( \\varphi \\left( {mn}\\right) = \\varphi \\left( m\\right) \\varphi \\left( n\\right) \), and so the Euler phi function is multiplicative. If \( {m}_{1},\\ldots ,{m}_{k} \) are pairwise relatively prime positive integers, then \( \\varphi \\left( {{m}_{1}\\cdots {m}_{k}}\\right) = \\varphi \\left( {m}_{1}\\right) \\cdots \\varphi \\left( {m}_{k}\\right) \) . In particular, if \( m = {p}_{1}^{{r}_{1}}\\cdots {p}_{k}^{{r}_{k}} \) is the standard factorization of \( m \), where \( {p}_{1},\\ldots ,{p}_{k} \) are distinct primes and \( {r}_{1},\\ldots ,{r}_{k} \) are positive integers, then\n\n\[ \n\\varphi \\left( m\\right) = \\mathop{\\prod }\\limits_{{i = 1}}^{k}\\varphi \\left( {p}_{i}^{{r}_{i}}\\right) = \\mathop{\\prod }\\limits_{{i = 1}}^{k}{p}_{i}^{{r}_{i}}\\left( {1 - \\frac{1}{{p}_{i}}}\\right) = m\\mathop{\\prod }\\limits_{{p \\mid m}}\\left( {1 - \\frac{1}{p}}\\right) .\n\]\n\nThis completes the proof.
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Yes
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Theorem 2.8 For every positive integer \( m \) , \n\n\[ \n\mathop{\sum }\limits_{{d \mid m}}\varphi \left( d\right) = m \n\]
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Proof. We first consider the case where \( m = {p}^{t} \) is a power of a prime \( p \) . The divisors of \( {p}^{t} \) are \( 1, p,{p}^{2},\ldots ,{p}^{t} \), and \n\n\[ \n\mathop{\sum }\limits_{{d \mid {p}^{t}}}\varphi \left( d\right) = \mathop{\sum }\limits_{{r = 0}}^{t}\varphi \left( {p}^{r}\right) = 1 + \mathop{\sum }\limits_{{r = 1}}^{t}\left( {{p}^{r} - {p}^{r - 1}}\right) = {p}^{t}. \n\] \n\nNext we consider the general case where \( m \) has the standard factorization \n\n\[ \nm = {p}_{1}^{{t}_{1}}{p}_{2}^{{t}_{2}}\cdots {p}_{k}^{{t}_{k}} \n\] \n\nwhere \( {p}_{1},\ldots ,{p}_{k} \) are distinct prime numbers and \( {t}_{1},\ldots ,{t}_{k} \) are positive integers. Every divisor \( d \) of \( m \) is of the form \n\n\[ \nd = {p}_{1}^{{r}_{1}}{p}_{2}^{{r}_{2}}\cdots {p}_{k}^{{r}_{k}} \n\] \n\nwhere \( 0 \leq {r}_{i} \leq {t}_{i} \) for \( i = 1,\ldots, k \) . By Theorem 2.7, \( \varphi \left( d\right) \) is multiplicative, and so \n\n\[ \n\varphi \left( d\right) = \varphi \left( {p}_{1}^{{r}_{1}}\right) \varphi \left( {p}_{2}^{{r}_{2}}\right) \cdots \varphi \left( {p}_{k}^{{r}_{k}}\right) . \n\] \n\nTherefore, \n\n\[ \n\mathop{\sum }\limits_{{d \mid m}}\varphi \left( d\right) = \mathop{\sum }\limits_{{{r}_{1} = 0}}^{{t}_{1}}\cdots \mathop{\sum }\limits_{{{r}_{k} = 0}}^{{t}_{k}}\varphi \left( {{p}_{1}^{{r}_{1}}\cdots {p}_{k}^{{r}_{k}}}\right) \n\] \n\n\[ \n= \mathop{\sum }\limits_{{{r}_{1} = 0}}^{{t}_{1}}\cdots \mathop{\sum }\limits_{{{r}_{k} = 0}}^{{t}_{k}}\varphi \left( {p}_{1}^{{r}_{1}}\right) \varphi \left( {p}_{2}^{{r}_{2}}\right) \cdots \varphi \left( {p}_{k}^{{r}_{k}}\right) \n\] \n\n\[ \n= \mathop{\prod }\limits_{{i = 1}}^{k}\mathop{\sum }\limits_{{{r}_{i} = 0}}^{{t}_{i}}\varphi \left( {p}_{i}^{{r}_{i}}\right) \n\] \n\n\[ \n= \mathop{\prod }\limits_{{i = 1}}^{k}{p}_{i}^{{t}_{i}} \n\] \n\n\[ \n= m\text{.} \n\] \n\nThis completes the proof.
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Yes
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Theorem 2.9 Let \( m \) and \( n \) be positive integers. For any integers \( a \) and \( b \) there exists an integer \( x \) such that\n\n\[ x \equiv a\;\left( {\;\operatorname{mod}\;m}\right) \]\n\nand\n\n\[ x \equiv b\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nif and only if\n\n\[ a \equiv b\;\left( {\;\operatorname{mod}\;\left( {m, n}\right) }\right) .
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Proof. If \( x \) is a solution of congruence (2.5), then \( x = a + {mu} \) for some integer \( u \) . If \( x \) is also a solution of congruence (2.6), then\n\n\[ x = a + {mu} \equiv b\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nthat is,\n\n\[ a + {mu} = b + {nv} \]\n\nfor some integer \( v \) . It follows that\n\n\[ a - b = {nv} - {mu} \equiv 0\;\left( {\;\operatorname{mod}\;\left( {m, n}\right) }\right) .\n\nConversely, if \( a - b \equiv 0\;\left( {\;\operatorname{mod}\;\left( {m, n}\right) }\right) \), then by Theorem 1.15 there exist integers \( u \) and \( v \) such that\n\n\[ a - b = {nv} - {mu}.\n\nThen\n\n\[ x = a + {mu} = b + {nv} \]\n\nis a solution of the two congruences.\n\nAn integer \( y \) is another solution of the congruences if and only if\n\n\[ y \equiv a \equiv x\;\left( {\;\operatorname{mod}\;m}\right) \]\n\nand\n\n\[ y \equiv b \equiv x\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nthat is, if and only if \( x - y \) is a common multiple of \( m \) and \( n \), or, equivalently, \( x - y \) is divisible by the least common multiple \( \left\lbrack {m, n}\right\rbrack \) . This completes the proof. \( ▱ \)
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Yes
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Theorem 2.10 (Chinese remainder theorem) Let \( k \geq 2 \) . If \( {a}_{1},\ldots ,{a}_{k} \) are integers and \( {m}_{1},\ldots ,{m}_{k} \) are pairwise relatively prime positive integers, then there exists an integer \( x \) such that\n\n\[ \nx \equiv {a}_{i}\;\left( {\;\operatorname{mod}\;{m}_{i}}\right) \;\text{ for all }i = 1,\ldots, k.\n\]\n\nIf \( x \) is any solution of this set of congruences, then the integer \( y \) is also a solution if and only if\n\n\[ \nx \equiv y\;\left( {{\;\operatorname{mod}\;{m}_{1}}\cdots {m}_{k}}\right) \n\]
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Proof. We prove the theorem by induction on \( k \) . If \( k = 2 \), then \( \left\lbrack {{m}_{1},{m}_{2}}\right\rbrack = \) \( {m}_{1}{m}_{2} \), and this is a special case of Theorem 2.9.\n\nLet \( k \geq 3 \), and assume that the statement is true for \( k - 1 \) congruences. Then there exists an integer \( z \) such that \( z \equiv {a}_{i}\;\left( {\;\operatorname{mod}\;{m}_{i}}\right) \) for \( i = 1,\ldots, k - \) 1. Since \( {m}_{1},\ldots ,{m}_{k} \) are pairwise relatively prime integers, we have\n\n\[ \n\left( {{m}_{1}\cdots {m}_{k - 1},{m}_{k}}\right) = 1 \n\]\n\nand so, by the case \( k = 2 \), there exists an integer \( x \) such that\n\n\[ \nx \equiv z\;\left( {{\;\operatorname{mod}\;{m}_{1}}\cdots {m}_{k - 1}}\right) , \n\]\n\n\[ \nx \equiv {a}_{k}\;\left( {\;\operatorname{mod}\;{m}_{k}}\right) \n\]\n\nThen\n\n\[ \nx \equiv z \equiv {a}_{i}\;\left( {\;\operatorname{mod}\;{m}_{i}}\right) \n\]\n\nfor \( i = 1,\ldots, k - 1 \) .\n\nIf \( y \) is another solution of the system of \( k \) congruences, then \( x - y \) is divisible by \( {m}_{i} \) for all \( i = 1,\ldots, k \) . Since \( {m}_{1},\ldots ,{m}_{k} \) are pairwise relatively prime, it follows that \( x - y \) is divisible by \( {m}_{1}\cdots {m}_{k} \) . This completes the proof. \( ▱ \)
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Yes
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Theorem 2.11 Let\n\n\\[ \nm = {p}_{1}^{{r}_{1}}\\cdots {p}_{k}^{{r}_{k}}\n\\]\n\nbe the standard factorization of the positive integer \\( m \\) . Let \\( f\\left( x\\right) \\) be a polynomial with integral coefficients. The congruence\n\n\\[ \nf\\left( x\\right) \\equiv 0\\;\\left( {\\;\\operatorname{mod}\\;m}\\right)\n\\]\n\nis solvable if and only if the congruences\n\n\\[ \nf\\left( x\\right) \\equiv 0\\;\\left( {\\;\\operatorname{mod}\\;{p}_{i}^{{r}_{i}}}\\right)\n\\]\n\nare solvable for all \\( i = 1,\\ldots, k \\) .
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Proof. If \\( f\\left( x\\right) \\equiv 0\\;\\left( {\\;\\operatorname{mod}\\;m}\\right) \\) has a solution in integers, then there exists an integer \\( a \\) such that \\( m \\) divides \\( f\\left( a\\right) \\) . Since \\( {p}_{i}^{{r}_{i}} \\) divides \\( m \\), it follows that \\( {p}_{i}^{{r}_{i}} \\) divides \\( f\\left( a\\right) \\), and so the congruences \\( f\\left( x\\right) \\equiv 0\\;\\left( {\\;\\operatorname{mod}\\;{p}_{i}^{{r}_{i}}}\\right) \\) are solvable for \\( i = 1,\\ldots, k \\) .\n\nConversely, suppose that the congruences \\( f\\left( x\\right) \\equiv 0\\;\\left( {\\;\\operatorname{mod}\\;{p}_{i}^{{r}_{i}}}\\right) \\) are solvable for \\( i = 1,\\ldots, k \\) . Then for each \\( i \\) there exists an integer \\( {a}_{i} \\) such that\n\n\\[ \nf\\left( {a}_{i}\\right) \\equiv 0\\;\\left( {\\;\\operatorname{mod}\\;{p}_{i}^{{r}_{i}}}\\right) .\n\\]\n\nSince the prime powers \\( {p}_{1}^{{r}_{1}},\\ldots ,{p}_{k}^{{r}_{k}} \\) are pairwise relatively prime, the Chinese remainder theorem tells us that there exists an integer \\( a \\) such that\n\n\\[ \na \\equiv {a}_{i}\\;\\left( {\\;\\operatorname{mod}\\;{p}_{i}^{{r}_{i}}}\\right)\n\\]\n\nfor all \\( i \\) . Then\n\n\\[ \nf\\left( a\\right) \\equiv f\\left( {a}_{i}\\right) \\equiv 0\\;\\left( {\\;\\operatorname{mod}\\;{p}_{i}^{{r}_{i}}}\\right)\n\\]\n\nfor all \\( i \\) . Since \\( f\\left( a\\right) \\) is divisible by each of the prime powers \\( {p}_{i}^{{r}_{i}} \\), it is also divisible by their product \\( m \\), and so \\( f\\left( a\\right) \\equiv 0\\left( {\\;\\operatorname{mod}\\;m}\\right) \\) . This completes the proof. \\( ▱ \\)
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Yes
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Theorem 2.12 (Euler) Let \( m \) be a positive integer, and let a be an integer relatively prime to \( m \) . Then\n\n\[ \n{a}^{\varphi \left( m\right) } \equiv 1\;\left( {\;\operatorname{mod}\;m}\right) \n\]
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Proof. Let \( \left\{ {{r}_{1},\ldots ,{r}_{\varphi \left( m\right) }}\right\} \) be a reduced set of residues modulo \( m \) . Since \( \left( {a, m}\right) = 1 \), we have \( \left( {a{r}_{i}, m}\right) = 1 \) for \( i = 1,\ldots ,\varphi \left( m\right) \) . Consequently, for every \( i \in \{ 1,\ldots ,\varphi \left( m\right) \} \) there exists \( \sigma \left( i\right) \in \{ 1,\ldots ,\varphi \left( m\right) \} \) such that\n\n\[ \na{r}_{i} \equiv {r}_{\sigma \left( i\right) }\;\left( {\;\operatorname{mod}\;m}\right) \n\]\n\nMoreover, \( a{r}_{i} \equiv a{r}_{j}\;\left( {\;\operatorname{mod}\;m}\right) \) if and only if \( i = j \), and so \( \sigma \) is a permutation of the set \( \{ 1,\ldots ,\varphi \left( m\right) \} \) and \( \left\{ {a{r}_{1},\ldots, a{r}_{\varphi \left( m\right) }}\right\} \) is also a reduced set of residues modulo \( m \) . It follows that\n\n\[ \n{a}^{\varphi \left( m\right) }{r}_{1}{r}_{2}\cdots {r}_{\varphi \left( m\right) }\; \equiv \;\left( {a{r}_{1}}\right) \left( {a{r}_{2}}\right) \cdots \left( {a{r}_{\varphi \left( m\right) }}\right) {\;(\operatorname{mod}\;m)} \n\]\n\n\[ \n\equiv \;{r}_{\sigma \left( 1\right) }{r}_{\sigma \left( 2\right) }\cdots {r}_{\sigma \left( {\varphi \left( m\right) }\right) }{\;(\operatorname{mod}\;m)} \n\]\n\n\[ \n\equiv \;{r}_{1}{r}_{2}\cdots {r}_{\varphi \left( m\right) }{\;(\operatorname{mod}\;m)}. \n\]\n\nDividing by \( {r}_{1}{r}_{2}\cdots {r}_{\varphi \left( m\right) } \), we obtain\n\n\[ \n{a}^{\varphi \left( m\right) } \equiv 1\;\left( {\;\operatorname{mod}\;m}\right) \n\]\n\nThis completes the proof.
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Yes
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Theorem 2.13 (Fermat) Let \( p \) be a prime number. If the integer \( a \) is not divisible by \( p \), then\n\n\[ \n{a}^{p - 1} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\nMoreover,\n\n\[ \n{a}^{p} \equiv a\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\nfor every integer \( a \) .
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Proof. If \( p \) is prime and does not divide \( a \), then \( \left( {a, p}\right) = 1,\varphi \left( p\right) = p - 1 \) , and\n\n\[ \n{a}^{p - 1} = {a}^{\varphi \left( p\right) } \equiv 1\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\nby Euler’s theorem. Multiplying this congruence by \( a \), we obtain\n\n\[ \n{a}^{p} \equiv a\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\nIf \( p \) divides \( a \), then this congruence also holds for \( a \) .
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Yes
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Theorem 2.14 Let \( m \) be a positive integer and a an integer relatively prime to \( m \) . If \( d \) is the order of a modulo \( m \), then \( {a}^{k} \equiv {a}^{\ell }\;\left( {\;\operatorname{mod}\;m}\right) \) if and only if \( k \equiv \ell \;\left( {\;\operatorname{mod}\;d}\right) \) . In particular, \( {a}^{n} \equiv 1\;\left( {\;\operatorname{mod}\;m}\right) \) if and only if \( d \) divides \( n \), and so \( d \) divides \( \varphi \left( m\right) \) .
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Proof. Since \( a \) has order \( d \) modulo \( m \), we have \( {a}^{d} \equiv 1\left( {\;\operatorname{mod}\;m}\right) \) . If \( k \equiv \ell \;\left( {\;\operatorname{mod}\;d}\right) \), then \( k = \ell + {dq} \), and so\n\n\[ \n{a}^{k} = {a}^{\ell + {dq}} = {a}^{\ell }{\left( {a}^{d}\right) }^{q} \equiv {a}^{\ell }\;\left( {\;\operatorname{mod}\;m}\right) .\n\]\n\nConversely, suppose that \( {a}^{k} \equiv {a}^{\ell }\;\left( {\;\operatorname{mod}\;m}\right) \) . By the division algorithm, there exist integers \( q \) and \( r \) such that\n\n\[ \nk - \ell = {dq} + r\;\text{ and }\;0 \leq r \leq d - 1.\n\]\n\nThen\n\n\[ \n{a}^{k} = {a}^{\ell + {dq} + r} = {a}^{\ell }{\left( {a}^{d}\right) }^{q}{a}^{r} \equiv {a}^{k}{a}^{r}\;\left( {\;\operatorname{mod}\;m}\right) .\n\]\n\nSince \( \left( {{a}^{k}, m}\right) = 1 \), we can divide this congruence by \( {a}^{k} \) and obtain\n\n\[ \n{a}^{r} \equiv 1\;\left( {\;\operatorname{mod}\;m}\right)\n\]\nSince \( 0 \leq r \leq d - 1 \), and \( d \) is the order of \( a \) modulo \( m \), it follows that \( r = 0 \) , and so \( k \equiv \ell \;\left( {\;\operatorname{mod}\;d}\right) \) .\n\nIf \( {a}^{n} \equiv 1 \equiv {a}^{0}\left( {\;\operatorname{mod}\;m}\right) \), then \( d \) divides \( n \) . In particular, \( d \) divides \( \varphi \left( m\right) \), since \( {a}^{\varphi \left( m\right) } \equiv 1\;\left( {\;\operatorname{mod}\;m}\right) \) by Euler’s theorem. \( ▱ \)
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Yes
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Theorem 2.15 (Lagrange’s theorem) If \( G \) is a finite group and \( H \) is a subgroup of \( G \), then the order of \( H \) divides the order of \( G \) .
