problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
9. Let $f(x)=x-[x]-\frac{1}{2}$, prove:
$$\sum_{k=0}^{n-1} f\left(x+\frac{k}{n}\right)=f(n x)$$ | 9. Proof:
$$\begin{array}{l}
\sum_{k=0}^{n-1} f\left(x+\frac{k}{n}\right)=\sum_{k=0}^{n-1}\left\{\left(x+\frac{k}{n}-\frac{1}{2}\right)-\left[x+\frac{k}{n}\right]\right\} \\
\quad=n\left(x-\frac{1}{2}\right)+\frac{1}{n} \cdot \frac{1}{2} n(n-1)-\sum_{k=0}^{n-1}\left[x+\frac{k}{n}\right] \\
\quad=n x-\frac{1}{2}-\sum_{k... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,052 |
10. Prove: $d(n)$ is odd if and only if $n$ is a perfect square. | 10. Proof: Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{m}^{a_{m}}$.
By Lemma 6, we know
$$d(n)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{m}+1\right)$$
If $d(n)$ is odd, then all $\alpha_{i}(1 \leqslant i \leqslant m)$ must be even.
Let
$$\alpha_{i}=2 \beta_{i}, \quad 1 \leqslan... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,053 |
11. Prove: $\prod_{t \mid n} t=n^{\alpha(n) / 2}$,
where $\prod_{t \mid n} t$ denotes the product of all divisors of $n$. | 11. Proof: Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{m^{\prime}}^{\alpha_{n}}$. If $m=1$, then $n=p_{1}^{\alpha_{1}}$. Its all divisors are $1, p_{1}, \cdots, p_{1}^{a_{1}}$, thus
$$\prod_{t \mid n} t=\prod_{j=0}^{a_{1}} p_{j}^{j}=p_{1}^{\frac{1}{2} \alpha_{1}\left(\alpha_{1}+1\right)}=\left(p_{1}^{\alpha_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,054 |
12. Prove: $\sum_{d^{2} \mid n} \mu(d)=\mu^{2}(n)$. | 12. Proof: By the definition of $\mu(n)$, we have
$$\mu^{2}(n)=\left\{\begin{array}{ll}
1, & n \text { is square-free, } \\
0, & n \text { has a square factor. }
\end{array}\right.$$
When $n$ is square-free, $\sum_{d^{2} \mid n} \mu(d)=\mu(1)=1$. When $n$ has a square factor, let $n=n_{0}^{2} m$, where $n_{0}>1$ and $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,055 |
13. If $F(n)=\sum_{d \mid n} f(d) ,$
then $f(n)=\sum_{d \mid n} F\left(\frac{n}{d}\right) \mu(d)$, and vice versa. | 13. Proof: By the assumption,
$$\begin{array}{l}
\sum_{d \mid n} F\left(\frac{n}{d}\right) \mu(d)=\sum_{i \mid n} \mu(d) \sum_{d \left\lvert\, \frac{n}{d}\right.} f(c) \\
\quad=\sum_{c \mid n} \mu\left(d^{\prime}\right) f(c)=\sum_{c \mid n} f(c) \sum_{d \left\lvert\, \frac{n}{c}\right.} \mu(d)
\end{array}$$
and
$$\sum... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,056 |
14. Let $n>0$ be an integer, prove:
(i) $n=\sum_{d \mid n} \varphi(d)$.
(ii) $\varphi(n)=n \sum_{d \mid n} \frac{\mu(d)}{d}$ | 14. Let $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}$.
(i) Proof:
$$\begin{aligned}
\sum_{d i p_{s}^{\alpha_{s}}} \varphi(d)= & \varphi(1)+\varphi\left(p_{s}\right)+\varphi\left(p_{s}^{2}\right)+\cdots+\varphi\left(p_{s}^{a_{s}}\right) \\
= & 1+\left(p_{s}-1\right)+\left(p_{s}^{2}-p_{s}\right)+\cdots \\
& +\left... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,057 |
15. Prove that an even perfect number must have the form $2^{n-1}\left(2^{n}-1\right)$, and $2^{n}-1$ is a prime. | 15. Proof: Let $N$ be an even perfect number, which can be written as $N=2^{n-1} b, n>1, b$ is an odd number. By Lemma 11 and Lemma 9, we have
$$\sigma(N)=\sigma\left(2^{n-1}\right) \sigma(b)=\left(2^{n}-1\right) \sigma(b)$$
Since $N$ is a perfect number, we have
$$\sigma(N)=2 N=2^{n} b$$
Therefore,
i.e., $\square$
$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,058 |
Example 1 Given $(5 x+\sqrt{3})-i=3+(\sqrt{2}-y) i, x$ and $y$ are real numbers, find $x$ and $y$. | Solve the system based on the condition of equality of complex numbers:
$$\left\{\begin{array}{l}
5 x+\sqrt{3}=3 \\
-1=\sqrt{2}-y
\end{array}\right.$$
Therefore, $x=\frac{3-\sqrt{3}}{5}, y=1+\sqrt{2}$. | x=\frac{3-\sqrt{3}}{5}, y=1+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,060 |
Example 2 Calculate $(1+2 i) \div(3-4 i)$.
.
The translation maintains the original text's line breaks and format, as requested. | \[
\begin{array}{l}
(1+2 i) \div(3-4 i)=\frac{1+2 i}{3-4 i}=\frac{(1+2 i)(3+4 i)}{(3-4 i)(3+4 i)} \\
=\frac{3+4 i+6 i+8 i^{2}}{9+12 i-12 i-16 i^{2}}=\frac{(3-8)+(4+6) i}{9+16} \\
=\frac{-5+10 i}{25}=\frac{-1+2 i}{5}
\end{array}
\] | \frac{-1+2 i}{5} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,061 |
Lemma 10 Let $l$ be a positive integer, $p$ be a prime, then we have
$$\varphi\left(p^{l}\right)=p^{l-1}(p-1)$$ | Since any integer in $1,2, \cdots, p-1$ is coprime with $p$, we have $\varphi(p)=p-1$. When $l=1$, we have $p^{l-1}=p^{0}=1$, so the lemma holds when $l=1$. Now suppose $l>1$. The positive integers not greater than 4 and coprime with 4 are 1,3, a total of 2, so we have $\varphi(4)=2$. The positive integers not greater ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,062 |
Example 3 Let $n$ be an integer $\geqslant 2$ and $z_{1}, z_{2}, \cdots, z_{n}$ be $n$ complex numbers, then we have
$$\left|z_{1} z_{2} \cdots z_{n}\right|=\left|z_{1}\right| \cdot\left|z_{2}\right| \cdots\left|z_{n}\right|$$ | Let $z_{1}=a+b i, z_{2}=c+d i$ be two complex numbers. From equation (10), we have
$$\begin{array}{c}
\left|z_{1} z_{2}\right|=\sqrt{(a c-b d)^{2}+(a d+b c)^{2}}, \quad \left|z_{1}\right|=\sqrt{a^{2}+b^{2}} \\
\left|z_{2}\right|=\sqrt{c^{2}+d^{2}}
\end{array}$$
where $\left|z_{1} z_{2}\right|, \left|z_{1}\right|, \lef... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,063 |
Side 4 Let $z_{1}$ and $z_{2}$ be two complex numbers, then we have $\left|z_{1}+z_{2}\right| \leqslant$ $\left|x_{1}\right|+\left|z_{2}\right|$ | Let $z_{1}=a+b i, z_{2}=c+d i$, where $a, b, c, d$ are all real numbers, then we have
$$\begin{array}{l}
\left(\left|z_{1}\right|+\left|z_{2}\right|\right)^{2}-\left|z_{1}+z_{2}\right|^{2} \\
=\left(\sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}}\right)^{2}-(a+c)^{2}-(b+d)^{2} \\
= a^{2}+b^{2}+c^{2}+d^{2}+2 \sqrt{a^{2}+b^{2}} \c... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 741,064 |
Example 6 Convert $\frac{3}{5} \pi$ radians to degrees. | Solve $\frac{3}{5} \pi$ radians $=\frac{180^{\circ}}{\pi} \times \frac{3}{5} \pi=108^{\circ}$. | 108^{\circ} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,066 |
Lemma 1 Let $\theta_{1}$ and $\theta_{2}$ be two real numbers, then we have
$$e^{i\left(\theta_{1}+\theta_{2}\right)}=e^{i \theta_{1}} \cdot e^{i \theta_{2}}$$ | Proof: From equation (24), we have
$$\begin{aligned}
e^{i \theta_{1}} \cdot & e^{i \theta_{2}}=\left(\cos \theta_{1}+i \sin \theta_{1}\right)\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\
= & \cos \theta_{1} \cdot \cos \theta_{2}+i^{2} \sin \theta_{1} \cdot \sin \theta_{2}+i \sin \theta_{1} \\
& \quad \cdot \cos \th... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,067 |
Lemma 2 Let $n$ be a positive integer, $\theta$ be a real number, then we have
