problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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Theorem 4 When $m \geqslant 1$, equation (23) must hold, i.e., there must be a prime $p$ satisfying
$$m<p \leqslant 2 m$$ | Let us first calculate $\alpha(p, 2 m)-2 \alpha(p, m)$ more precisely. When $m \geqslant 5$, we have
$$p^{2}>4 m^{2} / 9>2 m \text {, when } 2 m / 3<p \leqslant m \text {, }$$
Therefore, when $2 m / 3<p \leqslant m$, we have
$$\alpha(p, 2 m)-2 \alpha(p, m)=[2 m / p]-2[m / p]=0$$
Furthermore, from equation (6), we kno... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,482 |
Theorem 6 Let $x \geqslant 2$, then there exists a positive constant $c_{7}$, such that
$$\left(\ln x-c_{7}\right) \pi(x)<\theta(x)<(\ln x) \pi(x)$$
and
$$\theta(x) \leqslant \psi(x) \leqslant \theta(x)+x^{1 / 2} \ln x .$$ | To prove formula (35). We have
$$\begin{aligned}
\theta(x) & =\sum_{p \leq x} \ln p=\sum_{k \leq x} \ln k(\pi(k)-\pi(k-1)) \\
& =-\sum_{k=2}^{[x]-1} \pi(k)(\ln (k+1)-\ln k)+\pi([x]) \ln [x] .
\end{aligned}$$
Using formula (14), we get
$$\frac{1}{y+1}\ln (x-1)=\ln x+\ln (1-1 / x)>\ln x-1 /(x-1),$$
the last step uses f... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,484 |
Lemma 8 Let the integer $n \geqslant 1$. We have
$$\sum_{d \mid n} \Delta(d)=\ln n .$$ | Prove that $n=1$ is obviously true. If $n>1$, let the prime factorization of $n$ be
$$n=p_{1}^{a_{1}} \cdots p_{r}^{a_{r}}$$
By the definition of $\Lambda(d)$ (Equation (34)), we have
$$\begin{aligned}
\sum_{d \mid n} \Lambda(d) & =\sum_{d \mid p_{1}^{a_{1}}} \Lambda(d)+\cdots+\sum_{d \mid p_{r}^{a_{r}}} \Lambda(d) \\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,486 |
Lemma 9 Let $x \geqslant 1$. We have
$$\sum_{m \leq x} \psi(x / m)=\ln ([x]!) .$$ | Proof: By the definition of $\psi(x)$ (Equation (33)), we have
$$\sum_{m \leq x} \psi(x / m)=\sum_{m \leq x} \sum_{k \leq x / m} \Lambda(k)=\sum_{m \leq x} \sum_{k m \leqslant x} \Lambda(k),$$
By making the integer variable substitution $k m=d, k=k$, the above equation becomes
$$\sum_{m \leqslant x} \psi(x / m)=\sum_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,487 |
Theorem 10 Let $x \geqslant 2$. We have
$$(1 / 4)(\ln 2) x<\psi(x)<(4 \ln 2) x .$$ | Let $m$ be a positive integer. From equation (39), we have
$$\begin{aligned}
\ln (2 m)!-2 \ln m! & =\sum_{k \leqslant 2 m} \psi(2 m / k)-2 \sum_{d \leq m} \psi(m / d) \\
& =\sum_{k \leq 2 m} \psi(2 m / k)-2 \sum_{d \leq m} \psi(2 m /(2 d)) \\
& =\sum_{k \leqslant 2 m}(-1)^{k-1} \psi(2 m / k)
\end{aligned}$$
Therefore,... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 739,488 |
1. Let $\pi_{2}(x)$ denote the number of positive integers not exceeding $x$ that are exactly the product of two primes. Prove: there exist positive constants $c_{1}, c_{2}$, such that
$$c_{1} \frac{x \ln \ln x}{\ln x}<\pi_{2}(x)<c_{2} \frac{x \ln \ln x}{\ln x}, \quad x \geqslant 4 .$$ | 1. From the problem, we have
$$\begin{aligned}
\pi_{2}(x) & =\sum_{\substack{p_{1} p_{2}<x \\
p_{1} \leqslant p_{2}}} 1=\sum_{p_{1} \leqslant \sqrt{x}} \sum_{p_{1} \leqslant p_{2}<x / p_{1}} 1 \\
& =\sum_{p_{1} \leqslant \sqrt{x}}\left\{\pi\left(x / p_{1}\right)-\pi\left(p_{1}\right)+1\right\} .
\end{aligned}$$
Then, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,490 |
2. Prove: When $m \geqslant 6$, there must be two primes $p, q$ such that $m<p<q<2 m$.
| 2. When $m \geqslant 128$, $((\ln 2) / 6) m > 2 \ln (2 m)$, from this and equation (31), we derive that the conclusion holds for $m \geqslant 128$. Other cases are verified directly. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,491 |
3. Prove: There exists a positive constant $c$ such that
$$\pi(2 x)-\pi(x)>c x / \ln x, \quad x \geqslant 2 .$$ | 3. From equation (31), we know that when $m \geqslant 128$, $\pi(2 m)-\pi(m)>c_{1} m / \ln (2 m)$, which can lead to the desired conclusion, $c_{1}=(\ln 2) / 6$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,492 |
4. Prove:
(i) $\psi(x)=\sum_{n=1}^{\infty} \theta\left(x^{1 / n}\right)$;
(ii) $\theta(x)=\sum_{n=1}^{\infty} \mu(n) \psi\left(x^{1 / n}\right)$. | 4. (i) $\psi(x)=\sum_{p^{k} \leqslant x} \Lambda\left(p^{k}\right)=\sum_{k=1}^{\infty} \sum_{p \leqslant x^{1 / k}} \ln p$.
(ii) $\theta(x)=\sum_{n=1}^{\infty} \theta\left(x^{1 / n}\right) \sum_{d \mid n} \mu(d)$, by interchanging the order of summation, we get the result from (i). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,493 |
5. Prove: There exists a positive constant $c$, such that
$$\psi(x)<\theta(x)+c x^{1 / 2} .$$ | 5. In proving formula (36), we obtain
$$\psi(x) \leqslant \theta(x)+\ln x \cdot \sum_{p \leqslant \sqrt{x}} 1 .$$
From this and formula (1), the desired conclusion is obtained. | proof | Inequalities | proof | Yes | Yes | number_theory | false | 739,494 |
6. Prove: The prime number theorem $\lim _{x \rightarrow \infty} \pi(x)(\ln x) / x=1$ is equivalent to
$$\lim _{n \rightarrow \infty} p_{n} /(n \ln n)=1$$
Here $p_{n}$ denotes the $n$-th prime number. | 6. From equation (2), we can obtain $\lim _{n \rightarrow \infty} \ln p_{n} / \ln n=1$. If $\lim _{x \rightarrow \infty} \pi(x)(\ln x) / x=1$, taking $x=p_{n}$ leads to $\lim _{n \rightarrow \infty} p_{n} /(n \ln n)=1$. If the latter limit holds, then it must be that $p_{n} \leqslant x<p_{n+1}$, hence
$$\left(n \ln p_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,495 |
7. Let \( T(x) = \ln ([x]!) \). Prove: When \( x \geqslant 1 \),
\[
\psi(x) = \sum_{n \leqslant x} \mu(n) T(x / n)
\] | 7. $\psi(x)=\sum_{m=1}^{\infty} \psi(x / m) \sum_{d \mid m} \mu(d)$, interchange the summation signs, and use equation (39) to obtain the desired result. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,496 |
9. Prove: There exist positive constants $A_{1}, A_{2}$, such that
(i) $\sum_{k<x} \Lambda(k) / k=\ln x+\Delta_{1}(x),\left|\Delta_{1}(x)\right| \leqslant A_{1}, x \geqslant 1$;
(ii) $\sum_{p \leq x}(\ln p) / p=\ln x+\Delta_{2}(x),\left|\Delta_{2}(x)\right| \leqslant A_{2}, x \geqslant 2$. | 9. (i) From equations (39) and (41), we have
$$\ln ([x]!)=x \sum_{k \leqslant x} \Lambda(k) / k-\sum_{k \leqslant x} \Lambda(k)\{x / k\}$$
From this, using equations (16) and (44), the desired result follows; (ii) follows from (i). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,498 |
10. Let $f(x)$ be a non-negative and increasing function on the interval $[a, b]$. Prove:
$$\left|\sum_{a<n \leqslant b} f(n)-\int_{a}^{b} f(t) \mathrm{d} t\right| \leqslant f(b) .$$ | 10. $\int_{a}^{b} f(t) \mathrm{d} t=\left(\int_{a}^{\lfloor a \rfloor+1}+\int_{\lfloor a \rfloor+1}^{\lfloor a \rfloor+2}+\cdots+\int_{\lfloor b \rfloor-1}^{\lfloor b \rfloor}+\int_{\lfloor b \rfloor}^{b}\right) f(t) \mathrm{d} t$. | proof | Calculus | proof | Yes | Yes | number_theory | false | 739,499 |
20. Let $k \geqslant 1$. Prove:
(i) If $2^{k} \leqslant n<2^{k+1}$, and $1 \leqslant a \leqslant n, a \neq 2^{k}$, then $2^{k} \nmid a$;
(ii) If $3^{k} \leqslant 2 n-1<3^{k+1}, 1 \leqslant l \leqslant n, 2 l-1 \neq 3^{k}$, then $3^{k} \nmid 2 l-1$. | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,500 |
15. Let $f(n)$ be a function of a positive integer variable, and $F(n)=\sum_{d \mid n} f(d)$. Using
$$f(n)=\sum_{d \mid n} f(d)\left\{\sum_{l \mid n / d} \mu(l)\right\},$$
prove: $f(n)=\sum_{d \mid n} \mu(d) F(n / d)$. Conversely, if a function $F(n)$ of a positive integer variable is given and we set $f(n)=\sum_{d \m... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,505 |
Lemma 1 When the real number $s>1$, the infinite product
$$\prod_{p}\left(1-\frac{1}{p^{s}}\right)^{-1}$$
converges and is greater than 1, where the product sign indicates the product over all prime numbers. The product (1) is called the Euler product. | Proof: From § 2 equation (14), we have
