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7. Let $n$ be a positive integer, $n \mid a b, n \nmid a, n \nmid b$. Let $a=d(a, a b / n)$. Prove: $d \mid n, 1<d<n$. Explain the significance of this problem. | 7. $a=(d a, d a b / n), a \mid d a b / n$, so $n \mid d b$. Therefore, we have
$a=(d a, a(d b) / n)=a(d, d b / n) \quad(d, d b / n)=1$, which means $d(n, b)=n$. From $n \nmid b, n \mid d b$ we deduce $d>1$. From $n \mid a b, n \nmid a$ we deduce $(n, b)>1$, and from this and $d(n, b)=n$ we deduce $d<n$. This conclusion... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,923 |
8. Let $(a, b)=1$. Prove:
(i) $(d, a b)=(d, a)(d, b)$;
(ii) $d$ is a positive divisor of $a b$ if and only if $d$ can be expressed as $d_{1} d_{2}$, where $d_{1}$ is a positive divisor of $a$, $d_{2}$ is a positive divisor of $b$, and this representation is unique. | 8. (i) $(d, a b)=(d, a)(d /(d, a), b a /(d, a))=(d, a)(d /(d, a), b)=$ $(d, a)(d, b)$; (ii) Using (i). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,924 |
10. Let $a, b, c$ be positive integers. Prove:
(i)
$$\begin{array}{l}
{[a, b, c](a b, b c, c a)=(a, b, c)[a b, b c, c a]} \\
\quad=(a, b, c)[a, b, c][(a, b),(b, c),(c, a)]=a b c
\end{array}$$
(ii) $[a, b, c]=a b c$ if and only if $(a, b)=(b, c)=(c, a)=1$. | 10. (i) $[a, b, c]=[[a, b], c]=[a b /(a, b), c]=a b c /(a b,(a, b) c)$. (ii) From the condition, we get $(a b, b c, c a)=1$. Use $(\mathrm{i})$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,926 |
Example 2 Let $a=2 t-1$. (i) If $a \mid 2 n$, then $a \mid n$; (ii) If $2 \mid a b$, then $2 \mid b$. | Prove that from $a \mid 2 t n$ and $2 t n = a n + n$, we get $a \mid (2 t n - a n)$, which means $a \mid n$. This proves (i). Since $a b = 2 t b - b$, $b = 2 t b - a b$, so $2 \mid b$. This proves (ii). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,927 |
12. 证明: $(a, b, c)(a b, b c, c a)=(a, b)(b, c)(c, a)$. | 12. $\begin{aligned}(a, b)(b, c)(c, a) & =(a(a, b)(b, c), b(b, c)(c, a), c(c, a)(a, b)) \\ & =(a, b, c)(a b, b c, c a) .\end{aligned}$ | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,929 |
13. 证明: $(a,[b, c])=[(a, b),(a, c)]$. | $$\text { 13. } \begin{array}{l}
{[(a, b),(a, c)] }=(a, b)(a, c) /((a, b),(a, c)) \\
=(a, b)(a, c) /(a, b, c) . \\
(a,[b, c])=(a, b c /(b, c))=(a b, b c, c a) /(b, c) .