$$\left(e^{i \theta}\right)^{n}=e^{i n \theta}$$ | Proof: When $n=1$, by $\left(e^{i \theta}\right)^{1}=e^{i \theta}$, the lemma holds. Now assume $k \geqslant 2$, and the lemma holds for $n$ equal to $1,2, \cdots, k-1$, then by Lemma 1 we have
$$\left(e^{i \theta}\right)^{k}=\left(e^{i \theta}\right)^{k-1} \cdot e^{i \theta}=e^{i(k-1) \theta} \cdot e^{i \theta}=e^{i k... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,068 |
Example 8 Prove $(\sin x+i \cos x)^{n}=e^{i n\left(\frac{\pi}{2}-x\right)}$. | Proof: By Lemma 1 and equation (25), we have
$$\begin{aligned}
e^{i n\left(\frac{\pi}{2}-x\right)} & =e^{i n \frac{\pi}{2}} \cdot e^{-i n x}=\left(e^{\left.\frac{i}{2}\right)^{n}} e^{-i n x}=\left(i e^{-i x}\right)^{n}\right. \\
& =\left(i \cos x+i^{2} \sin (-x)\right)^{n}=(\sin x+i \cos x)^{n}
\end{aligned}$$
Thus, E... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,070 |
Example 9 Prove the identity
$$\begin{array}{l}
\sin (\alpha-\beta) \sin (\gamma-\delta)=\sin (\alpha-\delta) \sin (\gamma-\beta) \\
\quad+\sin (\alpha-\gamma) \sin (\beta-\delta)
\end{array}$$ | To prove that
$$\begin{aligned}
\left(a^{2}-\right. & \left.d^{2}\right)\left(c^{2}-b^{2}\right)+\left(a^{2}-c^{2}\right)\left(b^{2}-d^{2}\right) \\
= & a^{2} c^{2}-a^{2} b^{2}-c^{2} d^{2}+b^{2} d^{2}+a^{2} b^{2}-a^{2} d^{2} \\
& -b^{2} c^{2}+c^{2} d^{2} \\
= & a^{2} c^{2}-a^{2} d^{2}+b^{2} d^{2}-b^{2} c^{2} \\
= & \le... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,071 |
Lemma 3 Let $n$ be a positive integer and $z=a+b i$ be a complex number, then when $z \neq 1$ we have
$$\sum_{m=0}^{n} z^{m}=\frac{1-z^{n+1}}{1-z} .$$ | Proof When $n=1$, we have
$$1+z=\frac{(1+z)(1-z)}{1-z}=\frac{1-z^{2}}{1-z}$$
Thus, the lemma holds when $n=1$. Now assume $k \geqslant 2$, and the lemma holds for $n$ equal to $1,2, \cdots, k-1$, then we have
$$\sum_{m=0}^{k} z^{m}=\sum_{m=0}^{k-1} z^{m}+z^{k}=\frac{1-z^{k}}{1-z}+z^{k}=\frac{1-z^{k+1}}{1-z}$$
Thus, t... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,072 |
Example 6 Prove that $4,8,16,28,32,44,52,56$ is a reduced residue system modulo 15. | To prove that the positive integers less than 15 and coprime with 15 are 8 in total, which are
$$1,2,4,7,8,11,13,14$$
We have
$$\begin{aligned}
4 & \equiv 4(\bmod 15), \quad 8 \equiv 8(\bmod 15), \quad 16 \equiv 1(\bmod 15) \\
28 & \equiv 13(\bmod 15), \quad 32 \equiv 2(\bmod 15), \quad 44 \equiv 14(\bmod 15) \\
52 & ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,073 |
Lemma 4 We have
$$\sum_{m=0}^{n-1} e^{i(\theta+m \varphi)}=e^{i\left(\theta+\frac{n-1 \varphi}{2}\right)} \cdot \frac{\sin \frac{n \varphi}{2}}{\sin \frac{\varphi}{2}}$$
where $n$ is a positive integer, $\varphi \neq 2 l \pi$ where $l$ is any integer, i.e., $\left\{\frac{\varphi}{2 x}\right\} \neq 0(\{x\}$ denotes $x-... | Given that $\left\{\frac{\varphi}{2 \pi}\right\} \neq 0$, we have $e^{i \varphi} \neq 1$. By Lemma 3, we have
$$\begin{array}{l}
\sum_{m=0}^{n-1} e^{i m \varphi}=\frac{1-e^{i n \varphi}}{1-e^{i \varphi}}=\frac{e^{\frac{i n \varphi}{2}}\left(e^{\frac{i n \varphi}{2}}-e^{-\frac{i n \varphi}{2}}\right)}{e^{\frac{i \varphi... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,074 |
Let $n$ be a positive integer and $\left\{\frac{\varphi}{2 \pi}\right\} \neq 0$, then we have
$$\begin{aligned}
\sum_{m=0}^{n-1} \cos (\theta+m \varphi)= & \frac{\sin \frac{n \varphi}{2}}{\sin \frac{\varphi}{2}} \cdot \cos \left(\theta+\frac{n-1}{2} \varphi\right) \\
\sum_{m=0}^{n-1} \sin (\theta+m \varphi)= & \frac{\s... | Proof: By Lemma 4, we have
$$\begin{array}{l}
\sum_{m=0}^{n-1} \cos (\theta+m \varphi)+i \sum_{m=0}^{n-1} \sin (\theta+m \varphi) \\
=\frac{\sin \frac{n \varphi}{2}}{\sin \frac{\varphi}{2}} \cdot \cos \left(\theta+\frac{n-1}{2} \varphi\right) \\
\quad+i \frac{\sin \frac{n \varphi}{2}}{\sin \frac{\varphi}{2}} \cdot \sin... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,075 |
Lemma 6 Let $\alpha$ be any real number and $n$ be a positive integer, then we have
$$\left|\sum_{m=1}^{n} e^{2 \pi i \alpha m}\right| \leqslant \min \left(n, \frac{1}{|\sin \pi \alpha|}\right),$$
where $\min \left(n, \frac{1}{|\sin \pi \alpha|}\right)$ denotes the smaller of the two numbers $n$ and $\frac{1}{|\sin \p... | Assume $\alpha$ is not an integer, then $e^{2 \pi i \alpha} \neq 1$. By Lemma 2 and Lemma 3, we have
$$\begin{aligned}
\sum_{m=1}^{n} e^{2 \pi i \alpha m} & =e^{2 x i \alpha} \sum_{m=0}^{n-1} e^{2 x i \alpha m}=e^{2 \pi i \alpha} \sum_{m=0}^{n-1}\left(e^{2 \pi i a}\right)^{m} \\
& =\frac{e^{2 x i a}\left(1-e^{2 \pi i n... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 741,077 |
Lemma 8 Let $n$ be an integer $>2$, then we have
$$\begin{array}{c}
\sum_{m=1}^{n-1} m \cos \frac{2 \pi m}{n}=-\frac{n}{2} \\
\sum_{m=1}^{n-1} m \sin \frac{2 \pi m}{n}=-\frac{n\left(1+\cos \frac{2 \pi}{n}\right)}{2 \sin \frac{2 \pi}{n}}
\end{array}$$ | Prove that we have
$$\begin{aligned}
\frac{1}{e^{2 \pi i \frac{1}{n}}-1} & =\frac{e^{-2 \pi i \frac{1}{n}}+1}{\left(e^{2 \pi i \frac{1}{n}}-1\right)\left(e^{-2 \pi i \frac{1}{n}}+1\right)} \\
& =\frac{1+\cos \left(-\frac{2 \pi}{n}\right)+i \sin \left(-\frac{2 \pi}{n}\right)}{e^{2 \pi i \frac{1}{n}}-e^{-2 \pi i \frac{1}... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,079 |
Lemma 9 Let $n$ and $m$ be two positive integers and $(n, m)=1$. In this case, we call the trigonometric sum $S(n, m)=\sum_{x=0}^{m-1} e^{2 x i \frac{n x^{2}}{m}}$ a Gauss sum. Suppose $(n, m)=1$, when $m$ is odd, we have
$$|S(n, m)|=\sqrt{m}$$
Suppose $(n, m)=1$, when $m=4 k$ (where $k$ is a positive integer), we hav... | From (24), (15), and (16) we have
$$\begin{array}{l}
|S(n, m)|^{2}=\left(\sum_{x=0}^{m-1} \cos \frac{2 \pi n x^{2}}{m}\right)^{2}+\left(\sum_{x=0}^{m-1} \sin \frac{2 \pi n x^{2}}{m}\right)^{2} \\
=\left(\sum_{x=0}^{m-1} \cos \frac{2 \pi n x^{2}}{m}+i \sum_{x=0}^{m-1} \sin \frac{2 \pi n x^{2}}{m}\right)\left(\sum_{x=0}^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,080 |
Lemma 11 Let $m$ be an integer greater than 1, and $b_{1}, b_{2}, \cdots, b_{\varphi(m)}$ be a reduced residue system modulo $m$. If any two integers are taken from $b_{1}, b_{2}, \cdots, b_{\varphi(m)}$, then these two integers are incongruent modulo $m$. If any one integer is taken from $b_{1}, b_{2}, \cdots, b_{\var... | Let $1<a_{2}<\cdots<a_{\varphi(m)}$ be all positive integers not greater than $m$ and coprime with $m$. Let $r_{i}$ (where $i=1,2, \cdots, m$) be an integer satisfying the condition
$$b_{i} \equiv r_{i}(\bmod m), \quad 0 \leqslant r_{i} \leqslant m-1,$$
then we have
$$b_{1} \equiv r_{1}(\bmod m), b_{2} \equiv r_{2}(\b... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,084 |
1. Find the values of $x$ and $y$ that satisfy the following equations:
(i) $-4 x + 8 y i + 7 = 2 x - 3 y i + 7 i$.