$$00,$$
thus we have $\square$
$$\begin{aligned}
\sum_{p} \frac{1}{p^{s}} & 1
\end{aligned}$$
Here the summation $\sum_{p}$ denotes the sum over all prime numbers $p$. Since the series $\sum_{n=1}^{\infty} n^{-s}$ converges when $s>1$, it follows from equation (3) that the posit... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,506 |
1. Let $f(n)$ be a function defined on the set of positive integers, and let $s$ be a real number. The series $\sum_{n=1}^{\infty} f(n) n^{-s(1)}$ converges absolutely and is not equal to zero. Prove:
(i) If $f(m n)=f(m) f(n)$ whenever $(m, n)=1$, then
$$\sum_{n=1}^{\infty} f(n) n^{-s}=\prod_{p}\left(1+f(p) p^{-s}+f\le... | 1. (i) For any given positive integer $k$ and $x<2^{k}$, by the fundamental theorem of arithmetic and the multiplicative condition, we have
$$\begin{array}{c}
\left|\prod_{p \leqslant x}\left(1+f(p) p^{-s}+f\left(p^{2}\right) p^{-2 s}+\cdots+f\left(p^{k}\right) p^{-k s}\right)-\sum_{n \leqslant x} f(n) n^{-s}\right| \\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,509 |
2. Let $s$ be a real number, and the series $\sum_{n=1}^{\infty} f(n) n^{-s}$ and $\sum_{m=1}^{\infty} g(m) m^{-s}$ both converge absolutely. Prove:
$$\sum_{n=1}^{\infty} f(n) n^{-s} \cdot \sum_{m=1}^{\infty} g(m) m^{-s}=\sum_{l=1}^{\infty} h(l) l^{-s}$$
where
$$\begin{aligned}
h(l) & =\sum_{n \mid l} f(n) g(l / n)=\s... | 2. The product of absolutely convergent series can be arbitrarily grouped.
The above text has been translated into English, preserving the original text's line breaks and format. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,510 |
3. (i) Prove: $\sum_{n=1}^{\infty} \mu(n) n^{-s}=\prod_{p}\left(1-p^{-s}\right), s>1$;
(ii) Prove: $\sum_{n=1}^{\infty} \mu(n) n^{-s}=1 / \zeta(s), s>1$;
(iii) Use (ii) to give another proof of Lemma 3 in §1. | 3. (i) follows from Question 1 (i); (ii) follows from (i) and equation (9); (iii) $1=\zeta(s) \sum_{n=1}^{\infty} \mu(n) n^{-s}$, $s>1$, then use the result from Question 2, and compare the coefficients on both sides. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,512 |
7. Prove that when $s>2$,
$$\sum_{n=1}^{\infty} \varphi(n) n^{-s}=\zeta(s-1) / \zeta(s) .$$
Using this, provide new proofs for $\sum_{d \mid n} \varphi(d)=n$ and $\sum_{d \mid n} \mu(d) / d=\varphi(n) / n$. | 7. Using the result from Question 1(i), and
$$\begin{aligned}
1+ & (p-1) p^{-s}+p(p-1) p^{-2 s}+\cdots+p^{m}(p-1) p^{-(m+1) s}+\cdots \\
& =1+(p-1) p^{-s} /\left(1-p^{-s+1}\right)=\left(1-p^{-s}\right) /\left(1-p^{-s+1}\right)
\end{aligned}$$
From this proof, using the result from Question 2, we can provide new proofs... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,516 |
8. Let $s>1$. Prove: (i) $-\zeta^{\prime}(s) / \zeta(s)=\sum_{n=1}^{\infty} \Lambda(n) n^{-s}$;
(ii) Use (i) to give new proofs of $\sum_{d \mid n} \Lambda(n)=\ln n$ and $\Lambda(n)=-\sum_{d \mid n} \mu(d) \ln d$. | 8. (i) Differentiating both sides of $\ln \zeta(s)=-\sum_{p} \ln \left(1-p^{-s}\right)$ yields the result;
(ii) Use the result from Question 2. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,517 |
3. If the multiplicative function $f(n)$ is defined at $n=-1$, then $f(-1)= \pm 1$.
| 3. $(f(-1))^{2}=f((-1) \cdot(-1))=f(1)=1$.
The translation is as follows:
3. $(f(-1))^{2}=f((-1) \cdot(-1))=f(1)=1$. | f(-1) = \pm 1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,521 |
4. Let $T$ be a set composed of integers. If $T$ has a lower bound, that is, there exists an integer $a$ such that for all $t \in T$, $t \geqslant a$, then there must be a $t_{0} \in T$, such that for all $t \in T$, $t \geqslant t_{0}$. | 4. Consider the set $T^{*}=\left\{t^{*}=t-a+1: t \in T\right\}$. Apply Theorem 2 of $\S 1$ to $T^{*}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,522 |
4. Let $k$ be a given positive integer, and $P_{k}(n)$ be a function defined on the set of positive integers:
$$P_{k}(n)=\left\{\begin{array}{ll}
1, & n \text { is a } k \text {th power, } \\
0, & \text { otherwise. }
\end{array}\right.$$
Prove: $P_{k}(n)$ is multiplicative, and is completely multiplicative only when ... | 4. Let $\left(n_{1}, n_{2}\right)=1, n=n_{1} n_{2}$, then $n=m^{k} \Longleftrightarrow n_{1}=m_{1}^{k}, n_{2}=m_{2}^{k}$. Therefore, $P_{k}(n)$ is multiplicative. When $k=1$, it is evident that $P_{1}(n) \equiv 1$ is completely multiplicative; if $P_{k}(n)$ is completely multiplicative, then it must be that $k=1$, othe... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,524 |
5. Let $k$ be a given positive integer, and $Q_{k}(n)$ be a function defined on the set of positive integers:
$$Q_{k}(n)=\left\{\begin{array}{ll}
1, & n \text { has no } k \text { th power factor greater than } 1, \\
0, & \text { otherwise. }
\end{array}\right.$$
Prove: $Q_{k}(n)$ is multiplicative, and is completely ... | 5. Let $\left(n_{1}, n_{2}\right)=1, n=n_{1} n_{2}$, then $n$ has a $k$-th power divisor greater than 1 if and only if $n_{1}$ or $n_{2}$ has a $k$-th power divisor greater than 1, so it is multiplicative. When $k=1$, it is evident that $Q_{1}(n)=[1 / n]$, which is completely multiplicative. If $Q_{k}(n)$ is completely... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,525 |
6. Let $l$ be a given positive integer. Let $\tau_{l}(n)$ denote the number of distinct representations of a positive integer $n$ as a product of $l$ positive integers $d_{1}, d_{2}, \cdots, d_{l}$. For example, $\tau_{1}(n) \equiv 1, \tau_{2}(n)=\tau(n)$. Prove that $\tau_{l}(n)$ is a multiplicative function, and that... | 6. Let $\left(n_{1}, n_{2}\right)=1, n=n_{1} n_{2}$, then $n=d_{1} \cdots d_{l}$ if and only if $n_{1}=d_{11} \cdots d_{l 1}, n_{2}=d_{12} \cdots d_{l 2}, d_{j 1}=\left(d_{j}, n_{1}\right), d_{j 2}=\left(d_{j}, n_{2}\right)$. This implies that $\tau_{l}(n)$ is multiplicative. $\tau_{l}\left(p^{a}\right)$ $(l \geqslant ... | \frac{(\alpha+l-1)!}{\alpha!(l-1)!} | Number Theory | proof | Yes | Yes | number_theory | false | 739,526 |
7. Let $l$ be a given positive integer. Let $\tau_{i}^{*}(n)$ denote the number of representations: $n=d_{1} \cdots d_{l}$, $\left(d_{i}, d_{j}\right)=1, i \neq j, d_{i}$ are positive integers. Prove: $\tau_{i}(n)$ is a multiplicative function, and when $l \geqslant 2$ it is not completely multiplicative. Try to find t... | 7. Similar to the previous question, prove multiplicativity. $\tau_{1}^{*}(n) \equiv 1 . \tau_{i}^{*}\left(p^{e}\right)(l \geqslant 2)$ equals the number of solutions to the indeterminate equation $x_{1}+\cdots+x_{l}=\alpha, x_{j} \geqslant 0$ and not more than one of them can be $\geqslant 1$, hence it equals $l . \ta... | \tau_{i}^{*}(n)=l^{\omega(n)} | Number Theory | proof | Yes | Yes | number_theory | false | 739,527 |
Lemma 2 Let $(m, n)=1, k$ be a given positive integer, then for each positive integer $d$ the necessary and sufficient condition for $d^{k} \mid m n$ to hold is that there exists a unique pair of positive integers $d_{1}, d_{2}$ satisfying
$$d=d_{1} d_{2}, \quad d_{1}^{k}\left|m, \quad d_{2}^{k}\right| n$$ | We prove this using the conclusion from Example 4 (ii) in Section 4 of Chapter 1 (i.e., Corollary 5 in Section 5 of Chapter 1). From \((m, n) = 1\), we know that \(d^k \mid mn\) if and only if
\[ d^k = (d^k, mn) = (d^k, m)(d^k, n). \]
It is evident that \(( (d^k, m), (d^k, n) ) = 1\). Therefore, by Example 4 (ii) in S... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,530 |
Example 1 Find the Möbius transform of the Liouville function $\lambda(n)$. | $$\text{Solve } \begin{aligned}
\sum_{d \mid \varepsilon_{e}} \lambda(d) & =(-1)^{0}+(-1)^{1}+\cdots+(-1)^{\alpha} \\
& =\left\{\begin{array}{ll}
1, & 2 \mid \alpha, \\
0, & 2 \nmid \alpha .