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Proof. Let \( G \) be a group, written multiplicatively, and let \( X \) be a nonempty subset of \( G \) . For every \( a \in G \) we define the set\n\n\[ \n{aX} = \{ {ax} : x \in X\} .\n\]\n\nThe map \( f : X \rightarrow {aX} \) defined by \( f\left( x\right) = {ax} \) is a bijection, and so \( \left| X\right| = \) \( \left| {aX}\right| \) for all \( a \in G \) . If \( H \) is a subgroup of \( G \), then \( {aH} \) is called a coset of \( H \) . Let \( {aH} \) and \( {bH} \) be cosets of the subgroup \( H \) . If \( {aH} \cap {bH} \neq \varnothing \) , then there exist \( x, y \in H \) such that \( {ax} = {by} \), or, since \( H \) is a subgroup, \( b = {ax}{y}^{-1} = {az} \), where \( z = x{y}^{-1} \in H \) . Then \( {bh} = {azh} \in {aH} \) for all \( h \in H \) , and so \( {bH} \subseteq {aH} \) . By symmetry, \( {aH} \subseteq {bH} \), and so \( {aH} = {bH} \) . Therefore, cosets of a subgroup \( H \) are either disjoint or equal. Since every element of \( G \) belongs to some coset of \( H \) (for example, \( a \in {aH} \) for all \( a \in G \) ), it follows that the cosets of \( H \) partition \( G \) . We denote the set of cosets by \( G/H \) . If \( G \) is a finite group, then \( H \) and \( G/H \) are finite, and\n\n\[ \n\left| G\right| = \left| H\right| \left| {G/H}\right| \n\]\n\nIn particular, we see that \( \left| H\right| \) divides \( \left| G\right| \) . \( ▱ \)
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Yes
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Theorem 2.16 Let \( G \) be a finite group, and \( a \in G \) . Then the order of the element a divides the order of the group \( G \) .
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Proof. This follows immediately from Theorem 2.15, since the order of \( a \) is the order of the cyclic subgroup that \( a \) generates. \( ▱ \)
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Yes
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Theorem 2.17 Let \( G \) be a cyclic group of order \( m \), and let \( H \) be a subgroup of \( G \). If \( a \) is a generator of \( G \), then there exists a unique divisor \( d \) of \( m \) such that \( H \) is the cyclic subgroup generated by \( {a}^{d} \), and \( H \) has order \( m/d \).
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Proof. Let \( S \) be the set of all integers \( u \) such that \( {a}^{u} \in H \). If \( u, v \in S \), then \( {a}^{u},{a}^{v} \in H \). Since \( H \) is a subgroup, it follows that \( {a}^{u}{a}^{v} = {a}^{u + v} \in H \) and \( {a}^{u}{\left( {a}^{v}\right) }^{-1} = {a}^{u - v} \in H \). Therefore, \( u \pm v \in S \), and \( S \) is a subgroup of \( \mathbf{Z} \). By Theorem 1.3, there is a unique nonnegative integer \( d \) such that \( S = d\mathbf{Z} \), and so \( H \) is the cyclic subgroup generated by \( {a}^{d} \). Since \( {a}^{m} = 1 \in H \), we have \( m \in S \), and so \( d \) is a positive divisor of \( m \). It follows that \( H \) has order \( m/d \). \( ▱ \)
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Yes
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Theorem 2.18 Let \( G \) be a cyclic group of order \( m \), and let \( a \) be a generator of \( G \) . For every integer \( k \), the cyclic subgroup generated by \( {a}^{k} \) has order \( m/d \), where \( d = \left( {m, k}\right) \), and \( \left\langle {a}^{k}\right\rangle = \left\langle {a}^{d}\right\rangle \) . In particular, \( G \) has exactly \( \varphi \left( m\right) \) generators.
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Proof. Since \( d = \left( {k, m}\right) \), there exist integers \( x \) and \( y \) such that \( d = \) \( {kx} + {my} \) . Then\n\n\[ \n{a}^{d} = {a}^{{kx} + {my}} = {\left( {a}^{k}\right) }^{x}{\left( {a}^{m}\right) }^{y} = {\left( {a}^{k}\right) }^{x}, \n\]\n\nand so \( {a}^{d} \in \left\langle {a}^{k}\right\rangle \) and \( \left\langle {a}^{d}\right\rangle \subseteq \left\langle {a}^{k}\right\rangle \) . Since \( d \) divides \( k \), there exists an integer \( z \) such that \( k = {dz} \) . Then\n\n\[ \n{a}^{k} = {\left( {a}^{d}\right) }^{z} \n\]\n\nand so \( {a}^{k} \in \left\langle {a}^{d}\right\rangle \) and \( \left\langle {a}^{k}\right\rangle \subseteq \left\langle {a}^{d}\right\rangle \) . Therefore, \( \left\langle {a}^{k}\right\rangle = \left\langle {a}^{d}\right\rangle \) and \( {a}^{k} \) has order \( m/d \) . In particular, \( {a}^{k} \) generates \( G \) if and only if \( d = 1 \) if and only if \( \left( {m, k}\right) = 1 \), and so \( G \) has exactly \( \varphi \left( m\right) \) generators. This completes the proof. \( ▱ \)
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Yes
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Theorem 2.19 Let \( m \) be an integer that is the product of two prime numbers. The prime divisors of \( m \) are the roots of the quadratic equation\n\n\[ {x}^{2} - \left( {m + 1 - \varphi \left( m\right) }\right) x + m = 0, \]\n\nand so \( \varphi \left( m\right) \) determines the prime factors of \( m \) .
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Proof. If \( m = {pq} \), then\n\n\[ \varphi \left( m\right) = \left( {p - 1}\right) \left( {q - 1}\right) = {pq} - p - q + 1 = m - p - \frac{m}{p} + 1, \]\n\nand so\n\n\[ p - \left( {m + 1 - \varphi \left( m\right) }\right) + \frac{m}{p} = 0. \]\n\nEquivalently, \( p \) and \( q \) are the solutions of the quadratic equation\n\n\[ {x}^{2} - \left( {m + 1 - \varphi \left( m\right) }\right) x + m = 0. \]\n\nThis completes the proof.
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Yes
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Theorem 3.1 (Division algorithm for polynomials) Let \( F \) be a field. If \( f\left( x\right) \) and \( d\left( x\right) \) are polynomials in \( F\left\lbrack x\right\rbrack \) and if \( d\left( x\right) \neq 0 \), then there exist unique polynomials \( q\left( x\right) \) and \( r\left( x\right) \) such that \( f\left( x\right) = d\left( x\right) q\left( x\right) + r\left( x\right) \) and either \( r\left( x\right) = 0 \) or the degree of \( r\left( x\right) \) is strictly smaller than the degree of \( d\left( x\right) \) .
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Proof. Let \( d\left( x\right) = {b}_{m}{x}^{m} + \cdots + {b}_{1}x + {b}_{0} \), where \( {b}_{m} \neq 0 \) and \( \deg \left( d\right) = \) \( m \) . If \( d\left( x\right) \) does not divide \( f\left( x\right) \), then \( f - {dq} \neq 0 \) and \( \deg \left( {f - {dq}}\right) \) is a nonnegative integer for every polynomial \( q\left( x\right) \in F\left\lbrack x\right\rbrack \) . Choose \( q\left( x\right) \) such that \( \ell = \deg \left( {f - {dq}}\right) \) is minimal, and let\n\n\[ r\left( x\right) = f\left( x\right) - d\left( x\right) q\left( x\right) = {c}_{\ell }{x}^{\ell } + \cdots + {c}_{1}x + {c}_{0} \in F\left\lbrack x\right\rbrack ,\]\n\nwhere \( {c}_{\ell } \neq 0 \) . We shall prove that \( \ell < m \) .\n\nSince \( F \) is a field, \( {b}_{m}^{-1} \in F \) . If \( \ell \geq m \), then\n\n\[ d\left( x\right) {b}_{m}^{-1}{c}_{\ell }{x}^{\ell - m} \]\n\nis a polynomial of degree \( \ell \) with leading coefficient \( {c}_{\ell } \) . Then\n\n\[ Q\left( x\right) = q\left( x\right) + {b}_{m}^{-1}{c}_{\ell }{x}^{\ell - m} \in F\left\lbrack x\right\rbrack ,\]\nand\n\n\[ R\left( x\right) = f\left( x\right) - d\left( x\right) Q\left( x\right) \]\n\n\[ = f\left( x\right) - d\left( x\right) \left( {q\left( x\right) + {b}_{m}^{-1}{c}_{\ell }{x}^{\ell - m}}\right) \]\n\n\[ = r\left( x\right) - d\left( x\right) {b}_{m}^{-1}{c}_{\ell }{x}^{\ell - m} \]\n\nis a polynomial of degree at most \( \ell - 1 \) . This contradicts the minimality of \( \ell \), and so \( \ell < m \) .\n\nNext we prove that the polynomials \( q\left( x\right) \) and \( r\left( x\right) \) are unique. Suppose that\n\n\[ f\left( x\right) = d\left( x\right) {q}_{1}\left( x\right) + {r}_{1}\left( x\right) = d\left( x\right) {q}_{2}\left( x\right) + {r}_{2}\left( x\right) ,\]\n\nwhere \( {q}_{1}\left( x\right) ,{q}_{2}\left( x\right) ,{r}_{1}\left( x\right) ,{r}_{2}\left( x\right) \) are polynomials in \( F\left\lbrack x\right\rbrack \) such that \( {r}_{i}\left( x\right) = 0 \) or \( \deg \left( {r}_{i}\right) < \deg \left( d\right) \) for \( i = 1,2 \) . Then\n\n\[ d\left( x\right) \left( {{q}_{1}\left( x\right) - {q}_{2}\left( x\right) }\right) = {r}_{2}\left( x\right) - {r}_{1}\left( x\right) .\n\nIf \( {q}_{1}\left( x\right) \neq {q}_{2}\left( x\right) \), then\n\n\[ \deg \left( d\right) \leq \deg \left( {d\left( {{q}_{1} - {q}_{2}}\right) }\right) = \deg \left( {{r}_{2} - {r}_{1}}\right) < \deg \left( d\right) \]\n\nwhich is absurd. Therefore, \( {q}_{1}\left( x\right) = {q}_{2}\left( x\right) \), and so \( {r}_{1}\left( x\right) = {r}_{2}\left( x\right) \) . This completes the proof. \( ▱ \)
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Yes
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Theorem 3.2 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack, f\left( x\right) \neq 0 \), and let \( {N}_{0}\left( f\right) \) denote the number of distinct zeros of \( f\left( x\right) \) in \( F \) . Then \( {N}_{0}\left( f\right) \) does not exceed the degree of \( f\left( x\right) \), that is,\n\n\[ \n{N}_{0}\left( f\right) \leq \deg \left( f\right) \n\]
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Proof. We use the division algorithm for polynomials. Let \( \alpha \in F \) . Dividing \( f\left( x\right) \) by \( x - \alpha \), we obtain\n\n\[ \nf\left( x\right) = \left( {x - \alpha }\right) q\left( x\right) + r\left( x\right) \n\]\n\nwhere \( r\left( x\right) = 0 \) or \( \deg \left( r\right) < \deg \left( {x - \alpha }\right) = 1 \), that is, \( r\left( x\right) = {r}_{0} \) is a constant. Letting \( x = \alpha \), we see that \( {r}_{0} = f\left( \alpha \right) \), and so\n\n\[ \nf\left( x\right) = \left( {x - \alpha }\right) q\left( x\right) + f\left( \alpha \right) \n\]\n\nfor every \( \alpha \in F \) . In particular, if \( \alpha \) is a zero of \( f\left( x\right) \), then \( x - \alpha \) divides \( f\left( x\right) \) .\n\nWe prove the theorem by induction on \( n = \deg \left( f\right) \) . If \( n = 0 \), then \( f\left( x\right) \) is a nonzero constant and \( {N}_{0}\left( f\right) = 0 \) . If \( n = 1 \), then \( f\left( x\right) = {a}_{0} + {a}_{1}x \) with \( {a}_{1} \neq 0 \), and \( {N}_{0}\left( f\right) = 1 \) since \( f\left( x\right) \) has the unique zero \( \alpha = - {a}_{1}^{-1}{a}_{0} \) . Suppose that \( n \geq 2 \) and the theorem is true for all polynomials of degree\n\nat most \( n - 1 \) . If \( {N}_{0}\left( f\right) = 0 \), we are done. If \( {N}_{0}\left( f\right) \geq 1 \), let \( \alpha \in F \) be a zero of \( f\left( x\right) \) . Then\n\n\[ \nf\left( x\right) = \left( {x - \alpha }\right) q\left( x\right) \n\]\n\nand\n\n\[ \n\deg \left( q\right) = n - 1 \n\]\n\nIf \( \beta \) is a zero of \( f\left( x\right) \) and \( \beta \neq \alpha \), then\n\n\[ \n0 = f\left( \beta \right) = \left( {\beta - \alpha }\right) q\left( \beta \right) \n\]\n\nand so \( \beta \) is a zero of \( q\left( x\right) \) . Since \( \deg \left( q\right) = n - 1 \), the induction hypothesis implies that\n\n\[ \n{N}_{0}\left( f\right) \leq 1 + {N}_{0}\left( q\right) \leq 1 + \deg \left( q\right) = n. \n\]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 3.3 Let \( G \) be a finite subgroup of the multiplicative group of a field. Then \( G \) is cyclic.
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Proof. Let \( \left| G\right| = m \) . By Theorem 2.15, if \( a \in G \), then the order of \( a \) is a divisor of \( m \) . For every divisor \( d \) of \( m \), let \( \psi \left( d\right) \) denote the number of elements of \( G \) of order \( d \) . If \( \psi \left( d\right) \neq 0 \), then there exists an element \( a \) of order \( d \), and every element of the cyclic subgroup \( \langle a\rangle \) generated by \( a \) satisfies \( {a}^{d} = 1 \) . By Theorem 3.2, the polynomial \( f\left( x\right) = {x}^{d} - 1 \in F\left\lbrack x\right\rbrack \) has at most \( d \) zeros, and so every zero of \( f\left( x\right) \) belongs to the cyclic subgroup \( \langle a\rangle \) . In particular, every element of \( G \) of order \( d \) must belong to \( \langle a\rangle \) . By Theorem 2.18, a cyclic group of order \( d \) has exactly \( \varphi \left( d\right) \) generators, where \( \varphi \left( d\right) \) is the Euler phi function. Therefore, \( \psi \left( d\right) = 0 \) or \( \psi \left( d\right) = \varphi \left( d\right) \) for every divisor \( d \) of \( m \) . Since every element of \( G \) has order \( d \) for some divisor \( d \) of \( m \), it follows that\n\n\[\n\mathop{\sum }\limits_{{d \mid m}}\psi \left( d\right) = m\n\]\n\nBy Theorem 2.8,\n\n\[\n\mathop{\sum }\limits_{{d \mid m}}\varphi \left( d\right) = m\n\]\n\nand so \( \psi \left( d\right) = \varphi \left( d\right) \) for every divisor \( d \) of \( m \) . In particular, \( \psi \left( m\right) = \varphi \left( m\right) \geq \) 1, and so \( G \) is a cyclic group of order \( m \) . \( ▱ \)
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Yes
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For every prime \( p \), the multiplicative group of the finite field \( \mathbf{Z}/p\mathbf{Z} \) is cyclic. This group has \( \varphi \left( {p - 1}\right) \) generators. Equivalently, for every prime \( p \), there exist \( \varphi \left( {p - 1}\right) \) pairwise incongruent primitive roots modulo \( p \) .
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This follows immediately from Theorem 3.3, since \( \left| {\left( \mathbf{Z}/p\mathbf{Z}\right) }^{ \times }\right| = \) \( p - 1 \) . \( ▱ \)
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No
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Theorem 3.5 Let \( m \) be a positive integer that is not a power of 2 . If \( m \) has a primitive root, then \( m = {p}^{k} \) or \( 2{p}^{k} \), where \( p \) is an odd prime and \( k \) is a positive integer.
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Proof. Let \( a \) and \( m \) be integers such that \( \left( {a, m}\right) = 1 \) and \( m \geq 3 \) . Suppose that\n\n\[ m = {m}_{1}{m}_{2},\;\text{ where }\left( {{m}_{1},{m}_{2}}\right) = 1\text{ and }{m}_{1} \geq 3,{m}_{2} \geq 3. \]\n\n(3.2)\n\nThen \( \left( {a,{m}_{1}}\right) = \left( {a,{m}_{2}}\right) = 1 \) . The Euler phi function \( \varphi \left( m\right) \) is even for \( m \geq 3 \) (Exercise 4 in Section 2.2). Let\n\n\[ n = \frac{\varphi \left( m\right) }{2} = \frac{\varphi \left( {m}_{1}\right) \varphi \left( {m}_{2}\right) }{2}. \]\n\nBy Euler's theorem,\n\n\[ {a}^{\varphi \left( {m}_{1}\right) } \equiv 1\;\left( {\;\operatorname{mod}\;{m}_{1}}\right) \]\n\nand so\n\n\[ {a}^{n} = {\left( {a}^{\varphi \left( {m}_{1}\right) }\right) }^{\varphi \left( {m}_{2}\right) /2} \equiv 1\;\left( {\;\operatorname{mod}\;{m}_{1}}\right) . \]\n\nSimilarly,\n\n\[ {a}^{n} = {\left( {a}^{\varphi \left( {m}_{2}\right) }\right) }^{\varphi \left( {m}_{1}\right) /2} \equiv 1\;\left( {\;\operatorname{mod}\;{m}_{2}}\right) . \]\n\nSince \( \left( {{m}_{1},{m}_{2}}\right) = 1 \) and \( m = {m}_{1}{m}_{2} \), we have\n\n\[ {a}^{n} \equiv 1\;\left( {\;\operatorname{mod}\;m}\right) \]\n\nand so the order of \( a \) modulo \( m \) is strictly smaller than \( \varphi \left( m\right) \) . Consequently, if we can factor \( m \) in the form (3.2), then there does not exist a primitive root modulo \( m \) . In particular, if \( m \) is divisible by two distinct odd primes, then \( m \) does not have a primitive root. Similarly, if \( m = {2}^{\ell }{p}^{k} \), where \( \ell \geq 2 \) , then \( m \) does not have a primitive root. Therefore, the only moduli \( m \neq {2}^{\ell } \) for which primitive roots can exist are of the form \( m = {p}^{k} \) or \( m = 2{p}^{k} \) for some odd prime \( p \) .
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No
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Theorem 3.6 Let \( p \) be an odd prime, and let \( a \neq \pm 1 \) be an integer not divisible by \( p \) . Let \( d \) be the order of a modulo \( p \) . Let \( {k}_{0} \) be the largest integer such that \( {a}^{d} \equiv 1\left( {\;\operatorname{mod}\;{p}^{{k}_{0}}}\right) \) . Then the order of a modulo \( {p}^{k} \) is \( d \) for \( k = 1,\ldots ,{k}_{0} \) and \( d{p}^{k - {k}_{0}} \) for \( k \geq {k}_{0} \) .