(ii) $x + y i = \sqrt{a + b i}$. | 1. (i) Solution:
$$-4 x+8 y i+7=2 x-3 y i+7 i$$
Rearranging terms
$$-6 x+11 y i=-7+7 i$$
Thus
$$x=\frac{7}{6}, \quad y=\frac{7}{11}$$
(ii) Solution:
$$x+y i=\sqrt{a+b i}$$
Squaring both sides gives
$$x^{2}-y^{2}+2 x y i=a+b i,$$
Therefore
$$\left\{\begin{array}{l}
x^{2}-y^{2}=a \\
2 x y=b
\end{array}\right.$$
Squar... | x= \pm \sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}, y= \pm \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,089 |
2. Let $z_{1}$ and $z_{2}$ be any two complex numbers, prove:
$$\left|z_{1}-z_{2}\right| \geqslant|| z_{1}|-| z_{2}||$$ | 2. Proof: Let $z_{1}=a+b i, z_{2}=c+d i$, then
$$\begin{aligned}
\mid z_{1} & -\left.z_{2}\right|^{2}-\left(\left|z_{1}\right|-\left|z_{2}\right|\right)^{2} \\
& =(a-c)^{2}+(b-d)^{2}-\left(\sqrt{a^{2}+b^{2}}-\sqrt{c^{2}+d^{2}}\right)^{2} \\
& =2\left(-a c-b d+\sqrt{a^{2}+b^{2}} \cdot \sqrt{c^{2}+d^{2}}\right)
\end{alig... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 741,090 |
3. Find the cube root of 27. | 3. Solution: Let $\alpha$ be the cube root of 27, so
$$\begin{aligned}
a^{3} & =27=27 e^{2 k x i}, \\
\alpha & =3 e^{\frac{2 k \pi}{3}}
\end{aligned}$$
When $k$ takes all integer values, there are only three distinct roots:
$$\begin{array}{l}
\alpha_{1}=3, \alpha_{2}=3 e^{\frac{2 \pi}{3} i}=3\left(\cos 120^{\circ}+i \... | \alpha_{1}=3, \alpha_{2}=3 e^{\frac{2 \pi}{3} i}=3\left(\cos 120^{\circ}+i \sin 120^{\circ}\right), \alpha_{3}=3 e^{\frac{2 \pi}{3}}=3\left(\cos 240^{\circ}+i \sin 240^{\circ}\right) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,091 |
4. 证明下列三角恒等式:
(i) $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$
$$\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$$
(ii) $\sin 4 \alpha=4 \sin \alpha \cos ^{3} \alpha-4 \sin ^{3} \alpha \cos \alpha$,
$$\cos 4 \alpha=\cos ^{4} \alpha-6 \sin ^{2} \alpha \cos ^{2} \alpha+\sin ^{4} \alpha$$
(iii) $\cos ^{4} \alpha=\frac{... | 4. (i) 证: 由引理 2 和 (24) 式得
$$\begin{aligned}
e^{33 i} & =\left(e^{\alpha i}\right)^{3}=(\cos \alpha+i \sin \alpha)^{3} \\
& =\cos ^{3} \alpha+3 \cos ^{2} \alpha(i \sin \alpha)+3 \cos \alpha(i \sin \alpha)^{2}+(i \sin \alpha)^{3} \\
& =\left(\cos ^{3} \alpha-3 \cos \alpha \sin ^{2} \alpha\right)+i\left(3 \cos ^{2} \alpha... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,092 |
5. Prove
(i) $\sum_{k=1}^{n} \sin ^{2} k \alpha=\frac{1}{4 \sin \alpha}[(2 n+1) \sin \alpha-\sin (2 n+1) \alpha]$.
(ii)
$$\begin{array}{c}
\sum_{k=1}^{n} \cos ^{3} k \alpha=\frac{1}{4}\left[\frac{3 \sin \frac{n \alpha}{2}}{\sin \frac{\alpha}{2}} \cos \frac{(n+1) \alpha}{2}\right. \\
\left.+\frac{\sin \frac{3 n \alpha}{... | 5. (i) Proof: From equation (19) by setting $\beta=\alpha$ and from equation (20), we get
$$\cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha=1-2 \sin ^{2} \alpha$$
Therefore,
$$\sin ^{2} \alpha=\frac{-1-\cos 2 \alpha}{2}$$
Similarly,
$$\begin{array}{l}
\sin ^{2} 2 \alpha=\frac{1-\cos 4 \alpha}{2} \\
\sin ^{2} 3 \alpha... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,093 |
6. Prove: $\frac{1+\sin \theta+i \cos \theta}{1+\sin \theta-i \cos \theta}=\sin \theta+i \cos \theta$, and from this, deduce
$$\begin{aligned}
(1 & \left.+\sin \frac{\pi}{5}+i \cos \frac{\pi}{5}\right)^{5}+i\left(1+\sin \frac{\pi}{5}-i \cos \frac{\pi}{5}\right)^{5} \\
& =0
\end{aligned}$$ | 6. Proof:
$$\begin{array}{l}
\frac{1+\sin \theta+i \cos \theta}{1+\sin \theta-i \cos \theta} \\
=\frac{(1+\sin \theta+i \cos \theta)^{2}}{(1+\sin \theta)^{2}+\cos ^{2} \theta} \\
=\frac{(1+\sin \theta)^{2}-\cos ^{2} \theta+2 i(1+\sin \theta) \cos \theta}{1+2 \sin \theta+\sin ^{2} \theta+\cos ^{2} \theta} \\
=\frac{1+2... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,094 |
7. Find the sum:
$$\begin{array}{l}
A_{n}=1+r \cos \theta+r^{2} \cos 2 \theta+\cdots+r^{n-1} \cos (n-1) \theta, \\
B_{n}=r \sin \theta+r^{2} \sin 2 \theta+\cdots+r^{n-1} \sin (n-1) \theta .
\end{array}$$ | 7. Solution: Consider the complex number $A_{n}+i B_{n}$.
$$\begin{aligned}
A_{n}+i B_{n}=1 & +r(\cos \theta+i \sin \theta)+r^{2}(\cos 2 \theta \\
& +i \sin 2 \theta)+\cdots \\
& +r^{n-1}[\cos (n-1) \theta+i \sin (n-1) \theta] \\
=1 & +r e^{i \theta}+r^{2} e^{i 2 \theta}+\cdots+r^{n-1} e^{i(n-1) \theta}
\end{aligned}$$... | A_{n}=\frac{1-r \cos \theta-r^{n} \cos n \theta+r^{n+1} \cos (n-1) \theta}{1-2 r \cos \theta+r^{2}} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,096 |
8. Prove: when $\theta \neq m \pi$ ( $m$ is an integer),
$$\sum_{k=1}^{\infty} \cos ^{k-1} \theta \cos k \theta=0$$ | 8. Proof: In the previous question, take $r=\cos \theta$. Since $\theta \neq m \pi$, we have $|r|<1$. As $n \rightarrow \infty$, $r^{n} \rightarrow 0$, $r^{n+1} \rightarrow 0$, thus $r^{n} \cos n \theta \rightarrow 0$, $r^{n+1} \cos (n-1) \theta \rightarrow 0$. From the result of the previous question, as $n \rightarro... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,097 |
9. Prove: When $\alpha \neq \frac{k}{2} \pi$ ( $k$ is an integer),
$$\begin{array}{c}
\cos \alpha+\sin 3 \alpha+\cos 5 \alpha+\sin 7 \alpha+\cdots+\sin (4 n-1) \alpha \\
=\frac{\sin 2 n \alpha}{\sin 2 \alpha}(\cos 2 n \alpha+\sin 2 n \alpha)(\cos \alpha+\sin \alpha) .
\end{array}$$ | 9. Proof:
$$\begin{aligned}
\text { Left side } & =\sum_{k=1}^{n}[\cos (4 k-3) \alpha+\sin (4 k-1) \alpha] \\
& =\sum_{k=0}^{n-1} \cos (4 k+1) \alpha+\sum_{k=0}^{n-1} \sin (4 k+3) \alpha .
\end{aligned}$$
In equation (37), let $\theta=\alpha, \varphi=4 \alpha$, we get
$$\sum_{k=0}^{n-1} \cos (4 k+1) \alpha=\frac{\sin ... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,098 |
10. Prove:
$$\operatorname{tg} n \alpha=\frac{\sin \alpha+\sin 3 \alpha+\cdots+\sin (2 n-1) \alpha}{\cos \alpha+\cos 3 \alpha+\cdots+\cos (2 n-1) \alpha}$$ | 10. Proof: $\left\{\frac{\theta}{\pi}\right\}=0$ is obviously true. When $\left\{\frac{\theta}{\pi}\right\} \neq 0$, from (41) and (42) we get
$$\begin{array}{l}
\sin \alpha+\sin 3 \alpha+\cdots+\sin (2 n-1) \alpha \\
\quad=\frac{\sin n \alpha}{\sin \alpha} \sin n \alpha \\
\cos \alpha+\cos 3 \alpha+\cdots+\cos (2 n-1)... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,099 |
13. Let $S(n, m)=\sum_{x=0}^{m-1} e^{2 x i \frac{n x^{2}}{m}}$, if $\left(m, m^{\prime}\right)=1$, then $S\left(n, m m^{\prime}\right)=S\left(n m^{\prime}, m\right) S\left(n m, m^{\prime}\right)$ | 13. Proof:
$$\begin{array}{l}
S\left(n m^{\prime}, m\right) S\left(n m, m^{\prime}\right)=\left(\sum_{x=0}^{m-1} e^{2 \pi i \frac{n m x^{2}}{m}}\right)\left(\sum_{x^{\prime}=0}^{m m^{\prime}-1} e^{2 \pi i \frac{n m x^{\prime} i^{\prime}}{m^{\prime}}}\right) \\
\quad=\sum_{x=0}^{m-1} \sum_{x^{\prime}=0}^{m \prime-1} e^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,102 |
14. Let \( C_{q}(m)=\sum_{h} e^{2 \pi i \frac{h m}{q}} \), where \( h \) runs through the reduced residue system modulo \( q \). Prove:
(i) If \(\left(q, q^{\prime}\right)=1\), then
$$C_{q q^{\prime}}(m)=C_{q}(m) C_{q^{\prime}}(m)$$
(ii) \( C_{q}(m)=\sum_{d|q, d| m} u\left(\frac{q}{d}\right) d \).