\end{array}$$
再利用上题. | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,930 |
14. Prove: $[a,(b, c)]=([a, b],[a, c])$. | 14. $[a,(b, c)]=a(b, c) /(a, b, c)$. $([a, b],[a, c])=(a b /(a, b), a c /(a, c))=$ $a(a b, b c, c a) /(a, b, c)$. Then use the result from problem 12. | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,931 |
15. Prove: (i) $([a, b],[b, c],[c, a])=[(a, b),(b, c),(c, a)]$;
(ii) $(a, b)(b, c)(c, a)[a, b, c]^{2}=[a, b][b, c][c, a](a, b, c)^{2}$. | 15. (i) Using questions 13 and 14. $([a, b],[b, c],[c, a])=[([a, b],[b, c], c),([a, b]$, $[b, c], a)]=[(c,[a, b]),(a,[b, c])]=[(a, b),(b, c),(c, a)]$. (ii) Using question 10 (i) and question 12. | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,932 |
18. Let $p$ be a prime, $p \nmid a$. Prove: for any positive integer $k$, we have
(i) $\varphi\left(p^{k}\right)=(p-1) p^{k-1}, \varphi(n)$ is given by Exercise 18 of the second part of Exercise 2 in §2; $\square$
(ii) $p^{k} \mid a^{\varphi\left(p^{k}\right)}-1$. | 18. (i) $\left(a, p^{k}\right)=1 \Longleftrightarrow p \nmid a .1 \leqslant a \leqslant p^{k}$ contains exactly $p^{k-1}$ numbers $a$ that are divisible by $p$; (ii) Use Example 1 and (i). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,935 |
Example 3 Let $a, b$ be two given non-zero integers, and there exist integers $x, y$, such that $a x + b y = 1$. Prove: (i) If $a \mid n$ and $b \mid n$, then $a b \mid n$. (ii) If $a \mid b n$, then $a \mid n$. | Given $n=n(a x+b y)=(n a) x+(n b) y$, and $a b|n a, a b| n b$, it follows that $a b \mid n$, which proves (i). Note that $7 \cdot 1+3 \cdot(-2)=1$, which also proves Example 1. By $n=(1-a x) n, n=b y n+a x n$ it follows that $a \mid n$. This proves (ii). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,938 |
21. Let $m>1$. Prove: $m \nmid 2^{m}-1$.
| 21. It is evident that we only need to consider $2 \nmid m$. Let the smallest prime factor of $m$ be $p$, which must have $p>2$. Therefore, $p \mid 2^{m}-1$, $p \mid 2^{p-1}-1$, and consequently, $p \mid 2^{(m, p-1)}-1$. Since $p$ is the smallest prime factor of $m$, we have $(m, p-1)=1$, which implies $p \mid 1$, a co... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,939 |
22. Let $f(x)=a_{n} x^{n}+\cdots+a_{0}, g(x)=b_{m} x^{m}+\cdots+b_{0}$. Prove:
(i) If they are polynomials with integer coefficients and $h(x)=f(x) g(x)=c_{m+n} x^{m+n}+\cdots+c_{0}$, then
$$\left(a_{n}, \cdots, a_{0}\right)\left(b_{m}, \cdots, b_{0}\right)=\left(c_{m+n}, \cdots, c_{0}\right)$$
In particular, when $\l... | 22. (i) Without loss of generality, let $\left(a_{n}, \cdots, a_{0}\right)=\left(b_{m}, \cdots, b_{0}\right)=1$. Use proof by contradiction. If $d=\left(c_{m+n}, \cdots, c_{0}\right)>1$, then there exists a prime $p \mid d$. Let $i_{0}, j_{0}$ be the largest indices such that $p \nmid a_{i}, p \nmid b_{j}$, i.e., $p\le... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,940 |
22. Let $a, b, m$ be positive integers, $(a, b)=1$. Prove: In the arithmetic sequence
$$a+k b \quad(k=0,1,2, \cdots)$$
there are infinitely many numbers that are coprime to $m$. | 23. Let $c$ be the greatest positive divisor of $m$ such that $(c, a)=1$, prove that $(a+b c, m)=1$. Let $d=$ $(a+b c, m)$, from $(a, b c)=1, d \mid a+b c$ it follows that $(d, a)=(d, b c)=1$, from this and $d|m, c| m$ we get $d c \mid m$. Since $(a, d c)=1$, it follows from the maximality of $c$ that $d=1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,941 |
24. Let $a>b>0, n>1$. Prove: $a^{n}-b^{n} \nmid a^{n}+b^{n}$. | 24. Suppose $(a, b)=1, a>b$. If $a^{n}-b^{n} \mid a^{n}+b^{n}$, then $a^{n}-b^{n} \mid 2$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,942 |
25. Let $a>b \geqslant 1, n>1$. Prove:
$$\left(\left(a^{n}-b^{n}\right) /(a-b), a-b\right)=\left(n(a, b)^{n-1}, a-b\right) .$$ | 25. $\left(a^{n}-b^{n}\right) /(a-b)=n a^{n-1}+A(a-b)=n b^{n-1}+B(a-b), A, B$ are two integers.