\end{array}\right.
\end{aligned}$$
From this and the fact that $\lambda(n)$ is a multiplicative function, we get
$$\sum_{d \mid ... | \sum_{d \mid n} \lambda(d)=\left\{\begin{array}{ll}
1, & n \text{ is a perfect square, } \\
0, & \text{ otherwise. }
\end{array}\right.} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,531 |
Example 2 Find the Möbius transform of $\mu^{2}(n) / \varphi(n)$.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | Solve $\sum_{d \mid p^{e^{\circ}}} \mu^{2}(d) / \varphi(d)=1+1 /(p-1)=(1-1 / p)^{-1}$.
From this and the fact that $\mu^{2}(n) / \varphi(n)$ is a multiplicative function, we get
$$\sum_{d \mid n} \mu^{2}(d) / \varphi(d)=\prod_{p \mid n}(1-1 / p)^{-1}=n / \varphi(n)$$ | n / \varphi(n) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,532 |
Example 3 Find the Möbius transform $F(n)$ of $\Omega(n)$.
| Solve $\Omega(n)$ is not a multiplicative function. Only formula (2) can be used to calculate. $F(1)=0,1<n=$ $p_{1}^{a_{1}} \cdots p_{r}^{\sigma}$ when,
$$\begin{array}{l}
\quad F(n)=\sum_{e_{1}=0}^{\alpha_{1}} \cdots \sum_{e_{r}=0}^{\alpha_{r}} \Omega\left(p_{1}^{f_{1}} \cdots p_{r}^{e_{r}}\right) \\
=\sum_{e_{1}=0}^{... | \frac{1}{2} \Omega(n) \tau(n) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,533 |
Theorem 4 Let $f(m), F(n)$ be number-theoretic functions. Then the necessary and sufficient condition for equation (1) to hold is
$$f(n)=\sum_{d \mid n} \mu(d) F\left(\frac{n}{d}\right) .$$ | First, we prove the sufficiency. If equation (19) holds, then we have
$$\sum_{d \mid n} f(d)=\sum_{d \mid n}\left\{\sum_{k \mid d} \mu(k) F\left(\frac{d}{k}\right)\right\}=\sum_{k \mid n} \mu(k) \sum_{k|d, d| n} F\left(\frac{d}{k}\right).$$
Let $d=k l$, we get
$$\sum_{d \mid n} f(d)=\sum_{k \mid n} \mu(k) \sum_{l \mid... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,536 |
Theorem 5 Let $f(n), g(n)$ be number-theoretic functions,
$$h(n)=\sum_{d \mid n} f(d) g\left(\frac{n}{d}\right),$$
then, when $f(n), g(n)$ are both multiplicative functions, $h(n)$ is also a multiplicative function. | Given $f(1)=g(1)=1$, it follows that $h(1)=1$. If $(m, n)=1$, using Lemma 2 $(k=1)$, we have
$$\begin{aligned}
h(m n) & =\sum_{d \mid m n} f(d) g\left(\frac{m n}{d}\right)=\sum_{d_{1}\left|m, d_{2}\right| n} f\left(d_{1} d_{2}\right) g\left(\frac{m n}{d_{1} d_{2}}\right) \\
& =\sum_{d_{1} \mid m} f\left(d_{1}\right) g\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,537 |
Example 4 Find the Möbius inverse transform $f(n)$ of $F(n)=n^{t}$.
| Solve $n^{t}$ is multiplicative, from equation (27) we get
$$f\left(p^{a}\right)=p^{a t}-p^{(\sigma-1) t}=p^{a t}\left(1-p^{-t}\right) .$$
Therefore, we have
$$f(n)=n^{t} \prod_{p \mid n}\left(1-p^{-t}\right) .$$ | f(n)=n^{t} \prod_{p \mid n}\left(1-p^{-t}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,539 |
Example 5 Find the Möbius inverse transform of $F(n)=\varphi(n)$. | Since $\varphi(n)$ is multiplicative, from equation (27) we get
$$f\left(p^{\alpha}\right)=\varphi\left(p^{\alpha}\right)-\varphi\left(p^{\sigma-1}\right)=\left\{\begin{array}{ll}
p(1-2 / p), & \alpha=1 \\
p^{\alpha}\left(1-p^{-1}\right)^{2}, & \alpha \geqslant 2 .
\end{array}\right.$$
Therefore, we have
$$f(n)=n \pro... | f(n)=n \prod_{p \| n}\left(1-\frac{2}{p}\right) \prod_{p^{2} \mid n}\left(1-\frac{1}{p}\right)^{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,540 |
Example 6 Find the Möbius inverse transform $f(n)$ of $F(n)=\Lambda(n)$.
| $\Lambda(n)$ is not multiplicative, so we cannot use equation (27). However, using equation (19) to obtain a good expression for $f(n)$ is quite difficult. Noting equation (24), we have
$$\Lambda(n)=\left(\sum_{d \mid n} \mu(d)\right) \ln n-\sum_{d \mid n} \mu(d) \ln d$$
Using equation (18), we get
$$\Lambda(n)=-\sum_... | f(n) = -\mu(n) \ln n | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,541 |
Example 8 Let $P(x)$ be a polynomial with integer coefficients, and let $S(n ; P(x))$ denote the number of integers $d$ satisfying the following conditions:
$$(P(d), n)=1, \quad 1 \leqslant d \leqslant n$$ | Prove: $S(n)=S(n ; P(x))$ is a multiplicative function of $n$.
Proof Using property (18), we have
$$S(n)=\sum_{\substack{d=1 \\(P(d), n)=1}}^{n} 1=\sum_{d=1}^{n} \sum_{k \mid(P(d), n)} \mu(k)=\sum_{k \mid n} \mu(k) \sum_{\substack{d=1 \\ k \mid P(d)}}^{n} 1$$
Let $T(k)=T(k ; P(x))$ denote the number of solutions to th... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,543 |
3. Change the prime $p$ to an integer $K$, and find the expression for $\sum_{d \mid n} \mu(d) \mu((d, K))$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. is a multiplicative function, let the distinct prime factors of $K$ be $p_{1}, \cdots, p_{r}$. When $n=p_{1}^{a_{1}} \cdots p_{r}^{a_{r}}$, it equals $2^{a(n)}$, for other $n>1$ it equals zero. | null | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 739,547 |
4. Let $m$ be a given positive integer, find $\sum_{d \mid n} \mu(d) \ln ^{m} d$, and prove: when $n$ has more than $m$ distinct prime factors, the sum is zero. | 4. This is not a multiplicative function, so we directly use Theorem 1(i) to compute. When $n=1$, it equals zero; when $n=p_{1}^{a_{1}} \cdots p_{r}^{e_{r}}$, it equals
$$\sum_{\substack{k_{1}+\cdots+k_{k}=m \\ k_{j} \geqslant 1}}(-1)^{r} \frac{m!}{k_{1}!\cdots k_{r}!} \ln ^{k_{1}} p_{1} \cdots \ln ^{k_{r}} p_{r}$$
Wh... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,548 |
5. Prove: The Möbius transform of $f(n)$'s Möbius transform is
$$\sum_{d \mid n} f(d) \tau\left(\frac{n}{d}\right) .$$ | 5. $F_{1}(n)=\sum_{d \mid n} f(d), F_{2}(m)=\sum_{n \mid m} F_{1}(n)=\sum_{n \mid m} \sum_{d \mid n} f(d)$, interchange the summation signs. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,549 |
8. Find the Möbius transform and the Möbius inverse transform of $|\mu(n)|$.
| 8. $\sum_{d \mid n}|\mu(n)|=2^{\alpha(n)}, \sum_{d \mid n} \mu(d)|\mu(n / d)|=1$, when $n=1 ;=0$, when $n=p$ or $p^{\alpha}$, $\alpha>2 ;=-1$, when $n=p^{2}$, i.e.