(The right-hand side ... | 14. (i) Proof:
$$\begin{array}{l}
C_{q}(m) C_{q^{\prime}}(m)=\sum_{h} e^{2 \pi i \frac{h m}{q}} \cdot \sum_{h^{\prime}} e^{2 \pi i \frac{h^{\prime} m}{q^{\prime}}} \\
\quad=\sum_{h} \sum_{h^{\prime}} e^{2 \pi i m\left(\frac{h}{q}+\frac{h^{\prime}}{q^{\prime}}\right)} \\
= \sum_{h} \sum_{h^{\prime}} e^{2 \pi i m \frac{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,103 |
Lemma 13 Let $m$ be an integer greater than 1, and $a$ be an integer satisfying the condition $(a, m)=1$. If $b_{1}, b_{2}, \cdots, b_{q(m)}$ is a reduced residue system modulo $m$, then
$$a b_{1}, a b_{2}, \cdots, a b_{q(m)}$$
is also a reduced residue system modulo $m$. | Proof: By Lemma 11 and $b_{1}, b_{2}, \cdots, b_{q(m)}$ being a reduced residue system modulo $m$, we know that any integer taken from $b_{1}, b_{2}, \cdots, b_{q(m)}$ is coprime with $m$. Since $(a, m)=1$, we know that any integer taken from $a b_{1}, a b_{2}, \cdots, a b_{\psi(m)}$ is also coprime with $m$. Suppose t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,104 |
Theorem 1 (Euler) Let $m$ be an integer greater than 1, and let $a$ be an integer satisfying the condition $(a, m)=1$. Then we have
$$a^{\alpha(m)} \equiv 1(\bmod m)$$ | Let $1<a_{2}<\cdots<a_{q(m)}$ be all positive integers not exceeding $m$ and coprime with $m$. Let $r_{1}$ be an integer satisfying the condition
$$a \equiv r_{1}(\bmod m), \quad 0 \leqslant r_{1} \leqslant m-1$$
Let $r_{i}$ (where $i=2, \cdots, \varphi(m)$) be an integer satisfying the condition
$$a a_{i} \equiv r_{i... | a^{\varphi(m)} \equiv 1(\bmod m) | Number Theory | proof | Yes | Yes | number_theory | false | 741,105 |
Lemma 1 If $a, b, c$ are any three integers, $m$ is a positive integer, then when $a \equiv b(\bmod m), b \equiv c(\bmod m)$. are satisfied, we have
$$a \equiv c(\bmod m)$$ | Given $a-b=m q_{1}, b-c=m q_{2}$, where $q_{1}, q_{2}$ are two integers, we get $a-b+b-c=m q_{1}+m q_{2}$. Therefore, $a-c=$ $m\left(q_{1}+q_{2}\right)$, where $q_{1}+q_{2}$ is an integer. | a \equiv c(\bmod m) | Number Theory | proof | Yes | Yes | number_theory | false | 741,106 |
Theorem 2 (Fermat) If $p$ is a prime, $p \nmid a$, then we have $a^{p-1} \equiv 1(\bmod p)$. | By Lemma 10, we have $\varphi(p)=p-1$. Since $p$ is a prime number and $p \nmid a$, we get $(p, a)=1$. Taking $m=p$ in Theorem 1, we obtain
$$a^{p-1} \equiv 1(\bmod p)$$
Since $341=11 \times 31$, 341 is not a prime number. By $1024=341 \times 3 + 1$, we get $1024 \equiv 1(\bmod 341)$. By $2^{340}=(2^{10})^{34}=(1024)^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,107 |
Example 8 Find the remainder when $\left(12371^{55}+34\right)^{28}$ is divided by 111. | From $12371=111^{2}+50$, we get $12371 \equiv 50(\bmod 111)$. By Lemma 3, we have
$$12371^{56} \equiv 50^{56}(\bmod 111)$$
We also have $(50)^{28}=(125000)^{9}(50), 125000 \equiv 14(\bmod 111)$, so by Lemma 3, we get
$$(50)^{28} \equiv(14)^{9}(50)(\bmod 111)$$
Furthermore, $14^{3} \equiv 80(\bmod 111),(80)^{3} \equiv... | 70 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,109 |
Lemma 14 If $a=p_{1}^{a_{i}} \cdots p_{n}^{a_{n}}$, where $p_{1}, \cdots, p_{n}$ are all primes and $\alpha_{1}, \cdots, \alpha_{n}$ are all positive integers, then we have
$$\varphi(a)=p_{1}^{a_{1}-1}\left(p_{1}-1\right) \cdots p_{n}^{a_{n}-1}\left(p_{n}-1\right)$$ | Proof: When $n=1$, by Lemma 10, we know that this lemma holds. Now let $n \geqslant 2$. The multiples of $p_{1}$ not greater than $a$ are
$$p_{1}, 2 p_{1}, \cdots, \frac{a}{p_{1}} p_{1}$$
There are $\frac{a}{p_{1}}$ such numbers. Therefore, the number of positive integers not greater than $a$ and coprime with $p_{1}$ ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,110 |
Example 9 $3^{8833010}-59049$ can be divided by 24010000. | Given $24010000=245^{4} 7^{4}$ and Lemma 14, we have $\varphi(24010000)=2^{3} \times 5^{3} \times 4 \times 7^{3} \times 6=8232000$. Therefore, by Theorem 1 we have
$$3^{8232000} \equiv 1(\bmod 24010000)$$
By (28) we have
$$3^{8232010}-59049 \equiv 3^{10}-59049(\bmod 24010000)$$
By $3^{10}=59049$ and (29) we know that... | 3^{10}-59049 \equiv 0(\bmod 24010000) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,111 |
1. Let $m_{1}, m_{2}$ be two coprime positive integers, then when $x_{1}, x_{2}$ run through the complete residue systems modulo $m_{1}, m_{2}$ respectively, $m_{2} x_{1}+m_{1} x_{2}$ runs through the complete residue system modulo $m_{1} m_{2}$. | 1. Proof: Since $x_{1}, x_{2}$ pass through $m_{1}, m_{2}$ integers respectively, $m_{2} x_{1} + m_{1} x_{2}$ passes through exactly $m_{1} m_{2}$ integers. By Lemma 5, if we can prove that these $m_{1} m_{2}$ integers are incongruent modulo $m_{1} m_{2}$, then these $m_{1} m_{2}$ integers form a complete residue syste... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,113 |
2. Let $m_{1}, m_{2}, \cdots, m_{k}$ be $k$ pairwise coprime positive integers. Then when $x_{1}$, $x_{2}$, $\cdots$, $x_{k}$ run through the complete residue systems modulo $m_{1}$, $m_{2}$, $\cdots$, $m_{k}$ respectively, $M_{1} x_{1}+M_{2} x_{2}+\cdots+M_{k} x_{k}$ runs through the complete residue system modulo $m_... | 2. Proof: The case $k=2$ has been proven in the first question. Here we assume $k>2$. Clearly, $M_{1} x_{1}+M_{2} x_{2}+\cdots+M_{k} x_{k}$ passes through exactly $m_{1} m_{2} \cdots m_{k}$ integers. By Lemma 5, it suffices to prove that these $m_{1} m_{2} \cdots m_{k}$ integers are pairwise incongruent modulo $m_{1} m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,114 |
3. Let $m_{1}, m_{2}, \cdots, m_{k}$ be $k$ pairwise coprime positive integers, then when $x_{1}$, $x_{2}$, $\cdots$, $x_{k}$ respectively run through the complete residue systems modulo $m_{1}$, $m_{2}$, $\cdots$, $m_{k}$, $x_{1}+m_{1} x_{2}+m_{1} m_{2} x_{3}+\cdots+m_{1} m_{2} \cdots m_{k-1} x_{k}$ runs through the c... | 3. Proof: It is evident that $x_{1}+m_{1} x_{2}+m_{1} m_{2} x_{3}+\cdots+m_{1} m_{2} \cdots m_{k-1} x_{k}$ passes through exactly $m_{1} m_{2} \cdots m_{k}$ integers, so it suffices to prove that these integers are pairwise incongruent modulo $m_{1} m_{2} \cdots m_{k}$.
Assume
Here $x_{i}^{\prime}, x_{i}^{\prime \pri... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,115 |
4. Prove the following properties of the Euler's function:
(i) If $N>2$, then $\varphi(N)$ is necessarily even.