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Proof. There exists an integer \( {u}_{0} \) such that\n\n\[ \n{a}^{d} = 1 + {p}^{{k}_{0}}{u}_{0}\;\text{ and }\;\left( {{u}_{0}, p}\right) = 1.\n\]\n\n(3.3)\n\nLet \( 1 \leq k \leq {k}_{0} \), and let \( e \) be the order of \( a \) modulo \( {p}^{k} \) . If \( {a}^{e} \equiv 1\;\left( {\;\operatorname{mod}\;{p}^{k}}\right) \) , then \( {a}^{e} \equiv 1\left( {\;\operatorname{mod}\;p}\right) \), and so \( d \) divides \( e \) . By (3.3), we have \( {a}^{d} \equiv 1 \) \( \left( {\;\operatorname{mod}\;{p}^{k}}\right) \), and so \( e \) divides \( d \) . It follows that \( e = d \) .\n\nLet \( j \geq 0 \) . We shall show that there exists an integer \( {u}_{j} \) such that\n\n\[ \n{a}^{d{p}^{j}} = 1 + {p}^{j + {k}_{0}}{u}_{j}\;\text{ and }\;\left( {{u}_{j}, p}\right) = 1.\n\]\n\n(3.4)\n\nThe proof is by induction on \( j \) . The assertion is true for \( j = 0 \) by (3.3). Suppose we have (3.4) for some integer \( j \geq 0 \) . By the binomial theorem, there exists an integer \( {v}_{j} \) such that\n\n\[ \n{a}^{d{p}^{j + 1}} = {\left( 1 + {p}^{j + {k}_{0}}{u}_{j}\right) }^{p}\n\]\n\n\[ \n= 1 + {p}^{j + 1 + {k}_{0}}{u}_{j} + \mathop{\sum }\limits_{{i = 2}}^{p}\left( \begin{matrix} p \\ i \end{matrix}\right) {p}^{i\left( {j + {k}_{0}}\right) }{u}_{j}^{i}\n\]\n\n\[ \n= 1 + {p}^{j + 1 + {k}_{0}}{u}_{j} + {p}^{j + 2 + {k}_{0}}{v}_{j}\n\]\n\n\[ \n= 1 + {p}^{j + 1 + {k}_{0}}\left( {{u}_{j} + p{v}_{j}}\right)\n\]\n\n\[ \n= 1 + {p}^{j + 1 + {k}_{0}}{u}_{j + 1},\n\]\n\nand the integer \( {u}_{j + 1} = {u}_{j} + p{v}_{j} \) is relatively prime to \( p \) . Thus,(3.4) holds for all \( j \geq 0 \) .\n\nLet \( k \geq {k}_{0} + 1 \) and \( j = k - {k}_{0} \geq 1 \) . Suppose that the order of \( a \) modulo \( {p}^{k - 1} \) is \( d{p}^{j - 1} \) . Let \( {e}_{k} \) denote the order of \( a \) modulo \( {p}^{k} \) . The congruence\n\n\[ \n{a}^{{e}_{k}} \equiv 1\;\left( {\;\operatorname{mod}\;{p}^{k}}\right)\n\]\n\nimplies that\n\n\[ \n{a}^{{e}_{k}} \equiv 1\;\left( {\;\operatorname{mod}\;{p}^{k - 1}}\right)\n\]\n\nand so \( d{p}^{j - 1} \) divides \( {e}_{k} \) . Since\n\n\[ \n{a}^{d{p}^{j - 1}} = 1 + {p}^{k - 1}{u}_{j - 1} ≢ 1\;\left( {\;\operatorname{mod}\;{p}^{k}}\right) ,\n\]\n\nit follows that \( d{p}^{j - 1} \) is a proper divisor of \( {e}_{k} \) . On the other hand,\n\n\[ \n{a}^{d{p}^{j}} = 1 + {p}^{k}{u}_{j} \equiv 1\;\left( {\;\operatorname{mod}\;{p}^{k}}\right) ,\n\]\n\nand so \( {e}_{k} \) divides \( d{p}^{j} \) . It follows that the order of \( a \) modulo \( {p}^{k} \) is exactly \( {e}_{k} = d{p}^{j} = d{p}^{k - {k}_{0}} \) . This completes the proof. \( ▱ \)
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Yes
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Theorem 3.7 Let \( p \) be an odd prime. If \( g \) is a primitive root modulo \( p \) , then either \( g \) or \( g + p \) is a primitive root modulo \( {p}^{k} \) for all \( k \geq 2 \) . If \( g \) is a primitive root modulo \( {p}^{k} \) and \( {g}_{1} \in \left\{ {g, g + {p}^{k}}\right\} \) is odd, then \( {g}_{1} \) is a primitive root modulo \( 2{p}^{k} \) .
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Proof. Let \( g \) be a primitive root modulo \( p \) . The order of \( g \) modulo \( p \) is \( p - 1 \) . Let \( {k}_{0} \) be the largest integer such that \( {p}^{{k}_{0}} \) divides \( {g}^{p - 1} - 1 \) . By Theorem 3.6, if \( {k}_{0} = 1 \), then the order of \( g \) modulo \( {p}^{k} \) is \( \left( {p - 1}\right) {p}^{k - 1} = \varphi \left( {p}^{k}\right) \) , and \( g \) is a primitive root modulo \( {p}^{k} \) for all \( k \geq 1 \) . If \( {k}_{0} \geq 2 \), then \[ {g}^{p - 1} = 1 + {p}^{2}v \] for some integer \( v \) . By the binomial theorem, \[ {\left( g + p\right) }^{p - 1} = \mathop{\sum }\limits_{{i = 0}}^{{p - 1}}\left( \begin{matrix} p - 1 \\ i \end{matrix}\right) {g}^{p - 1 - i}{p}^{i} \] \[ \equiv {g}^{p - 1} + \left( {p - 1}\right) {g}^{p - 2}p\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) \] \[ \equiv 1 + {p}^{2}v + {g}^{p - 2}{p}^{2} - {g}^{p - 2}p\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) \] \[ \equiv 1 - {g}^{p - 2}p\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) \] \[ ≢ 1\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) . \] Then \( g + p \) is a primitive root modulo \( p \) such that \[ {\left( g + p\right) }^{p - 1} = 1 + p{u}_{0}\;\text{ and }\;\left( {{u}_{0}, p}\right) = 1. \] Therefore, \( g + p \) is a primitive root modulo \( {p}^{k} \) for all \( k \geq 1 \) . Next we prove that primitive roots exist for all moduli of the form \( 2{p}^{k} \) . If \( g \) is a primitive root modulo \( {p}^{k} \), then \( g + {p}^{k} \) is also a primitive root modulo \( {p}^{k} \) . Since \( {p}^{k} \) is odd, it follows that one of the two integers \( g \) and \( g + {p}^{k} \) is odd, and the other is even. Let \( {g}_{1} \) be the odd integer in the set \( \left\{ {g, g + {p}^{k}}\right\} \) . Since \( \left( {g + {p}^{k},{p}^{k}}\right) = \left( {g,{p}^{k}}\right) = 1 \), it follows that \( \left( {{g}_{1},2{p}^{k}}\right) = 1 \) . The order of \( {g}_{1} \) modulo \( 2{p}^{k} \) is not less than \( \varphi \left( {p}^{k}\right) \), which is the order of \( {g}_{1} \) modulo \( {p}^{k} \) , and not greater than \( \varphi \left( {2{p}^{k}}\right) \) . However, since \( p \) is an odd prime, we have \[ \varphi \left( {2{p}^{k}}\right) = \varphi \left( {p}^{k}\right) \] and so \( {g}_{1} \) has order \( \varphi \left( {2{p}^{k}}\right) \) modulo \( 2{p}^{k} \), that is, \( {g}_{1} \) is a primitive root modulo \( 2{p}^{k} \) . This completes the proof. \( ▱ \)
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Yes
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Theorem 3.8 There exists a primitive root modulo \( m = {2}^{k} \) if and only if \( m = 2 \) or 4 .
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Proof. We note that 1 is a primitive root modulo 2, and 3 is a primitive root modulo 4 . We shall prove that if \( k \geq 3 \), then there is no primitive root modulo \( {2}^{k} \) . Since \( \varphi \left( {2}^{k}\right) = {2}^{k - 1} \), it suffices to show that\n\n\[ \n{a}^{{2}^{k - 2}} \equiv 1\;\left( {\;\operatorname{mod}\;{2}^{k}}\right) \n\]\n\n(3.5)\n\nfor \( a \) odd and \( k \geq 3 \) . We do this by induction on \( k \) . The case \( k = 3 \) is congruence (3.1). Let \( k \geq 3 \), and suppose that (3.5) is true. Then\n\n\[ \n{a}^{{2}^{k - 2}} - 1 \n\]\n\nis divisible by \( {2}^{k} \) . Since \( a \) is odd, it follows that\n\n\[ \n{a}^{{2}^{k - 2}} + 1 \n\]\n\nis even. Therefore,\n\n\[ \n{a}^{{2}^{k - 1}} - 1 = \left( {{a}^{{2}^{k - 2}} - 1}\right) \left( {{a}^{{2}^{k - 2}} + 1}\right) \n\]\n\nis divisible by \( {2}^{k + 1} \), and so\n\n\[ \n{a}^{{2}^{k - 1}} \equiv 1\;\left( {\;\operatorname{mod}\;{2}^{k + 1}}\right) \n\]\n\nThis completes the induction and the proof of theorem. \( ▱ \)
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Yes
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Theorem 3.9 For every positive integer \( k \) ,\n\n\[ \n{5}^{{2}^{k}} \equiv 1 + 3 \cdot {2}^{k + 2}\;\left( {\;\operatorname{mod}\;{2}^{k + 4}}\right) .\n\]
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Proof. The proof is by induction on \( k \) . For \( k = 1 \) we have\n\n\[ \n{5}^{{2}^{1}} = {25} \equiv 1 + 3 \cdot {2}^{3}\;\left( {\;\operatorname{mod}\;{2}^{5}}\right) .\n\]\n\nSimilarly, for \( k = 2 \) we have\n\n\[ \n{5}^{{2}^{2}} = {625} = 1 + {48} + {576} \equiv 1 + 3 \cdot {2}^{4}\;\left( {\;\operatorname{mod}\;{2}^{6}}\right) .\n\]\n\nIf the theorem holds for \( k \geq 1 \), then there exists an integer \( u \) such that\n\n\[ \n{5}^{{2}^{k}} = 1 + 3 \cdot {2}^{k + 2} + {2}^{k + 4}u = 1 + {2}^{k + 2}\left( {3 + {4u}}\right) .\n\]\n\nSince \( {2k} + 4 \geq k + 5 \), we have\n\n\[ \n{5}^{{2}^{k + 1}} = {\left( {5}^{{2}^{k}}\right) }^{2}\n\]\n\n\[ \n= {\left( 1 + {2}^{k + 2}\left( 3 + 4u\right) \right) }^{2}\n\]\n\n\[ \n\equiv \;1 + {2}^{k + 3}\left( {3 + {4u}}\right) {\;(\operatorname{mod}\;{2}^{{2k} + 4})}\n\]\n\n\[ \n\equiv \;1 + 3 \cdot {2}^{k + 3}{\;(\operatorname{mod}\;{2}^{k + 5})}.\n\]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 3.10 If \( k \geq 3 \), then 5 has order \( {2}^{k - 2} \) modulo \( {2}^{k} \) . If \( a \equiv 1 \) \( \left( {\;\operatorname{mod}\;4}\right) \), then there exists a unique integer \( i \in \left\{ {0,1,\ldots ,{2}^{k - 2} - 1}\right\} \) such that\n\n\[ a \equiv {5}^{i}\;\left( {\;\operatorname{mod}\;{2}^{k}}\right) \]\n\nIf \( a \equiv 3\;\left( {\;\operatorname{mod}\;4}\right) \), then there exists a unique integer \( i \in \left\{ {0,1,\ldots ,{2}^{k - 2} - 1}\right\} \) such that\n\n\[ a \equiv - {5}^{i}\;\left( {\;\operatorname{mod}\;{2}^{k}}\right) \]
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Proof. In the case \( k = 3 \), we observe that 5 has order 2 modulo 8, and\n\n\[ 1 \equiv {5}^{0}\;\left( {\;\operatorname{mod}\;8}\right) \]\n\n\[ 3 \equiv - {5}^{1}\;\left( {\;\operatorname{mod}\;8}\right) \]\n\n\[ 5 \equiv {5}^{1}\;\left( {\;\operatorname{mod}\;8}\right) \]\n\n\[ 7 \equiv - {5}^{0}\left( {\;\operatorname{mod}\;8}\right) .\n\nLet \( k \geq 4 \) . By Theorem 3.9, we have\n\n\[ {5}^{{2}^{k - 2}} \equiv 1 + 3 \cdot {2}^{k}\;\left( {\;\operatorname{mod}\;{2}^{k + 2}}\right) \]\n\n\[ \equiv 1\;\left( {\;\operatorname{mod}\;{2}^{k}}\right) \]\n\nand\n\n\[ {5}^{{2}^{k - 3}} \equiv 1 + 3 \cdot {2}^{k - 1}\;\left( {\;\operatorname{mod}\;{2}^{k + 1}}\right) \]\n\n\[ \equiv \;1 + 3 \cdot {2}^{k - 1}{\;(\operatorname{mod}\;{2}^{k})} \]\n\n\[ ≢ 1\;\left( {\;\operatorname{mod}\;{2}^{k}}\right) \text{.} \]\n\nTherefore,5 has order exactly \( {2}^{k - 2} \) modulo \( {2}^{k} \), and so the integers \( {5}^{i} \) are pairwise incongruent modulo \( {2}^{k} \) for \( i = 0,1,\ldots ,{2}^{k - 2} - 1 \) . Since \( {5}^{i} \equiv 1 \) (mod 4) for all \( i \), and since exactly half, that is, \( {2}^{k - 2} \), of the \( {2}^{k - 1} \) odd numbers between 0 and \( {2}^{k} \) are congruent to 1 modulo 4, it follows that the congruence\n\n\[ {5}^{i} \equiv a\;\left( {\;\operatorname{mod}\;{2}^{k}}\right) \]\n\nis solvable for every \( a \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . If \( a \equiv 3\left( {\;\operatorname{mod}\;4}\right) \), then \( - a \equiv 1 \) \( \left( {\;\operatorname{mod}\;4}\right) \) and so the congruence\n\n\[ - a \equiv {5}^{i}\;\left( {\;\operatorname{mod}\;{2}^{k}}\right) \]\n\nor, equivalently,\n\n\[ a \equiv - {5}^{i}\;\left( {\;\operatorname{mod}\;{2}^{k}}\right) \]\n\nis solvable. This completes the proof. \( ▱ \)
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Yes
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Theorem 3.11 Let \( p \) be prime, \( k \geq 2 \), and \( d = \left( {k, p - 1}\right) \) . Let a be an integer not divisible by \( p \) . Let \( g \) be a primitive root modulo \( p \), Then \( a \) is a kth power residue modulo \( p \) if and only if\n\n\[{\operatorname{ind}}_{g}\left( a\right) \equiv 0\;\left( {\;\operatorname{mod}\;d}\right)\]\n\nif and only if\n\n\[{a}^{\left( {p - 1}\right) /d} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right)\]
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Proof. Let \( \ell = {\operatorname{ind}}_{g}\left( a\right) \), where \( g \) is a primitive root modulo \( p \) . Congruence (3.7) is solvable if and only if there exists an integer \( y \) such that\n\n\[{g}^{y} \equiv x\;\left( {\;\operatorname{mod}\;p}\right)\]\n\nand\n\n\[{g}^{ky} \equiv {x}^{k} \equiv a \equiv {g}^{\ell }\;\left( {\;\operatorname{mod}\;p}\right) .\]\n\nThis is equivalent to\n\n\[{ky} \equiv \ell \;\left( {{\;\operatorname{mod}\;p} - 1}\right)\]\n\n(3.8)\n\nThis linear congruence in \( y \) has a solution if and only if\n\n\[{\operatorname{ind}}_{g}\left( a\right) = \ell \equiv 0\;\left( {\;\operatorname{mod}\;d}\right)\]\n\nwhere \( d = \left( {k, p - 1}\right) \) . Thus, the \( k \) th power residues modulo \( p \) are precisely the integers in the \( \left( {p - 1}\right) /d \) congruence classes \( {g}^{id} + p\mathbf{Z} \) for \( i = 0,1,\ldots ,(p - \) 1)/ \( d - 1 \) . Moreover,\n\n\[{a}^{\left( {p - 1}\right) /d} \equiv {g}^{\left( {p - 1}\right) \ell /d} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right)\]\n\nif and only if\n\n\[\frac{\left( {p - 1}\right) \ell }{d} \equiv 0\;\left( {{\;\operatorname{mod}\;p} - 1}\right)\]\n\nif and only if\n\n\[{\operatorname{ind}}_{g}\left( a\right) = \ell \equiv 0\;\left( {\;\operatorname{mod}\;d}\right)\]\n\nFinally, if the linear congruence (3.8) is solvable, then by Theorem 2.2 it has exactly \( d \) solutions \( y \) that are pairwise incongruent modulo \( p - 1 \) , and so (3.7) has exactly \( d \) solutions \( x = {g}^{y} \) that are pairwise incongruent modulo \( p \) . This completes the proof.
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Yes
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Theorem 3.12 Let \( p \) be an odd prime. For every integer \( a \) ,\n\n\[ \left( \frac{a}{p}\right) \equiv {a}^{\left( {p - 1}\right) /2}\;\left( {\;\operatorname{mod}\;p}\right) \]
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Proof. If \( p \) divides \( a \), then both sides of the congruence are 0 . If \( p \) does not divide \( a \), then, by Fermat’s theorem,\n\n\[ {\left( {a}^{\left( {p - 1}\right) /2}\right) }^{2} \equiv {a}^{p - 1} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \]\n\nand so\n\n\[ {a}^{\left( {p - 1}\right) /2} \equiv \pm 1\;\left( {\;\operatorname{mod}\;p}\right) \]\n\nApplying Theorem 3.11 with \( k = 2 \), we have\n\n\[ {a}^{\left( {p - 1}\right) /2} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \;\text{ if and only if }\;\left( \frac{a}{p}\right) = 1, \]\n\nand so\n\n\[ {a}^{\left( {p - 1}\right) /2} \equiv - 1\;\left( {\;\operatorname{mod}\;p}\right) \;\text{ if and only if }\;\left( \frac{a}{p}\right) = - 1. \]\nThis completes the proof.
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Yes
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Theorem 3.13 Let \( p \) be an odd prime, and let \( a \) and \( b \) be integers. Then\n\n\[ \left( \frac{ab}{p}\right) = \left( \frac{a}{p}\right) \left( \frac{b}{p}\right) \]
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Proof. If \( p \) divides \( a \) or \( b \), then \( p \) divides \( {ab} \), and\n\n\[ \left( \frac{ab}{p}\right) = 0 = \left( \frac{a}{p}\right) \left( \frac{b}{p}\right) . \]\n\nIf \( p \) does not divide \( {ab} \), then, by Theorem 3.12,\n\n\[ \left( \frac{ab}{p}\right) \equiv {\left( ab\right) }^{\left( {p - 1}\right) /2}\;\left( {\;\operatorname{mod}\;p}\right) \]\n\n\[ \equiv \;{a}^{\left( {p - 1}\right) /2}{b}^{\left( {p - 1}\right) /2}\;\left( {\;\operatorname{mod}\;p}\right) \]\n\n\[ \equiv \left( \frac{a}{p}\right) \left( \frac{b}{p}\right) \;\left( {\;\operatorname{mod}\;p}\right) \]\n\nThe result follows immediately from the observation that each side of this congruence is \( \pm 1 \) .