$$\sum_{d \mid n} \mu(d)|\mu(n / d)|=\left\{\begin{array}{ll}
1, & n=1 \\
(-1)^{r}, & n=p_{1}^{2} \cdots p_{r}^{2}, \\
0, & \text { otherwise. }
\end{array}... | \sum_{d \mid n} \mu(d)|\mu(n / d)|=\left\{\begin{array}{ll}
1, & n=1 \\
(-1)^{r}, & n=p_{1}^{2} \cdots p_{r}^{2}, \\
0, & \text { otherwise. }
\end{array}\right.} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,552 |
9. Find the Möbius transform and the Möbius inverse transform of $P_{k}(n)$ (see Exercise 1, Question 4). Prove: The Möbius inverse transform of $P_{2}(n)$ is $\lambda(n)$. | 9. Let $n=p_{1}^{\varepsilon_{1}} \cdots p_{r}^{R_{r}} \cdot \sum_{d \mid n} P_{k}(d)=\prod_{i=1}^{r}\left(1+\left[a_{i} / k\right]\right)$. Also, let $\rho(r)=1, r=0 ; \rho(r)=$ $-1, r=1$; and $\rho(r)=0, r \geqslant 2$.
$$\sum_{d \mid \pi} \mu(n / d) P_{k}(d)=\prod_{i=1}^{r} \rho\left(\alpha_{i}-\left[\alpha_{i} / k\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,553 |
11. Let $f(n)$ denote the number of $d$ satisfying $1 \leqslant d \leqslant n,(d, n)=(d+1, n)=1$. Prove: $f(n)$ is a multiplicative function, and $f(n)=n \prod_{p / n}\left(1-\frac{2}{p}\right)$. | 11. Let $n_{1}^{-1} n_{1} \equiv 1\left(\bmod n_{2}\right), n_{2}^{-1} n_{2} \equiv 1\left(\bmod n_{1}\right), n=n_{1} n_{2},\left(n_{1}, n_{2}\right)=1$. By the Chinese Remainder Theorem, we know: $d=n_{2}^{-1} n_{2} d_{1}+n_{1}^{-1} n_{1} d_{2}, d$ traverses a complete residue system modulo $n$ if and only if $d_{1},... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,555 |
25. Let $n>1$. Prove: $n$ can be expressed as the sum of two or more consecutive positive integers if and only if $n \neq 2^{k}$. | 25. Let $m \geqslant 1, r \geqslant 1, m+(m+1)+\cdots+(m+r)=(r+1)(2 m+r) / 2=n . r+1$ and $2 m+r$ have opposite parity, which proves the necessity. When $n=2^{k} \cdot n^{\prime}, 2 \nmid n^{\prime}>1$, if $2^{k+1}>n^{\prime}$, take $r=n^{\prime}-1,2 m=2^{k+1}-r$; if $2^{k+1}<n^{\prime}$, take $r=2^{k+1}-1,2 m=n^{\prim... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,556 |
12. Let $k$ be a given positive integer. Let $\varphi_{k}(n)$ denote the number of arrays $\left\{d_{1}, d_{2}, \cdots, d_{k}\right\}$ satisfying the following conditions:
$$1 \leqslant d_{j} \leqslant n, \quad 1 \leqslant j \leqslant k \quad \text { and } \quad\left(d_{1}, \cdots, d_{k}, n\right)=1$$
(It is evident th... | 12. (i) $\varphi_{k}(n)=\sum_{1 \leqslant d_{1} \leqslant n} \cdots \sum_{1 \leqslant d_{k} \leqslant n d \mid\left(d_{1}, \cdots, d_{k}, n\right)} \mu(d)=\sum_{d \mid n} \mu(d)(n / d)^{k}$. Another proof is
$$\begin{aligned}
\sum_{d \mid n} \varphi_{k}(d) & =\sum_{d \mid n} \sum_{\substack{\left(d_{1}, \cdots, d_{k}, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,557 |
13. Let $S_{k}(n)=n^{-k} \sum_{j=1}^{n} j^{k}, S_{k}^{*}(n)=n^{-k} \sum_{\substack{j=1 \\(j, n)=1}}^{n} j^{k}$.
(i) Prove that the Möbius transform of $S_{k}^{*}(n)$ is $S_{k}(n)$ (prove using two methods);
(ii) Find the values of $S_{1}^{*}(n), S_{2}^{*}(n)$. | 13. (i) Using the method from the previous problem, from $S_{k}(n)=n^{-k} \sum_{d \mid n} \sum_{\substack{j=1 \\(j, n)=d}} j^{k}$ or
$$\sum_{d \mid n} S_{k}^{*}(d)=\sum_{d \mid n} d^{-k} \sum_{j=1}^{d} j^{2}\left(\sum_{l \mid(d, j)} \mu(l)\right)$$
we can deduce;
(ii) When $n>1$,
$$\begin{aligned}
S_{1}^{\prime}(n) & ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,558 |
17. Let $f_{1}, \cdots, f_{r}$ be number-theoretic functions. Prove:
(i) $f_{1} * f_{2}=f_{2} * f_{1}$;
(ii) $\left(f_{1} * f_{2}\right) * f_{3}=f_{1} *\left(f_{2} * f_{3}\right)$;
(iii) $\left(f_{1}+f_{2}\right) * f_{3}=\left(f_{1} * f_{3}\right)+\left(f_{2} * f_{3}\right)$, where the "+" sign denotes the addition of ... | 17. Regardless of the order in which the convolution $f_{1} * f_{2} * \cdots * f_{r}$ is performed, we have
$$\left(f_{1} * \cdots * f_{r}\right)(n)=\sum_{d_{1} \cdots d_{r}=n} f_{1}\left(d_{1}\right) \cdots f_{r}\left(d_{r}\right) .$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,562 |
19. Prove:
(i) $\mu * U=I$;
(ii) $\tau=U * U$;
(iii) $\varphi=\mu * E$;
(iv) $\sigma=U * E$;
(v) $\sigma=\varphi * \tau$;
(vi) $\sigma * \varphi=E * E$;
(vii) $\tau^{2} * \mu=\tau * \mu^{2}$. | 19. Direct verification, or using the result of problem 18, or using the convolution property to derive. For example: from (i) (iv) we can deduce that $\sigma=U * E=U *(U * \varphi)=\tau * \varphi$, which means (v) holds. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,564 |
26. Let $m>n$ be positive integers. Prove: $2^{n}-1 \mid 2^{m}-1$ if and only if $n \mid m$. Does the conclusion still hold if 2 is replaced by any positive integer $a>2$? | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 739,567 |
23. Prove: (i) $\mu^{-1}=U$; (ii) $\tau^{-1}=\mu^{*} \mu$; (iii) Find $E^{-1}$;
(iv) Find $\sigma^{-1}, \varphi^{-1}$; (v) Find the inverse of the Liouville function $\lambda$. | 23. (i), (ii) follow from 19 (i), (ii). (iii) $E^{-1}(1)=1, E^{-1}(p)=-p$, $E^{-1}\left(p^{\alpha}\right)=0, \alpha \geqslant 2$, i.e., $E^{-1}=\mu E$. (iv) $\sigma^{-1}=\mu * E^{-1}, \sigma^{-1}\left(p^{\alpha}\right)=E^{-1}\left(p^{\alpha}\right)-$ $E^{-1}\left(p^{\alpha-1}\right)=-p-1, \alpha=1 ; p, \alpha=2 ; 0, \a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,569 |
26. Let $f$ be a completely multiplicative function, $g$ be a number-theoretic function, $g(1) \neq 0$. Prove:
(i) $(f g)^{-1}=f g^{-1}$;
(ii) $f^{-1}=\mu f$. | 26. (i) Directly verify $\left(f g^{-1}\right) *(f g)^{-1}=I$; (ii) Take $g=U$ in (i). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,572 |
28. Let $\varphi_{1}=\mu^{2} * E$. Prove: $\varphi_{1}(n)=\sum_{d^{2} \mid n} \mu(d)_{\sigma}\left(\frac{n}{d^{2}}\right)$. | 28. Direct verification. $\left(\mu^{2} \times E\right)\left(p^{a}\right)=\sum_{d \mid \rho^{\alpha}} \mu^{2}(d)\left(p^{\alpha} / d\right)=p^{\alpha}+p^{\alpha-1}$.
$$\sum_{d^{2} \mid \rho^{a}} \mu(d) \sigma\left(p^{a} / d^{2}\right)=\left\{\begin{array}{ll}
\sigma(p)=p+1, & \text { when } \alpha=1, \\
\sigma\left(p^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,574 |
29. Let $f$ be a mapping from a set $S$ to itself, and its $n$-th iteration is denoted as
$$f^{(n)}(x)=\underbrace{f(f(\cdots(f}_{n \uparrow}(x)) \cdots)) .$$
Assume that for each positive integer $n$, $f^{(n)}$ has a finite number of fixed points, i.e., there are a finite number of $x \in S$ such that $f^{(n)}(x)=x$.... | 29. (i) Use division with remainder. (ii) Deduce from (i) and the definitions of $T(n), P(d)$. (iii) First prove: if $x \in P(n)$, then $x_{j}=f^{(j)}(x)(0 \leqslant j \leqslant n-1)$ are all distinct and belong to $T(n)$. Then prove that $x_{j}(0 \leqslant j \leqslant n-1)$ all belong to $P(n)$. Therefore, the points ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,575 |
30. Let $m \geqslant 3$. Prove that the arithmetic sequence $1 + lm (l=0,1, \cdots)$ contains infinitely many primes by the following steps:
(i) The original proposition is equivalent to: For any $m \geqslant 3$, the above arithmetic sequence must contain at least one prime.