(ii) If $(a, b)=1$, then $\varphi(a b)=\varphi(a) \cdot \varphi(b)$. | 4. (i) Proof: If $N$ does not contain any odd prime factors, then since $N>2$, we have $N=2^{\alpha}, \alpha \geqslant 2$. By Lemma 14, we get $\varphi(N)=2^{\alpha-1}, \alpha \geqslant 2$, hence $2 \mid \varphi(N)$.
If $N$ contains an odd prime factor, let $p$ be an odd prime factor of $N$. By Lemma 14, we know that ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,116 |
Lemma 2 If $a, b, c$ are any three integers, $m$ is a positive integer and $(m, c)=1$, then when $a c \equiv b c(\bmod m)$, we have
$$a \equiv b(\bmod m)$$ | Proof: Since $c(a-b)=ac-bc=mq$, where $\boldsymbol{g}$ is an integer, and $(m, c)=1$, we have $a-b=mq_{1}$, where $q_{1}$ is an integer. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,117 |
5. Let $N>1$, prove: the sum of all positive integers not greater than $N$ and coprime to $N$ is $\frac{1}{2} N \cdot p(N)$. | 5. Proof: If $N=2$, then $\varphi(N)=\varphi(2)=1$, so
$$\frac{1}{2} N \cdot \varphi(N)=\frac{1}{2} \cdot 2 \cdot 1=1$$
and the only number not greater than 2 and coprime with 2 is 1, thus the case $N=2$ is proved.
Now assume $N>2$. The number of positive integers not greater than $N$ and coprime with $N$ is $\varphi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,118 |
6. Suppose $m>1$ is a positive integer, $(a, m)=1$, and assume $b_{1}, b_{2}, \cdots$, $b_{q(m)}$ is a reduced residue system modulo $m$, and $a b_{i}=r_{i}(\bmod m)(0 \leqslant$ $r_{i}<m, 1 \leqslant i \leqslant \varphi(m)$ ), then
$$\frac{1}{m}\left(r_{1}+r_{2}+\cdots+r_{q(m)}\right)=\frac{1}{2} \varphi(m)$$ | 6. Proof: Let $1<a_{2}<\cdots<a_{\varphi(m)}$ be all positive integers not greater than $m$ and coprime with $m$. Since $b_{1}, b_{2}, \cdots, b_{p(m)}$ is a reduced residue system modulo $m$, and $(a, m)=1$, by Lemma 13, we know that $a b_{1}, a b_{2}, \cdots, a b_{q(m)}$ is also a reduced residue system modulo $m$. A... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,119 |
7. Let $m_{1}, m_{2}, \cdots, m_{k}$ be $k$ pairwise coprime positive integers, then when $x_{1}, x_{2}, \cdots, x_{k}$ respectively run through the reduced residue systems modulo $m_{1}, m_{2}, \cdots, m_{k}$, $M_{1} x_{1}+M_{1} x_{2}+\cdots+M_{k} x_{k}$ runs through the reduced residue system modulo $m_{1} m_{2} \cdo... | 7. Proof: Since $x_{1}, x_{2}, \cdots, x_{k}$ pass through $\varphi\left(m_{1}\right), \varphi\left(m_{2}\right), \cdots, \varphi\left(m_{k}\right)$ numbers respectively, $M_{1} x_{1}+M_{2} x_{2}+\cdots+M_{k} x_{k}$ passes through $\varphi\left(m_{1}\right) \varphi\left(m_{2}\right) \cdots \varphi\left(m_{k}\right)$ nu... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,120 |
8. (i) Let $N=9450$, find $\varphi(N)$.
(ii) Find the sum of all positive integers not greater than 9450 and coprime with 9450. | 8. (i) Solution: Given $9450=2 \cdot 3^{3} \cdot 5^{2} \cdot 7$, and by Lemma 14, we have
$$\begin{aligned}
\varphi(N) & =9450\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \\
& =\frac{2 \cdot 3^{3} \cdot 5^{2} \cdot 7 \cdot 2 \cdot 4 \cdot 6}{2 \cdot 3 \cdot 5 ... | 10206000 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,121 |
9. (i) Determine whether $121^{6}-1$ is divisible by 21.
(ii) Find the remainder when $8^{4965}$ is divided by 13.
(iii) Let $p$ be any prime number other than 2 and 5. Prove that: $p \mid \underbrace{99 \cdots 9}_{(p-1) k \uparrow}, k$ is any positive integer. | 9. (i) Solution: Since $\varphi(21)=(3-1)(7-1)=12$, and $121 = 11^{12}$, and $(11,21)=1$. By Theorem 1, $11^{12} \equiv 1(\bmod 21)$, so $21 \mid\left(121^{6}-1\right)$.
(ii) Solution: Since $\varphi(13)=12$, and $4965=413 \times 12+9$. By Theorem 2, $8^{12} \equiv 1(\bmod 13)$, so $8^{4955} \equiv 8^{9}(\bmod 13)$. Al... | 21 \mid (121^{6}-1), 8^{4965} \equiv 8(\bmod 13), proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,122 |
10. We call $F_{n}=2^{2^{n}}+1$ Fermat numbers, try to prove $641 \mid F_{5}$ | 10. Proof: $F_{5}=2^{2^{5}}+1=2^{32}+1$. Since $640=5 \cdot 2^{7}$, we have
$$5 \cdot 2^{7} \equiv -1 \pmod{641}$$
By Lemma 3, we get
$$5^{4} \cdot 2^{28} \equiv 1 \pmod{641}.$$
But
$$5^{4}=625 \equiv -2^{4} \pmod{641}$$
Therefore,
i.e. $\square$
$$\begin{array}{c}
-2^{4} \cdot 2^{28} \equiv 1 \pmod{641} \\
641 \mid... | 641 \mid (2^{32}+1) | Number Theory | proof | Yes | Yes | number_theory | false | 741,123 |
11. Suppose $p$ is a prime, $a$ and $b$ are any two integers, then
$$(a+b)^{p} \equiv a^{p}+b^{p}(\bmod p)$$ | 11. Proof: If $p|a, p| b$, the conclusion is obviously true. If one of $a$ and $b$ can divide $p$, without loss of generality, assume $p \mid b$, then
\[
(a+b)^{p}=a^{p}+b q_{1} \equiv a^{p}(\bmod p), \quad q_{1} \text{ is an integer.}
\]
And $a^{p}+b^{p} \equiv a^{p}(\bmod p)$, so the conclusion also holds. Now assume... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,124 |
12. Find positive integers $\boldsymbol{n}$ and $m, n>m \geqslant 1$, such that the last three digits of $1978^{n}$ and $1978^{m}$ are equal, and make $n+m$ as small as possible. (20th International Mathematical Olympiad Problem) | 12. Solution: $1978^{n}-1978^{m}=1978^{m}\left(1978^{n-m}-1\right)$
$$=2^{m} \cdot 989^{m}\left(1978^{n-m}-1\right)$$
Since the last three digits of $1978^{n}$ and $1978^{m}$ are the same, the last three digits of $1978^{n}-1978^{m}$ are all 0. Therefore, $1978^{n}-1978^{m}$ is divisible by 1000. And $1000=2^{3} \cdot... | 106 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,125 |
Lemma 1 Let $a, b$ be positive integers, $a<b$ and $(a, b)=1$. If there exists a prime $p$, such that $p \mid b$ but $p \nmid 10$, then $\frac{a}{b}$ - cannot definitely be expressed as a finite decimal. If $b=2^{\alpha} 5^{\beta}$, where $\alpha, \beta$ are non-negative integers, then $\frac{a}{b}$ can be expressed as... | If there exists a prime $p$, such that $p \mid b$ but $p \nmid 10$, and $\frac{a}{b}$ can be expressed as a finite decimal. Since $a0$. We also have
$$10^{n} a=\left(10^{n-1} a_{1}+\cdots+a_{n}\right) b^{-}$$
Since $(a, b)=1,10^{n-1} a_{1}+\cdots+a_{n}$ is a positive integer and (1) leads to $b \mid 10^{n}$. Since we ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,127 |
Lemma 3 If $a, b$ are any two integers, and $m, n$ are two positive integers, then when $a \equiv b(\bmod m)$, we have
$$a^{n} \equiv b^{n}(\bmod m)$$ | Given $a-b=m q$, where $q$ is an integer, we have
$$a^{n}=(b+m q)^{n}=b^{n}+\cdots+(m q)^{n}=b^{n}+m q_{1}$$
where $q_{1}$ is an integer. Therefore, $a^{n}-b^{n}=m q_{1}$, which means
$$a^{n} \equiv b^{n}(\bmod m)$$ | a^{n} \equiv b^{n}(\bmod m) | Number Theory | proof | Yes | Yes | number_theory | false | 741,128 |
Example 1 Prove that $\frac{4}{9}=0 . \dot{4}, \frac{8}{15}=0.5 \dot{3}, \frac{3}{14}=0.2 \dot{2} 4285 \dot{7}$. | Prove that since $\frac{4}{9}=0.44444 \cdots$, we have $\frac{4}{9}=0 . \dot{4}$. Since $\frac{80}{15}=\frac{16}{3}=5+\frac{1}{3}=5.33333 \cdots$, we have $\frac{8}{15}=0.533333 \cdots$, i.e., $\frac{8}{15}=0.53$. Since $\frac{30}{14}=\frac{15}{7}=2+\frac{1}{7}=2.1428571428571 \cdots$, we have $\frac{3}{14}=0.214285714... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,129 |
Lemma 2 Let $0<a<b$, and $(a, b)=1$. If $\frac{a}{b}$ can be expressed as a pure repeating decimal, then we have $(b, 10)=1$. | Proof: If $\frac{a}{b}$ can be expressed as a pure repeating decimal, then by $0<\frac{a}{b}<1$ and Definition 2, we have
$$\frac{a}{b}=0 . a_{1} \cdots a_{t} a_{1} \cdots a_{t} a_{1} \cdots a_{t} \cdots$$
where $a_{1}, \cdots, a_{t}$ are non-negative integers no greater than 9, but at least one $a_{i} \geqslant 1$ in... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,130 |
Lemma 3 Let $0<a<b$ and $(a, b)=1$. Let $h$ be the smallest positive integer such that
$$10^{h} \equiv 1(\bmod b)$$
holds, then $\frac{a}{b}$ can be expressed as a pure repeating decimal $0 . d_{1} \cdots \boldsymbol{a}_{\boldsymbol{h}}$. | Prove that from $10^{h} \equiv 1(\bmod b)$ we get $10^{h} a \equiv a(\bmod b)$. From $10^{h} a - a = b m$, where $m$ is an integer, we get $\frac{10^{h} a}{b} - \frac{a}{b} = m$. Let $\frac{a}{b} = 0 . a_{1} a_{2} \cdots a_{h} a_{h+1} a_{h+2} \cdots$; then we have $\frac{10^{h} a}{b} = 10^{h-1} a_{1} + \cdots + a_{h} +... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,131 |
Lemma 4 Let $b$ be a positive integer and $(10, b)=1$. Let $h$ be the smallest positive integer such that
$$10^{A} \equiv 1(\bmod b)$$
holds, then $h \mid p(b)$. | Given that $(10, b)=1$ and Theorem 1 of Chapter 5, we have
$$10^{\varphi(b)} \equiv 1(\bmod b)$$
By the definition of $h$ and equation (8), we have $0<h \leqslant \varphi(b)$. Let $\varphi(b)=h m + r$, where $r$ is a non-negative integer less than $h$ and $m$ is a positive integer. By equation (7) and Lemma 3 of Chapt... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,132 |
Let $a$ be a positive integer not greater than 6. Please express $\frac{a}{7}$ as a pure repeating decimal.