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Yes
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Theorem 3.14 Let \( p \) be an odd prime number. Then\n\n\[ \left( \frac{-1}{p}\right) = \begin{cases} 1 & \text{ if } & p \equiv 1 & \left( {\;\operatorname{mod}\;4}\right) , \\ - 1 & \text{ if } & p \equiv 3 & \left( {\;\operatorname{mod}\;4}\right) . \end{cases} \]\n\nEquivalently,\n\n\[ \left( \frac{-1}{p}\right) = {\left( -1\right) }^{\left( {p - 1}\right) /2} \]
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Proof. We observe that\n\n\[ {\left( -1\right) }^{\left( {p - 1}\right) /2} = \begin{cases} 1 & \text{ if } & p \equiv 1 & \left( {\;\operatorname{mod}\;4}\right) , \\ - 1 & \text{ if } & p \equiv 3 & \left( {\;\operatorname{mod}\;4}\right) . \end{cases} \]\n\nApplying Theorem 3.12 with \( a = - 1 \), we obtain\n\n\[ \left( \frac{-1}{p}\right) \equiv {\left( -1\right) }^{\left( {p - 1}\right) /2}\;\left( {\;\operatorname{mod}\;p}\right) \]\n\nAgain, the theorem follows immediately from the observation that both sides of this congruence are \( \pm 1 \) . \( ▱ \)
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Yes
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Theorem 3.15 (Gauss's lemma) Let \( p \) be an odd prime, and a an integer not divisible by \( p \) . Let \( S \) be a Gaussian set modulo \( p \) . For every \( s \in S \) there exist unique integers \( {u}_{a}\left( s\right) \in S \) and \( {\varepsilon }_{a}\left( s\right) \in \{ 1, - 1\} \) such that\n\n\[ \n{as} \equiv {\varepsilon }_{a}\left( s\right) {u}_{a}\left( s\right) \;\left( {\;\operatorname{mod}\;p}\right) .\n\]\n\nMoreover,\n\n\[ \n\left( \frac{a}{p}\right) = \mathop{\prod }\limits_{{s \in S}}{\varepsilon }_{a}\left( s\right) = {\left( -1\right) }^{m}\n\]\n\nwhere \( m \) is the number of \( s \in S \) such that \( {\varepsilon }_{a}\left( s\right) = - 1 \) .
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Proof. Since \( S \) is a Gaussian set, for every \( s \in S \) there exist unique integers \( {u}_{a}\left( s\right) \in S \) and \( {\varepsilon }_{a}\left( s\right) \in \{ 1, - 1\} \) such that\n\n\[ \n{as} \equiv {\varepsilon }_{a}\left( s\right) {u}_{a}\left( s\right) \;\left( {\;\operatorname{mod}\;p}\right) .\n\]\n\nLet \( s,{s}^{\prime } \in S \) . If \( {u}_{a}\left( s\right) = {u}_{a}\left( {s}^{\prime }\right) \), then\n\n\[ \na{s}^{\prime } \equiv {\varepsilon }_{a}\left( {s}^{\prime }\right) {u}_{a}\left( {s}^{\prime }\right) \equiv {\varepsilon }_{a}\left( {s}^{\prime }\right) {u}_{a}\left( s\right) \;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\n\[ \n\equiv \;{\varepsilon }_{a}\left( {s}^{\prime }\right) {\varepsilon }_{a}\left( s\right) {\varepsilon }_{a}\left( s\right) {u}_{a}\left( s\right) {\;(\operatorname{mod}\;p)}\n\]\n\n\[ \n\equiv \pm {as}\;\left( {\;\operatorname{mod}\;p}\right) \text{. }\n\]\n\nDividing by \( a \), we obtain\n\n\[ \n{s}^{\prime } \equiv \pm s\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\nand so \( {s}^{\prime } = s \) . It follows that the map \( {u}_{a} : S \rightarrow S \) is a permutation of \( S \) , and so\n\n\[ \n\mathop{\prod }\limits_{{s \in S}}s = \mathop{\prod }\limits_{{s \in S}}{u}_{a}\left( s\right)\n\]\n\nTherefore,\n\n\[ \n{a}^{\left( {p - 1}\right) /2}\mathop{\prod }\limits_{{s \in S}}s \equiv \mathop{\prod }\limits_{{s \in S}}{as}\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\n\[ \n\equiv \mathop{\prod }\limits_{{s \in S}}{\varepsilon }_{a}\left( s\right) {u}_{a}\left( s\right) \;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\n\[ \n\equiv \mathop{\prod }\limits_{{s \in S}}{\varepsilon }_{a}\left( s\right) \mathop{\prod }\limits_{{s \in S}}{u}_{a}\left( s\right) \;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\n\[ \n\equiv \mathop{\prod }\limits_{{s \in S}}{\varepsilon }_{a}\left( s\right) \mathop{\prod }\limits_{{s \in S}}s\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\nDividing by \( \mathop{\prod }\limits_{{s \in S}}s \), we obtain\n\n\[ \n\left( \frac{a}{p}\right) \equiv {a}^{\left( {p - 1}\right) /2} \equiv \mathop{\prod }\limits_{{s \in S}}{\varepsilon }_{a}\left( s\right) \;\left( {\;\operatorname{mod}\;p}\right) .\n\]\n\nThe proof is completed by the observation that the right and left sides of this congruence are \( \pm 1 \) . \( ▱ \)
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Yes
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Theorem 3.16 Let \( p \) be an odd prime. Then\n\n\[ \left( \frac{2}{p}\right) = \left\{ \begin{array}{lll} 1 & \text{ if }p \equiv \pm 1 & \left( {\;\operatorname{mod}\;8}\right) , \\ - 1 & \text{ if }p \equiv \pm 3 & \left( {\;\operatorname{mod}\;8}\right) . \end{array}\right. \]\n\nEquivalently,\n\n\[ \left( \frac{2}{p}\right) = {\left( -1\right) }^{\left( {{p}^{2} - 1}\right) /8}. \]
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Proof. We apply Gauss's lemma (Theorem 3.15) to the Gaussian set \( S = \{ 1,2,3,\ldots ,\left( {p - 1}\right) /2\} \) . Then\n\n\[ \{ {2s} : s \in S\} = \{ 2,4,6,\ldots, p - 1\} \]\n\nand\n\n\[ \left( \frac{2}{p}\right) = {\left( -1\right) }^{m} \]\n\nwhere \( m \) is the number of integers \( s \in S \) such that \( {\varepsilon }_{2}\left( s\right) = - 1 \) . If \( 1 \leq {2s} \leq \) \( \left( {p - 1}\right) /2 \), then \( {2s} \in S \), and so \( {u}_{2}\left( s\right) = {2s} \) and \( {\varepsilon }_{2}\left( s\right) = 1 \) . If \( \left( {p + 1}\right) /2 \leq \) \( {2s} \leq p - 1 \), then \( 1 \leq p - {2s} \leq \left( {p - 1}\right) /2 \), and so \( p - {2s} \in S \) . Since\n\n\[ {2s} \equiv - \left( {p - {2s}}\right) \;\left( {\;\operatorname{mod}\;p}\right) \]\n\nit follows that \( {u}_{2}\left( s\right) = p - {2s} \) and \( {\varepsilon }_{2}\left( s\right) = - 1 \) . Therefore, \( m \) is the number of integers \( s \in S \) such that \( \left( {p + 1}\right) /2 \leq {2s} \leq p - 1 \), or, equivalently,\n\n\[ \frac{p + 1}{4} \leq s \leq \frac{p - 1}{2} \]\n\n(3.10)\n\nSince every odd prime \( p \) is congruent to \( 1,3,5 \), or 7 modulo 8, there are four cases to consider.\n\n(i) If \( p \equiv 1\left( {\;\operatorname{mod}\;8}\right) \), then \( p = {8k} + 1 \), and \( s \in S \) satisfies (3.10) if and only if\n\n\[ {2k} + \frac{1}{2} \leq s \leq {4k} \]\n\nand so \( m = {2k} \) and \( \left( \frac{2}{p}\right) = {\left( -1\right) }^{2k} = 1 \) .\n\n(ii) If \( p \equiv 3\left( {\;\operatorname{mod}\;8}\right) \), then \( p = {8k} + 3 \), and \( s \in S \) satisfies (3.10) if and\nonly if\n\n\[ {2k} + 1 \leq s \leq {4k} + 1 \]\n\nand so \( m = {2k} + 1 \) and \( \left( \frac{2}{p}\right) = {\left( -1\right) }^{{2k} + 1} = - 1 \) .\n\n(iii) If \( p \equiv 5\left( {\;\operatorname{mod}\;8}\right) \), then \( p = {8k} + 5 \), and \( s \in S \) satisfies (3.10) if and only if\n\n\[ {2k} + 1 + \frac{1}{2} \leq s \leq {4k} + 2 \]\n\nand so \( m = {2k} + 1 \) and \( \left( \frac{2}{p}\right) = {\left( -1\right) }^{{2k} + 1} = - 1 \) .\n\n(iv) If \( p \equiv 7\left( {\;\operatorname{mod}\;8}\right) \), then \( p = {8k} + 7 \), and \( s \in S \) satisfies (3.10) if and\nonly if\n\n\[ {2k} + 2 \leq s \leq {4k} + 3 \]\n\nand so \( m = {2k} + 2 \) and \( \left( \frac{2}{p}\right) = {\left( -1\right) }^{{2k} + 2} = 1 \) .\n\nFinally, we observe that\n\n\[ \frac{{p}^{2} - 1}{8} \equiv 0\;\left( {\;\operatorname{mod}\;2}\right) \;\text{ if }p \equiv 1\text{ or }7\;\left( {\;\operatorname{mod}\;8}\right) \]\n\nand\n\n\[ \frac{{p}^{2} - 1}{8} \equiv 1\;\left( {\;\operatorname{mod}\;2}\right) \;\text{ if }p \equiv 3\text{ or }7\;\left( {\;\operatorname{mod}\;8}\right) . \]\n\nThis completes the proof.
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Yes
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Theorem 3.18 Let \( R \) be a ring and \( f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} \) a polynomial with coefficients in \( R \) . Then\n\n\[ f\left( {x + h}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) h + r\left( {x, h}\right) {h}^{2}. \]\n\nwhere \( r\left( {x, h}\right) \) is a polynomial in the two variables \( x \) and \( h \) with coefficients in \( R \) .
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Proof. This is a standard calculation. Expanding \( f\left( {x + h}\right) \) by the binomial theorem, we obtain\n\n\[ f\left( {x + h}\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{\left( x + h\right) }^{i} \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}\mathop{\sum }\limits_{{j = 0}}^{i}\left( \begin{array}{l} i \\ j \end{array}\right) {x}^{i - j}{h}^{j} \]\n\n\[ = \mathop{\sum }\limits_{{j = 0}}^{n}\mathop{\sum }\limits_{{i = j}}^{n}\left( \begin{array}{l} i \\ j \end{array}\right) {a}_{i}{x}^{i - j}{h}^{j} \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} + \mathop{\sum }\limits_{{i = 1}}^{n}i{a}_{i}{x}^{i - 1}h + \mathop{\sum }\limits_{{j = 2}}^{n}\mathop{\sum }\limits_{{i = j}}^{n}\left( \begin{array}{l} i \\ j \end{array}\right) {a}_{i}{x}^{i - j}{h}^{j} \]\n\n\[ = f\left( x\right) + {f}^{\prime }\left( x\right) h + r\left( {x, h}\right) {h}^{2}, \]\n\nwhere\n\n\[ r\left( {x, h}\right) = \mathop{\sum }\limits_{{j = 2}}^{n}\mathop{\sum }\limits_{{i = j}}^{n}\left( \begin{array}{l} i \\ j \end{array}\right) {a}_{i}{x}^{i - j}{h}^{j - 2} \]\n\nis a polynomial in \( x \) and \( h \) with coefficients in \( R \) .
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Yes
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Theorem 3.20 Let \( p \) be an odd prime, and let \( a \) be an integer not divisible by \( p \) . If \( a \) is a quadratic residue modulo \( p \), then \( a \) is a quadratic residue modulo \( {p}^{k} \) for every \( k \geq 1 \) .
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Proof. Consider the polynomial \( f\left( x\right) = {x}^{2} - a \) and its derivative \( {f}^{\prime }\left( x\right) = \) \( {2x} \) . If \( a \) is a quadratic residue modulo \( p \), then there exists an integer \( {x}_{1} \) such that \( {x}_{1} ≢ 0\;\left( {\;\operatorname{mod}\;p}\right) \) and \( {x}_{1}^{2} \equiv a\;\left( {\;\operatorname{mod}\;p}\right) \) . Then \( f\left( {x}_{1}\right) \equiv 0\;\left( {\;\operatorname{mod}\;p}\right) \) and \( {f}^{\prime }\left( {x}_{1}\right) ≢ 0\;\left( {\;\operatorname{mod}\;p}\right) \) . By Hensel’s lemma, the polynomial congruence \( f\left( x\right) \equiv 0\left( {\;\operatorname{mod}\;{p}^{k}}\right) \) is solvable for every \( k \geq 1 \), and so \( a \) is a quadratic residue modulo \( {p}^{k} \) for every \( k \geq 1 \) . \( ▱ \)
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Yes
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Theorem 4.1 Let \( G \) be a finite abelian group, written additively, and let \( \left| G\right| = m \) . For every prime number \( p \), let \( G\left( p\right) \) be the set of all elements \( g \in G \) whose order is a power of \( p \) . Then\n\n\[ G = {\bigoplus }_{p \mid m}G\left( p\right) \]
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Proof. Let \( m = \mathop{\prod }\limits_{{i = 1}}^{k}{p}_{i}^{{r}_{i}} \) be the standard factorization of \( m \), and let \( {m}_{i} = m{p}_{i}^{-{r}_{i}} \) for \( i = 1,\ldots, k \) . Then \( \left( {{m}_{1},\ldots ,{m}_{k}}\right) = 1 \) by Exercise 15 in Section 1.4, and so there exist integers \( {u}_{1},\ldots ,{u}_{k} \) such that\n\n\[ {m}_{1}{u}_{1} + \cdots + {m}_{k}{u}_{k} = 1 \]\n\nLet \( g \in G \), and define \( {g}_{i} = {m}_{i}{u}_{i}g \in G \) for \( i = 1,\ldots, k \) . Since \( {p}_{i}^{{r}_{i}}{g}_{i} = \) \( m{u}_{i}g = 0 \), it follows that \( {g}_{i} \in G\left( p\right) \) . Moreover,\n\n\[ g = \left( {{m}_{1}{u}_{1} + \cdots + {m}_{k}{u}_{k}}\right) g = {m}_{1}{u}_{1}g + \cdots + {m}_{k}{u}_{k}g \]\n\n\[ = {g}_{1} + \cdots + {g}_{k} \in G\left( {p}_{1}\right) + \cdots + G\left( {p}_{k}\right) ,\]\n\nand so\n\n\[ G = G\left( {p}_{1}\right) + \cdots + G\left( {p}_{k}\right) \]\n\nSuppose that\n\n\[ {g}_{1} + \cdots + {g}_{k} = 0 \]\n\nwhere \( {g}_{i} \in G\left( {p}_{i}\right) \) for \( i = 1,\ldots, k \) . There exist nonnegative integers \( {r}_{1},\ldots ,{r}_{k} \) such that \( {g}_{i} \) has order \( {p}_{i}^{{r}_{i}} \) for \( i = 1,\ldots, k \) . Let\n\n\[ {d}_{j} = \mathop{\prod }\limits_{\substack{{i = 1} \\ {i \neq j} }}^{k}{p}_{i}^{{r}_{i}} \]\n\nIf \( {g}_{j} \neq 0 \), then \( {d}_{j}{g}_{j} \neq 0 \) . Since \( {d}_{j}{g}_{i} = 0 \) for \( i = 1,\ldots, k, i \neq j \), it follows that\n\n\[ 0 = {d}_{j}\left( {{g}_{1} + \cdots + {g}_{k}}\right) = {d}_{j}{g}_{j} \]\n\nand so \( {g}_{j} = 0 \) for all \( j = 1,\ldots, k \) . Thus,0 has no nontrivial representation in \( G = G\left( {p}_{1}\right) + \cdots + G\left( {p}_{k}\right) \) . By Exercise 4, we conclude that \( G \) is the direct sum of the subgroups \( G\left( {p}_{i}\right) \).
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No
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Lemma 4.1 Let \( G \) be a finite abelian p-group. Let \( {g}_{1} \in G \) be an element of maximum order \( {p}^{{r}_{1}} \), and let \( {G}_{1} = \left\langle {g}_{1}\right\rangle \) be the cyclic subgroup generated by \( {g}_{1} \) . Consider the quotient group \( G/{G}_{1} \) . Let \( h \in G \) . If \( h + {G}_{1} \in G/{G}_{1} \) has order \( {p}^{r} \), then there exists an element \( g \in G \) such that \( g + {G}_{1} = h + {G}_{1} \) and \( g \) has order \( {p}^{r} \) in \( G \) .
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Proof. If \( h + {G}_{1} \) has order \( {p}^{r} \) in \( G/{G}_{1} \), then the order of \( h \) in \( G \) is at most \( {p}^{{r}_{1}} \) (since \( {p}^{{r}_{1}} \) is the maximum order in \( G \) ) and at least \( {p}^{r} \) (by Exercise 7). Since \( {G}_{1} = {p}^{r}\left( {h + {G}_{1}}\right) = {p}^{r}h + {G}_{1} \), it follows that \( {p}^{r}h \in {G}_{1} \) , and so \( {p}^{r}h = u{g}_{1} \) for some positive integer \( u \leq {p}^{{r}_{1}} \) (since \( {g}_{1} \) has order \( {p}^{{r}_{1}} \) ). Write \( u = {p}^{s}v \), where \( \left( {p, v}\right) = 1 \) and \( 0 \leq s \leq {r}_{1} \) . Then \( v{g}_{1} \) also has order \( {p}^{{r}_{1}} \), and so \( {p}^{s}v{g}_{1} \) has order \( {p}^{{r}_{1} - s} \) in \( G \) . Then \( {p}^{r}h = {p}^{s}v{g}_{1} \) has order \( {p}^{{r}_{1} - s} \) in \( G \), and so \( h \) has order \( {p}^{{r}_{1} + r - s} \leq {p}^{{r}_{1}} \) . It follows that \( r \leq s \), and \[ {p}^{r}h = {p}^{s}v{g}_{1} = {p}^{r}\left( {{p}^{s - r}v{g}_{1}}\right) = {p}^{r}{g}_{1}^{\prime }, \] where \[ {g}_{1}^{\prime } = {p}^{s - r}v{g}_{1} \in {G}_{1} \] Let \[ g = h - {g}_{1}^{\prime } \] Then \[ g + {G}_{1} = h + {G}_{1} \] Moreover, \( {p}^{r}g = {p}^{r}h - {p}^{r}{g}_{1}^{\prime } = 0 \), and so the order of \( g \) is at most \( {p}^{r} \) . On the other hand, \( g + {G}_{1} \) has order \( {p}^{r} \) in the quotient group \( G/{G}_{1} \), and so the order of \( g \) is at least \( {p}^{r} \) . Therefore, \( g \) has order \( {p}^{r} \) . \( ▱ \)
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Yes
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Theorem 4.3 Every finite abelian group is a direct sum of cyclic groups.