So $\left(\left(a^{n}-b^{n}\right) /(a-b), a-b\right)=\left(n a^{n-1}, a-b\right)=\left(n b^{n-1}, a-b\right)$
$$=\left(n a^{n-1}, n b^{n-1}, a-b\right)=\left(n(a, b)^{n-1}, a-b\right) .$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,943 |
26. (i) If $n \mid 2^{n}-2$, is $n$ necessarily a prime number? Consider $n=341$. Numbers with this property are called pseudoprimes.
(ii) If $n \mid 2^{n}-2$, then $m \mid 2^{m}-2$, where $m=2^{n}-1$.
(iii) Let $n=161038$, verify that $n \mid 2^{n}-2$. | 26. (i) Not necessarily. It can be directly verified that $341=11 \cdot 31=n$ satisfies.
(ii) Let $2^{n}-2=n k .2^{2^{n}-1}-2=2\left(2^{n k}-1\right)=2 A\left(2^{n}-1\right)$.
(iii) $161038=2 \cdot 73 \cdot 1103,161038-1=3^{2} \cdot 29 \cdot 617,73 \mid 2^{9}-1$, $1103 \mid 2^{29}-1$ | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,944 |
27. (i) A composite number $n$ is called an absolute pseudoprime if for any integer $a$ it must be that $n \mid a^{n}-a$. Prove: 561 is an absolute pseudoprime, but 341 is not.
(ii) Let $m \geqslant 1$. If $q_{1}=6 m+1, q_{2}=12 m+1, q_{3}=18 m+1$ are all primes, then $n=q_{1} q_{2} q_{3}$ is an absolute pseudoprime. G... | 27. (i) $31 \nmid 11^{301}-1$.
(ii) Let $n=q_{1} \cdots q_{k}, q_{1}, \cdots, q_{k}$ be distinct primes. If $q_{i}-1 \mid n-1,1 \leqslant i \leqslant k$, then $n$ is an absolute pseudoprime. This implies that $561=3 \cdot 11 \cdot 17$ is an absolute pseudoprime and (ii). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,945 |
28. Prove: There exist infinitely many $n$ such that $n \mid 2^{n}+1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 28. Prove by induction that $3^{k} \mid 2^{3^{k}}+1, k=1,2, \cdots$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,946 |
29. Prove: (i) If $n\left|2^{n}+2, n-1\right| 2^{n}+1$, then $m \mid 2^{m}+2$, $m-1 \mid 2^{m}+1$, where $m=2^{n}+2$;
(ii) There are infinitely many $n$ such that $n \mid 2^{n}+2$. | 29. (i) $m\left|2^{m}+2 \Longleftrightarrow 2^{n-1}+1\right| 2^{2^{n}+1}+1=2^{k(n-1)}+1, k$ is odd.
$$m-1\left|2^{m}+1 \Longleftrightarrow 2^{n}+1\right| 2^{2^{n}+2}+1=2^{n n}+1,$$
Since $n$ must be even, $h$ is odd.
(ii) $n=2$ satisfies both conditions in (i). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,947 |
30. Prove: There exist infinitely many composite numbers $n$, such that for any integer $a$ we have
$$n \mid a^{n-1}-a$$ | 30. $n=2 p, p$ is an odd prime, all satisfy.
The text has been translated while preserving the original line breaks and format. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,948 |
Example 4 Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} \in \boldsymbol{Z}[x]$, where $\boldsymbol{Z}[x]$ denotes the set of all monic polynomials with integer coefficients. If $d \mid b-c$, then
$$d \mid f(b)-f(c) \text {. }$$
In particular, we have
$$b-c \mid f(b)-f(c) .$$ | Prove that we have
$$\begin{aligned}
f(b)-f(c)= & a_{n}\left(b^{n}-c^{n}\right)+a_{n-1}\left(b^{n-1}-c^{n-1}\right) \\
& +\cdots+a_{1}(b-c),
\end{aligned}$$
From this and $(b-c) \mid b^{j}-c^{j}$, we can deduce the desired conclusion.
