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Let $g$ be a primitive root modulo $m$, and $(a, m)=1$. We have
$$\delta_{m}(a)=\varphi(m) /\left(\gamma_{m, g}(a), \varphi(m)\right)$$
From this, we can deduce that when $m$ has a primitive root, for each positive divisor $d \mid \varphi(m)$, there are exactly $\varphi(d)$ elements in a reduced residue system modulo ... | In §1 Property VI, taking $a=g$ and $k=\gamma_{m, g}(a)$, by $\delta_{m}(g)=\varphi(m)$, §1 formula (3), and §1 Property I, we obtain formula (8). When the modulus $m$ has a primitive root, we take the reduced residue system given by formula (1). From formula (8), we know that the element $g^{j}$ in this reduced residu... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,610 |
Example 1 Construct the index table for modulo 23 with primitive root 5.
保留源文本的换行和格式，直接输出翻译结果。
(注意：最后一句为中文，可能是注释或说明，翻译时可忽略或根据上下文调整。)
Example 1 Construct the index table for modulo 23 with primitive root 5.
(Keep the original text's line breaks and format, and output the translation directly.) | From Example 1 in §2, we know that 5 is a primitive root modulo 23, and $\varphi(23)=22$. First, list Table 1 in the order of indices, as it is easier to compute the absolute least residues of $5^{j}$ modulo 23. Arranging according to the size order of the absolute least reduced residue system, Table 1 becomes Table 2.... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,611 |
Property VI Let $\alpha \geqslant 3,1 \leqslant d \mid 2^{\alpha-2}$, and let $\psi(d)$ denote the number of elements of order $d$ in a reduced residue system modulo $2^{\alpha}$. We have
$$\psi(d)=\left\{\begin{array}{ll}
1, & d=1 \\
3, & d=2 \\
2 \varphi(d), & 2<d \mid 2^{\alpha-2}
\end{array}\right.$$ | We take the reduced residue system modulo $2^{\circ}$ given by formula (4) $\left(g_{0}=5\right)$. From formula (17), we know that the sufficient and necessary condition for $\delta_{2^{a}}(a)=1$ is $\gamma^{(0)}(a)=\gamma^{(-1)}(a)=0$, i.e., $a \equiv 1\left(\bmod 2^{a}\right)$. Therefore, formula (19) holds. From for... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,616 |
Example 2 Construct the table of indices modulo $2^{8}=64$ with $(\alpha=6)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Table 3 is arranged according to the size order of the indicator $\gamma^{(0)}(a)$, respectively with
$$\gamma^{(-1)}(a)=0, \quad \gamma^{(-1)}(a)=1$$
From this, we can obtain the indicator table arranged according to the absolute least residue system modulo $2^{6}$ (see Table 4).
Table 3
\begin{tabular}{|c|ccccccccc... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,617 |
1. Using the properties of indices and Example 1 in §3, construct a table of indices modulo 23 with base 11. | 1. From $\gamma_{23,11}(5) \cdot \gamma_{23,5}(11) \equiv 1(\bmod 22)$, we get $\gamma_{23,11}(5)=5$, hence $\gamma_{23,11}(a) \equiv 5 \gamma_{23,5}(a)(\bmod 22)$.
\begin{tabular}{|c|rrrrrrrrrrr|}
\hline$a$ & -11 & -10 & -9 & -8 & -7 & -6 & -5 & -4 & -3 & -2 & -1 \\
\hline$\gamma_{23.11}(a)$ & 12 & 4 & 17 & 19 & 18 & ... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,626 |
4. Find the reduced residue system modulo $m=7^{2} \cdot 11^{2}$ in the form of equation (25) and equation (30) of §3. | $$\text { 4. } \begin{array}{l}
-17 \cdot 121 \cdot 3^{\gamma^{(1)}}+42 \cdot 49 \cdot 7^{(2)}, 0 \leqslant \gamma^{(1)} \leqslant 41,0 \leqslant \gamma^{(2)} \leqslant 109 . \\
\widetilde{g}_{1}=-17 \cdot 121 \cdot 3+42 \cdot 49=-4113 \\
\widetilde{g}_{2}=-17 \cdot 121+42 \cdot 49 \cdot 7=12349 .
\end{array}$$
Since ... | 1816^{r^{(1)}} \cdot 491^{r^{(2)}}, \quad 0 \leqslant \gamma^{(1)} \leqslant 41, \quad 0 \leqslant \gamma^{(2)} \leqslant 109 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,629 |
5. Find the reduced residue systems modulo $m=2^{6} \cdot 43$ of the form similar to equation (25) and equation (30). | $$\begin{array}{l}
\text { 5. } 3 \cdot 43(-1)^{r^{(-1)}} 5^{r^{(0)}}+(-2) \cdot 64 \cdot 3^{r^{(1)}} \text {, } \\
0 \leqslant \gamma^{(-1)} \leqslant 1,0 \leqslant \gamma^{(0)} \leqslant 15,0 \leqslant \gamma^{(1)} \leqslant 41 . \\
\tilde{g}_{-1}=3 \cdot 43 \cdot(-1)+(-2) \cdot 64=-257 \text {, } \\
\tilde{g}_{0}=3 ... | (-257)^{r^{(-1)}} 517^{r^{(0)}}(-255)^{r^{(1)}} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,630 |
8. 求 3 对模 $m$ 的指标组 (选定相应的原根):
(i) $m=2^{5} \cdot 29 \cdot 41^{2}$ ；
(ii) $m=2 \cdot 13 \cdot 23 \cdot 41 \cdot 47$. | 8. (i) 对 $2^{5}$ 取 $-1,5$; 对 29 取原根 2 ; 对 $41^{2}$ 取原根 7.3 的指标组为
$$\{1,3 ; 5,825\} \text {. }$$
(ii) 对 2 取 $-1,5$; 对 $13,23,41,47$ 依次取原根 $2,5,6,5.3$ 的指标组为
$$\{0,0 ; 4,16,15,20\} .$$ | \{1,3 ; 5,825\} \text{ 和 } \{0,0 ; 4,16,15,20\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,633 |
Theorem 1 Let $m \geqslant 2, (a, m)=1$, and $g$ be a primitive root modulo $m$. Then the congruence equation (1) has a solution, i.e., $a$ is an $n$-th power residue modulo $m$ if and only if
$$(n, \varphi(m)) \mid \gamma(a),$$
where $\gamma(a)=\gamma_{m, g}(a)$ is the index of $a$ modulo $m$ with respect to the base... | Prove that if $x \equiv x_{1}(\bmod m)$ is a solution to (1), then by $(a, m)=1$ we know $\left(x_{1}, m\right)=1$. Therefore, by Theorem 5 in §2, there must be a $y_{1}$ such that
$$x_{1} \equiv g^{y_{1}}(\bmod m)$$
Thus, we have
$$g^{n y_{1}} \equiv a(\bmod m)$$
Furthermore, by Property I in §3, we have
$$n y_{1} \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,634 |
Example 1 Solve the congruence equation $x^{8} \equiv 41(\bmod 23)$. | From Example 1 in §2, we know that 5 is a primitive root modulo 23. Since \(41 \equiv -5 \pmod{23}\), we find from Table 2 in Example 1 of §3 that \(\gamma_{23,5}(41) = 12\). Therefore, we need to solve the congruence equation:
$$8 y \equiv 12 \pmod{22}$$
Since \((8, 22) = 2 \mid 12\), the above congruence equation ha... | x \equiv -6, 6 \pmod{23} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,635 |
1. Let $k_{0}$ be a given integer, and $P(n)$ be a property or proposition concerning the integer $n$. If
(i) $P\left(k_{0}\right)$ holds when $n=k_{0}$;
(ii) $P(n)$ holding implies that $P(n+1)$ holds,
then $P(n)$ holds for all integers $n \geqslant k_{0}$. | 1. Make a variable substitution: $n=m+k_{0}-1 . P(n)=P\left(m+k_{0}-1\right)=P^{*}(m)$. Apply Theorem 1 of §1 to $P^{*}(m)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,636 |
Example 1 Let $a \geqslant 2$ be a given positive integer, $j=0,1, \cdots, a-1$. For a given $j$, all integers that leave a remainder of $j$ when divided by $a$ are
$$k a+j, \quad k=0, \pm 1, \pm 2, \cdots$$ | The set of integers composed of these integers is denoted as $j \bmod a$. When $0 \leqslant j \neq j^{\prime} \leqslant a-1$, the sets $j \bmod a$ and $j^{\prime} \bmod a$ are disjoint, and the union
$$0 \bmod a \cup 1 \bmod a \cup \cdots \cup(a-1) \bmod a=\boldsymbol{Z}$$
That is, all integers are classified accordin... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,637 |
Theorem 2 If the modulus $m$ has a primitive root, $n \geqslant 2$, then, in a reduced residue system modulo $m$, the $n$-th residues modulo $m$ are exactly $\varphi(m) /(n, \varphi(m))$ in number. | For example, there are 11 eighth residues modulo 23, which are
$$a \equiv 5^{2 t}(\bmod 23), \quad 0 \leqslant t<11$$
From Table 1 in Example 1 of §3, we can find
$$a \equiv 1,2,4,8,-7,9,-5,-10,3,6,-11(\bmod 23)$$
When the modulus $m$ has a primitive root, there is a relationship between the exponent and the index (P... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,638 |
Theorem 3 Suppose that $m$ has a primitive root, $n \geqslant 2$, then $a$ is an $n$-th residue modulo $m$, i.e., the binomial congruence equation $((a, m)=1)$ has a solution if and only if equation (9) (i.e., equation (8)) holds, and when there is a solution, there are $(n, \varphi(m))$ solutions. | When $m=23, n=8$, $\varphi(23) /(8, \varphi(23))=11$. From Table 1 in Example 1 of $\S 3$, we know that all $a$ satisfying $\delta_{23}(a) \mid 11$ are exactly given by formula (7).