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Proof. This follows immediately from Theorem 4.1 and Theorem 4.2. \( ▱ \)
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No
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Lemma 4.2 The dual of a cyclic group of order \( n \) is also a cyclic group of order \( n \) .
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Proof. We introduce the exponential functions\n\n\[ e\left( x\right) = {e}^{2\pi ix} \]\n\nand\n\n\[ {e}_{n}\left( x\right) = e\left( {x/n}\right) = {e}^{{2\pi ix}/n}. \]\n\nThe \( n \) th roots of unity are the complex numbers \( {e}_{n}\left( a\right) \) for \( a = 0,1,\ldots, n - 1 \) . Let \( G \) be a finite cyclic group of order \( n \) with generator \( {g}_{0} \) . Then \( G = \) \( \left\{ {j{g}_{0} : j = 0,1,\ldots, n - 1}\right\} \) . For every integer \( a \), we define \( {\psi }_{a} \in \widehat{G} \) by\n\n\[ {\psi }_{a}\left( {j{g}_{0}}\right) = {e}_{n}\left( {aj}\right) \]\n\n(4.1)\n\nBy Exercise 3, we have \( {\psi }_{a}{\psi }_{b} = {\psi }_{a + b},{\psi }_{a}^{-1} = {\psi }_{-a},{\psi }_{a} = {\psi }_{b} \) if or only if \( a \equiv b\;\left( {\;\operatorname{mod}\;n}\right) \) . It follows that\n\n\[ {\psi }_{a} = {\psi }_{1}^{a} \]\n\nfor every integer \( a \) . If \( \chi \) is a character in \( \widehat{G} \), then \( \chi \) is completely determined by its value on \( {g}_{0} \) . Since \( \chi \left( {g}_{0}\right) \) is an \( n \) th root of unity, we have \( \chi \left( {g}_{0}\right) = {e}_{n}\left( a\right) \) for some integer \( a = 0,1,\ldots, n - 1 \), and so \( \chi \left( {j{g}_{0}}\right) = {e}_{n}\left( {aj}\right) \) for every integer \( j \) . Therefore, \( \chi = {\psi }_{a} \) and\n\n\[ \widehat{G} = \left\{ {{\psi }_{a} : a = 0,1,\ldots, n - 1}\right\} = \left\{ {{\psi }_{1}^{a} : a = 0,1,\ldots, n - 1}\right\} \]\n\nis also a cyclic group of order \( n \), that is, \( G \cong \widehat{G} \) . \( ▱ \)
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No
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Lemma 4.3 Let \( G \) be a finite abelian group and let \( {G}_{1},\ldots ,{G}_{k} \) be subgroups of \( G \) such that \( G = {G}_{1} \oplus \cdots \oplus {G}_{k} \) . For every character \( \chi \in \widehat{G} \) there exist unique characters \( {\chi }_{i} \in \widehat{{G}_{i}} \) such that if \( g \in G \) and \( g = {g}_{1} + \cdots + {g}_{k} \) with \( {g}_{i} \in {G}_{i} \) for \( i = 1,\ldots, k \), then\n\n\[ \n\chi \left( g\right) = {\chi }_{1}\left( {g}_{1}\right) \cdots {\chi }_{k}\left( {g}_{k}\right) \n\]
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Proof. If \( {\chi }_{i} \in \widehat{{G}_{i}} \) for \( i = 1,\ldots, k \), then we can construct a map \( \chi : G \rightarrow \) \( {\mathbf{C}}^{ \times } \) as follows. Let \( g \in G \) . There exist unique elements \( {g}_{i} \in {G}_{i} \) such that \( g = {g}_{1} + \cdots + {g}_{k} \) . Define\n\n\[ \n\chi \left( g\right) = \chi \left( {{g}_{1} + \cdots + {g}_{k}}\right) = {\chi }_{1}\left( {g}_{1}\right) \cdots {\chi }_{k}\left( {g}_{k}\right) .\n\]\n\nThen \( \chi \) is a character in \( \widehat{G} \), and this construction induces a map\n\n\[ \n\Psi : \widehat{{G}_{1}} \times \cdots \times \widehat{{G}_{k}} \rightarrow \widehat{G}\n\]\n\nBy Exercise 5, the map \( \Psi \) is a one-to-one homomorphism. We shall show that the map \( \Psi \) is onto. Let \( \chi \in \widehat{G} \) . We define the function \( {\chi }_{i} \) on \( {G}_{i} \) by\n\n\[ \n{\chi }_{i}\left( {g}_{i}\right) = \chi \left( {g}_{i}\right) \;\text{ for all }{g}_{i} \in {G}_{i}.\n\]\n\nThen \( {\chi }_{i} \) is a character in \( \widehat{{G}_{i}} \) . If \( g \in G \) and \( g = {g}_{1} + \cdots + {g}_{k} \) with \( {g}_{i} \in {G}_{i} \) , then\n\n\[ \n\chi \left( g\right) = \chi \left( {{g}_{1} + \cdots + {g}_{k}}\right) = \chi \left( {g}_{1}\right) \cdots \chi \left( {g}_{k}\right) = {\chi }_{1}\left( {g}_{1}\right) \cdots {\chi }_{k}\left( {g}_{k}\right) .\n\]\n\nIt follows that\n\n\[ \n\Psi \left( {{\chi }_{1},\ldots ,{\chi }_{k}}\right) = \chi\n\]\n\nand so \( \Psi \) is onto.
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Yes
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Theorem 4.5 A finite abelian group \( G \) is isomorphic to its dual, that is,\n\n\[ G \cong \widehat{G} \]
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Proof. By Lemma 4.2, the dual of a finite cyclic group of order \( n \) is also a finite cyclic group of order \( n \) . By Theorem 4.3, a finite abelian group \( G \) has cyclic subgroups \( {G}_{1},\ldots ,{G}_{k} \) such that\n\n\[ G = {G}_{1} \oplus \cdots \oplus {G}_{k} \]\n\nBy Lemma 4.3 and Exercise 5 in Section 4.1,\n\n\[ \widehat{G} \cong \widehat{{G}_{1}} \times \cdots \times \widehat{{G}_{k}} \cong {G}_{1} \times \cdots \times {G}_{k} \cong {G}_{1} \oplus \cdots \oplus {G}_{k} = G. \]\n\nThis completes the proof.
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No
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Theorem 4.6 (Orthogonality relations) Let \( G \) be a finite abelian group of order \( n \), and let \( \widehat{G} \) be its dual group. If \( \chi \in \widehat{G} \), then\n\n\[ \mathop{\sum }\limits_{{a \in G}}\chi \left( a\right) = \left\{ \begin{array}{ll} n & \text{ if }\chi = {\chi }_{0} \\ 0 & \text{ if }\chi \neq {\chi }_{0} \end{array}\right. \]\n\nIf \( a \in G \), then\n\n\[ \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\chi \left( a\right) = \left\{ \begin{array}{ll} n & \text{ if }a = 0 \\ 0 & \text{ if }a \neq 0 \end{array}\right. \]
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Proof. For \( \chi \in \widehat{G} \), let\n\n\[ S\left( \chi \right) = \mathop{\sum }\limits_{{a \in G}}\chi \left( a\right) \]\n\nIf \( \chi = {\chi }_{0} \), then \( S\left( {\chi }_{0}\right) = \left| G\right| = n \) . If \( \chi \neq {\chi }_{0} \), then \( \chi \left( b\right) \neq 1 \) for some \( b \in G \) , and\n\n\[ \chi \left( b\right) S\left( \chi \right) = \chi \left( b\right) \mathop{\sum }\limits_{{a \in G}}\chi \left( a\right) \]\n\n\[ = \mathop{\sum }\limits_{{a \in G}}\chi \left( {ba}\right) \]\n\n\[ = \mathop{\sum }\limits_{{a \in G}}\chi \left( a\right) \]\n\n\[ = S\left( \chi \right) \]\n\nand so \( S\left( \chi \right) = 0 \) .\n\nFor \( a \in G \), let\n\n\[ T\left( a\right) = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\chi \left( a\right) \]\n\nIf \( a = 0 \), then \( T\left( a\right) = \left| \widehat{G}\right| = n \) . If \( a \neq 0 \), then \( {\chi }^{\prime }\left( a\right) \neq 1 \) for some \( {\chi }^{\prime } \in \widehat{G} \) (by Theorem 4.4), and\n\n\[ {\chi }^{\prime }\left( a\right) T\left( a\right) = {\chi }^{\prime }\left( a\right) \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\chi \left( a\right) \]\n\n\[ = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}{\chi }^{\prime }\chi \left( a\right) \]\n\n\[ = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\chi \left( a\right) \]\n\n\[ = T\left( a\right) \]\n\nand so \( T\left( a\right) = 0 \) . This completes the proof.
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Yes
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Theorem 4.7 (Orthogonality relations) Let \( G \) be a finite abelian group of order \( n \), and let \( \widehat{G} \) be its dual group. If \( {\chi }_{1},{\chi }_{2} \in \widehat{G} \), then \[ \mathop{\sum }\limits_{{a \in G}}{\chi }_{1}\left( a\right) \overline{{\chi }_{2}}\left( a\right) = \left\{ \begin{array}{ll} n & \text{ if }{\chi }_{1} = {\chi }_{2}, \\ 0 & \text{ if }{\chi }_{1} \neq {\chi }_{2} \end{array}\right. \] If \( a, b \in G \), then \[ \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\chi \left( a\right) \bar{\chi }\left( b\right) = \left\{ \begin{array}{ll} n & \text{ if }a = b, \\ 0 & \text{ if }a \neq b. \end{array}\right. \]
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Proof. These identities follow immediately from Theorem 4.6, since \[ {\chi }_{1}\left( a\right) \overline{{\chi }_{2}}\left( a\right) = {\chi }_{1}{\chi }_{2}^{-1}\left( a\right) \] and \[ \chi \left( a\right) \bar{\chi }\left( b\right) = \chi \left( {a - b}\right) \] This completes the proof.
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Yes
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Theorem 4.8 (Fourier inversion) Let \( G \) be a finite abelian group of order \( n \) with dual group \( \widehat{G} \) . If \( f \in {L}^{2}\left( G\right) \), then\n\n\[ f = \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{f}\left( \chi \right) \chi \]\n\nand (4.7) is the unique representation of \( f \) as a linear combination of characters of \( G \) .
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Proof. This is a straightforward calculation. Let \( a \in G \) . Defining the Fourier transform by (4.6), we have\n\n\[ \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{f}\left( \chi \right) \chi \left( a\right) = \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\left( {\mathop{\sum }\limits_{{b \in G}}f\left( b\right) \bar{\chi }\left( b\right) }\right) \chi \left( a\right) \]\n\n\[ = \mathop{\sum }\limits_{{b \in G}}f\left( b\right) \left( {\frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\chi \left( a\right) \bar{\chi }\left( b\right) }\right) \]\n\n\[ = f\left( a\right) ,\]\n\nby the orthogonality relations (Theorem 4.7). This proves (4.7). The uniqueness of the series (4.7) is Exercise 2.\n\nTo prove (4.8), we have\n\n\[ \widehat{\widehat{f}}\left( {\Delta \left( a\right) }\right) = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{f}\left( \chi \right) \overline{\Delta \left( a\right) }\left( \chi \right) \]\n\n\[ = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\mathop{\sum }\limits_{{g \in G}}f\left( g\right) \bar{\chi }\left( g\right) \bar{\chi }\left( a\right) \]\n\n\[ = \mathop{\sum }\limits_{{g \in G}}f\left( g\right) \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\bar{\chi }\left( {g + a}\right) \]\n\n\[ = {nf}\left( {-a}\right) \text{.} \]\n\nThis completes the proof. \( ▱ \)
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No
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Theorem 4.9 (Plancherel’s formula) If \( G \) is a finite abelian group of order \( n \) and \( f \in {L}^{2}\left( G\right) \), then \[ \parallel \widehat{f}{\parallel }_{2} = \sqrt{n}\parallel f{\parallel }_{2} \]
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Proof. We have \[ \parallel \widehat{f}{\parallel }_{2}^{2} = \left( {\widehat{f},\widehat{f}}\right) \] \[ = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{f}\left( \chi \right) \overline{\widehat{f}\left( \chi \right) } \] \[ = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\left( {\mathop{\sum }\limits_{{b \in G}}f\left( b\right) \overline{\chi \left( b\right) }}\right) \left( {\mathop{\sum }\limits_{{a \in G}}\overline{f\left( a\right) }\chi \left( a\right) }\right) \] \[ = \mathop{\sum }\limits_{{a \in G}}\mathop{\sum }\limits_{{b \in G}}\overline{f\left( a\right) }f\left( b\right) \left( {\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\chi \left( a\right) \overline{\chi \left( b\right) }}\right) \] \[ = n\mathop{\sum }\limits_{{a \in G}}{\left| f\left( a\right) \right| }^{2} \] \[ = n\parallel f{\parallel }_{2}^{2}\text{.} \] This completes the proof. \( ▱ \)
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Yes
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Theorem 4.10 (Uncertainty principle) If \( G \) is a finite abelian group and \( f \in {L}^{2}\left( G\right), f \neq 0 \), then\n\n\[ \left| {\operatorname{supp}\left( f\right) }\right| \left| {\operatorname{supp}\left( \widehat{f}\right) }\right| \geq \left| G\right| \text{.} \]
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Proof. Let \( a \in G \) . By Theorem 4.8,\n\n\[ f\left( a\right) = \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{f}\left( \chi \right) \chi \left( a\right) \]\n\nSince \( \left| {\chi \left( a\right) }\right| = 1 \) for all \( \chi \in \widehat{G} \), it follows that\n\n\[ \left| {f\left( a\right) }\right| \leq \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\left| {\widehat{f}\left( \chi \right) }\right| = \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \operatorname{supp}\left( \widehat{f}\right) }}\left| {\widehat{f}\left( \chi \right) }\right| \]\n\nand so\n\n\[ \parallel f{\parallel }_{\infty } \leq \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \operatorname{supp}\left( \widehat{f}\right) }}\left| {\widehat{f}\left( \chi \right) }\right| \]\n\nApplying the Cauchy-Schwarz inequality (4.5) with \( {f}_{1} = \widehat{f}\left( \chi \right) \) and with \( {f}_{2} \) the characteristic function of the set \( \operatorname{supp}\left( \widehat{f}\right) \), we have\n\n\[ {\left( \mathop{\sum }\limits_{{\chi \in \operatorname{supp}\left( \widehat{f}\right) }}\left| \widehat{f}\left( \chi \right) \right| \right) }^{2} = \mathop{\sum }\limits_{{\chi \in \operatorname{supp}\left( \widehat{f}\right) }}{\left| \widehat{f}\left( \chi \right) \right| }^{2}\left| {\operatorname{supp}\left( \widehat{f}\right) }\right| \]\n\nUsing Plancherel's formula (Theorem 4.9), and inequality (4.9), we obtain\n\n\[ \parallel f{\parallel }_{\infty }^{2} \leq \frac{1}{{n}^{2}}{\left( \mathop{\sum }\limits_{{\chi \in \operatorname{supp}\left( \widehat{f}\right) }}\left| \widehat{f}\left( \chi \right) \right| \right) }^{2} \]\n\n\[ \leq \frac{1}{{n}^{2}}\mathop{\sum }\limits_{{\chi \in \operatorname{supp}\left( \widehat{f}\right) }}{\left| \widehat{f}\left( \chi \right) \right| }^{2}\left| {\operatorname{supp}\left( \widehat{f}\right) }\right| \]\n\n\[ = \frac{1}{{n}^{2}}\parallel \widehat{f}{\parallel }_{2}^{2}\left| {\operatorname{supp}\left( \widehat{f}\right) }\right| \]\n\n\[ = \frac{1}{n}\parallel f{\parallel }_{2}^{2}\left| {\operatorname{supp}\left( \widehat{f}\right) }\right| \]\n\n\[ \leq \frac{1}{n}\parallel f{\parallel }_{\infty }^{2}\left| {\operatorname{supp}\left( f\right) }\right| \left| {\operatorname{supp}\left( \widehat{f}\right) }\right| . \]\n\nSince \( f \neq 0 \), we have \( \parallel f{\parallel }_{\infty } > 0 \) and so\n\n\[ \left| {\operatorname{supp}\left( f\right) }\right| \left| {\operatorname{supp}\left( \widehat{f}\right) }\right| \geq n = \left| G\right| . \]\n\nThis completes the proof. \( ▱ \)
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Yes
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Lemma 4.4 Let \( G \) be a finite abelian group with subgroup \( H \) . Then\n\n\[ \n{\widehat{G}}^{H} = \widehat{G} \cap {L}^{2}{\left( G\right) }^{H} \n\]
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Proof. If \( \chi \in {\widehat{G}}^{H} \subseteq \widehat{G} \), then \( \chi \left( {x + h}\right) = \chi \left( x\right) \chi \left( h\right) = \chi \left( x\right) \) for all \( x \in G \) and \( h \in H \), and so \( \chi \in \widehat{G} \cap {L}^{2}\left( {G/H}\right) \) . Conversely, if \( \chi \in \widehat{G} \cap {L}^{2}\left( {G/H}\right) \) , then \( \chi \left( h\right) = \chi \left( {0 + h}\right) = \chi \left( 0\right) = 1 \) for all \( h \in H \), and \( \chi \in {\widehat{G}}^{H} \) . \( ▱ \)
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Yes
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Lemma 4.5 Let \( G \) be a finite abelian group with subgroup \( H \), and let \( \pi \) : \( G \rightarrow G/H \) be the natural map onto the quotient group. For \( {f}^{\sharp } \in {L}^{2}\left( {G/H}\right) \) , define the map \( {\pi }^{\sharp }\left( {f}^{\sharp }\right) \in {L}^{2}\left( G\right) \) by\n\n\[ \n{\pi }^{\sharp }\left( {f}^{\sharp }\right) \left( x\right) = {f}^{\sharp }\pi \left( x\right) = {f}^{\sharp }\left( {x + H}\right) \n\]\n\nfor all \( x \in G \) . Then \( {\pi }^{\sharp } \) is a vector space isomorphism from \( {L}^{2}\left( {G/H}\right) \) onto \( {L}^{2}{\left( G\right) }^{H} \) . Moreover,\n\n\[ \n{\pi }^{\sharp }\left( \widehat{G/H}\right) \subseteq {\widehat{G}}^{H} \n\]\n\nand the map\n\n\[ \n{\pi }^{\sharp } : \widehat{G/H} \rightarrow {\widehat{G}}^{H} \n\]\n\nis a group isomorphism.