By definition, for an integer $a \neq 0$, its multiples are
$$q a, \quad q=0, \pm 1,... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,949 |
31. Let $m \geqslant 2, (a, m)=1$. Suppose there exists a positive integer $d$ such that $m \mid a^{d}+1$, and denote the smallest such $d$ by $\delta_{m}^{-}(a)$. Additionally, denote $\delta_{m}(a)$ from Examples 1, 4, and 5 as $\delta_{m}^{+}(a)$. Prove:
(i) If $m \mid a^{h}-1$ or $m \mid a^{h}+1$, then $\delta_{m}^... | 31. (i) follows from (ii), (iii), and (iv). (iii) First, prove by contradiction: $\delta_{m}^{+}(a)>\delta_{m}^{-}(a)$. From Example 5, we get $\delta_{m}^{+}(a) \mid 2 \delta_{m}^{-}(a)$. (iv) Let $h=q \delta_{m}^{-}(a)+r, 0 \leqslant r<\delta_{m}^{-}(a)$, and then deduce that $m \mid a^{r}+(-1)^{q}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,950 |
32. (i) For any positive integer $k$, it must be true that $3^{k} \mid 2^{3^{k-1}}+1, 3^{k+1} \nmid 2^{k^{-1}}+1$.
(ii) Let $k, s$ be positive integers. Prove: $3^{k} \mid 2^{s}+1$ if and only if
$$2 \nmid s, \quad 3^{k-1} \mid s .$$ | 32. (i) Use induction; (ii) Sufficiency follows from (i). Note that when $2 \mid s$, $3 \nmid 2^{s}+1$, set $s=3^{t}$. $f,(f, 6)=1$, use (i) to prove necessity. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,951 |
3. Let $a_{1}<a_{2}<a_{3}<\cdots$ be an infinite sequence of positive integers. Prove: In this sequence, there must exist two numbers $a_{s}, a_{t}$, such that there are infinitely many $a_{n}$ that can be expressed as
$$a_{n}=x a_{s}+y a_{t},$$
where $x, y$ are integers. | 3. If there exists $\left(a_{s}, a_{t}\right)=1$, then the conclusion holds. If for any $s, t$, there is always $\left(a_{s}, a_{t}\right)>1$, consider $d_{i}=\left(a_{1}, a_{i}\right), i>1 . d_{i} \mid a_{1}$, so $d_{i}>1$ can only take a finite number of values. Therefore, there must be an infinite subsequence $a_{i_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,954 |
4. In Example 8 of $\S 4$, when $(n, k)=d>1$, if the numbers in set $M$ are colored according to conditions (i), (ii), how many different colors can the numbers in $M$ be painted at most? | 4. $[d / 2]+1$.
The above text has been translated into English, retaining the original text's line breaks and format. | [d / 2]+1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,955 |
6. Let $1 \leqslant a<b,(a, b)=1$. Prove:
(i) The reduced fraction $a / b$ is a pure repeating decimal if and only if $(b, 10)=1$;
(ii) If $a / b$ is a pure repeating decimal, and the smallest repeating cycle is $t_{0}$ (i.e.,
$$a / b=0 . \quad d_{1} \cdots d_{t_{0}} d_{1} \cdots d_{t_{0}} \cdots$$
is the pure repeati... | 6. (i) If $a / b=0 . a_{1} a_{2} \cdots a_{k} a_{1} a_{2} \cdots a_{k} \cdots$ is a pure repeating decimal ${ }^{(1)}$, then $a / b=a_{1} \cdots a_{k} /\left(10^{k}-1\right)$, so $(b, 10)=1$. Conversely, if $(b, 10)=1$, then there must be a $k$ such that $b \mid 10^{k}-1$. Therefore,
$$a / b=a \cdot A /\left(10^{k}-1\r... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,957 |