(ii) Let $a > 1, m > 1$, and $q \geqslant 1$... | 30. (ii) $f(\bar{x})=a \bar{x}-x_{1}\left(a^{m}-1\right)$. If $\bar{x} \in S, f^{\left(m / p_{i}\right)}(\bar{x})=\bar{x}$, let $m_{i}=m / p_{i}$, then it must be that $\bar{x}=\left(x_{1} a^{m_{i}-1}+\cdots+x_{m_{i}-1} a+x_{m_{i}}\right)\left(a^{m}-1\right) /\left(a^{m_{i}}-1\right)$. From $q \mid D$ it follows that $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,576 |
Theorem 1 Let $x \geqslant 1$. We have
$$D(x)=\sum_{n \leqslant x} \tau(n)=x \ln x+r_{1}(x),$$
where
$$\left|r_{1}(x)\right| \leqslant x$$ | Prove that using $\tau(n)=\sum_{d \mid n} 1$, i.e., $\tau(n)$ is the Möbius transform of the number-theoretic function $U(n) \equiv 1$, we can obtain (by making the integer transformation $n=d l$)
$$\begin{aligned}
D(x) & =\sum_{n \leq x} \sum_{d \mid n} 1=\sum_{d \leq x} \sum_{d \mid n \leq x} 1=\sum_{d \leq x} \sum_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,577 |
27. Let $a, b$ be positive integers, $b>2$. Prove: $2^{b}-1 \nmid 2^{a}+1$. | 27. For $a<b$ and $a \geqslant b$ two cases. When $a<b$, it is only possible for $2^{b}-1 \mid 2^{a}+1$ when $a=1, b=2$; when $a \geqslant b$, let $a=q b+r, 0 \leqslant r<b$. It is easy to prove that $2^{b}-1\left|2^{a}+1 \Longleftrightarrow 2^{b}-1\right| 2^{r}+1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,578 |
Lemma 2 Let $y \geqslant 1$. We have
$$\sum_{1 \leqslant n \leqslant y} \frac{1}{n}=\ln y+\gamma+\Delta_{1}(y),$$
where
$$\begin{array}{l}
\left|\Delta_{1}(y)\right| \leqslant 1 / y \\
\gamma=\sum_{n=1}^{\infty} \int_{n}^{n+1}\left(\frac{1}{n}-\frac{1}{t}\right) \mathrm{d} t=\int_{1}^{\infty} \frac{t-[t]}{t[t]} \mathr... | Prove that when $y \geqslant 1$, we have
$$\begin{aligned}
\sum_{1 \leqslant n \leqslant y} \frac{1}{n}-\ln y & =\sum_{1 \leqslant n \leqslant[y]} \frac{1}{n}-\int_{1}^{y} \frac{1}{t} \, \mathrm{d} t \\
& =\sum_{n=1}^{[y]} \int_{n}^{n+1} \frac{1}{n} \, \mathrm{d} t-\int_{1}^{y} \frac{1}{t} \, \mathrm{d} t \\
& =\sum_{n... | proof | Calculus | proof | Yes | Yes | number_theory | false | 739,579 |
Theorem 3 Let $x \geqslant 1$. We have
$$D(x)=\sum_{n \leqslant x} \tau(n)=x \ln x+(2 \gamma-1) x+r_{2}(x),$$
where $\gamma$ is the Euler constant given by equation (14), and
$$\left|r_{2}(x)\right|<4 \sqrt{x} .$$ | Proof: From (11) and Lemma 2, we have
$$\begin{aligned}
D(x)= & 2 x\left(\ln \sqrt{x}+\gamma+\Delta_{1}(\sqrt{x})\right)-x \\
& +2 \sqrt{x}\{x\}-\{x\}^{2}-2 \sum_{1 \leqslant d \leqslant \sqrt{x}}\left\{\frac{x}{d}\right\} \\
= & x \ln x+(2 \gamma-1) x+r_{2}(x)
\end{aligned}$$
where
$$r_{2}(x)=2 x \Delta_{1}(\sqrt{x})... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,580 |
Lemma $4^{(1)}$ Let $y \geqslant 1$. We have
$$\ln ([y]!)=\sum_{1 \leqslant n<y} \ln n=y \ln y-y+\Delta_{2}(y),$$
where
$$\left|\Delta_{2}(y)\right|<2+\ln y .$$ | Prove that when $1 \leqslant y$,
\[
\begin{aligned}
& y(y-1) \ln (y-1) \\
& =(y-1) \ln y+(y-1) \ln \left(1-\frac{1}{y}\right) \\
& >y \ln y-\ln y-1
\end{aligned}
\]
This leads to
\[
\begin{aligned}
2+\int_{1}^{[y]} \frac{t-[t]}{t} \mathrm{~d} t> & \Delta_{2}(y) \\
= & {[y] \ln [y]-y \ln y+y-[y]+1 } \\
& +\int_{1}^{[y]... | 2+\ln y>\Delta_{2}(y)>-\ln y | Calculus | proof | Yes | Yes | number_theory | false | 739,581 |
Theorem 5 Let $x \geqslant 1$. We have
$$C(x)=\sum_{n \leq x} N(n)=\pi x+r_{3}(x),$$
where
$$-12 \sqrt{x}<r_{3}(x)<12 \sqrt{x}$$ | By Theorem 5 and formula (23) of Chapter 6, §2 (taking \( g(n) = h(n), k(n) \equiv 1 \)), we have
\[
\begin{aligned}
\frac{1}{4} C(x) = & \sum_{n \leqslant x} \frac{1}{4} N(n) = \sum_{\substack{d \leqslant x \\ d \geqslant 1, l \geqslant 1}} h(d) \\
= & \sum_{1 \leqslant d \leqslant \sqrt{x}} h(d) \sum_{1 \leqslant l \... | \pi x / 4 - 3 \sqrt{x} < C(x) / 4 < \pi x / 4 + 3 \sqrt{x} | Number Theory | proof | Yes | Yes | number_theory | false | 739,582 |
Theorem 6 Let $x \geqslant 1$. We have
$$\Phi(x)=\sum_{n \leqslant x} \varphi(n)=\frac{1}{2}\left(\sum_{d=1}^{\infty} \frac{\mu(d)}{d^{2}}\right) x^{2}+r_{4}(x),$$
where
$$\left|r_{4}(x)\right|<3 x \ln x+4 x .$$ | Prove that using equation (17) and equation (24) (taking \( s=x, t=1 \)) we can obtain
$$\begin{aligned}
\Phi(x) & =\sum_{n \leqslant x} \sum_{d \mid n} \mu(d) \frac{n}{d}=\sum_{\substack{d l \leqslant x \\
d \geqslant 1, l \geqslant 1}} \mu(d) l \\
& =\sum_{1 \leqslant d \leqslant x} \mu(d) \sum_{1 \leqslant l \leqsla... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,583 |
Theorem 7 Let $x \geqslant 1$. We have
$$\left|\sum_{n \leqslant x} \frac{\mu(n)}{n}\right| \leqslant 1$$ | Proof: From equation (18) in §2, we have
$$\begin{aligned}
1 & =\sum_{n \leqslant x}\left[\frac{1}{n}\right]=\sum_{n \leqslant x} \sum_{d \mid n} \mu(d)=\sum_{\substack{d \leqslant x \\
d \geqslant 1, l \geqslant 1}} \mu(d) \\
& =\sum_{d \leqslant x} \mu(d) \sum_{1 \leqslant l \leqslant x / d} 1=\sum_{d \leqslant x} \m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,584 |
Theorem 8 Let $x \geqslant 1$. We have
$$\sum_{n \leq x} \frac{\Lambda(n)}{n}=\ln x+r_{5}(x),$$
where
$$\left|r_{5}(x)\right|<4 \ln 2+2$$ | Proof: By Lemma 8 of Chapter 8, §2, we have
$$\begin{aligned}
\ln ([x]!) & =\sum_{n \leqslant x} \ln n=\sum_{n \leqslant x} \sum_{d \mid n} \Lambda(d)=\sum_{\substack{d \leqslant x \\
d \geqslant 1}} \Lambda(d) \\
& =\sum_{d \leqslant x} \Lambda(d) \sum_{1 \leqslant l \leqslant x / d} 1=\sum_{d \leqslant x} \Lambda(d)\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,585 |
Theorem 9 Let $x \geqslant 1$. We have
$$\sum_{p \leqslant x} \frac{\ln p}{p}=\ln x+r_{6}(x),$$
where
$$\left|r_{6}(x)\right|<5 \ln 2+3 .$$ | Proof: From the definition of $\Lambda(n)$ (let $n=p^{m}$)
$$\begin{aligned}
\sum_{n \leqslant x} \frac{\Lambda(n)}{n}-\sum_{p \leqslant x} \frac{\ln p}{p} & =\sum_{p^{m} \leqslant x, m \geqslant 2} \frac{\ln p}{p^{m}}<\sum_{p}\left(\frac{1}{p^{2}}+\frac{1}{p^{3}}+\cdots\right) \ln p \\
& =\sum_{p} \frac{\ln p}{p(p-1)}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,586 |
Lemma 10 Let $x \geqslant 1, b(n)$ be an arithmetic function,
$$B(x)=\sum_{n \leq x} b(n)$$
Suppose $\alpha(x)$ is a continuously differentiable function on the interval $\left[x_{1}, x_{2}\right]$, $x_{2}>x_{1} \geqslant 0$. Then, we have
$$\sum_{x_{1}<n \leqslant x_{2}} b(n) \alpha(n)=B\left(x_{2}\right) \alpha\left... | Let $n_{1}=\left[x_{1}\right], n_{2}=\left[x_{2}\right]$. We have (with the convention $B(0)=0$)
$$\begin{aligned}
\sum_{x_{1}<n \leq x_{2}} b(n) \alpha(n)= & \sum_{n_{1}<n \leq n_{2}} b(n) \alpha(n) \\
= & \sum_{n=n_{1}+1}^{n_{2}}\{B(n)-B(n-1)\} \alpha(n) \\