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Proof. Let \( {f}^{\sharp } \in {L}^{2}\left( {G/H}\right) \) . If \( x \in G \) and \( h \in H \), then\n\n\[ \n{\pi }^{\sharp }\left( {f}^{\sharp }\right) \left( {x + h}\right) = {f}^{\sharp }\pi \left( {x + h}\right) = {f}^{\sharp }\pi \left( x\right) = {\pi }^{\sharp }\left( {f}^{\sharp }\right) \left( x\right) , \n\]\n\nand so \( {\pi }^{\sharp } \) maps \( {L}^{2}\left( {G/H}\right) \) into \( {L}^{2}{\left( G\right) }^{H} \) . It is easy to check that \( {\pi }^{\sharp } \) is linear. Moreover, \( {\pi }^{\sharp } \) is onto, since if \( f \in {L}^{2}{\left( G\right) }^{H} \), then there is a well-defined map \( {f}^{\sharp } \in {L}^{2}\left( {G/H}\right) \) given by \( {f}^{\sharp }\left( {x + H}\right) = f\left( x\right) \), and \( {\pi }^{\sharp }\left( {f}^{\sharp }\right) \left( x\right) = {f}^{\sharp }\left( {x + H}\right) = \) \( f\left( x\right) \) for all \( x \in G \) . Finally, \( {\pi }^{\sharp } \) is one-to-one since \( {\pi }^{\sharp }\left( {f}^{\sharp }\right) \left( x\right) = 0 \) for all \( x \in G \) if and only if \( {f}^{\sharp }\left( {x + H}\right) = 0 \) for all \( x + H \in G/H \), that is, if and only if \( {f}^{\sharp } = 0 \) . This proves that \( {\pi }^{\sharp } \) is an isomorphism.\n\nIf \( {\chi }^{\sharp } \in \widehat{G/H} \), then\n\n\[ \n{\pi }^{\sharp }\left( {\chi }^{\sharp }\right) \left( {x + y}\right) = {\chi }^{\sharp }\left( {\pi \left( {x + y}\right) }\right) \n\]\n\n\[ \n= {\chi }^{\sharp }\left( {x + y + H}\right) \n\]\n\n\[ \n= {\chi }^{\sharp }\left( {x + H}\right) {\chi }^{\sharp }\left( {y + H}\right) \n\]\n\n\[ \n= {\pi }^{\sharp }\left( {\chi }^{\sharp }\right) \left( x\right) {\pi }^{\sharp }\left( {\chi }^{\sharp }\right) \left( y\right) \n\]\n\nand so\n\n\[ \n{\pi }^{\sharp }\left( {\chi }^{\sharp }\right) \in \widehat{G} \cap {L}^{2}{\left( G\right) }^{H} = {\widehat{G}}^{H}. \n\]\n\nIt is left as an exercise to prove that \( {\pi }^{\sharp } : \widehat{G/H} \rightarrow {\widehat{G}}^{H} \) is a group isomorphism (Exercise 2). \( ▱ \)
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No
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Theorem 4.11 (Poisson summation formula) Let \( G \) be a finite abelian group and \( H \) a subgroup of \( G \) . If \( f \in {L}^{2}\left( G\right) \), then\n\n\[ \frac{1}{\left| H\right| }\mathop{\sum }\limits_{{y \in H}}f\left( y\right) = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{\chi \in {\widehat{G}}^{H}}}\widehat{f}\left( \chi \right) \]
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Proof. Let \( f \in {L}^{2}\left( G\right) \) and \( \chi \in {\widehat{G}}^{H} \) . We define the function \( {f}^{\sharp } \in \) \( {L}^{2}\left( {G/H}\right) \) by\n\n\[ {f}^{\sharp }\left( {x + H}\right) = \mathop{\sum }\limits_{{y \in H}}f\left( {x + y}\right) \]\n\nWe define the character \( {\chi }^{\sharp } \in \widehat{G/H} \) by \( {\chi }^{\sharp }\left( {x + H}\right) = \chi \left( x\right) \) . If \( {\pi }^{\sharp } : \widehat{G/H} \rightarrow {\widehat{G}}^{H} \) is the isomorphism constructed in Lemma 4.5, then \( {\pi }^{\sharp }\left( {\chi }^{\sharp }\right) = \chi \), and the Fourier transform of \( {f}^{\sharp } \) is\n\n\[ {\widehat{f}}^{\sharp }\left( {\chi }^{\sharp }\right) = \mathop{\sum }\limits_{{x + H \in G/H}}{f}^{\sharp }\left( {x + H}\right) \overline{{\chi }^{\sharp }}\left( {x + H}\right) \]\n\n\[ = \mathop{\sum }\limits_{{x + H \in G/H}}\mathop{\sum }\limits_{{y \in H}}f\left( {x + y}\right) \bar{\chi }\left( x\right) \]\n\n\[ = \mathop{\sum }\limits_{{x + H \in G/H}}\mathop{\sum }\limits_{{y \in H}}f\left( {x + y}\right) \bar{\chi }\left( {x + y}\right) \]\n\n\[ = \mathop{\sum }\limits_{{x \in G}}f\left( x\right) \bar{\chi }\left( x\right) \]\n\n\[ = \widehat{f}\left( \chi \right) \]\n\nIt follows that the Fourier series for \( {f}^{\sharp } \) is\n\n\[ {f}^{\sharp }\left( {x + H}\right) = \frac{1}{\left| G/H\right| }\mathop{\sum }\limits_{{{\chi }^{\sharp } \in \widehat{G/H}}}\widehat{{f}^{\sharp }}\left( {\chi }^{\sharp }\right) \overline{{\chi }^{\sharp }}\left( {x + H}\right) \]\n\n\[ = \frac{\left| H\right| }{\left| G\right| }\mathop{\sum }\limits_{{\chi \in {\widehat{G}}^{H}}}\widehat{f}\left( \chi \right) \bar{\chi }\left( x\right) \]\n\nEquivalently, for \( x \in G \) ,\n\n\[ \frac{1}{\left| H\right| }\mathop{\sum }\limits_{{y \in H}}f\left( {x + y}\right) = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{\chi \in {\widehat{G}}^{H}}}\widehat{f}\left( \chi \right) \bar{\chi }\left( x\right) . \]\n\nThis is the Poisson summation formula.
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Yes
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Theorem 4.13 Let \( G = \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) be an abelian group of order \( n \) . Let \( K \in {L}^{2}\left( {G \times G}\right) \) and let \( {\Phi }_{K} \) be the associated integral operator on \( {L}^{2}\left( G\right) \) . The matrix of \( {\Phi }_{K} \) with respect to the orthonormal basis \( \left\{ {{\delta }_{{x}_{i}} : i = 1,\ldots, n}\right\} \) is \( \left( {K\left( {{x}_{i},{x}_{j}}\right) }\right) \), and the trace of \( {\Phi }_{K} \) is\n\n\[ \n\operatorname{tr}\left( {\Phi }_{K}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}K\left( {{x}_{i},{x}_{i}}\right) \n\]
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Proof. The matrix of the operator \( {\Phi }_{K} \) is \( \left( {c}_{ij}\right) \), where \( {c}_{ij} \) is defined by\n\n\[ \n{\Phi }_{K}\left( {\delta }_{{x}_{j}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{c}_{ij}{\delta }_{{x}_{i}} \n\]\n\nThen\n\n\[ \n{c}_{ij} = {\Phi }_{K}\left( {\delta }_{{x}_{j}}\right) \left( {x}_{i}\right) = \mathop{\sum }\limits_{{y \in G}}K\left( {{x}_{i}, y}\right) {\delta }_{{x}_{j}}\left( y\right) = K\left( {{x}_{i},{x}_{j}}\right) . \n\]\n\nThis completes the proof.
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Yes
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Theorem 4.15 (Trace formula) For \( h \in {L}^{2}\left( G\right) \), let \( {C}_{h} \) be the convolution operator on \( {L}^{2}\left( G\right) \), that is, \( {C}_{h}\left( f\right) = h * f \) for \( f \in {L}^{2}\left( G\right) \). The dual group \( \widehat{G} \) is a basis of eigenvectors for \( {C}_{h} \). If \( \chi \) is a character in \( \widehat{G} \), then \( \chi \) has eigenvalue \( \widehat{h}\left( \chi \right) \), that is,\n\n\[ \n{C}_{h}\left( \chi \right) = \widehat{h}\left( \chi \right) \chi \n\]\n\nand\n\n\[ \n{nh}\left( 0\right) = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{h}\left( \chi \right) \n\]
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Proof. This is a straightforward calculation. For \( x \in G \), we have\n\n\[ \n{C}_{h}\left( \chi \right) \left( x\right) = h * \chi \left( x\right) = \chi * h\left( x\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{{y \in G}}\chi \left( {x - y}\right) h\left( y\right) \n\]\n\n\[ \n= \left( {\mathop{\sum }\limits_{{y \in G}}h\left( y\right) \bar{\chi }\left( y\right) }\right) \chi \left( x\right) \n\]\n\n\[ \n= \widehat{h}\left( \chi \right) \chi \left( x\right) \n\]\n\nand so \( \chi \) is an eigenvector of the convolution \( {C}_{h} \) with eigenvalue \( \widehat{h}\left( \chi \right) \). By Theorem 4.12, since \( \widehat{G} \) is a basis for \( {L}^{2}\left( G\right) \), the trace of \( {C}_{h} \) is the sum of the eigenvalues, that is,\n\n\[ \n\operatorname{tr}\left( {C}_{h}\right) = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{h}\left( \chi \right) \n\]\n\nBy Theorem 4.14, we also have\n\n\[ \n\operatorname{tr}\left( {C}_{h}\right) = {nh}\left( 0\right) \n\]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 4.17 If \( p \) is an odd prime and \( \left( {a, p}\right) = 1 \), then\n\n\[ \n\tau \left( {{\ell }_{p}, a}\right) = \mathop{\sum }\limits_{{x = 0}}^{{p - 1}}{e}_{p}\left( {a{x}^{2}}\right) .\n\]\n\nIn particular,\n\n\[ \n\tau \left( p\right) = \mathop{\sum }\limits_{{x = 0}}^{{p - 1}}{e}^{{2\pi i}{x}^{2}/p}.\n\]
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Proof. The set \( R = \left\{ {k \in \{ 1,\ldots, p - 1\} : {\ell }_{p}\left( {k + p\mathbf{Z}}\right) = 1}\right\} \) is a set of representatives of the congruence classes of quadratic residues modulo \( p \) , and \( N = \left\{ {k \in \{ 1,\ldots, p - 1\} : {\ell }_{p}\left( {k + p\mathbf{Z}}\right) = - 1}\right\} \) is a set of representatives of the congruence classes of quadratic nonresidues modulo \( p \) . We have \( \left| R\right| = \) \( \left| N\right| = \left( {p - 1}\right) /2 \) . If \( {x}^{2} \equiv k\left( {\;\operatorname{mod}\;p}\right) \), then also \( {\left( p - x\right) }^{2} \equiv k\left( {\;\operatorname{mod}\;p}\right) \) . Let \( x ≢ 0\;\left( {\;\operatorname{mod}\;p}\right) \) . Since \( p \) is odd, \( x ≢ p - x\;\left( {\;\operatorname{mod}\;p}\right) \), and\n\n\[ \n\mathop{\sum }\limits_{{x = 1}}^{{p - 1}}{e}_{p}\left( {a{x}^{2}}\right) = 2\mathop{\sum }\limits_{{k \in R}}{e}_{p}\left( {ak}\right) .\n\]\n\nIt follows that\n\n\[ \n\tau \left( {{\ell }_{p}, a}\right) = \mathop{\sum }\limits_{{k = 1}}^{{p - 1}}\left( \frac{k}{p}\right) {e}_{p}\left( {ak}\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{{k \in R}}{e}_{p}\left( {ak}\right) - \mathop{\sum }\limits_{{k \in N}}{e}_{p}\left( {ak}\right) \n\]\n\n\[ \n= 2\mathop{\sum }\limits_{{k \in R}}{e}_{p}\left( {ak}\right) - \mathop{\sum }\limits_{{k \in R \cup N}}{e}_{p}\left( {ak}\right) \n\]\n\n\[ \n= 1 + 2\mathop{\sum }\limits_{{k \in R}}{e}_{p}\left( {ak}\right) - \mathop{\sum }\limits_{{k = 0}}^{{p - 1}}{e}_{p}\left( {ak}\right) \n\]\n\n\[ \n= 1 + \mathop{\sum }\limits_{{x = 1}}^{{p - 1}}{e}_{p}\left( {a{x}^{2}}\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{{x = 0}}^{{p - 1}}{e}_{p}\left( {a{x}^{2}}\right) \n\]\n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 4.18 If \( p \) is prime and \( \left( {a, p}\right) = 1 \), then\n\n\[ \tau {\left( {\ell }_{p}, a\right) }^{2} = \left( \frac{-1}{p}\right) p = {\left( -1\right) }^{\frac{p - 1}{2}}p. \]
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Proof. If \( p \) does not divide \( a \), then\n\n\[ \tau {\left( {\ell }_{p}, a\right) }^{2} = \mathop{\sum }\limits_{{x = 1}}^{{p - 1}}\left( \frac{x}{p}\right) {e}_{p}\left( {ax}\right) \mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{y}{p}\right) {e}_{p}\left( {ay}\right) \]\n\n\[ = \mathop{\sum }\limits_{{x = 1}}^{{p - 1}}\mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{xy}{p}\right) {e}_{p}\left( {a\left( {x + y}\right) }\right) . \]\n\nLet \( \left( {x, p}\right) = 1 \) . Then \( \{ x,{2x},\ldots ,\left( {p - 1}\right) x\} \) is a reduced set of residues modulo \( p,\left( \frac{{x}^{2}}{p}\right) = 1 \), and\n\n\[ \mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{xy}{p}\right) {e}_{p}\left( {-a\left( {x + y}\right) }\right) = \mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{x\left( {xy}\right) }{p}\right) {e}_{p}\left( {-a\left( {x + {xy}}\right) }\right) \]\n\n\[ = \mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{{x}^{2}y}{p}\right) {e}_{p}\left( {-{ax}\left( {1 + y}\right) }\right) \]\n\n\[ = \mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{{x}^{2}}{p}\right) \left( \frac{y}{p}\right) {e}_{p}\left( {-{ax}\left( {1 + y}\right) }\right) \]\n\n\[ = \mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{y}{p}\right) {e}_{p}\left( {-{ax}\left( {1 + y}\right) }\right) . \]\n\nSince\n\n\[ \mathop{\sum }\limits_{{x = 1}}^{{p - 1}}{e}_{p}\left( {-{ax}\left( {1 + y}\right) }\right) = \left\{ \begin{array}{lll} p - 1 & \text{ if }y \equiv p - 1 & \left( {\;\operatorname{mod}\;p}\right) , \\ - 1 & \text{ if }y ≢ p - 1 & \left( {\;\operatorname{mod}\;p}\right) , \end{array}\right. \]\n\nit follows that\n\n\[ \tau {\left( {\ell }_{p}, a\right) }^{2} = \mathop{\sum }\limits_{{x = 1}}^{{p - 1}}\mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{xy}{p}\right) {e}_{p}\left( {a\left( {x + y}\right) }\right) \]\n\n\[ = \mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{y}{p}\right) \mathop{\sum }\limits_{{x = 1}}^{{p - 1}}{e}_{p}\left( {-{ax}\left( {1 + y}\right) }\right) \]\n\n\[ = \left( \frac{-1}{p}\right) \left( {p - 1}\right) - \mathop{\sum }\limits_{{y = 1}}^{{p - 2}}\left( \frac{y}{p}\right) \]\n\n\[ = \left( \frac{-1}{p}\right) p - \mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{y}{p}\right) \]\n\n\[ = \left( \frac{-1}{p}\right) p \]\n\n\[ = {\left( -1\right) }^{\frac{p - 1}{2}}p \]\n\nby Theorem 3.14. \( ▱ \)
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Yes
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Theorem 4.20 If \( p \) and \( q \) are distinct odd primes, then\n\n\[ \left( \widehat{{\widehat{{\ell }_{p}}}^{q}}\right) \left( {\Delta \left( {-q + p\mathbf{Z}}\right) }\right) = {p\tau }{\left( p\right) }^{q - 1}\left( \frac{q}{p}\right) .
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Proof. The function on the left side of the equation is a bit complicated. Let \( G = \mathbf{Z}/p\mathbf{Z} \) . Since \( {\ell }_{p} \in {L}^{2}\left( G\right) \), it follows that the Fourier transform \( {\ell }_{p} \in {L}^{2}\left( \widehat{G}\right) \), and also its \( q \) th power \( {\widehat{\ell }}_{p}^{q} \in {L}^{2}\left( \widehat{G}\right) \) . The Fourier transform of this function is \( \widehat{{\ell }_{p}^{q}} \in {L}^{2}\left( \widehat{\widehat{G}}\right) \), and so its domain is \( \widehat{\widehat{G}} = \{ \Delta \left( {a + p\mathbf{Z}}\right) \) : \( a + p\mathbf{Z} \in G\} \) . We have\n\n\[ \left( \widehat{{\widehat{{\ell }_{p}}}^{q}}\right) \left( {\Delta \left( {-q + p\mathbf{Z}}\right) }\right) = \mathop{\sum }\limits_{{x = 0}}^{{p - 1}}{\widehat{{\ell }_{p}}}^{q}\left( {\psi }_{x}\right) \overline{\Delta \left( {-q + p\mathbf{Z}}\right) }\left( {\psi }_{x}\right) \]\n\n\[ = \mathop{\sum }\limits_{{x = 0}}^{{p - 1}}{\left( \widehat{{\ell }_{p}}\left( {\psi }_{x}\right) \right) }^{q}\overline{\Delta \left( {-q + p\mathbf{Z}}\right) \left( {\psi }_{x}\right) } \]\n\n\[ = \mathop{\sum }\limits_{{x = 0}}^{{p - 1}}\tau {\left( {\ell }_{p}, - x\right) }^{q}\overline{{\psi }_{x}\left( {-q + p\mathbf{Z}}\right) } \]\n\n\[ = \mathop{\sum }\limits_{{x = 1}}^{{p - 1}}{\left( \left( \frac{-x}{p}\right) \tau \left( p\right) \right) }^{q}{\psi }_{x}\left( {q + p\mathbf{Z}}\right) \]\n\n\[ = \tau {\left( p\right) }^{q}\mathop{\sum }\limits_{{x = 1}}^{{p - 1}}\left( \frac{-x}{p}\right) {e}_{p}\left( {qx}\right) \]\n\n\[ = \left( \frac{-q}{p}\right) \tau {\left( p\right) }^{q}\mathop{\sum }\limits_{{x = 1}}^{{p - 1}}\left( \frac{qx}{p}\right) {e}_{p}\left( {qx}\right) \]\n\n\[ = \left( \frac{-q}{p}\right) \tau {\left( p\right) }^{q}\mathop{\sum }\limits_{{x = 1}}^{{p - 1}}\left( \frac{x}{p}\right) {e}_{p}\left( x\right) \]\n\n\[ = \left( \frac{-q}{p}\right) \tau {\left( p\right) }^{q + 1} \]\n\n\[ = \left( \frac{-q}{p}\right) \left( \frac{-1}{p}\right) {p\tau }{\left( p\right) }^{q - 1} \]\n\n\[ = {p\tau }{\left( p\right) }^{q - 1}\left( \frac{q}{p}\right) \]\n\nby Theorem 4.18. This completes the proof.