1. Use the Euclidean algorithm to find the greatest common divisor (GCD) of the following sets of numbers, and express it as an integer linear combination of these numbers: (i) $15, 21, -35$; (ii) $210, -330, 1155$. | (i) $(15,21)=3=3 \cdot 15-2 \cdot 21, (3,-35)=1=3 \cdot 12-35$. Therefore, $(15,21,-35)=1=12(3 \cdot 15-2 \cdot 21)-35=36 \cdot 15-24 \cdot 21+(-35)$;
(ii) $(210,-330)=30(7,-11)=30, 1=-3 \cdot 7-2 \cdot(-11), 30=-3 \cdot 210-$
$$\begin{aligned}
2 \cdot(-330) \cdot(30,1155)=15(2,77) & =15, 15=-38 \cdot 30+1155 . \text{ ... | (i) 1=36 \cdot 15-24 \cdot 21+(-35); (ii) 15=114 \cdot 210+76 \cdot(-330)+1155 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,959 |
Theorem 2 Let $b \neq 0, d_{1}, d_{2}, \cdots, d_{k}$ be all the divisors of the integer $b$, then $b / d_{1}$, $b / d_{2}, \cdots, b / d_{k}$ are also all the divisors of $b$. That is, as $d$ runs through all the divisors of $b$, $b / d$ also runs through all the divisors of $b$. Moreover, if $b>0$, then as $d$ runs t... | Prove that when $d_{j} \mid b$, $b / d_{j}$ is an integer, $b=d_{j}\left(b / d_{j}\right)$, so $b / d_{j}$ is also a divisor of $b$, and when $d_{i} \neq d_{j}$, $b / d_{i} \neq b / d_{j}$. Thus, $b / d_{1}, \cdots, b / d_{k}$ are $k$ distinct divisors of $b$. Since the number of divisors of $b$ is fixed, this proves t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,960 |
2. Let $a>1$. Prove:
(i) $\left(a^{m}-1, a^{n}-1\right)=a^{(m, n)}-1$;
(ii) $\left(a^{m}-(-1)^{m /(m, n)}, a^{n}-(-1)^{n /(m, n)}\right)=a^{(m, n)}+1$;
(iii) Except for the cases in (i), (ii), we always have $\left(a^{m} \pm 1, a^{n} \pm 1\right)=\left\{\begin{array}{ll}1, & 2 \mid a, \\ 2, & 2 \nmid a,\end{array}\righ... | 2. Let's assume $(m, n)=1$. When the parity of $m, n$ is chosen arbitrarily (there are three possibilities) and the signs are chosen arbitrarily, the form of $\left(a^{m} \pm 1, a^{n} \pm 1\right)$ has 12 possible cases. Then, using the property $(u, v)=(u, v \pm u)$ to transform $\left(a^{m} \pm 1, a^{n} \pm 1\right)$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,961 |
3. Let $a>b \geqslant 1,(a, b)=1$. Prove:
$$\left(a^{m}-b^{m}, a^{n}-b^{n}\right)=a^{(m, n)}-b^{(m, n)}$$ | 3. Following the method of Example 7 in $\S 3$, or using the following method. Let $d=(m, n) . A=a^{d}-b^{d}, B=\left(a^{m}-b^{m}, a^{n}-b^{n}\right)$. It is evident that $A \mid B$. Without loss of generality, assume $d=m x-n y, x>0, y>0$ (if necessary, $m$ and $n$ can be swapped).
$$\begin{aligned}
a^{m x} & =a^{n y}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,962 |
4. Let $m, n$ be positive integers, satisfying $m n \mid m^{2}+n^{2}+1$. Prove:
$$m^{2}+n^{2}+1=3 m n$$ | 4. Let's assume $m \geqslant n$. Clearly, $(m, n)=1$. Let $m=q n+r, -n / 2 \leqslant r<n / 2$. Except for the obvious case, it must be that $n r \mid n^{2}+r^{2}+1, -n / 2 \leqslant r<0$. Then, by using the Euclidean algorithm, we can obtain the result. This problem can also be solved as follows: Let $n_{1} m=n^{2}+1$.... | m^{2}+n^{2}+1=3 m n | Number Theory | proof | Yes | Yes | number_theory | false | 739,963 |