When $m$ is a prime $p$, Theorem 2 and Theorem 3 have already been proven in Theorems 7, 8, and 9 of Chapter 4, $\S 8$. Here, these concl... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 740,639 |
Example 2 Solve the congruence equation $x^{12} \equiv 17\left(\bmod 2^{6}\right)$. | From Table 4 in Example 2 of §3, we find that the indicator set of 17 is $\gamma^{(-1)}(17)=0, \gamma^{(0)}(17)=12$. Therefore, we need to solve two linear congruences:
$$\begin{array}{ll}
12 u \equiv 0(\bmod 2), & 0 \leqslant u<1 \\
12 v \equiv 12\left(\bmod 2^{4}\right), & 0 \leqslant v<2^{4}
\end{array}$$
It is eas... | x \equiv 5,-11,-27,21,-5,11,27,-21\left(\bmod 2^{6}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,641 |
Example 3 Solve the congruence equation $x^{11} \equiv 27\left(\bmod 2^{6}\right)$. | Solve Table 4 of Example 2 in §3 to get $\gamma^{(-1)}(27)=1, \gamma^{(0)}(27)=9$. Therefore, we need to solve two linear congruences:
$$\begin{array}{ll}
11 u \equiv 1(\bmod 2), & 0 \leqslant u<1, \\
11 v \equiv 9\left(\bmod 2^{4}\right), & 0 \leqslant v<2^{4} .
\end{array}$$
It is easy to solve and get $u=1, v=11$. ... | x \equiv -29 \pmod{2^6} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,642 |
Theorem 5 Let $m=2^{a}, \alpha \geqslant 3$, then when $2 \nmid n$, all elements in a reduced residue system modulo $2^{a}$ are $n$-th residues modulo $2^{a}$; when $2 \mid n$, there are $2^{a-2} /\left(n, 2^{a-2}\right)$ elements in a reduced residue system modulo $2^{a}$ that are $n$-th residues modulo $2^{a}$. | Proof When $2 \nmid n$, condition (10) always holds, so the conclusion holds. When $2 \mid n$, condition (10) holds if and only if $\gamma^{(-1)}(a)=0$ and $\left(n, 2^{a^{-2}}\right) \mid \gamma^{(0)}(a), 0 \leqslant \gamma^{(0)}(a)< 2^{a-2}$, so there are exactly $2^{a-2} /\left(n, 2^{a-2}\right)$ such index groups, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,643 |
Theorem 7 Let $m=2^{a}, \alpha \geqslant 3,2 \nmid a$, then $a$ is an $n$-th residue modulo $2^{a}$, i.e., the binomial congruence equation (1) has a solution if and only if
$$(a-1) / 2 \equiv 0(\bmod (n, 2)), \quad \delta_{2^{a}}(a) \mid 2^{a-2} /\left(n, 2^{a-2}\right),$$
which is equivalent to the conditions
$$(a-1... | Proof: By Property VII of §3, we have
$$\begin{array}{ll}
2^{a-2}=\left(2^{a-2}, \gamma^{(0)}(a)\right) \delta_{2^{a}}(a), & \gamma^{(0)}(a) \neq 0 \\
2=\left(2, \gamma^{(-1)}(a)\right) \delta_{2^{a}}(a), & \gamma^{(0)}(a)=0
\end{array}$$
We will prove the necessity and sufficiency of condition (18). The equivalence b... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,645 |
1. Solve the following congruences using the index table:
(i) $3 x^{6} \equiv 5(\bmod 7)$;
(ii) $x^{12} \equiv 16(\bmod 17)$;
(iii) $5 x^{11} \equiv-6(\bmod 17)$;
(iv) $3 x^{14} \equiv 2(\bmod 23)$;
(v) $x^{15} \equiv 14(\bmod 41)$;
(vi) $7 x^{7} \equiv 11(\bmod 41)$;
(vii) $3^{x} \equiv 2(\bmod 23)$;
(viii) $13^{x} \e... | 1. (i) no solution;
(ii) $x \equiv 1,13,16,4(\bmod 17)$;
(iii) $x \equiv 2(\bmod 17)$;
(iv) $x \equiv-9(\bmod 23)$;
(v) $x \equiv 29,3,30,13,7(\bmod 41)$;
(vi) $x \equiv 4(\bmod 41)$;
(vii) $x \equiv 7,18(\bmod 22)$;
(viii) no solution. | no solution | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,646 |
2. For which integers $a$, is the congruence equation $a x^{5} \equiv 3(\bmod 19)$ solvable? | 2. All $a,(a, 19)=1$.
| All\ a,\ (a, 19)=1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,647 |
Example 2 (i) $0 \bmod 2 \cap 0 \bmod 3=0 \bmod 6$;
(ii) $1 \bmod 2 \cap 1 \bmod 3=$ $1 \bmod 6$; (iii) $0 \bmod 2 \cap 1 \bmod 3=4 \bmod 6$. | To prove (i). That is, to prove that the necessary and sufficient condition for $a=2k$ and $a=3h$ is $a=6d$. The sufficiency is obvious. From $2k=3h$, we know $h=2(k-h)$, so $a=6(k-h)$. This proves the necessity.
(ii) Is to prove that the necessary and sufficient condition for $a=2k+1$ and $a=3h+1$ is $a=6d+1$, which m... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,648 |
3. For which integers $b$, is the congruence equation $7 x^{8} \equiv b(\bmod 41)$ solvable? | 3. $b \equiv 29,3,30,13,7(\bmod 41)$. | b \equiv 29,3,30,13,7(\bmod 41) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,649 |
4. Find all solutions to the congruence equation $x^{x} \equiv x(\bmod 19)$ that satisfy $(x, 19)=1$. | 4. $x \equiv a(\bmod 18), x^{a-1} \equiv 1(\bmod 19), 1 \leqslant a \leqslant 18,(x, 19)=1$. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,650 |
5. Find all solutions to the congruence equation $5^{x} \equiv x(\bmod 23)$. | 5. $x \equiv a(\bmod 22), x \equiv 5^{a}(\bmod 23), 0 \leqslant a \leqslant 21$. | x \equiv a(\bmod 22), x \equiv 5^{a}(\bmod 23), 0 \leqslant a \leqslant 21 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,651 |
6. Let $p>2$ be a prime. Prove: The congruence equation $x^{4} \equiv-1(\bmod p)$ has a solution if and only if $p \equiv 1(\bmod 8)$. From this, deduce that there are infinitely many primes of the form $p \equiv 1(\bmod 8)$. | 6. The index of -1 to any primitive root modulo $p$ is $(p-1) / 2$. Therefore,
$$x^{4} \equiv-1(\bmod p)$$
has a solution if and only if $(4, p-1) \mid(p-1) / 2$, i.e., $p \equiv 1(\bmod 8)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,652 |
7. 解同余方程:
(i) $x^{6} \equiv-15(\bmod 64)$;
(ii) $x^{12} \equiv 7(\bmod 128)$. | 7. (i) $x \equiv \pm 9, \pm 23(\bmod 64)$; (ii) 无解. | x \equiv \pm 9, \pm 23(\bmod 64); 无解 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,653 |
12. Let $p$ be a prime. Prove: The congruence equation $x^{8} \equiv 16(\bmod p)$ always has a solution. | 12. Let $g$ be a primitive root of $p$, $\gamma=\gamma_{p, g}$ (2). The original congruence equation is equivalent to
$$8 y \equiv 4 \gamma(\bmod p-1)$$
When $p \equiv \pm 1(\bmod 8)$, $2 \mid \gamma$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,658 |
13. Let $p$ be a prime, $2 \nmid \delta_{p}(a)$. Prove: The congruence equation $a^{x}+1 \equiv 0(\bmod p)$ has no solution. | 13. Let $p-1=2^{l} \cdot c, 2 \nmid c$, by property IV of $\S 3$ and $2 \nmid \delta_{p}(a)$, we know that $2^{l}$ must divide the index of $a$ (with any primitive root $g$ as the base). From this and the fact that the index of -1 is $(p-1) / 2$, the desired conclusion can be derived. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,660 |
15. 对哪些整数 $a$, 同余方程 $10^{x} \equiv a(\bmod 41)$ 有解? | 15. $a \equiv 1,10,16,18,37(\bmod 41)$. | a \equiv 1,10,16,18,37(\bmod 41) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,662 |
16. Prove: 2 is an 8th power residue modulo 73. | 16. $\delta_{73}(2)=9$. By Theorem 3 and equation (8), we deduce. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,663 |
17. Solve the congruence equations using methods that both utilize and do not utilize primitive roots:
(i) $x^{4} \equiv 41(\bmod 37) ;$
(ii) $x^{4} \equiv 37(\bmod 41)$. | 17. (i) i.e., $\left(x^{2}-2\right)\left(x^{2}+2\right) \equiv 0(\bmod 37),\left(\frac{-2}{37}\right)=\left(\frac{2}{37}\right)=-1$, no solution;
(ii) $x \equiv 10,33,31,8(\bmod 41)$. | x \equiv 10,33,31,8(\bmod 41) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,664 |
18. Solve the congruence equation $\left(x^{2}+1\right)(x+1) x \equiv -1(\bmod 41)$. | 18. The original equation is $x^{4}+x^{3}+x^{2}+x+1 \equiv 0(\bmod 41), x \equiv 1(\bmod 41)$ is not a solution, so the solutions to the original equation are the solutions to $x^{5} \equiv 1(\bmod 41)$ excluding $x \equiv 1(\bmod 41)$, which are
$$x \equiv 10,18,16,37(\bmod 41)$$ | x \equiv 10,18,16,37(\bmod 41) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,665 |
19. Let the prime $p \equiv 3(\bmod 4)$. Prove: $a$ is a fourth power residue modulo $p$ if and only if $\left(\frac{a}{p}\right)=1$, i.e., $a$ is a quadratic residue modulo $p$. Solve the congruence equation
$$x^{4} \equiv 3(\bmod 11)$$ | 19. Necessity is obvious. If $\left(\frac{a}{p}\right)=1$, then $a \equiv b^{2}(\bmod p)$.