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Yes
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Theorem 4.21 If \( p \) and \( q \) are distinct odd primes, then\n\n\[ \left( \widehat{{\widehat{{\ell }_{p}}}^{q}}\right) \left( {\Delta \left( {-q + p\mathbf{Z}}\right) }\right) = p\mathop{\sum }\limits_{\substack{{{x}_{1} + \cdots + {x}_{q} \equiv q\;\left( {\;\operatorname{mod}\;p}\right) } \\ {1 \leq {x}_{i} \leq p - 1} }}\left( \frac{{x}_{1}\cdots {x}_{q}}{p}\right) .
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Proof. Let \( k \) be a positive integer. By Exercise 10 in Section 4.3, a product of Fourier transforms is the Fourier transform of the convolution,\n\nand so\n\[ {\widehat{{\ell }_{p}}}^{k} = \underset{k\text{ times }}{\underbrace{\widehat{{\ell }_{p}} * \cdots * \widehat{{\ell }_{p}}}} = \underset{k\text{ times }}{\underbrace{\underline{{\ell }_{p}} * \cdots * {\ell }_{p}}}. \]\n\nBy (4.8) of Theorem 4.8, for every integer \( a \) we have\n\n\[ \left( \widehat{{\ell }_{p}^{k}}\right) \left( {\Delta \left( {-a + p\mathbf{Z}}\right) }\right) = \underset{k\text{ times }}{\underbrace{{\ell }_{p} * \cdots * {\ell }_{p}}}\left( {\Delta \left( {-a + p\mathbf{Z}}\right) }\right) \]\n\n\[ = p\underset{k\text{ times }}{\underbrace{{\ell }_{p} * \cdots * {\ell }_{p}}}\left( {a + p\mathbf{Z}}\right) . \]\n\nBy Exercise 12 in Section 4.3,\n\n\[ \underset{k\text{ times }}{\underbrace{{\ell }_{p} * \cdots * {\ell }_{p}}}\left( {a + p\mathbf{Z}}\right) = \mathop{\sum }\limits_{\substack{{{x}_{1} + \cdots + {x}_{k} \equiv a\;\left( {\;\operatorname{mod}\;p}\right) } \\ {1 \leq {x}_{i} \leq p - 1} }}\left( \frac{{x}_{1}\cdots {x}_{k}}{p}\right) . \]\n\nIf \( k = a = q \), then\n\n\[ \left( \widehat{{\widehat{{\ell }_{p}}}^{q}}\right) \left( {\Delta \left( {-q + p\mathbf{Z}}\right) }\right) = p\mathop{\sum }\limits_{\substack{{{x}_{1} + \cdots + {x}_{q} \equiv q\;\left( {\;\operatorname{mod}\;p}\right) } \\ {1 \leq {x}_{i} \leq p - 1} }}\left( \frac{{x}_{1}\cdots {x}_{q}}{p}\right) . \]\nThis completes the proof.
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Yes
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Theorem 4.22 For all functions \( f \in {L}^{2}\left( {\mathbf{Z}/n\mathbf{Z}}\right) \) , \[ {\mathcal{F}}^{2}\left( f\right) \left( {a + n\mathbf{Z}}\right) = {nf}\left( {-a + n\mathbf{Z}}\right) . \]
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Proof. This is similar to the proof of (4.8) in Theorem 4.8. Writing \( \mathcal{F}\left( f\right) = g \), we have \[ g\left( {x + n\mathbf{Z}}\right) = \mathop{\sum }\limits_{{y = 0}}^{{n - 1}}f\left( {y + n\mathbf{Z}}\right) {\omega }^{-{xy}} \] and \[ {\mathcal{F}}^{2}\left( f\right) \left( {a + n\mathbf{Z}}\right) = \mathcal{F}\left( g\right) \left( {a + n\mathbf{Z}}\right) \] \[ = \mathop{\sum }\limits_{{x = 0}}^{{n - 1}}g\left( {x + n\mathbf{Z}}\right) {\omega }^{-{ax}} \] \[ = \mathop{\sum }\limits_{{x = 0}}^{{n - 1}}\mathop{\sum }\limits_{{y = 0}}^{{n - 1}}f\left( {y + n\mathbf{Z}}\right) {\omega }^{-{xy}}{\omega }^{-{ax}} \] \[ = \mathop{\sum }\limits_{{y = 0}}^{{n - 1}}f\left( {y + n\mathbf{Z}}\right) \mathop{\sum }\limits_{{x = 0}}^{{n - 1}}{\omega }^{-x\left( {a + y}\right) } \] \[ = {nf}\left( {-a + n\mathbf{Z}}\right) \text{.} \] This completes the proof. \( ▱ \)
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Yes
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Theorem 5.1 The spectrum of the ring of integers is\n\n\[ \operatorname{Spec}\left( \mathbf{Z}\right) = \{ p\mathbf{Z} : p\text{ is prime or }p = 0\} . \]
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Proof. Since \( \mathbf{Z} \) is principal, every ideal is of the form \( d\mathbf{Z} \) for some nonnegative integer \( d \) . If \( d = 0 \), then \( d\mathbf{Z} = \{ 0\} \), and the zero ideal is prime, since \( {ab} = 0 \) if and only if \( a = 0 \) or \( b = 0 \) . Let \( d \geq 1 \) . If \( d = p \) is prime and \( {ab} \in p\mathbf{Z} \), then \( p \) divides \( {ab} \) . By Euclid’s lemma, \( p \) divides \( a \) or \( p \) divides \( b \) , and so \( a \in p\mathbf{Z} \) or \( b \in p\mathbf{Z} \) . Therefore, \( p\mathbf{Z} \) is a prime ideal for every prime number \( p \) .\n\nIf \( d \) is composite, then we can write \( d = {ab} \), where \( 1 < a \leq b < d \) . If \( a \in d\mathbf{Z} \), then \( a = {dk} = {abk} \) for some positive integer \( k \), and so \( 1 = {bk} \), which is absurd. Therefore, \( a \notin d\mathbf{Z} \) and, similarly, \( b \notin d\mathbf{Z} \) . Since \( d = {ab} \in d\mathbf{Z} \), it follows that \( d\mathbf{Z} \) is not a prime ideal. Thus, the prime ideals in the ring \( \mathbf{Z} \) are the ideals of the form \( p\mathbf{Z} \), where \( p \) is a prime number or \( p = 0 \) .
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Yes
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Theorem 5.2 For \( m \geq 2 \), let \( \mathbf{Z}/m\mathbf{Z} \) be the ring of congruence classes modulo \( m \) . Then\n\n(i) \( \mathbf{Z}/m\mathbf{Z} \) is principal, and the ideals of \( \mathbf{Z}/m\mathbf{Z} \) are the ideals generated by the congruence classes \( d + m\mathbf{Z} \), where \( d \) is a divisor of \( m \) ;\n\n(ii) the prime ideals of \( \mathbf{Z}/m\mathbf{Z} \) are the ideals generated by the congruence classes \( p + m\mathbf{Z} \), where \( p \) is a prime divisor of \( m \) ; and\n\n(iii) the radical of \( \mathbf{Z}/m\mathbf{Z} \) is the ideal generated by the congruence class \( \operatorname{rad}\left( m\right) + m\mathbf{Z} \) .
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Proof. Let \( J \) be an ideal in the ring \( R = \mathbf{Z}/m\mathbf{Z} \) . Consider the union of congruence classes\n\n\[ I = \mathop{\bigcup }\limits_{{a + m\mathbf{Z} \in J}}\left( {a + m\mathbf{Z}}\right) \]\n\nThe set \( I \) is an ideal in \( \mathbf{Z} \) . Since \( \mathbf{Z} \) is principal, \( I = d\mathbf{Z} \) for some positive integer \( d \in I \) . Since \( m \in m\mathbf{Z} \subseteq I \), it follows that \( d \) is a divisor of \( m \) . Moreover, \( d + m\mathbf{Z} \in J \), and so the principal ideal generated by \( d + m\mathbf{Z} \) in \( \mathbf{Z}/m\mathbf{Z} \) is contained in \( J \) . If \( a + m\mathbf{Z} \in J \), then \( a \in a + m\mathbf{Z} \subseteq I \), and so \( a = {dr} \) for some integer \( r \) . It follows that \( a + m\mathbf{Z} = \left( {d + m\mathbf{Z}}\right) \left( {r + m\mathbf{Z}}\right) \) belongs to the principal ideal generated by \( d + m\mathbf{Z} \) . Therefore, \( J \) is the principal ideal generated by \( d + m\mathbf{Z} \), and \( a + m\mathbf{Z} \in J \) if and only if \( d \) divides \( a \) . (See Exercise 3 for a different proof.)\n\nNext we compute the spectrum of the ring \( \mathbf{Z}/m\mathbf{Z} \) . Let \( J \) be the principal ideal generated by \( d + m\mathbf{Z} \), where \( d \) divides \( m \) and \( d \geq 2 \) . If \( d = p \) is prime and\n\n\[ \left( {a + m\mathbf{Z}}\right) \left( {b + m\mathbf{Z}}\right) = {ab} + m\mathbf{Z} \in J \]\n\nthen \( p \) divides \( {ab} \) and so \( p \) divides \( a \) or \( p \) divides \( b \), that is, \( a + m\mathbf{Z} \in J \) or \( b + m\mathbf{Z} \in J \), and \( J \) is a prime ideal.\n\nIf \( d = {ab} \) is composite, where \( 1 < a \leq b < d \), then \( a + m\mathbf{Z} \notin J, b + m\mathbf{Z} \notin J \) , but \( \left( {a + m\mathbf{Z}}\right) \left( {b + m\mathbf{Z}}\right) = d + m\mathbf{Z} \in J \), and so \( J \) is not a prime ideal. Thus, the prime ideals of the ring \( \mathbf{Z}/m\mathbf{Z} \) are the ideals of the form \( p + m\mathbf{Z} \), where \( p \) is a prime divisor of \( m \) .\n\nFinally, the congruence class \( a + m\mathbf{Z} \) is nilpotent in \( R \) if and only if \( {\left( a + m\mathbf{Z}\right) }^{k} = {a}^{k} + m\mathbf{Z} = m\mathbf{Z} \) for some positive integer \( k \) . Equivalently, \( a + m\mathbf{Z} \) is nilpotent if and only if \( m \) divides \( {a}^{k} \) for some positive integer \( k \) . By Theorem 1.13, this is possible if and only if \( a \) is divisible by \( \operatorname{rad}\left( m\right) \), and so \( \mathcal{N}\left( {\mathbf{Z}/m\mathbf{Z}}\right) \) is the ideal generated by the congruence class \( \operatorname{rad}\left( m\right) + m\mathbf{Z} \) .
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Yes
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Theorem 5.3 The ring \( \mathbf{C}\left\lbrack t\right\rbrack \) of polynomials with coefficients in the field \( \mathbf{C} \) of complex numbers is a principal ring.
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Proof. This is a special case of Exercise 18 in Section 3.1. \( ▱ \)
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No
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Theorem 5.4 Let \( f\left( t\right) \in \mathbf{C}\left\lbrack t\right\rbrack \) and \( R = \mathbf{C}\left\lbrack t\right\rbrack /I \), where \( I = \langle f\left( t\right) \rangle \) is the principal ideal generated by \( f\left( t\right) \) . The radical of \( R \) is the principal ideal generated by \( \operatorname{rad}\left( f\right) + I \) .
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Proof. This follows immediately from the observation that if \( f\left( t\right) \) and \( g\left( t\right) \) are polynomials with complex coefficients, then there exists a positive integer \( k \) such that \( f\left( t\right) \) divides \( g{\left( t\right) }^{k} \) if and only if \( \operatorname{rad}\left( f\right) \) divides \( g\left( t\right) \) . \( ▱ \)
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Yes
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Theorem 5.5 Let \( R \) be a ring and \( R\\left\\lbrack t\\right\\rbrack \) the ring of polynomials with coefficients in \( R \) . Define \( D : R\\left\\lbrack t\\right\\rbrack \\rightarrow R\\left\\lbrack t\\right\\rbrack \) by\n\n\[ D\\left( {\\mathop{\\sum }\\limits_{{i = 0}}^{m}{a}_{i}{t}^{i}}\\right) = \\mathop{\\sum }\\limits_{{i = 1}}^{m}i{a}_{i}{t}^{i - 1}. \]\n\nThen \( D \) is a derivation on \( R\\left\\lbrack t\\right\\rbrack \) .
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Proof. Let \( f = f\\left( t\\right) = \\mathop{\\sum }\\limits_{{i = 0}}^{m}{a}_{i}{t}^{i} \) and \( g = g\\left( t\\right) = \\mathop{\\sum }\\limits_{{j = 0}}^{n}{b}_{j}{t}^{j} \) . It is immediate that \( D\\left( {f + g}\\right) = D\\left( f\\right) + D\\left( g\\right) \), and so \( D \) is a homomorphism of the additive group of polynomials. Since\n\n\[ f\\left( t\\right) g\\left( t\\right) = \\mathop{\\sum }\\limits_{{i = 0}}^{m}\\mathop{\\sum }\\limits_{{j = 0}}^{n}{a}_{i}{t}^{i}{b}_{j}{t}^{j} = \\mathop{\\sum }\\limits_{{k = 0}}^{{m + n}}\\mathop{\\sum }\\limits_{{i + j = k}}{a}_{i}{b}_{j}{t}^{k}, \]\n\nwe have\n\n\[ D\\left( {fg}\\right) = \\mathop{\\sum }\\limits_{{k = 1}}^{{m + n}}k\\mathop{\\sum }\\limits_{{i + j = k}}{a}_{i}{b}_{j}{t}^{k - 1} \]\n\n\[ = \\mathop{\\sum }\\limits_{{k = 1}}^{{m + n}}\\mathop{\\sum }\\limits_{{i + j = k}}\\left( {i + j}\\right) {a}_{i}{b}_{j}{t}^{i + j - 1} \]\n\n\[ = \\mathop{\\sum }\\limits_{{k = 1}}^{{m + n}}\\mathop{\\sum }\\limits_{{i + j = k}}i{a}_{i}{t}^{i - 1}{b}_{j}{t}^{j} + \\mathop{\\sum }\\limits_{{k = 1}}^{{m + n}}\\mathop{\\sum }\\limits_{{i + j = k}}{a}_{i}{t}^{i}j{b}_{j}{t}^{j - 1} \]\n\n\[ = \\mathop{\\sum }\\limits_{{i = 1}}^{m}\\mathop{\\sum }\\limits_{{j = 0}}^{n}i{a}_{i}{t}^{i - 1}{b}_{j}{t}^{j} + \\mathop{\\sum }\\limits_{{i = 0}}^{m}\\mathop{\\sum }\\limits_{{j = 1}}^{n}{a}_{i}{t}^{i}j{b}_{j}{t}^{j - 1} \]\n\n\[ = D\\left( f\\right) g + {fD}\\left( g\\right) \\text{.} \]\n\nTherefore, \( D \) is a derivation on \( R\\left\\lbrack t\\right\\rbrack \) .
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Yes
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Theorem 5.6 Let \( R \) be an integral domain with quotient field \( F \), and let \( D \) be a derivation on \( R \) . There exists a unique derivation \( {D}_{F} \) on \( F \) such that \( {D}_{F}\left( x\right) = D\left( x\right) \) for all \( x \in R \) .
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Proof. Suppose that there exists a derivation \( {D}_{F} \) on \( F \) such that \( {D}_{F}\left( a\right) = \) \( D\left( a\right) \) for all \( a \in R \) . Let \( x \in F, x \neq 0 \) . There exist \( a, b \in R \) with \( b \neq 0 \) and \( x = a/b \) . Since \( a = {bx} \in R \), it follows that\n\n\[ D\left( a\right) = {D}_{F}\left( a\right) = {D}_{F}\left( {bx}\right) = {D}_{F}\left( b\right) x + b{D}_{F}\left( x\right) = D\left( b\right) x + b{D}_{F}\left( x\right) ,\]\n\nand so\n\n\[ {D}_{F}\left( \frac{a}{b}\right) = {D}_{F}\left( x\right) = \frac{D\left( a\right) - D\left( b\right) x}{b} = \frac{D\left( a\right) b - {aD}\left( b\right) }{{b}^{2}}. \]\n\n(5.3)\n\nThus, the derivation \( {D}_{F} \) on \( \mathrm{F} \) is uniquely determined by the derivation \( D \) on \( R \) . In Exercise 3 we prove that (5.3) defines a derivation on the quotient field \( {R}_{F} \) .
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No
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Theorem 5.8 If \( n \geq 3 \), then the Fermat equation \( {x}^{n} + {y}^{n} = {z}^{n} \) has no solution in nonzero, relatively prime polynomials, not all constant.
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Proof. Let \( n \geq 3 \), and suppose that \( x, y \), and \( z \) are nonzero, relatively prime polynomials, not all constant, such that \( {x}^{n} + {y}^{n} = {z}^{n} \) . We apply Mason’s theorem with \( a = {x}^{n}, b = {y}^{n} \), and \( c = {z}^{n} \) . Then\n\n\[ \operatorname{rad}\left( {abc}\right) = \operatorname{rad}\left( {{x}^{n}{y}^{n}{z}^{n}}\right) = \operatorname{rad}\left( {xyz}\right) .\n\]\n\nSince \( \deg \left( {x}^{n}\right) = n\deg \left( x\right) \), we obtain\n\n\[ n\deg \left( x\right) \leq n\max \left( {\deg \left( x\right) ,\deg \left( y\right) ,\deg \left( z\right) }\right)\n\]\n\n\[ = \max \left( {\deg \left( {x}^{n}\right) ,\deg \left( {y}^{n}\right) ,\deg \left( {z}^{n}\right) }\right)\n\]\n\n\[ = \max \left( {\deg \left( a\right) ,\deg \left( b\right) ,\deg \left( c\right) }\right)\n\]\n\n\[ \leq \deg \left( {\operatorname{rad}\left( {abc}\right) }\right) - 1\n\]\n\n\[ = \deg \left( {\operatorname{rad}\left( {xyz}\right) }\right) - 1\n\]\n\n\[ \leq \deg \left( {xyz}\right) - 1\n\]\n\n\[ = \deg \left( x\right) + \deg \left( y\right) + \deg \left( z\right) - 1\text{.} \]\n\nIt follows that\n\n\[ n\left( {\deg \left( x\right) + \deg y + \deg \left( z\right) }\right) \leq 3\left( {\deg \left( x\right) + \deg y + \deg \left( z\right) }\right) - 3\n\]\n\n\[ \leq n\left( {\deg \left( x\right) + \deg y + \deg \left( z\right) }\right) - 3. \]\n\nThis is impossible.
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Yes
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Theorem 5.9 (Asymptotic Fermat theorem) The abc conjecture implies that there exists an integer \( {n}_{0} \) such that the Fermat equation has no solution in relatively prime integers for any exponent \( n \geq {n}_{0} \) .