= & -B\left(n_{1}\right) \alpha\left(n_{1}+1\right) \\
& -\s... | proof | Calculus | proof | Yes | Yes | number_theory | false | 739,587 |
Theorem 11 Let $x \geqslant 2$. We have
$$\sum_{p \leqslant x} \frac{1}{p}=\ln \ln x+A_{1}+r_{7}(x)$$
where the constant
$$A_{1}=1-\ln \ln 2+\int_{2}^{\infty} r_{6}(t)\left(t \ln ^{2} t\right)^{-1} \mathrm{~d} t,$$
and
$$\left|r_{7}(x)\right|<2(5 \ln 2+3)(\ln x)^{-1}$$ | Let $x_{1}=2, x_{2}=x, b(n)$ be given by equation (43), and $\alpha(x)=(\ln x)^{-1}$. From equation (42), we have
$$\begin{aligned}
\sum_{p \leqslant x} \frac{1}{p} & =\frac{1}{2}+\sum_{2<n \leqslant x} b(n) \alpha(n) \\
& =\frac{1}{2}+\frac{1}{\ln x}\left\{\sum_{p \leqslant x} \frac{\ln p}{p}\right\}-\frac{1}{2}+\int_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,588 |
1. Prove: $\gamma=1-\int_{1}^{\infty}(t-[t]) t^{-2} \mathrm{~d} t$, where $\gamma$ is the Euler constant. | 1. From equation (14), we know that it is sufficient to prove:
$$\begin{array}{l}
I=\int_{1}^{\infty}\left\{(t-[t]) / t[t]+(t-[t]) / t^{2}\right\} \mathrm{d} t=1 \\
I=\sum_{n=1}^{\infty} \int_{n}^{n+1}\left(1 / n-n / t^{2}\right) \mathrm{d} t=\sum_{n=1}^{\infty}(1 / n-1 /(n+1))=1
\end{array}$$
We can also use Lemma 10... | proof | Calculus | proof | Yes | Yes | number_theory | false | 739,590 |
5. Let $D(x)$ be given by Theorem 1 of $\S 3$. Prove:
$$\sum_{n \leqslant x} 2^{\omega(n)}=\sum_{n \leqslant \sqrt{x}} \mu(n) D\left(\frac{x}{n^{2}}\right) .$$ | 5. $\begin{array}{l}2^{\omega(n)}=\sum_{\substack{n=d_{1} d_{2} \\\left(d_{1}, d_{2}\right)=1}} 1=\sum_{n=d_{1} d_{2}} \sum_{l \mid\left(d_{1}, d_{2}\right)} \mu(l) . \\ \sum_{n \leqslant x} 2^{\alpha(n)}=\sum_{l \leqslant \sqrt{x}} \mu(l) \sum_{k_{1} k_{2} \leqslant x / l^{2}} 1=\sum_{l \leqslant \sqrt{x}} \mu(l) D\le... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,595 |
9. (i) Prove: When $n \geqslant 2$, $\sigma(n) / n < n / \varphi(n) < \left(\pi^{2} / 6\right) \sigma(n) / n$;
(ii) There exists a positive constant $A$, such that $\sum_{n \leqslant x} \frac{n}{\varphi(n)} \leqslant A x, x \geqslant 1$;
(iii) There exists a positive constant $B$, such that $\sum_{n \leqslant x} \frac{... | 9. (i) $\sigma(n)=\prod_{p \mid n}\left(p^{a+1}-1\right) /(p-1), \frac{n^{2}}{\varphi(n)}=n \prod_{p \mid n} \frac{p}{p-1}$. Therefore,
$$n^{2} /(\sigma(n) \varphi(n))=\prod_{\rho \mid n}\left(1-1 / p^{\alpha+1}\right) ;$$
(ii) Follows from (i) and part (ii) of problem 7;
(iii) Follows from (i) and part (ii) of problem... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,599 |
29. Find: (i) the smallest positive integer $d$ such that $7 \mid 2^{d}-1$;
(ii) the smallest positive integer $d$ such that $11 \mid 3^{d}-1$;
(iii) the smallest non-negative remainder and the absolute smallest remainder when $2^{d}$ is divided by 7;
(iv) the smallest non-negative remainder and the absolute smallest r... | 29. (i) 3 ;
(ii) 5 ;
(iii) $1,2,4 ; 1,2,-3$;
(iv) $1,3,9,5,4 ; 1,3,-2,5,4$. | 3, 5, 1,2,4, 1,2,-3, 1,3,9,5,4, 1,3,-2,5,4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,600 |
16. Let $x \geqslant 2, \sigma_{t}(n)$ be as in the previous problem. Prove:
(i) $\sum_{n \leq x} \sigma_{-1}(n)=\zeta(2) x+r_{1}(x),\left|r_{1}(x)\right| \leqslant A_{1} \ln x, A_{1}$ is a positive constant;
(ii) When $t>0, t \neq 1$,
$$\sum_{n \leqslant x} \sigma_{-t}(n)=\zeta(t+1) x+r_{2}(x), \quad\left|r_{2}(x)\rig... | 16. $\sum_{n \leq x} \sigma_{-t}(n)=\sum_{d \leq x} d^{-t}[x / d]$, then use problem 13. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,607 |
22. Let $x \geqslant 4$. Prove:
$$\sum_{n \leqslant x} \omega^{2}(n)=x(\ln \ln x)^{2}+A_{3} \ln \ln x+r_{3}(x),$$
where $A_{3}$ is a positive constant, $\left|r_{3}(x)\right| \leqslant B_{1} x, B_{1}$ is a positive constant. | 22. Let $p, q$ be prime variables.
$$\begin{aligned}
\sum_{n \leq x} \omega^{2}(n) & =\sum_{n \leq x}\left(\sum_{p \mid n} 1\right)\left(\sum_{q \mid n} 1\right)=\sum_{p \leq x} \sum_{q \leq x} \sum_{p|n, q| n, n \leq x} 1 \\
& =\sum_{p=q}+\sum_{p \neq q}=S_{1}+S_{2}
\end{aligned}$$
It is clear that $S_{1}=\sum_{n \le... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,614 |
Property II Let $k=k_{1} k_{2},\left(k_{1}, k_{2}\right)=1, \chi(n ; k)$ be a character modulo $k$, then there exists a unique character $\chi\left(n ; k_{1}\right)$ modulo $k_{1}$ such that
$$\chi(n ; k)=\chi\left(n ; k_{1}\right), \quad n \equiv 1\left(\bmod k_{2}\right)$$ | Define the number-theoretic function $f(m)$ as follows: for any integer $m$, by the Chinese Remainder Theorem, there exists a unique $n$ modulo $k$ satisfying
$$\left\{\begin{array}{l}
n \equiv m\left(\bmod k_{1}\right) \\
n \equiv 1\left(\bmod k_{2}\right)
\end{array}\right.$$
We define
$$f(m)=\chi(n ; k)$$
We will ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,623 |
Theorem 1 Let $k=k_{1} k_{2},\left(k_{1}, k_{2}\right)=1, \chi(n ; k)$ be a character modulo $k$, then there exists a unique pair of characters $\chi\left(n ; k_{1}\right)$ modulo $k_{1}$ and $\chi\left(n ; k_{2}\right)$ modulo $k_{2}$, such that for any integer $n$, we have
$$\chi(n ; k)=\chi\left(n ; k_{1}\right) \ch... | First, we prove the uniqueness. If equation (19) holds, then we have
$$\begin{array}{ll}
\chi(n ; k)=\chi\left(n ; k_{1}\right), & n \equiv 1\left(\bmod k_{2}\right), \\
\chi(n ; k)=\chi\left(n ; k_{2}\right), & n \equiv 1\left(\bmod k_{1}\right),
\end{array}$$
Thus, by property VII, we know that $\chi\left(n ; k_{1}\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,624 |
Theorem 6 Let $k>1,(a, k)=1$ and $a \neq 1(\bmod k)$, then there must exist a non-principal character $\chi$ modulo $k$, such that $\chi(a) \neq 1$. | Let $k$ be given by (23), and using the notation of Theorem 5, the conditions imply
$$a \not \equiv 1\left(\bmod 2^{a_{0}}\right), \quad a \not \equiv 1\left(\bmod p_{1}^{a_{1}}\right), \quad \cdots, \quad a \not \equiv 1\left(\bmod p_{s}^{a_{s}}\right)$$
at least one of these holds. If $a \neq 1\left(\bmod 2^{a_{0}}\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,630 |
Theorem 7 Let $k \geqslant 1$. We have
$$\sum_{\chi \bmod k} \chi(n)=\left\{\begin{array}{ll}
\varphi(k), & n \equiv 1(\bmod k) \\
0, & n \neq 1(\bmod k)
\end{array}\right.$$
Here the summation sign indicates the sum over all characters modulo $k$. | Prove that when $n \equiv 1(\bmod k)$, for any $\chi \bmod k$ we have $\chi(n)=1$, so the left side of equation (37) is the number of characters modulo $k$. By Theorem 5, this number is $\varphi(k)$. This proves the first part of equation (37).