$$x^{4}-a \equiv\left(x^{2}-b\right)\left(x^{2}+b\right)(\bmod p)$$
When $p \equiv 3(\bmod 4)$, $\left(\frac{b}{p}\right),\left(\frac{-b}{p}\right)$ must have one that is 1. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,666 |
Lemma 3 Let $p>2$ be a prime. The congruence equation
$$\left\{\begin{array}{l}
x^{2}+y^{2}+1 \equiv 0(\bmod p) \\
0 \leqslant x, y \leqslant(p-1) / 2
\end{array}\right.$$
has a solution. | It is easy to see that the following $(p+1) / 2$ numbers are pairwise incongruent modulo $p$:
$$a^{2}, \quad a=0,1, \cdots,(p-1) / 2 .$$
Similarly, the following $(p+1) / 2$ numbers are also pairwise incongruent modulo $p$:
$$-b^{2}-1, \quad b=0,1, \cdots,(p-1) / 2$$
But among these total $p+1$ numbers, there must be... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,669 |
Example 4 Let $a \geqslant 2$ be a given positive integer, then any positive integer $n$ can be uniquely represented as
$$n=r_{k} a^{k}+r_{k-1} a^{k-1}+\cdots+r_{1} a+r_{0}$$
where the integer $k \geqslant 0,0 \leqslant r_{j} \leqslant a-1(0 \leqslant j \leqslant k), r_{k} \neq 0$. This is the $a$-ary representation o... | Prove that for any positive integer $n$, there exists a unique $k \geqslant 0$ such that $a^{k} \leqslant n < a^{k+1}$ (why). By the division algorithm, there must exist unique $q_{0}, r_{0}$ satisfying
$$n = q_{0} a + r_{0}, \quad 0 \leqslant r_{0} < a.$$
If $k=0$, then we must have $q_{0}=0, 1 \leqslant r_{0} < a$, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,670 |
Lemma 4 Let $p>2$ be a prime. There exist integers $x_{0}, y_{0}$ and $m_{0}, 1 \leqslant m_{0}<p$, such that
$$m_{0} p=1+x_{0}^{2}+y_{0}^{2} .$$ | Prove that $x_{0}, y_{0}$ are solutions to the congruence equation in Lemma 3. We have
$$x_{0}^{2}+y_{0}^{2}+1=m_{0} p, \quad m_{0} \geqslant 1$$
But on the other hand
$$\begin{aligned}
x_{0}^{2}+y_{0}^{2}+1 & \leqslant\left(\frac{p-1}{2}\right)^{2}+\left(\frac{p-1}{2}\right)^{2}+1 \\
& =p^{2} / 2-p+3 / 2<p^{2} / 2
\e... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,671 |
Theorem $5^{\Phi}$ When $n$ is a positive integer of the form $4^{a}(8 k+7)(\alpha \geqslant 0, k \geqslant 0)$, $n$ cannot be expressed as the sum of three squares of integers. | Prove that for any integer $x$,
$$x^{2} \equiv 0,1 \text { or } 4(\bmod 8) .$$
Therefore, for any integers $x_{1}, x_{2}, x_{3}$, we must have
$$x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \not \equiv 7(\bmod 8)$$
From this, it follows that a positive integer $n$ of the form $8 k+7$ cannot be expressed as the sum of three squares,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,672 |
Theorem 7 Except for the following twelve numbers:
$$1,2,3,4,6,7,9,10,12,15,18,33$$
every positive integer is the sum of five positive squares. | The theorem can be proved as follows: The twelve numbers given by equation (8) can be directly verified, none of which can be expressed as the sum of five positive squares; when $n \leqslant 168$ and not equal to the above twelve numbers, it can be directly verified that they can all be expressed as the sum of five pos... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,674 |
4. Express (i) $23 \cdot 53$, (ii) $43 \cdot 197$, (iii) $47 \cdot 223$ as the sum of four squares in two different ways. | 4. (i) $23 \cdot 53=27^{2}+15^{2}+12^{2}+11^{2}=25^{2}+19^{2}+13^{2}+8^{2}$;
(ii) $43 \cdot 197=74^{2}+51^{2}+15^{2}+13^{2}=69^{2}+57^{2}+19^{2}+10^{2}$;
(iii) $47 \cdot 223=101^{2}+12^{2}+10^{2}+6^{2}=77^{2}+54^{2}+40^{2}+6^{2}$. | 23 \cdot 53=27^{2}+15^{2}+12^{2}+11^{2}=25^{2}+19^{2}+13^{2}+8^{2}; 43 \cdot 197=74^{2}+51^{2}+15^{2}+13^{2}=69^{2}+57^{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,678 |
6. Prove: except for a finite number of exceptional values, every positive integer is the sum of six positive squares. Find these exceptional values. | 6. When $N-1$ is not equal to $0,1,2,3,4,6,7,9,10,12,15,18,33$, by Theorem 7, $N$ can definitely be expressed as the sum of six positive squares. $34=3^{2}+3^{2}+2^{2}+2^{2}+2^{2}+2^{2}$. Direct verification shows that only when $N=$ $1,2,3,4,5,7,8,10,11,13,16,19$ cannot be expressed as the sum of six positive squares. | 1,2,3,4,5,7,8,10,11,13,16,19 | Number Theory | proof | Yes | Yes | number_theory | false | 740,680 |
Example 5 Let $a>2$ be an odd number. Prove:
(i) There must exist a positive integer $d \leqslant a-1$, such that $a \mid 2^{d}-1$;
(ii) Let $d_{0}$ be the smallest positive integer $d$ satisfying (i), then $a \mid 2^{h}-1(h \in \boldsymbol{N})$ if and only if $d_{0} \mid h$.