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Proof. Let \( x, y \), and \( z \) be relatively prime positive integers such that\n\n\[ \n{x}^{n} + {y}^{n} = {z}^{n} \n\] \n\nWe note that \n\n\[ \n\operatorname{rad}\left( {{x}^{n}{y}^{n}{z}^{n}}\right) = \operatorname{rad}\left( {xyz}\right) \leq {xyz} \leq {z}^{3}. \n\] \n\nIf \( n \geq 2 \), then \( z \geq 3 \) . Applying the \( {abc} \) conjecture with \( \varepsilon = 1 \) and \( {K}_{1} = \) \( \max \left( {1, K\left( 1\right) }\right) \), we obtain \n\n\[ \n{z}^{n} = \max \left( {{x}^{n},{y}^{n},{z}^{n}}\right) \leq {K}_{1}\operatorname{rad}{\left( {x}^{n}{y}^{n}{z}^{n}\right) }^{2} < {K}_{1}{z}^{6}, \n\] \n\nand so \n\[ \nn < 6 + \frac{\log {K}_{1}}{\log z} \leq 6 + \frac{\log {K}_{1}}{\log 3}. \n\] \n\nThis completes the proof. \( ▱ \)
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Yes
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Theorem 5.10 (Asymptotic Catalan theorem) The abc conjecture implies that the Catalan equation has only finitely many solutions.
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Proof. Let \( \left( {x, y, m, n}\right) \) be a solution of the Catalan equation with \( \min \left( {m, n}\right) \geq \) 3. Then \( x \) and \( y \) are relatively prime. It follows from the \( {abc} \) conjecture with \( \varepsilon = 1/4 \) that there exists a constant \( {K}_{2} = K\left( {1/4}\right) \) such that\n\n\[ \n{y}^{n} < {x}^{m} \leq {K}_{2}\operatorname{rad}{\left( {x}^{m}{y}^{n}\right) }^{5/4} = {K}_{2}\operatorname{rad}{\left( xy\right) }^{5/4} \leq {K}_{2}{\left( xy\right) }^{5/4}, \n\] \n\nand so \n\n\[ \nm\log x \leq \log {K}_{2} + \frac{5}{4}\left( {\log x + \log y}\right) \n\] \n\nand \n\n\[ \nn\log y < \log {K}_{2} + \frac{5}{4}\left( {\log x + \log y}\right) \n\] \n\nIt follows that \n\n\[ \nm\log x + n\log y < 2\log {K}_{2} + \frac{5}{2}\left( {\log x + \log y}\right) , \n\] \n\nand so \n\n\[ \n\left( {m - \frac{5}{2}}\right) \log x + \left( {n - \frac{5}{2}}\right) \log y < 2\log {K}_{2}. \n\] \n\n(5.7) \n\nSince \( x \geq 2 \) and \( y \geq 2 \), we have \n\n\[ \nm + n < \frac{2\log {K}_{2}}{\log 2} + 5 \n\] \n\nThus, there are only finitely many pairs of exponents \( \left( {m, n}\right) \) for which the Catalan equation is solvable. For fixed exponents \( m \geq 3 \) and \( n \geq 3 \) , equation (5.7) has only only finitely many solutions in positive integers \( x \) and \( y \) . This completes the proof. \( ▱ \)
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Yes
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Lemma 5.1 Let \( p \) be an odd prime. If there exists a positive integer \( n \) such that \( {2}^{n} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \) but \( {2}^{n} ≢ 1\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) \), then \( p \) is a Wieferich prime.
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Proof. Let \( d \) be the order of 2 modulo \( p \) . Then \( d \) divides \( n \) . Since \( {2}^{n} ≢ 1 \) \( \left( {\;\operatorname{mod}\;{p}^{2}}\right) \), it follows that \( {2}^{d} ≢ 1\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) . Then \( {2}^{d} = 1 + {kp} \), where \( \left( {k, p}\right) = 1 \) . Moreover, \( d \) divides \( p - 1 \), since \( {2}^{p - 1} \equiv 1\left( {\;\operatorname{mod}\;p}\right) \), and so \( p - 1 = {de} \) for some integer \( e \) such that \( 1 \leq e \leq p - 1 \) . Then \( \left( {{ek}, p}\right) = 1 \) and\n\n\[
{2}^{p - 1} = {\left( {2}^{d}\right) }^{e} = {\left( 1 + kp\right) }^{e} \equiv 1 + {ekp} ≢ 1\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) ,
\]\n\nand \( p \) is a Wieferich prime.
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Yes
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Theorem 5.11 The abc conjecture implies that there exist infinitely many Wieferich primes.
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Proof. Let \( W \) be the set of Wieferich primes. For every positive integer \( n \), we write\n\n\[ {2}^{n} - 1 = {u}_{n}{v}_{n} \]\n\nwhere \( {v}_{n} \) is the maximal powerful divisor of \( {2}^{n} - 1 \) . Then \( {u}_{n} \) is a square-free integer,\n\n\[ {u}_{n} = \mathop{\prod }\limits_{\substack{{p \mid n} \\ {{v}_{p}\left( n\right) = 1} }}p \]\n\nand\n\n\[ {v}_{n} = \mathop{\prod }\limits_{\substack{{p \mid n} \\ {{v}_{p}\left( n\right) \geq 2} }}{p}^{{v}_{p}\left( n\right) } \]\n\nIf \( p \) divides \( {u}_{n} \), then\n\n\[ {2}^{n} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \]\n\nbut\n\n\[ {2}^{n} ≢ 1\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) \]\n\nIt follows from Lemma 5.1 that \( p \in W \), and so \( {u}_{n} \) is a square-free integer divisible only by Wieferich primes.\n\nIf the set \( W \) is finite, then there exist only finitely many square-free integers whose prime divisors all belong to \( W \), and so the set \( \left\{ {{u}_{n} : n = }\right. \) \( 1,2,3,\ldots \} \) is finite. It follows that the set \( \left\{ {{v}_{n} : n = 1,2,3,\ldots }\right\} \) is infinite, and, consequently, unbounded. Since \( {v}_{n} \) is powerful, we have\n\n\[ \operatorname{rad}\left( {v}_{n}\right) \leq {v}_{n}^{1/2} \]\n\nLet \( 0 < \varepsilon < 1 \) . Applying the \( {abc} \) conjecture to the identity\n\n\[ \left( {{2}^{n} - 1}\right) + 1 = {2}^{n} \]\n\nwe obtain\n\n\[ {v}_{n} < {2}^{n} \]\n\n\[ \leq K\left( \varepsilon \right) \operatorname{rad}{\left( {2}^{n}\left( {2}^{n} - 1\right) \right) }^{1 + \varepsilon } \]\n\n\[ \leq K\left( \varepsilon \right) \operatorname{rad}{\left( 2{u}_{n}{v}_{n}\right) }^{1 + \varepsilon } \]\n\n\[ \leq K\left( \varepsilon \right) {\left( 2{u}_{n}\right) }^{1 + \varepsilon }\operatorname{rad}{\left( {v}_{n}\right) }^{1 + \varepsilon } \]\n\n\[ \ll {v}_{n}^{\left( {1 + \varepsilon }\right) /2}. \]\n\nThis implies that the numbers \( {v}_{n} \) are bounded, which is absurd. This completes the proof. \( ▱ \)
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Yes
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Lemma 5.2 Let \( a, b, c \) be relatively prime positive integers such that\n\n\[ a < b < c \]\n\nand\n\n\[ a + b = c.\]\n\nLet \( n \geq 2 \) . If \( c \) is odd, define\n\n\[ {A}_{n} = {\left( b - a\right) }^{n} \]\n\n\[ {B}_{n} = {c}^{n} - {\left( b - a\right) }^{n} \]\n\n\[ {C}_{n} = {c}^{n}. \]\n\nIf \( c \) is even, define\n\n\[ {A}_{n} = {\left( \frac{b - a}{2}\right) }^{n} \]\n\n\[ {B}_{n} = {\left( \frac{c}{2}\right) }^{n} - {\left( \frac{b - a}{2}\right) }^{n} \]\n\n\[ {C}_{n} = {\left( \frac{c}{2}\right) }^{n}. \]\n\nThen \( {A}_{n},{B}_{n},{C}_{n} \) are distinct, relatively prime positive integers such that\n\n\[ {A}_{n} + {B}_{n} = {C}_{n} \]
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Proof. It is left to the reader to show that \( {A}_{n},{B}_{n},{C}_{n} \) are distinct, relatively prime positive integers such that \( {A}_{n} + {B}_{n} = {C}_{n} \) (Exercises 1,2, and 3).\n\nLet \( m \geq 3 \) and \( n = \varphi \left( m\right) \) . Then \( n \geq 2 \) . We must prove that\n\n\[ {A}_{n}{B}_{n}{C}_{n} \equiv 0\;\left( {\;\operatorname{mod}\;m}\right) \]\n\nIt suffices to prove that if \( p \) is a prime and \( {p}^{r} \) divides \( m \), then\n\n\[ {A}_{n}{B}_{n}{C}_{n} \equiv 0\;\left( {\;\operatorname{mod}\;{p}^{r}}\right) . \]\n\n(5.8)\n\nNote that if \( p \) is a prime and \( {p}^{r} \) divides \( m \), then \( \left( {p - 1}\right) {p}^{r - 1} \) divides \( n \), and so\n\n\[ r \leq {2}^{r - 1} \leq \left( {p - 1}\right) {p}^{r - 1} \leq n. \]\n\nSuppose that \( p \) is an odd prime. If \( p \) divides \( c \), then \( {p}^{n} \) divides \( {c}^{n} \) and \( {p}^{n} \) divides \( {C}_{n} \) . Since \( r \leq n \), it follows that \( {C}_{n} \equiv 0\;\left( {\;\operatorname{mod}\;{p}^{r}}\right) \) . Similarly, if \( p \) divides \( b - a \), then \( {A}_{n} \equiv 0\;\left( {\;\operatorname{mod}\;{p}^{r}}\right) \) . If \( p \) divides neither \( c \) nor \( b - a \), then, by Theorem 2.12,\n\n\[ {c}^{\left( {p - 1}\right) {p}^{r - 1}} \equiv 1\;\left( {\;\operatorname{mod}\;{p}^{r}}\right) \]\n\nand\n\n\[ {\left( b - a\right) }^{\left( {p - 1}\right) {p}^{r - 1}} \equiv 1\;\left( {\;\operatorname{mod}\;{p}^{r}}\right) . \]\n\nSince \( \left( {p - 1}\right) {p}^{r - 1} \) divides \( n \), we have\n\n\[ {c}^{n} \equiv {\left( b - a\right) }^{n} \equiv 1\;\left( {\;\operatorname{mod}\;{p}^{r}}\right) ,\]\nand so \( {B}_{n} \equiv 0\;\left( {\;\operatorname{mod}\;{p}^{r}}\right) \) . This proves (5.8) for odd primes \( p \) .\n\nFinally, we consider the prime 2 . If \( {2}^{r} \) divides \( m \), then \( {2}^{r - 1} \) divides \( n \) and \( r \leq n \) . If \( c \) is even, then \( b - a \) is even and exactly one of the integers \( c \) and \( b - a \) is divisible by 4 (Exercise 4). It follows that either \( {c}^{n} \) or \( {\left( b - a\right) }^{n} \) is divisible by \( {4}^{n} \), and so either \( {C}_{n} \) or \( {A}_{n} \) is divisible by \( {2}^{n} \), which is divisible by \( {2}^{r} \) .\n\nIf \( c \) is odd, then \( b - a \) is odd and\n\n\[ {c}^{{2}^{r - 1}} \equiv {\left( b - a\right) }^{{2}^{r - 1}} \equiv 1\;\left( {\;\operatorname{mod}\;{2}^{r}}\right) . \]\n\nSince \( {2}^{r - 1} \) divides \( n \), we have\n\n\[ {B}_{n} = {c}^{n} - {\left( b - a\right) }^{n} \equiv 0\;\left( {\;\operatorname{mod}\;{2}^{r}}\right) . \]\n\nThis proves (5.8) for the prime 2.
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No
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Theorem 6.1 The set of all complex-valued arithmetic functions, with addition defined by pointwise sum and multiplication defined by Dirichlet convolution, is a commutative ring with additive identity \( 0\left( n\right) \) and multiplicative identity \( \delta \left( n\right) \) .
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Proof. It is easy to check that the set of arithmetic functions is an additive abelian group with the zero function as the additive identity.\n\nWe shall prove that Dirichlet convolution is commutative, associative, and distributes over addition, that is,\n\n\[ f * g = g * f \]\n\n\[ \left( {f * g}\right) * h = f * \left( {g * h}\right) \]\n\nand\n\n\[ f * \left( {g + h}\right) = f * g + f * h \]\n\nfor all arithmetic functions \( f, g \), and \( h \) . These are straightforward calculations. We have\n\n\[ f * g\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) g\left( {n/d}\right) = \mathop{\sum }\limits_{{d \mid n}}g\left( {n/d}\right) f\left( d\right) = \mathop{\sum }\limits_{{d \mid n}}g\left( d\right) f\left( {n/d}\right) = g * f\left( n\right) \]\n\nand\n\n\[ \left( {\left( {f * g}\right) * h}\right) \left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\left( {f * g}\right) \left( d\right) h\left( \frac{n}{d}\right) \]\n\n\[ = \mathop{\sum }\limits_{{{dm} = n}}\left( {f * g}\right) \left( d\right) h\left( m\right) \]\n\n\[ = \mathop{\sum }\limits_{{{dm} = n}}\mathop{\sum }\limits_{{k\ell = d}}f\left( k\right) g\left( \ell \right) h\left( m\right) \]\n\n\[ = \mathop{\sum }\limits_{{k\ell m = n}}f\left( k\right) g\left( \ell \right) h\left( m\right) \]\n\n\[ = \mathop{\sum }\limits_{{k \mid n}}f\left( k\right) \mathop{\sum }\limits_{{\ell m = n/k}}g\left( \ell \right) h\left( m\right) \]\n\n\[ = \mathop{\sum }\limits_{{k \mid n}}f\left( k\right) \mathop{\sum }\limits_{{\ell \mid \left( {n/k}\right) }}g\left( \ell \right) h\left( \frac{n}{k\ell }\right) \]\n\n\[ = \mathop{\sum }\limits_{{k \mid n}}f\left( k\right) \left( {g * h}\right) \left( \frac{n}{k}\right) \]\n\n\[ = \left( {f * \left( {g * h}\right) }\right) \left( n\right) \text{.} \]\n\nSimilarly,\n\n\[ f * \left( {g + h}\right) \left( n\right) = \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) \left( {g\left( {n/d}\right) + h\left( {n/d}\right) }\right) \]\n\n\[ = \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) g\left( {n/d}\right) + \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) h\left( {n/d}\right) \]\n\n\[ = f * g\left( n\right) + f * h\left( n\right) \text{.} \]\n\nFinally, we observe that\n\n\[ \delta * f\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\delta \left( d\right) f\left( {n/d}\right) = f\left( n\right) \]\n\nfor every arithmetic function \( f \), and so the arithmetic functions form a commutative ring with multiplicative identity \( \delta \left( n\right) \) . This completes the proof. \( ▱ \)
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Yes
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Theorem 6.2 Consider the arithmetic function \( L\left( n\right) \) defined by\n\n\[ L\left( n\right) = \log n\;\text{ for all }n \geq 1. \]\n\nPointwise multiplication by \( L\left( n\right) \) is a derivation on the ring of arithmetic functions.
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Proof. Observe that if \( d \) is a positive divisor of \( n \), then\n\n\[ L\left( n\right) = L\left( d\right) + L\left( {n/d}\right) \]\n\nWe must prove that\n\n\[ L \cdot \left( {f * g}\right) = \left( {L \cdot f}\right) * g + f * \left( {L \cdot g}\right) \]\n\nfor all arithmetic functions \( f \) and \( g \) . We have\n\n\[ L \cdot \left( {f * g}\right) \left( n\right) = L\left( n\right) \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) g\left( {n/d}\right) \]\n\n\[ = \mathop{\sum }\limits_{{d \mid n}}L\left( n\right) f\left( d\right) g\left( {n/d}\right) \]\n\n\[ = \mathop{\sum }\limits_{{d \mid n}}\left( {L\left( d\right) + L\left( {n/d}\right) }\right) f\left( d\right) g\left( {n/d}\right) \]\n\n\[ = \mathop{\sum }\limits_{{d \mid n}}L\left( d\right) f\left( d\right) g\left( {n/d}\right) + \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) L\left( {n/d}\right) g\left( {n/d}\right) \]\n\n\[ = \left( {L \cdot f}\right) * g + f * \left( {L \cdot g}\right) \text{.} \]\n\nThis completes the proof.
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Yes
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Theorem 6.3 Let \( a \) and \( b \) be integers with \( a < b \), and let \( f\left( t\right) \) be a function that is monotonic on the interval \( \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ \min \left( {f\left( a\right), f\left( b\right) }\right) \leq \mathop{\sum }\limits_{{n = a}}^{b}f\left( n\right) - {\int }_{a}^{b}f\left( t\right) {dt} \leq \max \left( {f\left( a\right), f\left( b\right) }\right) . \]
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Proof. If \( f\left( t\right) \) is increasing on \( \left\lbrack {n, n + 1}\right\rbrack \), then\n\n\[ f\left( n\right) \leq {\int }_{n}^{n + 1}f\left( t\right) {dt} \leq f\left( {n + 1}\right) \]\n\nIf \( f\left( t\right) \) is increasing on the interval \( \left\lbrack {a, b}\right\rbrack \), then\n\n\[ f\left( a\right) + {\int }_{a}^{b}f\left( t\right) {dt} \leq \mathop{\sum }\limits_{{n = a}}^{b}f\left( n\right) \leq f\left( b\right) + {\int }_{a}^{b}f\left( t\right) {dt}. \]\n\nSimilarly, if \( f\left( t\right) \) is decreasing on the interval \( \left\lbrack {n, n + 1}\right\rbrack \), then\n\n\[ f\left( {n + 1}\right) \leq {\int }_{n}^{n + 1}f\left( t\right) {dt} \leq f\left( n\right) . \]\n\nIf \( f\left( t\right) \) is decreasing on the interval \( \left\lbrack {a, b}\right\rbrack \), then\n\n\[ f\left( b\right) + {\int }_{a}^{b}f\left( t\right) {dt} \leq \mathop{\sum }\limits_{{n = a}}^{b}f\left( n\right) \leq f\left( a\right) + {\int }_{a}^{b}f\left( t\right) {dt}. \]\n\nThis proves (6.3).
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Yes
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Theorem 6.4 For \( x \geq 2 \) ,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\log n = x\log x - x + O\left( {\log x}\right) . \]
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Proof. The function \( f\left( t\right) = \log t \) is increasing on \( \left\lbrack {1, x}\right\rbrack \) . By Theorem 6.3,\n\n\[ {\int }_{1}^{x}\log {tdt} \leq \mathop{\sum }\limits_{{n \leq x}}\log n \leq {\int }_{1}^{x}\log {tdt} + \log x \]\n\nand so\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\log n = x\log x - x + O\left( {\log x}\right) . \]\n\nThis completes the proof. \( ▱ \)
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Yes
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