Express the above text in English, keeping the original text's line breaks and format, and output the translation result directly. | Since $(a, 7)=1, \varphi(7)=6$,
$$\begin{aligned}
10 & \equiv 3(\bmod 7) \\
10^{2} & \equiv 2(\bmod 7) \\
10^{3} & \equiv 6(\bmod 7) \\
10^{6} & \equiv 1(\bmod 7)
\end{aligned}$$
and by Lemma $\dot{4}$ we have $h=6$. Also, by Lemma 3 we have $\frac{a}{7}=0 . d_{1} a_{2} a_{3} a_{4}$ $a_{5} a_{69}$, hence we get
$$\beg... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,133 |
Example 3 Let $a$ be a positive integer not greater than 12, please express $\frac{a}{13}$ as a pure repeating decimal. | $$\text { Sol } \begin{array}{l}
(a, 13)=1, \varphi(13)=12, \text { since } \\
10 \equiv 10(\bmod 13), \quad 10^{2} \equiv 9(\bmod 13), \\
10^{3} \equiv 12(\bmod 13), \quad 10^{4} \equiv 3(\bmod 13), \\
10^{6} \equiv 1(\bmod 13),
\end{array}$$
and by Lemma 4 we have $h=6$. Also, by Lemma 3 we have $\frac{a}{13}=0 . a_... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,134 |
Example 4 Let $a$ be any positive integer not greater than 3988, prove that $\frac{a}{3989}$ is a pure repeating decimal, and its repeating cycle is no less than 997 digits (i.e., $\frac{a}{3989}=0 . a_{1}$ $a_{2} a_{3} \cdots a_{n}$, then $n \geqslant 997$ ). | Given that 3989 is a prime number and $a$ is a positive integer no greater than 3988, we have $(a, 3989)=1$. Let $n$ be the smallest positive integer such that
$$10^{n} \equiv 1(\bmod 3989)$$
holds. Since $\varphi(3989)=3988$ and by Lemma 4, we have $n \mid 3988$. Given that
$$\begin{array}{c}
10 \equiv 10(\bmod 3989)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,135 |
Example 5 Please convert $\frac{15}{308}$ into a decimal. | Since $308=4 \times 77=2^{2} \cdot 77, \varphi(77)=(11-1) \times (7-1)=60$.
We also have
$$\begin{array}{ll}
10 \equiv 10(\bmod 77), & 10^{2} \equiv 23(\bmod 77) \\
10^{3} \equiv 76(\bmod 77), & 10^{4} \equiv 67(\bmod 77) \\
10^{5} \equiv 54(\bmod 77), & 10^{6} \equiv 1(\bmod 77)
\end{array}$$
Therefore, in Lemma 5, w... | 0.04 \dot{8} 7012 \dot{9} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,137 |
Example 6 Please convert $\frac{1}{17408}$ into a decimal. | Since $17408=1024 \times 17=2^{10} \cdot 17, \varphi(17)=16$. We also have
$$\begin{array}{cc}
10 \equiv 10(\bmod 17), & 10^{2} \equiv 15(\bmod 17) \\
10^{4} \equiv 4(\bmod 17), & 10^{8} \equiv 16(\bmod 17) \\
10^{16} \equiv 1(\bmod 17)
\end{array}$$
Therefore, in Lemma 5, we can take $a=1, b=17408, a=10, \beta=0$, $h... | 0.0000574448 \dot{5} 29411764705882 \dot{3} | Other | math-word-problem | Yes | Yes | number_theory | false | 741,138 |
Prove that $-10, -6, -1, 2, 10, 12, 14$ is a complete residue system modulo 7.
Column 3 | To prove that since
$$\begin{array}{l}
-10 \equiv 4(\bmod 7), \quad-6 \equiv 1(\bmod 7), \quad-1 \equiv 6(\bmod 7) \\
2 \equiv 2(\bmod 7), \quad 10 \equiv 3(\bmod 7), \quad 12 \equiv 5(\bmod 7), \quad 14 \equiv \\
0(\bmod 7),
\end{array}$$
and $4,1,6,2,3,5,0$ and $0,1,2,3,4,5,6$ differ only in order, hence $-10,-6,-1,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,139 |
Lemma 7 Let $p$ be a prime, $m$ be a positive integer and $m=n \alpha+$ $\beta$, where $\boldsymbol{\alpha}$ is a non-negative integer and $\boldsymbol{\beta}$ is a non-negative integer no greater than $n-1$. Let $a=p^{m}$, when $\beta=0$, $\sqrt[n]{a}$ is an integer. When $1 \leqslant \beta \leqslant$ $n-1$, $\sqrt[n]... | Proof (i) When $\beta=0$. In this case, we have $m=n \alpha$ and $a=p^{n \alpha}$, hence we get $\sqrt[n]{a}=p^{\alpha}$ is an integer.
(ii) When $\alpha=0$ and $1 \leqslant \beta \leqslant n-1$. In this case, we have $m=\beta$ and $a=p^{\beta}$. If there exist two positive integers $b, c$ such that $\sqrt[n]{a} \Right... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,141 |
Lemma 8 Let $p$ be a prime number, $m$ be a positive integer, $m=n \alpha+\beta$, where $\alpha$ is a non-negative integer and $\beta$ is a non-negative integer no greater than $n-1$. Let $a=p^{m}$, when $\beta=0$, then $\sqrt[n]{a}$ is a positive integer. When $1 \leqslant \beta \leqslant$ $n-1$, then $\sqrt[n]{a}=b+c... | Proof: Let $1 \leqslant \beta \leqslant n-1$, and $c$ is a finite decimal, then $c$ can be converted into a fraction, i.e., $c=\frac{a_{1}}{b_{1}}$, where $a_{1}, b_{1}$ are positive integers. When $1 \leqslant \beta \leqslant n-1$, then $\sqrt[n]{a}=b+c=\frac{b_{1} b+a_{1}}{b_{1}}$. That is, at this time $\sqrt[n]{a}$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,142 |
Example 7 Find the cube root of 3.652264. | From (31) we have
$$\sqrt[3]{3.652264}=\frac{\sqrt[3]{3652264}}{10^{2}}$$
Since $3652264=2^{3} \times 7^{3} \times 11^{3}$, by (32) we get
$$\sqrt[3]{3.652264}=\frac{2 \times 7 \times 11}{100}=1.54$$ | 1.54 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,144 |
Example 8. Find the square root of 7.93. | Solving, from equation (31) we have
$$\sqrt{7.93}=\frac{\sqrt{793}}{10}$$
Since $793=61 \times 13$, where 61 and 13 are both prime numbers, from Table 1 we have
Table of square roots of prime numbers below 150
- (Table 1)
\begin{tabular}{c|c||c|c}
\hline$p$ & $\sqrt{p}$ & $p$ & $\sqrt{p}$ \\
\hline 2 & $1.41421356 \cd... | 2.81602556 \cdots | Other | math-word-problem | Yes | Yes | number_theory | false | 741,145 |
Example 10 Find the tenth root of 194400. | Since $194400=2^{2} \times 3^{5} \times 5^{2}$, we have
$$\sqrt[10]{194400}=\sqrt{2} \times \sqrt{3} \times \sqrt[5]{5}$$
From Table 1 we have
$$\sqrt{2}=1.41421356 \cdots, \quad \sqrt{3}=1.73205080 \cdots$$
From Table 2 we have
$$\sqrt[5]{5}=1.37972966 \cdots$$
From equations (37) to (39) we have
$$\begin{array}{l}... | 3.3796336 \cdots | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,147 |
Example 12 Find $\sum_{k=1}^{8} \frac{1}{k!}$ equals what. | Solve $\begin{aligned} \sum_{k=1}^{8} \frac{1}{k!}= & 1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!} \\ = & 1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\frac{1}{720}+\frac{1}{5040} \\ & +\frac{1}{40320}=1.71827877 \cdots\end{aligned}$ | 1.71827877 \cdots | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,149 |
Example 4 Prove that $6,9,12,15,18,21,24,27$ is a complete residue system modulo 8. | To prove that since
$$\begin{array}{rrr}
6 \equiv 6(\bmod 8), & 9 \equiv 1(\bmod 8), & 12 \equiv 4(\bmod 8), \\
15 \equiv 7(\bmod 8), & 18 \equiv 2(\bmod 8), & 21 \equiv 5(\bmod 8), \\
24 \equiv 0(\bmod 8), & 27 \equiv 3(\bmod 8), &
\end{array}$$
and $6,1,4,7,2,5,0,3$ and $0,1,2,3,4,5,6,7$ differ only in order, hence ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,150 |
Example 13 Find $\sum_{k=1}^{15} \frac{1}{k^{2}}$ equals what.