(iii) There must exist a positive integer $... | First, we prove (i). Consider the following $a$ numbers:
$$2^{0}, 2^{1}, 2^{2}, \cdots, 2^{a-1}$$
From Example 2 in §2, we know that $a \nmid 2^{j} (0 \leqslant j < a)$. From this and Theorem 1, we get: for each $j, 0 \leqslant j < a$,
$$2^{j} = q_{j} a + r_{j}, \quad 0 < r_{j} < a.$$
Thus, the $a$ remainders $r_{0},... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,681 |
7. Prove that $2^{k}(k \geqslant 0)$ cannot be expressed as the sum of three positive squares, and directly find all solutions to $2^{k}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$. | 7. Prove the first conclusion by induction. When $k=0$, the conclusion holds. Assume that the conclusion holds for $k=n(\geqslant 0)$. When $k=n+1$, if $2^{n+1}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}, x_{1}>0, x_{2}>0, x_{3}>0$, then $x_{1}, x_{2}, x_{3}$ must be two odd and one even. And the two odd numbers cannot be equal. Th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,682 |
Theorem 1 Let $p$ be a prime, then the indeterminate equation
$$x^{2}+y^{2}=p$$
has a solution if and only if $p=2$ or $\left(\frac{-1}{p}\right)=1$, i.e., $p=2$ or $p=4k+1$. | To prove the necessity: Since $p$ is a prime, if the indeterminate equation (1) has a solution $x_{0}, y_{0}$, it must satisfy
$$\left(x_{0}, y_{0}\right)=\left(x_{0} y_{0}, p\right)=1,$$
Thus, there must be a $y_{0}^{-1}$ such that $y_{0}^{-1} y_{0} \equiv 1(\bmod p)$. This leads to
$$\left(x_{0} y_{0}^{-1}\right)^{2... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,684 |
Theorem 2 Let the positive integer $n=d^{2} m, m$ be square-free, then the indefinite equation (6) has a solution if and only if $m$ has no prime factors of the form $4 k+3$.
| To prove the sufficiency, from the given conditions we have
$$m=2^{a} p_{1} \cdots p_{r}, \quad p_{j} \equiv 1(\bmod 4), \quad 1 \leqslant j \leqslant r, 0 \leqslant \alpha \leqslant 1 .$$
Equation (5) indicates that if the indeterminate equation (6) is solvable for \( n=n_{1} \) and \( n=n_{2} \), then it is also sol... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,685 |
Theorem 3 The indeterminate equation (6) has solutions satisfying the condition $(x, y)=1$ if and only if $n=1,2, n_{1}$ or $2 n_{1}$, where
$$n_{1}=p_{1}^{a_{1}} \cdots p_{r}^{a_{r}}, \quad p_{j} \equiv 1(\bmod 4), 1 \leqslant j \leqslant r$$ | To prove the necessity, we need to show that $4 \nmid n$ and $n$ has no prime factors of the form $4k+3$. In this case, we must have $(x, y) = (x, n) = (y, n) = 1$. If $4 \mid n$, then we have
$$x^{2} + y^{2} \equiv 0 \pmod{4}$$
This is impossible when $(x, 2) = (y, 2) = 1$. Therefore, we must have $4 \nmid n$. If the... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,686 |
Theorem 4 Let $p>3$ be a prime number, then the indeterminate equation
$$x^{2}+3 y^{2}=p$$
has a solution if and only if $\left(\frac{-3}{p}\right)=1$, i.e., $p$ is a prime of the form $6 k+1$. | To prove the necessity, if the indeterminate equation (17) has a solution \(x_{0}, y_{0}\), it is evident that
$$\left(x_{0}, 3 y_{0}\right)=\left(p, 3 x_{0} y_{0}\right)=1$$
Using Theorem 1 from Chapter 4, Section 6, we get from this and equation (17):
$$1=\left(\frac{x_{0}^{2}}{p}\right)=\left(\frac{-3 y_{0}^{2}}{p}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,687 |
Theorem 5 Let $n \geqslant 1$. The number of solutions to the indeterminate equation
$$x^{2}+y^{2}=n$$
is
$$N(n)=4 \sum_{d \mid n} h(d)$$
where the arithmetic function $h(d)$ is defined as follows: $h(1)=1$,
$$h(d)=\left\{\begin{array}{ll}
0, & 2 \mid d, \\
(-1)^{(d-1) / 2}, & 2 \nmid d,
\end{array}\right.$$
and two... | Theorem 5 Proof: From equations (25) and (41), we have
$$N(n)=4 \sum_{\left.d^{2}\right|_{n}} R\left(\frac{n}{d^{2}}\right)$$
Let $n=n_{1} n_{2},\left(n_{1}, n_{2}\right)=1$. If $d^{2} \mid n$, then there must be $\left(d^{2}, n_{1}\right)=d_{1}^{2}$, $\left(d^{2}, n_{2}\right)=d_{2}^{2}, d=d_{1} d_{2}{ }^{(1)}$; conv... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,688 |
Lemma 6 We have
$$\begin{array}{c}
N(1)=Q(1)=4, \quad P(1)=2, \quad P(2)=1 \\
Q(n)=4 P(n), \quad n>1
\end{array}$$
and
$$N(n)=\sum_{d^{2} \mid n} Q\left(\frac{n}{d^{2}}\right), \quad n \geqslant 1$$ | Prove that when $n=1$, all solutions of the indeterminate equation (20) are
$$\{ \pm 1,0\}, \quad\{0, \pm 1\} ;$$
When $n=2$, all solutions are
$$\{ \pm 1, \pm 1\} .$$
This leads to equation (23). The primitive solutions $x, y$ of the indeterminate equation (20) must satisfy
$$(x, y)=(n, x y)=1$$
Therefore, when $n>... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,689 |
Lemma 7 Let $n>1$, then for each primitive solution $x, y$ of the indeterminate equation (20), there must be an $s$ satisfying
$$s y \equiv x(\bmod n), \quad s^{2} \equiv-1(\bmod n)$$
Moreover, if $\left\{x_{1}, y_{1}\right\},\left\{x_{2}, y_{2}\right\}$ are two different non-negative primitive solutions of (20), and ... | Proof: Since the primitive solution must satisfy equation (26), the linear congruence equation \( y s \equiv x \pmod{n} \) must have a unique solution \( s \) modulo \( n \). Therefore, we have
\[ s^{2} y^{2} \equiv x^{2} \equiv -y^{2} \pmod{n} \]
From this and \((y, n) = 1\), we get \( s^{2} \equiv -1 \pmod{n} \). Th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,690 |
Lemma 9 Let $n>1$. If the quadratic congruence
$$s^{2} \equiv-1(\bmod n)$$
has a solution $s_{1} \bmod n$, then, the indefinite equation (20) has a primitive solution, and there must be a non-negative primitive solution $x_{1}, y_{1}$, satisfying
$$s_{1} y_{1} \equiv x_{1}(\bmod n)$$ | Proof: Clearly, we have $\left(s_{1}, n\right)=1$, hence by Lemma 4, there must exist $u_{0}, v_{0}$ satisfying
$$0<\left|u_{0}\right| \leqslant \sqrt{n}, \quad 0<\left|v_{0}\right|<\sqrt{n}$$
and
$$s_{1} u_{0} \equiv v_{0}(\bmod n)$$
From equation (37), we get
$$2 \leqslant u_{0}^{2}+v_{0}^{2}<2 n$$
Since $s_{1}$ s... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,693 |
Lemma 10 We have $R(1)=1$,
$$R(2)=1, \quad R\left(2^{a}\right)=0, \quad \alpha>1 \text {, }$$
and for an odd prime $p$ we have
$$R\left(p^{a}\right)=1+\left(\frac{-1}{p}\right)=\left\{\begin{array}{ll}
2, & p \equiv 1(\bmod 4), \\
0, & p \equiv 3(\bmod 4),
\end{array} \quad \alpha \geqslant 1\right.$$
Furthermore, if... | To prove $R(1)=1$ and formula (42) can be easily verified directly. By Corollary 3 and Theorem 1 of Chapter 4, §5, formula (43) holds when $\alpha=1$. Furthermore, since
$$2 s \equiv 0(\bmod p)$$
has no common solution with congruence equation (35) $(n=p)$, from Corollary 4 of Chapter 4, §4, it follows that formula (4... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,694 |
1. Find which numbers between $1 \sim 99$ can be expressed as the sum of two squares, and which cannot. | 1. $1,2,4,5,8,9,10,13,16,17,18,20,25,26,29,32,34,36,37,40,41,45,48$, $49,50,52,53,58,61,64,65,68,72,73,74,80,81,82,85,89,90,97,98$ can be expressed as the sum of two squares, while others cannot. | 1,2,4,5,8,9,10,13,16,17,18,20,25,26,29,32,34,36,37,40,41,45,48,49,50,52,53,58,61,64,65,68,72,73,7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,696 |