6. This question requires providing an algorithm to determine the greatest common divisor $g=\left(a_{1}, \cdots, a_{k}\right)$ by directly finding $x_{1,0}, \cdots, x_{k, 0}$ in formula (2) of §4. (i) Choose a set of integers $x_{1,1}, \cdots, x_{k, 1}$ such that the positive integer $a_{1} x_{1,1}+\cdots+a_{k} x_{k, ... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 739,965 |
Proposition 3 Let $a$ be given by (2), then $d$ is a positive divisor of $a$ if and only if
$$d=p_{1}^{c_{1}} \cdots p_{s}^{e_{s}}, \quad 0 \leqslant e_{j} \leqslant \alpha_{j}, 1 \leqslant j \leqslant s .$$ | The sufficiency is obvious. Now we prove the necessity. When $d=1$, $e_{j}=0(1 \leqslant j \leqslant s)$, the conclusion obviously holds. If $d>1$, then by $d \mid a$ and Theorem 1, we know that the prime divisors of $d$ must be among $p_{1}, \cdots, p_{s}$. Therefore, by Theorem 2, the standard factorization of $d$ mu... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,968 |
Inference 5 If $(a, b)=1, a b=c^{k}$, then $a=u^{k}, \quad b=v^{k}$. | Let $c=p_{1}^{\alpha_{1}} \cdots p_{s}^{\alpha_{s}}$, then
$$c^{k}=p_{1}^{k_{1}} \cdots p_{s}^{k_{z_{3}}}$$
By Corollary 4, we can assume
$$a=p_{1}^{\beta_{1}} \cdots p_{s}^{\beta_{s}}, \quad b=p_{1}^{\gamma_{1}} \cdots p_{s}^{\gamma_{s}} .$$
From the condition $a b=c^{k}$, we know that $\beta_{j}+\gamma_{j}=k \alpha... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,970 |
Example 1 Prove: $(a,[b, c])=[(a, b),(a, c)]$. | If $a=0$, the equation obviously holds. Therefore, we can assume $a, b, c$ are positive integers,
$$a=p_{1}^{q_{1}} \cdots p_{s}^{\sigma_{3}}, \quad b=p_{1}^{\beta_{1}} \cdots p_{3}^{\beta_{3}}, \quad c=p_{1}^{\gamma_{1}} \cdots p_{3}^{\gamma_{3}}.$$
By Corollary 4, we get
$$\begin{array}{l}
(a,[b, c])=p_{1}^{\eta_{1}... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,976 |
Example 3 Find $\sum_{d \mid a} \frac{1}{d}$. | From Theorem 2 in § 2, we know
$$\sum_{d \mid a} \frac{1}{d}=\sum_{d \mid a} \frac{1}{(a / d)}=\frac{1}{a} \sum_{d \mid a} d=\frac{1}{a} \sigma(a) .$$
From this and Example 2, we can obtain
$$\sum_{d \mid 180} \frac{1}{d}=\frac{1}{180} \sigma(180)=\frac{91}{30} .$$ | \frac{91}{30} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,978 |
Theorem 4 If $a$ is a composite number, then there must be an irreducible number $p \mid a$.
| Proof: By definition, $a$ must have a divisor $d \geqslant 2$. Let the set $T$ consist of all divisors $d \geqslant 2$ of $a$. By the principle of the smallest natural number, the set $T$ must have the smallest natural number, denoted as $p$. $p$ must be an irreducible number. Otherwise, if $p \geqslant 2$ is a composi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,983 |
4. Let $a, b, n$ be positive integers and $a>b$. Prove: If $n \mid\left(a^{n}-b^{n}\right)$, then
$$n \mid\left(a^{n}-b^{n}\right) /(a-b) .$$ | 4. Using the method from question 1. Let $p^{a} \| n$. Consider the cases:
(i) $p \mid a$;
(ii) $p \nmid a, p \mid a-b$;
(iii) $p \nmid a, p \nmid a-b$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,984 |