| Solve $\begin{aligned} \sum_{k=1}^{15} \frac{1}{k^{2}}= & 1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\frac{1}{5^{2}}+\frac{1}{6^{2}}+\frac{1}{7^{2}} \\ & +\frac{1}{8^{2}}+\frac{1}{9^{2}}+\frac{1}{10^{2}}+\frac{1}{11^{2}}+\frac{1}{12^{2}}+\frac{1}{13^{2}} \\ & +\frac{1}{14^{2}}+\frac{1}{15^{2}} \\ = & 1+\frac{1}{... | 1.58044028 \cdots | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,151 |
Column 14 Find $\sum_{k=1}^{11} \frac{1}{k!}$ equals what. | Solve $\begin{aligned} \sum_{k=1}^{11} \frac{1}{k!}= & \sum_{k=1}^{8} \frac{1}{k!}+\frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!} \\ = & 1.71827877 \cdots+\frac{1}{362880}+\frac{1}{3628800} \\ & +\frac{1}{39916800}=1.71828182 \cdots\end{aligned}$ | 1.71828182 \cdots | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,152 |
1. Convert the following fractions to decimals:
(i) $\frac{371}{6250}$,
(ii) $\frac{190}{37}$,
(iii) $\frac{13}{28}$,
(iv) $\frac{a}{875}, a=4,29,139,361$. | 1. (i) Solution: Since $6250=2 \times 5^{5}$, $\frac{371}{6250}$ is a finite decimal. After calculation, we get
$$\frac{371}{6250}=0.05936$$
(ii) Solution: $\frac{190}{37}=5+\frac{5}{37}$. Since $(10,37)=1$, $\frac{5}{37}$ is a pure repeating decimal. Also, $p(37)=36$, and $10^{2} \neq 1(\bmod 37), 10^{3} \equiv 1(\bmo... | \begin{array}{l}
\text{(i) } 0.05936 \\
\text{(ii) } 5 . \dot{1} 3 \dot{5} \\
\text{(iii) } 0.46 \dot{4} 2857 \dot{1} \\
\text{(iv) } \begin{array}{l}
\frac{4}{875}=0.00 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,153 |
4. Prove: For the equation with integer coefficients
$$x^{n}+a_{1} x^{n-1}+\cdots+a_{n}=0, \quad n \geqslant 1$$
if a real root is not an integer, then it must be irrational. | 4. Proof: We can assume that this equation has only non-zero roots, i.e., we can assume $a_{n} \neq 0$. Because if this equation has a zero root of multiplicity $m$, then we can divide both sides of the equation by $x^{m}$ to obtain an integer coefficient equation with a non-zero constant term.
Suppose the equation ha... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,156 |
5. Prove: $\log _{10} 2$ is an irrational number. | 5. Proof: If $\log _{10} 2=\frac{q}{r},(q, r)=1$, then by the definition of logarithms we get $10^{q / r}=2$, i.e., $10^{q}=2^{r}$. Therefore, $5^{q}=2^{r-q}$. Since $q$ and $r-q$ are positive integers, and $(5,2)=1$, the above equation cannot hold. Therefore, $\log _{10} 2$ is irrational. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,157 |
6. If positive integers $M$ and $N$ cannot be expressed as powers of the same base (the base is also a positive integer), $\log _{M} N$ is definitely an irrational number. | 6. Proof: Assume $\log _{M} N=\frac{q}{r},(q, r)=1$. By the definition of logarithms, we have: $M^{q / r}=N$, which implies $M^{q}=N^{r}$. Let the standard factorizations of $M$ and $N$ be:
$$M=u_{1}^{\alpha_{1}} u_{2}^{\alpha_{2}} \cdots u_{m}^{\alpha_{n}}, N=v_{1}^{\beta_{1}} v_{2}^{\beta_{2}} \cdots v_{n}^{\beta_{n}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,158 |
7. Prove: $e$ is an irrational number. | 7. Proof: Assume $e=\frac{q}{r},(q, r)=1$. Let $k \geqslant r$ be a positive integer, then $r \mid k!$. And for any positive integer $n \leq k$, we have $n! \mid k!$. Therefore, when $e=\frac{q}{r}$,
$$A=k!\left(e-1-\frac{1}{1!}-\frac{1}{2!}-\cdots-\frac{1}{k!}\right)$$
is an integer. By the definition of $e$,
$$\begi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,159 |
8. Prove:
$$J=1-\frac{1}{2^{2}}+\frac{1}{2^{2} \cdot 4^{2}}-\frac{1}{2^{2} \cdot 4^{2} \cdot 6^{2}}+\cdots$$
is an irrational number. | 8. Proof: Assume $J=\frac{q}{r},(q, r)=1$. Let $k \geqslant r$ be a positive integer, then $r \mid k!$. Therefore, when $J=\frac{q}{r}$,
$$A_{!}=\left(2^{k} \cdot k!\right)^{2}\left[J-\left(1-\frac{1}{2^{2}}+\frac{1}{2^{2} \cdot 4^{2}}-\cdots+\frac{(-1)^{k}}{\left(2^{k} \cdot k!\right)^{2}}\right)\right]$$
is an integ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,160 |
Lemma 4 Let $m$ be an integer greater than 1, and $a_{1}, a_{2}, \cdots, a_{m}$ be a complete residue system modulo $m$. If any two integers are taken from $a_{1}, a_{2}, \cdots, a_{m}$, then these two integers are not congruent modulo $m$. | To prove that for any integer modulo $m$, it must be congruent to one of the following $m$ integers:
$$0,1, \cdots, m-1$$
Let $r_{i}$ (where $i=1,2, \cdots, m$) be an integer satisfying the condition
$$a_{i} \equiv r_{i}(\bmod m), \quad 0 \leqslant r_{i} \leqslant m-1,$$
Then we have
$$a_{1} \equiv r_{1}(\bmod m), a_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,161 |
9. For any real number $\alpha$, we define $[\alpha]$ as the greatest integer not exceeding $\alpha$. Prove: For positive integers $a$ and $b$, the number of positive integers not exceeding $a$ and divisible by $b$ is $\left[\frac{a}{b}\right]$. | 9. Proof: When $a < b$, there does not exist a positive integer that is a multiple of $b$ and not greater than $a$. According to the definition, $\left[\frac{a}{b}\right]=0$, so the conclusion holds. Now assume $a \geqslant b$. Arrange all positive integers that are multiples of $b$ and not greater than $a$ as follows:... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,162 |
10. In the standard factorization of $n!$, the exponent of the prime $p$ is
$$S=\left[\frac{n}{p}\right]+\left[\frac{n}{p^{2}}\right]+\left[\frac{n}{p^{3}}\right]+\cdots=\sum_{r=1}^{\infty}\left[\frac{n}{p^{r}}\right] .$$ | 10. Proof: When $p^{k}>n$, $\left[\frac{n}{p^{k}}\right]=0$, so $s$ only contains a finite number of non-zero terms.
If $n<p$, then the standard factorization of $n!$ does not contain $p$, and by definition $S=0$, so the conclusion holds.