2. If the prime $p=a^{2}+b^{2}$, then this representation is actually unique, i.e., assuming $a \geqslant b>0$, if there is also $p=a_{1}^{2}+b_{1}^{2}, a_{1} \geqslant b_{1}>0$, then it must be that $a=a_{1}, b=b_{1}$. | 2. Let $c^{2} \equiv-1(\bmod p)$. It must be that $a \equiv c b$ or $-c b(\bmod p), a_{1} \equiv c b_{1}$ or $-c b_{1}(\bmod p)$. $p^{2}=\left(a a_{1} \pm b b_{1}\right)^{2}+\left(a b_{1} \mp b a_{1}\right)^{2}$, one of the terms on the right side must be zero. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,697 |
3. Find all possible representations as a sum of two squares for (i) $5 \cdot 13$, (ii) $17 \cdot 29$, (iii) $37 \cdot 41$, (iv) $5 \cdot 13 \cdot 17 \cdot 29$, (v) $7^{2} \cdot 13 \cdot 17$. | $\begin{array}{l}\text { 3. (i) } 5 \cdot 13=8^{2}+1^{2}=7^{2}+4^{2} \text {; (ii) } 17 \cdot 29=22^{2}+3^{2}=18^{2}+13^{2} \text {; (iii) } 37 \\ 41=34^{2}+19^{2}=29^{2}+26^{2} \text {; (iv) } 5 \cdot 13 \cdot 17 \cdot 29=179^{2}+2^{2}=173^{2}+46^{2}=178^{2}+ \\ 19^{2}=163^{2}+74^{2}=157^{2}+86^{2}=131^{2}+122^{2}=166... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,698 |
Theorem 5 Under the conditions and notation of Theorem 4, we have
(i) $u_{k+1}=\left(u_{0}, u_{1}\right)$, that is, the last non-zero remainder $u_{k+1}$ is the greatest common divisor of $u_{0}$ and $u_{1}$;
(ii) $d \mid u_{0}$ and $d \mid u_{1}$ if and only if $d \mid u_{k+1}$;
(iii) There exist integers $x_{0}, x_{1... | Prove: Using Theorem 8 (i) and (iv) of §2, starting from the last equation of (5) and proceeding upwards, we get
$$\begin{aligned}
u_{k+1} & =\left(u_{k+1}, u_{k}\right)=\left(u_{k}, u_{k-1}\right)=\left(u_{k-1}, u_{k-2}\right)=\cdots \\
& =\left(u_{4}, u_{3}\right)=\left(u_{3}, u_{2}\right)=\left(u_{2}, u_{1}\right)=\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,703 |
9. Let prime $p \equiv 1(\bmod 4),(k, p)=1$, and
$$S(k)=\sum_{j=0}^{p-1}\left(\frac{j\left(j^{2}+k\right)}{p}\right)$$
Prove: (i) $S(k)$ is even; (ii) for any integer $l$,
$$S\left(k l^{2}\right)=\left(\frac{l}{p}\right) S(k) .$$ | 9. (i) Let $p=4 m+1$,
$$S(k)=2 \sum_{j=1}^{2 m}\left(\frac{j\left(j^{2}+k\right)}{p}\right) ;$$
(ii)
$$\begin{array}{r}
\text { When } p \mid l \text {, } S\left(k l^{2}\right)=\sum_{j=0}^{p-1}\left(\frac{j}{p}\right)=0 \text {; when } p \nmid l \text {, } \\
S\left(k l^{2}\right)=\sum_{j=0}^{p-1}\left(\frac{j l\left((... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,705 |
11. Let the prime $p \equiv 1(\bmod 4), a, b$ be as in problem 10, and $q=(p-1) / 2$. Prove:
(i)
$$q(S(a))^{2}+q(S(b))^{2}=\sum_{x=1}^{p-1} \sum_{y=1}^{p-1} \sum_{k=1}^{p-1}\left(\frac{x y\left(x^{2}+k\right)\left(y^{2}+k\right)}{p}\right),$$
where $S(k)$ is defined as in problem 9;
(ii)
$$\sum_{k=1}^{p-1}\left(\frac{... | 11.
$$\text { (i) } \begin{aligned}
q(S(a))^{2}+q(S(b))^{2} & =\sum_{l=1}^{q}\left(S\left(a l^{2}\right)\right)^{2}+\sum_{l=1}^{q}\left(S\left(b l^{2}\right)\right)^{2} \\
& =\sum_{k=1}^{p-1}(S(k))^{2}
\end{aligned}$$
(ii) Use problem 8; (iii) From (ii), we get $q(S(a))^{2}+q(S(b))^{2}=4 p q$, then use problem 9 (i). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,707 |
14. Let $p>5$ be an odd prime. Is the statement "The indeterminate equation $p=x^{2}+5 y^{2}$ has a solution if and only if $\left(\frac{-5}{p}\right)=1$" correct? Can the proof method of Theorem 4 in § 2 be used here? Why? | 14. Not applicable, the last part of the sufficiency proof may not hold here. For example, when $p=7$, $\left(\frac{-5}{7}\right)=1$, but $7=x^{2}+5 y^{2}$ has no solution. | not found | Number Theory | proof | Yes | Yes | number_theory | false | 740,710 |
15. Let $d, x_{1}, y_{1}, x_{2}, y_{2}$ be integers. Prove: There exist integers $x, y$ such that:
$$\left(x_{1}^{2}-d y_{1}^{2}\right)\left(x_{2}^{2}-d y_{2}^{2}\right)=x^{2}-d y^{2}$$ | 15. Let $(x-\sqrt{d} y)=\left(x_{1}-\sqrt{d} y_{1}\right)\left(x_{2}-\sqrt{d} y_{2}\right)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,711 |
Example 6 Find the greatest common divisor of 198 and 252, and express it as an integer linear combination of 198 and 252. | \begin{tabular}{|c|c|c|c|}
\hline $252=1 \cdot 198+54$ & & & $18=-198+4(252-198)$ \\
\hline $198=3 \cdot 54+36$ & & & $=4 \cdot 252-5 \cdot 198$ \\
\hline $54=1 \cdot 36+18$ & $\gamma$ & $\uparrow$ & $18=54-(198-3 \cdot 54)$ \\
\hline $36=2 \cdot 18$ & & & $=-198+4 \cdot 54$ \\
\hline & & & $18=54-36$ \\
\hline
\end{ta... | 18=4 \cdot 252-5 \cdot 198 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,714 |
18. Let $p$ be a prime, $p>3$. Prove: the indeterminate equation $x^{2}-x y+y^{2}=p$ has a solution if and only if $p \equiv 1(\bmod 6)$. | 18. The original equation can be written as $4 p=(2 x-y)^{2}+3 y^{2}$, and then use Theorem 4. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,715 |
4. Prove directly that for any positive integer $n$, the number of positive divisors of the form $4k+1$ is no less than the number of positive divisors of the form $4k+3$.
| 4. Let $n=2^{\sigma_{0}} p_{1}^{\alpha_{1}} \cdots p_{r_{r}^{r}}^{\alpha_{r}} q_{1}^{\beta_{1}} \cdots q_{3}^{\beta_{3}}, p_{i} \equiv 1(\bmod 4), q_{i} \equiv 3(\bmod 4) . n$ and $n^{\prime}=n / 2^{\alpha_{0}}$ have the same number of odd positive divisors. Use induction or direct comparison on the number of different... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,720 |
8. Let $1<n \equiv 1(\bmod 4)$ be a given positive integer. Prove that we can narrow down the range of non-negative solutions $x, y$ for $4 x^{2}+y^{2}=n$ using the following method: (i) $0 \leqslant x \leqslant \sqrt{n} / 2$; (ii) Choose an odd prime $q$ and a quadratic non-residue $a$ modulo $q$, such that $-4 x^{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 | 740,724 |
Example 7 Let $m, n$ be positive integers. Prove that
$$\left(2^{m}-1,2^{n}-1\right)=2^{(m, n)}-1$$ | Assume $m \geqslant n$. By the division algorithm, we have
$$m=q_{1} n+r_{1}, \quad 0 \leqslant r_{1}<n$$
We then have
$$2^{m}-1=2^{q_{1} n+r_{1}}-2^{r_{1}}+2^{r_{1}}-1=2^{r_{1}}\left(2^{q_{1} n}-1\right)+2^{r_{1}}-1 .$$
From this and $2^{n}-1 \mid 2^{q_{1} n}-1$, we get
$$\left(2^{m}-1,2^{n}-1\right)=\left(2^{n}-1,2... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,725 |
10. Let $N_{4}(n)$ denote the total number of solutions to $n=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$ (different orders and signs are considered as different solutions), and let $\sigma(n)$ denote the sum of the positive divisors of $n$, i.e., $\sigma(n) = \sum_{d \mid n} d$ (for example, $\sigma(6) = 1 + 2 + 3 + 6 = ... | 10. (i) For the terms where $s \neq t$, pair them as $\{s, t, x, y\},\left\{s^{\prime}, t^{\prime}, x^{\prime}, y^{\prime}\right\}$: Let
$$\begin{array}{c}
l=[t /(s-t)], \quad x^{\prime}=-l s+(l+1) t \\
y^{\prime}=(l+1) s-(l+2) t, \quad s^{\prime}=(l+2) x+(l+1) y
\end{array}$$
and $t^{\prime}=(l+1) x+l y$, then $h(s t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,727 |
Theorem 1 Let $a, b, c$ be non-zero integers, and the product $abc$ be square-free. Then the necessary and sufficient condition for the indefinite equation (1) to have a solution is: the signs of $a, b, c$ are not all the same, and $-bc, -ca, -ab$ are quadratic residues modulo $a, b, c$ respectively, i.e., the quadrati... | Theorem 1 Proof: From the condition that $abc$ has no square factors, it can be deduced that $a, b, c$ are pairwise coprime.