When $n \not\equiv 1(\bmod k)$, we must have $k>1$. If $(n, k)>1$, then th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,631 |
Theorem 8 Let $k \geqslant 1, \chi$ be a character modulo $k$. We have
$$\sum_{n \bmod k}^{\prime} \chi(n)=\left\{\begin{array}{ll}
\varphi(k), & \chi=\chi^{0} \bmod k \\
0, & \chi \neq \chi^{0} \bmod k
\end{array}\right.$$
where the summation is over a reduced residue system modulo $k$. | Prove that when $\chi=\chi^{0} \bmod k$, for any $(n, k)=1$ we have $\chi(n)=1$. This proves the first equation of (38). By Theorem 5, there exists a set of $l_{-1}, \cdots, l_{s}$, such that
$$\chi(n)=\chi\left(n ; k, l_{-1}, l_{0}, \cdots, l_{s}\right)$$
When $\chi \neq \chi^{0}$, there must be an $h(-1 \leqslant h ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,632 |
5. Let $M$ be a set composed of integers. If $M$ has an upper bound, that is, there exists an integer $a$ such that for all $m \in M$ we have $m \leqslant a$, then there must be an $m_{0} \in M$ such that for all $m \in M$, we have $m \leqslant m_{0}$. | 5. Consider the set $M^{*}=\left\{m^{*}=-m: m \in M\right\}$. Apply the conclusion of question 4 to $M^{*}$. | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,633 |
32. Let the odd number $a>2, a \mid 2^{d}-1$ with the smallest positive integer $d=d_{0}$. Prove: The different smallest non-negative remainders that $2^{d}$ can take when divided by $a$ are $d_{0}$ in number. | 32. Prove: $2^{0}, 2^{1}, \cdots, 2^{d_{0}-1}$ when divided by $a$ yield distinct smallest non-negative remainders, and any $2^{k}$ when divided by $a$ yields a smallest non-negative remainder that must be the same as one of the $d_{0}$ remainders above. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,634 |
3. Let $k=2^{a_{0}} p_{1}^{a_{1}} \cdots p_{s}^{a_{s}}$. Prove: $\chi(n ; k)$ is a real character if and only if:
(i) When $\alpha_{0}=0$, $\chi(n ; k)=\left(\frac{n}{p_{1}}\right)^{\beta_{1}} \cdots\left(\frac{n}{p_{s}}\right)^{\beta_{3}}$, where $\beta_{j}=1$ or 2;
(ii) When $\alpha_{0} \geqslant 1$,
$$\chi(n ; k)=\l... | 3. Using expression (33). The principal character modulo $p^{\alpha}$ is $\chi\left(n ; p^{\alpha}, 0\right)=\chi(n ; p, 0)$, and the non-principal real character modulo $p^{\alpha}$ is $\chi\left(n ; p^{\alpha}, \varphi\left(p^{\alpha}\right) / 2\right)=\chi(n ; p,(p-1) / 2)=\left(\frac{n}{p}\right)$, where expression... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,641 |
4. A number-theoretic function $f(n)$ is called periodic if there exists a positive integer $q$ such that for any integer $n$, $f(n+q)=f(n)$. Let $f(n)$ be periodic. Prove:
(i) If $q_{0}$ is the smallest such $q$, then $q_{0} \mid q$. $q_{0}$ is called the smallest positive period of $f(n)$. $\square$
(ii) If $f(n)$ is... | 4. (i) Use division with remainder. (ii) It suffices to prove that when $\left(n, q_{0}\right)>1$, $f(n)=0$. That is, we need to prove that for any $p|q_{0}$, $f(p)=0$. If $f(p) \neq 0$, let $p^{a} \| q_{0}, q_{0}=p^{a} q_{1}$.
$$f\left(p^{\sigma}\right) f\left(n+q_{1}\right)=f\left(p^{\sigma} n+q_{0}\right)=f\left(p^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,642 |
6. (i) How to determine the smallest positive period of $\chi\left(n; p^{\alpha}\right)$ given by equation (28)?
(ii) How to determine the smallest positive period of $\chi\left(n ; 2^{\circ}\right)$ given by equation (31)?
(iii) Let $\chi(n ; k)$ be given by equation (24). Prove: its smallest positive period is equal ... | 6. (i) $p^{a} /\left(l, p^{\alpha-1}\right)$. (ii) $2^{a} /\left(\left(l_{-1}, 2\right) \cdot\left(l_{0}, 2^{a-3}\right)\right)$. (iii) Let the smallest positive period of $\chi(n ; k)$ be $q ; \chi\left(n ; 2^{a_{0}}\right), \chi\left(n ; p_{1}^{q_{1}}\right), \cdots, \chi\left(n ; p_{s}^{a_{s}}\right)$ have the small... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,644 |
1. Use the Euclidean algorithm from Theorem 4 in $\S 3$ to find the greatest common divisor (gcd) of the following sets of numbers, and express the gcd as an integer linear combination of these numbers:
(i) 1819, 3587;
(ii) 2947, 3997;
(iii) $-1109, 4999$. | 1. (i) $3587=1819+1768,1819=1768+51,1768=34 \cdot 51+34,51=$ $1 \cdot 34+17,34=2 \cdot 17$, so $(3587,1819)=17$.
$$\begin{aligned}
17 & =51-34=51-(1768-34 \cdot 51) \\
& =35 \cdot 51-1768=35(1819-1768)-1768 \\
& =35 \cdot 1819-36 \cdot 1768=35 \cdot 1819-36(3587-1819) \\
& =-36 \cdot 3587+71 \cdot 1819
\end{aligned}$$
... | 17 = -36 \cdot 3587 + 71 \cdot 1819 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,645 |
7. Prove: (i) If $\chi(n ; k)$ is a non-primitive character, let the smallest $k^{\prime}$ satisfying the definition be $k^{*}$, then it must be that $k^{*} \mid k$.
(ii) The definition of a non-primitive character is equivalent to the following four definitions: (a) There exists a positive integer $k^{\prime}<k$, such... | 7. (i) Use division with remainder; (ii) $\left(n_{1} n_{2}, k\right)=1, n_{1} \equiv n_{2}\left(\bmod k^{\prime}\right)$ is equivalent to
$$\left(n_{1} n_{2}, k\right)=1, \quad n_{1} n_{2}^{-1} \equiv 1\left(\bmod k^{\prime}\right)$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,646 |
8. Prove: (i) The modulus 1 character is the principal character;
(ii) There is no primitive character modulo 2;
(iii) $\chi(n ; 4,0)$ (see equation (8)) is not a primitive character, $\chi(n ; 4,1)$ (see equation (9)) is a primitive character;
(iv) $\chi\left(n ; p^{a}, l\right)$ (see equation (28)) is a primitive cha... | 8. (i), (ii), (iii) direct verification; (iv), (v) use the corresponding expressions (28), (32), the previous problem (i), the relationship between the indices of a given primitive root $g$ (when $k=p^{\alpha}, p$ is an odd prime, for all $\left.\alpha \geqslant 1\right), n$ modulo $p^{\alpha_{1}}$ and modulo $p^{a_{2}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,647 |