Now assume $n \geqslant p$. From the previous problem, we know that among the $... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,163 |
11. If $3^{k} \mid 1000!$ and $3^{k+1} \nmid 1000!$, find $k$. | 11. Solution: From the result of the previous problem, we have.
$$\begin{aligned}
k= & {\left[\frac{1000}{3}\right]+\left[\frac{1000}{9}\right]+\left[\frac{1000}{27}\right]+\left[\frac{1000}{81}\right] } \\
& +\left[\frac{1000}{243}\right]+\left[\frac{1000}{729}\right] \\
= & 333+111+37+12+4+1=498
\end{aligned}$$ | 498 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,164 |
12. Let $C_{m}^{n}=\frac{m!}{n!(m-n)!}, m, n$ are positive integers and $m>n$, prove:
(i) $C_{m}^{n}=C_{m}^{m-n}$;
(ii) $C_{m}^{n}$ is a positive integer;
(iii) The product of $k$ consecutive positive integers is divisible by $k!$;
(iv) Let $p$ be a prime, $k<p$, then $p \mid C_{p}^{k}$. | 12. (i) Proof: By the definition of $C_{m}^{n}$, we have
$$\begin{aligned}
C_{m}^{m-n} & =\frac{m!}{(m-n)![m-(m-n)]!} \\
& =\frac{m!}{n!(m-n)!}=C_{m}^{n}
\end{aligned}$$
(ii) Proof: $C_{m}^{n}=\frac{m!}{n!(m-n)!}$, since $n \leqslant m, m-n \leqslant m$,
if there is a prime $p \mid n!(m-n)!$, then $p \mid m!$. That is... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 741,165 |
13. Prove:
(i) Wilson's Theorem: If $p$ is a prime, then
$$(p-1)!\equiv-1(\bmod p)$$
(ii) If $(p-1)!\equiv-1(\bmod p)$, then $p$ is a prime. | 13. (i) Proof: When $p=2$, the conclusion is obviously true. Now assume $p$ is an odd prime. If we take $x_{0}$ as one of the numbers $1,2, \cdots, p-1$, then since $\left(x_{0}, p\right)=1$ and by Lemma 12 of Chapter 4, the congruence $x_{0} x \equiv 1(\bmod p)$ has a solution, and its smallest non-negative integer so... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,166 |
14. When $n$ and $n+2$ are both prime numbers, $n$ and $n+2$ are called a pair of twin primes. Prove: $n$ and $n+2$ are twin primes if and only if
$$4[(n-1)!+1]+n \equiv 0[\bmod n(n+2)], \quad n>1$$ | 14. Proof: Suppose
$$4[(n-1)!+1]+n \equiv 0[\bmod n(n+2)], n>1 \text { (1) }$$
is true. We can prove that $n$ must be odd, i.e., $(4, n)=1$. It is easy to verify that $n=2,4$ do not satisfy equation (1). When $n=2 m, m>2$, from equation (1) we get $n \mid 4[(n-1)!+1]$, thus $m \mid 2[(2 m-1)!+1]$. Since $m>2$ and $2 m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,167 |
15. Let $a^{m-1} \equiv 1(\bmod m)$, and for any divisor $n$ of $m-1$, when $0<n<m-1$, $a^{n} \neq 1(\bmod m)$, then $m$ is a prime. | 15. Proof: From $a^{m-1} \equiv 1(\bmod m)$, we know that $(a, m)=1$. By Theorem 1 of Chapter 5, we have
$$a^{\varphi(m)} \equiv 1(\bmod m)$$
Let $d$ be the smallest positive integer solution to the congruence $a^{x} \equiv 1(\bmod m)$, then it must be that $d \mid \varphi(m)$, because otherwise it can be written as $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,168 |
16. Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{n}^{a_{n}}$, $d(n)$ denote the number of all divisors of $n$, and $\sigma(n)$ denote the sum of all divisors of $n$. Prove:
(i) $d(n)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{n}+1\right)$.
(ii) $\sigma(n)=\left(\frac{p_{1}^{a_{1}+1... | 16. (i) Proof: Let $d$ be a divisor of $n$, and let $d=q_{1}^{r_{1}} q_{2}^{r_{2}} \cdots q_{m}^{r_{m}}$ be its standard factorization, then we have
$$q_{1}^{r} q_{2}^{r_{2}} \cdots q_{m}^{r_{m}} \mid p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{n}^{\alpha_{n}}$$
Since the standard factorization is unique, each $q_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,169 |
17. Let $n$ be a positive integer, $n \geqslant 2$. Prove that $n$ is a prime number if and only if
$$\varphi(n) \mid (n-1) \text{ and } (n+1) \mid \sigma(n) \text{ .}$$ | 17. Proof: If $n$ is a prime, then $\varphi(n)=n-1$, and $\sigma(n)=n+1$. Therefore, $\varphi(n) \mid (n-1)$ and $(n+1) \mid \sigma(n)$.
Conversely, if $\varphi(n) \mid (n-1)$ and $(n+1) \mid \sigma(n)$, if $n=2^{m}$, then $\varphi(n)=2^{m}-2^{m-1}=2^{m-1}$ and $n-1=2^{m}-1$. When $m>1$, $2^{m-1}$ is even, and $2^{m}-... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,170 |
Example 1 Prove
$$\sqrt{101}=10.0498756211 \cdots, \quad \sqrt{65}=8.0622577 \cdots$$ | In (11), taking $b=10$, we get
$$\begin{array}{l}
10.0498756211 \leqslant \frac{1620050}{161201} \leqslant \sqrt{101} \\
\leqslant \frac{80801}{8040} \leqslant 10.0498756219
\end{array}$$
Thus, we obtain $\sqrt{101}=10.0498756211 \cdots$. In (11), taking $b=8$, we have
$$\begin{array}{c}
8.062257747 \leqslant \frac{53... | \sqrt{101}=10.0498756211 \cdots, \quad \sqrt{65}=8.0622577 \cdots | Number Theory | proof | Yes | Yes | number_theory | false | 741,171 |
Lemma 5 Let $m$ be an integer greater than 1, and let $a_{1}, a_{2}, \cdots, a_{m}$ be $m$ integers. Suppose that any two integers taken from $a_{1}, a_{2}, \cdots, a_{m}$ are incongruent modulo $m$, then $a_{1}, a_{2}, \cdots, a_{m}$ form a complete residue system modulo $m$. | To prove that for any integer modulo $m$, it must be congruent to one of the following $m$ integers:
$$0,1, \cdots, m-1$$
Let $r_{i}$ (where $i=1,2, \cdots, m$) be an integer satisfying the condition
$$a_{i} \equiv r_{i}(\bmod m), \quad 0 \leqslant r_{i} \leqslant m-1,$$
Then we have
$$a_{1} \equiv r_{1}(\bmod m), a_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,172 |
Example 3 Prove that $n^{3}+5 n$ is a multiple of 6 (here $n$ is a positive integer). | To prove that the mathematical proposition here is that $n^{3}+5 n$ is a multiple of 6.
(1) When $n=1$, we have $n^{3}+5 n=6$, so the proposition holds for $n=1$.
(2) Let $k$ be an integer $\geqslant 2$. Assume the proposition holds for $n=k-1$, i.e., assume
$$(k-1)^{3}+5(k-1)=6 m$$
holds, where $m$ is an integer. Fro... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,174 |
Example 4 Let $n$ be a positive integer, and $x_{1}, \cdots, x_{n}, y_{1}, \cdots, y_{n}$ be real numbers, then
$$\left(x_{1} y_{1}+\cdots+x_{n} y_{n}\right)^{2} \leqslant\left(x_{1}^{2}+\cdots+x_{n}^{2}\right)\left(y_{1}^{2}+\cdots+y_{n}^{2}\right)$$
holds. | Prove that the mathematical proposition (18) is correct.
(1) When $n=1$, we have $x_{1}^{2} y_{1}^{2} \geqslant\left(x_{1} y_{1}\right)^{2}$, so (18) holds.
(2) Let $k$ be an integer $\geqslant 2$. Assume that the proposition holds for $n=k-1$. That is, assume
$$\begin{array}{l}
\left(x_{1} y_{1}+\cdots+x_{k-1} y_{k-1}... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 741,175 |
Lemma 1 Let $n \geqslant 3$ and the continued fraction $\left[a_{1}, a_{2}, \cdots, a_{n}\right]$ has convergents $\frac{p_{1}}{q_{1}}, \frac{p_{2}}{q_{2}}, \cdots, \frac{p_{n}}{q_{n}}$, then among these convergents, the following relations hold
$$p_{1}=a_{1}, q_{1}=1, p_{2}=a_{1} a_{2}+1, q_{2}=a_{2} .$$
and when $3 ... | Proof: Since
$$\frac{p_{1}}{q_{1}}=\frac{a_{1}}{1},$$
we have
$$p_{1}=a_{1}, q_{1}=1$$
Since
$$\frac{p_{2}}{q_{2}}=\frac{a_{1} a_{2}+1}{a_{2}}$$
we have
Since
$$\begin{array}{l}
p_{2}=a_{1} a_{2}+1, q_{2}=a_{2} \\
\frac{p_{3}}{q_{3}}=\frac{a_{3}\left(a_{1} a_{2}+1\right)+a_{1}}{a_{2} a_{3}+1}
\end{array}$$
we have
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,176 |
Lemma 2 If the $n$ convergents of the continued fraction $\left[a_{1}, a_{2}, \cdots, a_{n}\right]$ are $\frac{p_{k}}{q_{k}}$ (where $\left.k=1,2, \cdots, n\right)$, then for $k \geqslant 2$ we have
$$p_{k} q_{k-1}-p_{k-1} q_{k}=(-1)^{k},$$
and for $k \geqslant 3$ we have
$$p_{k} q_{k-2}-p_{k-2} q_{k}=(-1)^{k-1} a_{k}... | (i) Since $p_{2}=a_{1} a_{2}+1, q_{1}=1, p_{1}=a_{1}, q_{2}=a_{2}$, we have
$p_{2} q_{1}-p_{1} q_{2}=a_{1} a_{2}+1-a_{1} a_{2}=1$. Therefore, when $k=2$, equation (28) holds. Now assume $k \geqslant 3$, and suppose equation (28) holds for $k-1$, i.e., $p_{k-1} q_{k-2}-p_{k-2} q_{k-1}=(-1)^{k-1}$. Then, by equation (25)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,177 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.