Necessity: Suppose the indeterminate equation (1) has a solution $x_{1}, y_{1}, z_{1}$. It is evident that $a, b, c$ cannot have the same sign. Without loss of generality, assume $\left(x_{1}, y_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,728 |
Lemma 2 Let $m$ be a positive integer, $\alpha, \beta, \gamma$ be positive real numbers satisfying $\alpha \beta \gamma=m$. Then, for any integers $d, e, f$, the congruence equation
$$d x+e y+f z \equiv 0(\bmod m)$$
must have a non-zero solution $x_{1}, y_{1}, z_{1}$, satisfying
$$\left|x_{1}\right| \leqslant \alpha, ... | Consider the set of integers:
$$d u+e v+f w, \quad 0 \leqslant u \leqslant \alpha, 0 \leqslant v \leqslant \beta, 0 \leqslant w \leqslant \gamma$$
The number of elements in this set is equal to
$$(1+[\alpha])(1+[\beta])(1+[\gamma])>\alpha \beta \gamma=m$$
Therefore, there must be two different sets $\left\{u_{1}, v_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,729 |
Lemma 3 Let $\left(m_{1}, m_{2}\right)=1$. If
$$\begin{array}{l}
a x^{2}+b y^{2}+c z^{2} \\
\quad \equiv\left(d_{1} x+e_{1} y+f_{1} z\right)\left(d_{1}^{\prime} x+e_{1}^{\prime} y+f_{1}^{\prime} z\right)\left(\bmod m_{1}\right), \\
a x^{2}+b y^{2}+c z^{2} \\
\quad \equiv\left(d_{2} x+e_{2} y+f_{2} z\right)\left(d_{2}^{... | By the Chinese Remainder Theorem (Theorem 1, Section 3, Chapter 4), we know that there must exist $d, e, f, d^{\prime}, e^{\prime}, f^{\prime}$ satisfying
$$\begin{array}{l}
d \equiv d_{j}\left(\bmod m_{j}\right), e \equiv e_{j}\left(\bmod m_{j}\right), f \equiv f_{j}\left(\bmod m_{j}\right), j=1,2 \\
d^{\prime} \equiv... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,730 |
Example 1 Determine whether the following indeterminate equations have non-zero solutions, and find a set of solutions if they exist:
(i) $5 x^{2}-14 y^{2}-41 z^{2}=0$;
(ii) $3 x^{2}-14 y^{2}-19 z^{2}=0$. | (i) Here $a=5, b=-14, c=-41$. By Theorem 1, we need to verify whether the congruence equations (2), (3), (4) have solutions. We have
$$s^{2} \equiv-(-14) \cdot(-41) \equiv 1(\bmod 5),$$
so (2) has a solution; $\square$
$$s^{2} \equiv-(-41) \cdot(5) \equiv-5 \equiv 3^{2}(\bmod 14),$$
so (3) has a solution; Finally, we... | x_{1}=-5, y_{1}=2, z_{1}=-1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,731 |
1. Determine whether the following indeterminate equations have non-zero solutions. If solvable, find a set of solutions according to the method of Theorem 1:
(i) $3 x^{2}+7 y^{2}-5 z^{2}=0$;
(ii) $5 x^{2}-17 y^{2}+3 z^{2}=0$;
(iii) $19 x^{2}-7 y^{2}-11 z^{2}=0$;
(iv) $7 x^{2}-43 y^{2}-3 z^{2}=0$;
(v) $4 x^{2}+3 y^{2}-... | 1. (i) no solution; (ii) has solution, $x=1, y=1, z=2$; (iii) no solution; (iv) has solution, $x=5, y=$ $2, z=1$; (v) no solution; (vi) has solution, $x=y=1, z=2$. | (i) \text{no solution}; (ii) x=1, y=1, z=2; (iii) \text{no solution}; (iv) x=5, y=2, z=1; (v) \text{no solution}; (vi) x=y=1, z=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,732 |
3. Prove the proposition in question 2 by the following means:
(i) The proposition holds when $a=1$.
(ii) The proposition holds when $a=b$.
Hereafter, assume $a>b>1$.
(iii) There exist integers $T, c, |c| \leqslant a / 2$, such that
$$c^{2}-b=a T=a A m^{2}, \quad A, m \in \boldsymbol{Z},$$
where $A$ is a square-free ... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 740,734 |
Theorem 1 The indeterminate equation
$$x^{3}+y^{3}=z^{3}$$
has no solution for $x y z \neq 0$. | Theorem 1 is proved by contradiction. Suppose there is a solution $x y z \neq 0$. Let $x_{0}, y_{0}, z_{0}$ be such a solution, making $\left|x_{0} y_{0} z_{0}\right|$ the smallest. We will prove that in this case, there must be another solution $x_{1}, y_{1}, z_{1}$, such that $01$ (it has been pointed out that $\left... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,737 |
Lemma 2 If $2 \nmid s$, then
$$s^{3}=a^{2}+3 b^{2}, \quad(a, b)=1$$
is satisfied if and only if there exist $\alpha, \beta$, such that
$$s=\alpha^{2}+3 \beta^{2}, \quad(\alpha, 3 \beta)=1$$
and
$$a=\alpha^{3}-9 \alpha \beta^{2}, \quad b=3 \alpha^{2} \beta-3 \beta^{3} .$$ | Proof of Sufficiency: Suppose equations (3) and (4) hold. To make the derivation clear, we use complex number operations of the form \(x + y \sqrt{-3}\). From equation (3), we have
\[ s = (\alpha + \sqrt{-3} \beta)(\alpha - \sqrt{-3} \beta), \]
thus
\[ \begin{aligned}
s^{3} &= (\alpha + \sqrt{-3} \beta)^{3}(\alpha - \s... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,738 |
Example 1 If you don't have a square root table, nor a calculator, what simple method can be used to find the approximate value of $\sqrt{11}$? Of course, you can use the usual method for finding square roots, but the method below seems more convenient. | Repeating this process yields
One would immediately think of removing the irrational numbers $(\sqrt{11}-3)$ or $(\sqrt{11}-3) / 2$ from equations (2), (3), (4), (5), (6), (7), (8) to obtain the "fractional" values as approximations of $\sqrt{11}$. It is easy to calculate that these "approximations" are sequentially
$... | \sqrt{11}=3.31662479 \cdots | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,739 |
Example 3 So far, apart from calculating specific values, we have no method to solve the indeterminate equation
$$x^{2}-11 y^{2}=1, \quad x>0, y>0$$ | The indeterminate equation can be transformed into
$$\begin{array}{l}
x-\sqrt{11} y=\frac{1}{x+\sqrt{11} y}, \quad x>0, y>0 \\
\frac{x}{y}-\sqrt{11}=\frac{1}{y(x+\sqrt{11} y)}, \quad x>0, y>0
\end{array}$$
This indicates that the solutions \(x, y\) of the indeterminate equation (14) provide a very precise approximatio... | x=10, y=3 ; \quad x=199, y=60 ; \quad x=3970, y=1197 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,741 |
Theorem 2 Let $x_{0}, x_{1}, x_{2}, \cdots$ be an infinite sequence of real numbers, with $x_{j}>0, j \geqslant 1$; and let
$$P_{-2}=0, \quad P_{-1}=1, \quad Q_{-2}=1, \quad Q_{-1}=0,$$
and when $n \geqslant 0$, $P_{n}, Q_{n}$ are given by the recurrence relations (37). Then
$$\begin{array}{l}
\left\langle x_{0}, \cdo... | Proof: When $n=0$, $P_{0}=x_{0}, Q_{0}=1$, so equation (39) holds. Assume that when $n=k$ $(\geqslant 0)$, equation (39) holds. When $n=k+1$, from equation (25) we get
$$\left\langle x_{0}, \cdots, x_{k-1}, x_{k}, x_{k+1}\right\rangle=\left\langle x_{0}, \cdots, x_{k-1}, x_{k}+1 / x_{k+1}\right\rangle$$
By the assumpt... | proof | Algebra | proof | Yes | Yes | number_theory | false | 740,743 |
Example 4 Find the value of the finite continued fraction $\langle-2,1,2 / 3,2,1 / 2,3\rangle$. | We use equation (25) to calculate.
$$\begin{aligned}
\langle-2,1 & , 2 / 3,2,1 / 2,3\rangle=\langle-2,1,2 / 3,2,1 / 2+1 / 3\rangle \\
& =\langle-2,1,2 / 3,2,5 / 6\rangle=\langle-2,1,2 / 3,2+6 / 5\rangle \\
& =\langle-2,1,2 / 3,16 / 5\rangle=\langle-2,1,2 / 3+5 / 16\rangle \\
& =\langle-2,1,47 / 48\rangle=\langle-2,1+48... | -\frac{143}{95} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,744 |
Example 5 Find the successive convergents of the finite simple continued fraction $\langle 1,1,1,1,1,1,1,1,1,1\rangle$. | Sure, here is the translated text with the original formatting preserved:
```
Solution Of course, we can calculate one by one using the method in Example 1, but at this point, it is more convenient to use Theorem 2 to recursively calculate $P_{n}, Q_{n}$. According to formulas (38), (37), the following table can be li... | \frac{89}{55} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,745 |
1. Compute the value and the successive convergents of the following finite continued fractions:
(i) $\langle 1,2,3\rangle$;
(ii) $\langle 0,1,2,3\rangle$;
(iii) $\langle 3,2,1\rangle$;
(iv) $\langle 2,1,1,4,1,1\rangle$;
(v) $\langle-4,2,1,7,8\rangle$;
(vi) $\langle-1,1 / 2,1 / 3\rangle$;
(vii) $\langle 1 / 2,1 / 4,1 /... | 1. (i) $10 / 7$; Nearest fractions are: $1,3 / 2,10 / 7$.