9. Let $P(k)$ denote the number of primitive roots modulo $k$.
(i) Prove: $P(k)$ is a multiplicative function of $k$;
(ii) Prove: $\sum_{d \mid k} P(d)=\varphi(k)$;
(iii) Find the value of $P\left(p^{a}\right)$, where $p$ is a prime. | 9. (i) Use the result from (vi); (ii) Let $k$ be a given positive integer. For each character $\chi(n ; k)$ modulo $k$, there is a unique correspondence to a divisor $d \mid k$ and a primitive character $\chi^{*}(n ; d)$ modulo $d$, and vice versa (i.e., Question 12). From this and the fact that there are $\varphi(k)$ ... | P\left(p^{a}\right)=\left\{\begin{array}{ll}p-2, & \text { when } \alpha=1, \\ p^{a}-2 p^{\sigma-1}+p^{\sigma-2}, & \text { when } \alpha>1 .\end{array}\right.} | Number Theory | proof | Yes | Yes | number_theory | false | 739,648 |
13. When $k=p^{e}, \chi(n ; k)$ is not a principal character, the $k^{*}$ from the previous problem is the smallest positive period of $\chi(n ; k)$ (see problem 4). Provide an example to show that this conclusion does not hold when $k$ is not a prime power. | 13. $\chi\left(n ; p^{a}\right)$ is a non-principal character, the corresponding $k^{*}=p^{\lambda} \neq 1, \lambda \leqslant \alpha$. At this time, $\left(n, p^{a}\right)=1$ is the same as $(n, p)=1$. As long as a principal character appears on the right side of expression (24), it does not hold. | not found | Number Theory | proof | Yes | Yes | number_theory | false | 739,652 |
17. Let $G(a ; \chi)$ be given by equation (67), where $\chi$ is a primitive character modulo $k$, and $(a, k)=\lambda>1$. Let $a=\lambda a^{\prime}, k=\lambda k^{\prime}$. Prove:
(i) $G(a ; \chi)=\sum_{n=1}^{k^{\prime}} S(n) e\left(a^{\prime} \frac{n}{k^{\prime}}\right)$, where $e(\theta)=\mathrm{e}^{2 \pi i \theta}$,... | 17. (iii) There must be $(m, k)=1, m \equiv 1\left(\bmod k^{\prime}\right)$ such that $\chi(m) \neq 1$ (see Question 7 (ii)), prove
$$\chi(m) S(n)=S(n) .$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,657 |
18. (i) When $\chi$ is a primitive character modulo $k$, we always have $G(a ; \chi)=\bar{\chi}(a) G(1 ; \chi)$;
(ii) When $\chi$ is a primitive character modulo $k$, $|G(1 ; \chi)|=\sqrt{k}$;
(iii) When $\chi$ is a real primitive character modulo $k$, $G^{2}(1, \chi)=\chi(-1) k$;
(iv) When $\chi$ is the Legendre symbo... | 18. (i) Deduced from the previous question (iv) and Question 16 (v); (ii) Consider $\sum_{a=1}^{k}|G(a ; \chi)|^{2}$, use (i) and formula (58) to calculate this sum in two ways, compare to obtain; (iii) Use Question 16 (iii), from which (iv) is also deduced. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,658 |
20. Let $k \geqslant 3, \chi$ be a primitive character modulo $k$. Using the result from problem 18, prove that for any integer $M$ and positive integer $N$, we have
$$\left|\sum_{n=M+1}^{M+N} \chi(n)\right| \leqslant \frac{1}{\sqrt{k}} \sum_{l=1}^{k-1}\left(\sin \frac{\pi l}{k}\right)^{-1}$$
Furthermore, show that
$$... | 20. $\sum_{n=M+1}^{M+N} \chi(n)=(G(1 ; \bar{\chi}))^{-1} \sum_{n=M+1}^{M+N} G(n ; \bar{\chi})$ (here we use the result from problem 18). Substituting this into equation (58) and calculating directly yields the first inequality. Then, using the fact that when $0 \leqslant x \leqslant \pi / 2$, $2 x / \pi \leqslant \sin ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,660 |
21. Using the results of questions 12 and 20, deduce that for any integer $M$ and positive integer $N$, when $\chi$ is a non-principal character modulo $k$, we have $\left|\sum_{n=M+1}^{M+N} \chi(n)\right|<2 \sqrt{k} \ln k$. | 21. $\sum_{n=M+1}^{M+N} \chi(n)=\sum_{\substack{n=M+1 \\\left(n, k_{2}\right)=1}}^{M+N} \chi^{*}\left(n ; k^{*}\right)$, here $k=k_{1} k_{2}, k_{1}$ and $k^{*}$ have the same prime factors, $k^{*} \mid k_{1},\left(k_{1}, k_{2}\right)=1$. Then use $\sum_{d \mid n} \mu(d)=[1 / n]$ and the estimate from the previous probl... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,661 |
23. Let $p$ be an odd prime, $N_{p}$ denote the smallest positive quadratic non-residue modulo $p$, and $N(x)$ denote the number of positive quadratic non-residues modulo $p$ that are less than or equal to $x$. Prove:
(i) If $a$ is a positive quadratic non-residue modulo $p$, then there must be a prime $q \mid a, q \ge... | 23. (iii) Let the number of positive quadratic residues modulo $p$ not exceeding $x$ be $R(x)$.
$$R(x)-N(x)=\sum_{a=1}^{[x]}\left(\frac{a}{p}\right)=\theta \sqrt{p} \ln p, \quad|\theta| \leqslant 1$$
(This follows from Problem 20). Additionally, it is obvious that $R(x)+N(x)=[x]$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,663 |
26. Let $p$ be an odd prime, and let $T_{p}^{*}(r)$ denote the number of solutions to the congruence equation
$$x_{1}^{2}+\cdots+x_{r}^{2} \equiv 0(\bmod p), \quad 1 \leqslant x_{1} \leq c$$
where $c = (p-1)/2$. When $r > 5$, we have
$$\begin{aligned}
(3!) p T_{p}^{*}(3)= & \sum_{l=1}^{p} \sum_{x_{1}=1}^{c} \sum_{x_{2}... | 26. The number of solutions to the congruence equation $f\left(x_{1}, \cdots, x_{r}\right) \equiv 0(\bmod p), 1 \leqslant x_{j} \leqslant a_{j}, 1 \leqslant j \leqslant r$ is
$$p^{-1} \sum_{x_{1}=1}^{s_{1}} \cdots \sum_{x_{r}=1}^{a_{r}} \sum_{l=1}^{p} \mathrm{e}^{2 x_{i} i f\left(x_{1}, \cdots, x_{r}\right) / p} .$$
F... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 739,666 |
Theorem 1 For any $n \in \boldsymbol{N}$, we have $n \neq n^{+}$. | Let the subset of elements $n$ in $\boldsymbol{N}$ for which $n \neq n^{+}$ holds be denoted as $S$. By axiom (ii), we know $e \neq e^{+}$, so $e \in S$, and $S$ is non-empty. If $n \in S$, i.e., $n \neq n^{+}$, we will prove that it must be the case that $n^{+} \in S$. If not, then we would have $n^{+}=\left(n^{+}\rig... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,668 |
Theorem 2 Let $m \in \boldsymbol{N}, m \neq e$, then there must be $n \in \boldsymbol{N}$, such that $n^{+}=m$, that is, each element in $\boldsymbol{N}$ that is not equal to $e$ must be the successor of some element, $e$ is the only element without a successor. Moreover, this element $n$ is unique, denoted as $m^{-}$,... | Let set $A$ consist of all elements $a$ in $\boldsymbol{N}$ such that $a$ is the successor of some element. Since $e^{+} \in A$, $A$ is non-empty. Let the union $S=\{e\} \cup A$. Clearly, $e \in S$. If $n \in S$, by the definition of $A$, $n^{+} \in A$, and thus $n^{+} \in S$. By the induction axiom (iv), it follows th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,669 |
Theorem 3 (Principle of Mathematical Induction) Let $P(n)$ be a property or proposition about the natural number $n$. If $P(e)$ holds when $n=e$, and if $P(n)$ being true necessarily implies that $P\left(n^{+}\right)$ is true, then $P(n)$ holds for all $n \in \boldsymbol{N}$. | Proof: Let $S$ be the set of all $n$ for which $P(n)$ holds. By the condition, $e \in S$, and if $n \in S$, then $n^{+} \in S$. Therefore, by the induction axiom (iv), the theorem holds. Q.E.D. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,670 |
Theorem 1 There exists a unique binary operation $\sigma$ on the set of natural numbers $N$, satisfying the conditions:
(i) For any $n \in \boldsymbol{N}$, we have
$$n \sigma e=n^{+} ;$$
(ii) For any $n, m \in \boldsymbol{N}$, we have
$$n \sigma m^{+}=(n \sigma m)^{+} \text {. }$$ | Proof of Theorem 1: First, we prove the existence. For any $n \in \boldsymbol{N}$, let $f_{n}$ be the mapping determined in Lemma 2. Now define the binary operation $\sigma$:
$$n \sigma m=f_{n}(m), \quad n \in \mathbf{N}, m \in \mathbf{N}$$
From equation (3), we get
$$n \sigma e=f_{n}(e)=n^{+},$$
so condition (1) is ... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,671 |
Lemma 2 For any given $n \in \boldsymbol{N}$, there exists a unique mapping $f_{n}$ from $\boldsymbol{N}$ to itself, satisfying
$$f_{n}(e)=n^{+} ;$$
and for any $m \in \boldsymbol{N}$, we have
$$f_{n}\left(m^{+}\right)=\left(f_{n}(m)\right)^{+} .$$ | First, we prove the uniqueness. If there exists another such mapping $g_{n}$, then when $m=e$, by equation (3) we get $g_{n}(e)=f_{n}(e)=n^{+}$. Assuming for some $m \in N$, $g_{n}(m)=f_{n}(m)$, then by equation (4) we get
$$g_{n}\left(m^{+}\right)=\left(g_{n}(m)\right)^{+}=\left(f_{n}(m)\right)^{+}=f_{n}\left(m^{+}\ri... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,672 |
(1) Associative Law of Addition For any $a, b, c \in \boldsymbol{N}$, we have
$$(a+b)+c=a+(b+c) .$$ | Prove that when $c=e$, for any $a, b \in \boldsymbol{N}$, from equations (1) and (2) we get
$$(a+b)+e=(a+b)^{+}=a+b^{+}=a+(b+e),$$
so equation (8) holds. Assume that equation (8) holds for some $c=n$ and any $a, b \in \boldsymbol{N}$. When $c=n^{+}$, for any $a, b \in \boldsymbol{N}$, we have (using equation (2) and t... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,673 |
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