(ii) $7 / 10$; Nearest fractions are: $0,1,2 / 3,7 / 10$.
(iii) $10 / 3$; Nearest fractions are: $3,7 / 2,10 / 3$.
(iv) $51 / 20$; Nearest fractions are: $2,3,5 / 2,23 / 9,28 / 11,51 / 20$.
(v) $-683 / 187$; Nearest fractions are: $-4,-7 / 2,-11 / 3,-84 / 23,-68... | \frac{1193}{322} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,746 |
2. Let $a>0$. Prove: Among any $a$ consecutive integers, there is exactly one that is divisible by $a$.
| 2. Let the $a$ consecutive integers be: $m, m+1, \cdots, m+a-1$, and their smallest non-negative remainders when divided by $a$ are: $r_{0}, r_{1}, \cdots, r_{a-1}, 0 \leqslant r_{j}<a$. Thus, $r_{j}=r_{0}+j$, when $r_{0}+j<a ; r_{j}=r_{0}+j-$ $a$, when $r_{0}+j \geqslant a$. From this, it follows that either only $a \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,748 |
2. 把下面的有理数表为有限简单连分数, 并求各个渐近分数:
(i) $121 / 21$;
(ii) $-19 / 29$;
(iii) $177 / 292$;
(iv) $873 / 4867$. | 2. (i) $\langle 5,1,3,5\rangle$;
(ii) $\langle-1,2,1,9\rangle$;
(iii) $\langle 0,1,1,1,1,5,1,8\rangle$;
(iv) $\langle 0,5,1,1,2,1,4,1,21\rangle$. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,749 |
3. Find the successive convergents and their values of the finite simple continued fraction $\langle 2,1,2,1,1,4,1,1,6,1,1,8\rangle$, and compare them with the value of the base of the natural logarithm $e$. | 3. $2,3,2+2 / 3,2+3 / 4,2+5 / 7,2+23 / 32,2+28 / 39,2+51 / 71,2+334 / 465$, $2+385 / 536,2+719 / 1001,2+3799 / 5289 \approx 2.718283229 . \mathrm{e}$ The approximate value of e is: $2.718281828 \cdots$. This continued fraction is an approximant of the infinite simple continued fraction expansion of e. | 2.718283229 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,750 |
11. Prove: When $n \geqslant 1$,
$$\left(\begin{array}{ll}
P_{n} & P_{n-1} \\
Q_{n} & Q_{n-1}
\end{array}\right)=\left(\begin{array}{cc}
x_{0} & 1 \\
1 & 0
\end{array}\right)\left(\begin{array}{cc}
x_{1} & 1 \\
1 & 0
\end{array}\right) \cdots\left(\begin{array}{cc}
x_{n} & 1 \\
1 & 0
\end{array}\right) .$$ | 11. Using formulas (37), (38), prove by induction, where $P_{n}, Q_{n}$ are both considered as polynomials. | proof | Algebra | proof | Yes | Yes | number_theory | false | 740,758 |
Theorem 2 Let $\left\langle a_{0}, \cdots, a_{n}\right\rangle,\left\langle b_{0}, \cdots, b_{s}\right\rangle$ be two finite simple continued fractions, $a_{n}>1, b_{s}>1$. If
$$\left\langle a_{0}, \cdots, a_{n}\right\rangle=\left\langle b_{0}, \cdots, b_{s}\right\rangle$$
then it must be that $s=n, a_{j}=b_{j}, 0 \leq... | Proof: Without loss of generality, assume $s \geqslant n$. We use induction on $n$. When $n=0$, if $s \geqslant 1$, then by formula (24) in §1, we have
$$a_{0}=\left\langle b_{0}, b_{1}, \cdots, b_{s}\right\rangle=\left\langle b_{0},\left\langle b_{1}, \cdots, b_{s}\right\rangle\right\rangle=b_{0}+1 /\left\langle b_{1}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,762 |
Example 1 Find the finite simple continued fraction of $13 / 5$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | We solve according to formula (12).
$$\begin{aligned}
13 / 5 & =\langle 2+3 / 5\rangle=\langle 2,5 / 3\rangle=\langle 2,\langle 1+2 / 3\rangle\rangle \\
& =\langle 2,\langle 1,3 / 2\rangle\rangle=\langle 2,1,1+1 / 2\rangle \\
& =\langle 2,1,1,2\rangle .
\end{aligned}$$
The characteristic of this finite simple continue... | 13 / 5=\langle 2,1,1,2\rangle | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,764 |
Example 2 Find the finite simple continued fraction of $7700 / 2145$ and its successive convergents. | Solve
$$\begin{aligned}
7700 & / 2145=\langle 3+1265 / 2145\rangle=\langle 3,2145 / 1265\rangle \\
& =\langle 3,1+880 / 1265\rangle=\langle 3,1,1265 / 880\rangle \\
= & \langle 3,1,1+385 / 880\rangle=\langle 3,1,1,880 / 385\rangle \\
= & \langle 3,1,1,2+110 / 385\rangle=\langle 3,1,1,2,385 / 110\rangle \\
= & \langle 3... | 140 / 39 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,765 |
1. Let $a / b$ be a rational fraction, $\left\langle a_{0}, \cdots, a_{n}\right\rangle$ be its finite simple continued fraction, $b \geqslant 1$. Prove:
$$a k_{n-1}-b h_{n-1}=(-1)^{n+1}(a, b)$$ | 1. Derive from equations (4) and (5).
Translate the above text into English, preserving the original text's line breaks and format, and output the translation result directly.
However, it seems there's a misunderstanding in your request. The text you provided is already in English. If you meant to provide a Chinese ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,766 |
2. Specifically explain that the method given in question 1 provides a new way to find the greatest common divisor $(a, b)$ and solve the linear Diophantine equation $a x+b y=c$. Use this method to solve the following greatest common divisors and linear Diophantine equations: $\square$
(i) $205 x+93 y=1$;
(ii) $65 x-56... | 2. (i) $205 / 93=\langle 2,93 / 19\rangle=\langle 2,4,19 / 17\rangle=\langle 2,4,1,17 / 2\rangle=\langle 2,4,1,8,2\rangle$. $\langle 2,4,1,8\rangle=\langle 2,4,9 / 8\rangle=\langle 2,44 / 9\rangle=97 / 44$. From the first question, we know that $205 \cdot 44 - 93 \cdot 97 = -1 = -(205,93)$, so the solution is $x = -44 ... | x = -44 + 93t, y = 97 - 205t, t = 0, \pm 1, \pm 2, \cdots | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,767 |
3. 求有理分数 (i) $7 / 11$, (ii) $173 / 55$, (iii) $-43 / 1001$, (iv) $5391 / 3976$, (v) $-873 / 4867$ 的两种有限简单连分数表示式, 以及它们的各个渐近分数、渐近分数与有理分数的误差. | 3. (i) $\langle 0,1,1,1,3\rangle=\langle 0,1,1,1,2,1\rangle$;
(ii) $\langle 3,6,1,7\rangle=\langle 3,6,1,6,1\rangle$;
(iii) $\langle-1,1,22,3,1,1,2,2\rangle=\langle-1,1,22,3,1,1,2,1,1\rangle$;
(iv) $\langle 1,2,1,4,3,1,5,2,1,3\rangle=\langle 1,2,1,4,3,1,5,2,1,2,1\rangle$;
(v) $\langle-1,1,3,1,1,2,1,4,1,21\rangle=\langl... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,768 |
4. Let $r_{0}=\left\langle a_{0}, \cdots, a_{n}\right\rangle, s_{0}=\left\langle b_{0}, \cdots, b_{n}, b_{n+1}\right\rangle$ be two finite simple continued fractions. Try to find: (i) the necessary and sufficient condition for $r_{0}=s_{0}$; (ii) the necessary and sufficient condition for $r_{0}<s_{0}$. | 4. (i) $a_{i}=b_{i}, 0 \leqslant i \leqslant n-1, a_{n}=b_{n}+1, b_{n+1}=1$;
(ii) The necessary and sufficient condition is: (I) $a_{i}=b_{i}, 0 \leqslant i \leqslant n, 2 \mid n$; or (II) there exists $r, 0 \leqslant r \leqslant n$, such that $a_{i}=b_{i}, 0 \leqslant i < r, a_{r} > b_{r}$; (b) $r \leqslant n$, $2 \mi... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,769 |
5. Let the rational fraction $a / b((a, b)=1, a \geqslant b \geqslant 1)$ have the finite simple continued fraction $\left\langle a_{0}, a_{1}, \cdots, a_{n}\right\rangle$. Prove:
$$\left\langle a_{0}, a_{1}, \cdots, a_{n-1}, a_{n}\right\rangle=\left\langle a_{n}, a_{n-1}, \cdots, a_{1}, a_{0}\right\rangle$$
if and on... | 5. Using the result from Question 1 and Exercise 1, Question 6 (ii), we can obtain: if the conclusion holds, then $a k_{n-1} = b^2 + (-1)^{n+1}$. This establishes the necessity. When condition (i) or (ii) holds, from Question 1 we can deduce: $a \mid b - h_{n-1}$, i.e., $h_{n} \mid k_{n} - h_{n-1}$. This leads to $b = ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,771 |
Theorem 1 An infinite simple continued fraction $\left\langle a_{0}, a_{1}, a_{2}, \cdots\right\rangle$ is certainly convergent, that is, if $r^{(n)}=\left\langle a_{0}, \cdots, a_{n}\right\rangle$ is its $n$-th convergent, then there must exist a limit
$$\lim _{n \rightarrow \infty} r^{(n)}=\theta$$
Moreover,
$$r^{(0)... | Proof: From Theorem 1 of §1, equations (31), (32), and (33), we know:
(i) The sequence of rational numbers \( r^{(0)}, r^{(2)}, \cdots, r^{(2 t)}, \cdots \) is strictly increasing and has an upper bound \( r^{(1)} = a_{0} + 1 / a_{1} \). Therefore, it must have a limit:
\[ \lim _{t \rightarrow \infty} r^{(2 t)} = \the... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,774 |
Theorem 2 Let $\left\langle a_{0}, a_{1}, a_{2}, \cdots\right\rangle$ be an infinite simple continued fraction, and let
$$\theta_{n}=\left\langle a_{n}, a_{n+1}, \cdots\right\rangle, \quad n \geqslant 0$$
Then we have
$$a_{n}=\left[\theta_{n}\right], \quad \theta_{n+1}=\left\{\theta_{n}\right\}^{-1}, \quad n \geqslant... | By Theorem 1, we know that all infinite simple continued fractions $\left\langle a_{n}, a_{n+1}, \cdots\right\rangle(n \geqslant 0)$ are convergent. From Theorem 1 and equation (24) in §1, we have:
$$\begin{aligned}
\theta_{0} & =\lim _{n \rightarrow \infty}\left\langle a_{0}, a_{1}, \cdots, a_{n}\right\rangle \\
& =\l... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,775 |
Theorem 4 Let $\xi_{0}$ be an irrational number; and let
$$\left\{\begin{array}{ll}
a_{0}=\left[\xi_{0}\right], & \xi_{1}=1 /\left\{\xi_{0}\right\}, \\
a_{j}=\left[\xi_{j}\right], & \xi_{j+1}=1 /\left\{\xi_{j}\right\}, \quad j \geqslant 1 .
\end{array}\right.$$
Then, we have $a_{j} \geqslant 1, j \geqslant 1$ and
$$\x... | Proof: Since $\xi_{0}$ is an irrational number, $\xi_{j} (j \geqslant 1)$ are all irrational numbers. Therefore, $01, a_{j} \geqslant 1, j \geqslant 1$. By Theorem 1, the infinite simple continued fraction on the right side of equation (12) is convergent. Let
$$\theta_{0}=\left\langle a_{0}, a_{1}, a_{2}, \cdots\right\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,777 |
Example 3 Find the complete quotients of the continued fraction in Example 2.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve $\xi_{0}=\theta, \xi_{2}=\theta^{\prime}$, we only need to find $\xi_{1}=\left\langle 3, \theta^{\prime}\right\rangle, \xi_{3}=\left\langle 2,4, \theta^{\prime}\right\rangle, \xi_{4}=$ $\left\langle 4, \theta^{\prime}\right\rangle$, because for $n \geqslant 5$, $\xi_{n}=\xi_{n-3}$.
$$\begin{aligned}
\xi_{1} & =\l... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,780 |
Theorem 7 Let $\xi_{0}$ be an irrational number. If there is a rational fraction $a / b, b \geqslant 1$, such that
$$\left|\xi_{0}-a / b\right|<1 /\left(2 b^{2}\right),$$
then $a / b$ must be one of the convergents of $\xi_{0}$. | Solving according to formula (13) we get
$$\begin{aligned}
\sqrt{8} & =2+(\sqrt{8}-2)=2+4 /(\sqrt{8}+2) \\
& =\langle 2,(\sqrt{8}+2) / 4\rangle \\
& =\langle 2,1+(\sqrt{8}-2) / 4\rangle=\langle 2,1, \sqrt{8}+2\rangle \\
& =\langle 2,1,4+(\sqrt{8}-2)\rangle \\
& =\langle 2,1,4,(\sqrt{8}+2) / 4\rangle
\end{aligned}$$
Th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,783 |
Example 5 Find the first six digits of the infinite simple continued fraction for $\sqrt[3]{2}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | By Theorem 4, let $\xi_{0}=\sqrt[3]{2}$.
Therefore,
$$\sqrt[3]{2}=\left\langle 1,3,1,5,1,1, \xi_{6}\right\rangle .$$
$$\begin{array}{l}
a_{0}=\left[\xi_{0}\right]=1, \quad \xi_{1}=\left(\xi_{0}-a_{0}\right)^{-1}=(\sqrt[3]{2}-1)^{-1} \\
=\sqrt[3]{4}+\sqrt[3]{2}+1, \\
a_{1}=\left[\xi_{1}\right]=3, \quad \xi_{2}=\left(\xi... | \sqrt[3]{2}=\left\langle 1,3,1,5,1,1, \xi_{6}\right\rangle | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,784 |
Theorem 5 For $n \geqslant 0$, we have
$$\frac{1}{k_{n}\left(k_{n}+k_{n+1}\right)}<\left|\xi_{0}-r^{(n)}\right|=\left|\xi_{0}-\frac{h_{n}}{k_{n}}\right|<\frac{1}{k_{n} k_{n+1}} .$$ | Proof: From equation (13) and equation (39) of Theorem 2 in §1, we have
$$\xi_{0}=\left\langle a_{0}, \cdots, a_{n}, \xi_{n+1}\right\rangle=\frac{h_{n} \xi_{n+1}+h_{n-1}}{k_{n} \xi_{n+1}+k_{n-1}}, \quad n \geqslant 0 .$$
From this and Theorem 1 in §2, we get
$$\begin{aligned}
\xi_{0}-r^{(n)} & =\xi_{0}-\frac{h_{n}}{k_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,785 |
Theorem 8 Let $d>1$ not be a square number. If the indefinite equation
$$x^{2}-d y^{2}= \pm 1$$
has a solution $x=x_{0}>0, y=y_{0}>0$, then $x_{0} / y_{0}$ must be some convergent fraction $h_{n} / k_{n}$ of $\sqrt{d}$, and $x_{0}=h_{n}, y_{0}=k_{n}$. | Prove that in this case, there must be $x_{0} \geqslant y_{0}$. If not, from $y_{0}>x_{0}$ we can deduce
$$\pm 1=x_{0}^{2}-d y_{0}^{2}<y_{0}^{2}-d y_{0}^{2} \leqslant-y_{0}^{2} \leqslant-1$$
This is impossible. Given that $x_{0}, y_{0}$ are solutions and $x_{0} \geqslant y_{0}$, we can obtain
$$\left|x_{0} / y_{0}-\sq... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,787 |
Example 6 Find the simplest fraction (with the smallest denominator) that approximates $\sqrt{8}$, with an error $\leqslant 10^{-6}$. | In Example 4, we have found $\sqrt{8}=\langle 2,1,4,1,4,1,4, \cdots\rangle$. We first list to find $h_{n}$, $k_{n}$, and then estimate the error according to formula (16).
\begin{tabular}{|c|rrrrrrrrrr|}
\hline$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline$a_{n}$ & 2 & 1 & 4 & 1 & 4 & 1 & 4 & 1 & 4 & 1 \\
$h_{n}... | \frac{2786}{985} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,788 |
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