5. Find the smallest positive integer $n$ that satisfies $\tau(n)=6$.
| 5. $n=p_{1}^{a_{1}} \cdots p_{s}^{a_{s}}, p_{1}<\cdots<p_{3} . \tau(n)=\left(\alpha_{1}+1\right) \cdots\left(\alpha_{s}+1\right)=6$. It must be that $\alpha_{1}=1$, $\alpha_{2}=2$; or $\alpha_{1}=2, \alpha_{2}=1$. Therefore, the smallest $n=2^{2} \cdot 3^{1}=12$. | 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,985 |
6. (i) Find the smallest positive integer $a$ such that $\sigma(n)=a$ has no solution, exactly one solution, exactly two solutions, and exactly three solutions;
(ii) There exist infinitely many $a$ for which $\sigma(n)=a$ has no solution. | 6. (i) For $a=2,1,12,24$, the corresponding solutions are: $\sigma(1)=1, \sigma(6)=\sigma(11)=12$; $\sigma(14)=\sigma(15)=\sigma(23)=24$. To utilize the following conclusion: when $r \geqslant 3$, $\sigma\left(p_{1}^{\circ} p_{2}^{\alpha_{2}^{2}} p_{3}^{\alpha_{3}}\right) \geqslant 72$, and when $r \leqslant 2$ there a... | 2, 1, 12, 24 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,986 |
8. Prove: (i) $\tau(n)$ is odd if and only if $n$ is a perfect square;
(ii) $\prod_{d \mid n} d=n^{\tau(n) / 2}$ (prove using two methods). | 8. As in problem 7. When proving (ii), note that $\prod_{d \mid n} d=\prod_{d \mid n}(n / d)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,988 |
10. Let $t$ be a real number, $\sigma_{t}(n)=\sum_{d \mid n} d^{t}$. Prove: $\sigma_{t}(n)=n^{t} \sigma_{-t}(n)$, and find the calculation formula for $\sigma_{t}(n)$. | 10. $\sigma_{t}(n)=\prod_{j=1}^{s}\left(p_{j}^{t\left(\sigma_{j}+1\right)}-1\right) /\left(p_{j}^{t}-1\right), n=p_{1}^{q_{1}} \cdots p_{s}^{q^{\prime}}$. | \sigma_{t}(n)=\prod_{j=1}^{s}\left(\frac{p_{j}^{t(q_{j}+1)}-1}{p_{j}^{t}-1}\right) | Number Theory | proof | Yes | Yes | number_theory | false | 739,990 |
Theorem 5 Let the integer $a \geqslant 2$, then $a$ can certainly be expressed as a product of irreducible numbers (including the case where $a$ itself is an irreducible number), i.e.,
$$a=p_{1} p_{2} \cdots p_{s},$$
where $p_{j}(1 \leqslant j \leqslant s)$ are irreducible numbers. | We prove this using proof by contradiction and the principle of the least natural number. Assume the conclusion does not hold, then there exists a positive integer $\geqslant 2$ that cannot be expressed as a product of irreducible numbers. Let the set of all such positive integers be $T$, which is non-empty. Let $n_{0}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,994 |
15. If $2^{k}-1$ is a prime, then $2^{k-1}\left(2^{k}-1\right)$ is a perfect number. | 15. Find the sum of the divisors of $2^{k-1}\left(2^{k}-1\right)$ directly. | not found | Number Theory | proof | Yes | Yes | number_theory | false | 739,996 |
16. If $\sigma(n)=n+k<2 n$ and $k \mid n$, then $n$ is a prime. | 16. Prove that there must be $k=1$.
| proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,997 |
17. If $2 \mid m, m$ is a perfect number, then $m=2^{k-1}\left(2^{k}-1\right), 2^{k}-1$ is a prime. | 17. Let $m=2^{r-1} m_{1}, 2 \nmid m_{1}, r>1.2 m=\sigma(m)=\left(2^{r}-1\right) \sigma\left(m_{1}\right)$. Let $\sigma\left(m_{1}\right)=m_{1}+k$, then $m_{1}=\left(2^{r}-1\right) k$. Use the previous problem. | not found | Number Theory | proof | Yes | Yes | number_theory | false | 739,998 |
21. Let $n$ be an odd number. Find the number of ways $n$ can be expressed as the difference of two integer squares.
Expressing the above text in English, while retaining the original text's line breaks and format, results in the output provided. | 21. Let $n=x^{2}-y^{2}(x>y \geqslant 0)$, and the number of representations of $n$ be $T$. From
$$n=(x-y) \cdot(x+y), \quad x+y \geqslant x-y$$
we know that $T$ equals the number of positive divisors $d$ of $n$ that do not exceed $\sqrt{n}$, where we set $x-y=d \leqslant \sqrt{n}$. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,002 |
22. Prove: $\log _{2} 10, \log _{3} 7, \log _{15} 21$ are all irrational numbers. | 22. If $\log _{2} 10=a / b,(a, b)=1, a \geqslant 1, b \geqslant 1.2^{a}=2^{b} 5^{b}$. This is impossible. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,003 |
Theorem 1 Let $x, y$ be real numbers. We have
(i) If $x \leqslant y$, then $[x] \leqslant[y]$.
(ii) If $x=m+v, m$ is an integer, $0 \leqslant v<1$, then $m=[x], v=\{x\}$. In particular, when $0 \leqslant x<1$, $[x]=0,\{x\}=x$.
(iii) For any integer $m$, $[x+m]=[x]+m,\{x+m\}=\{x\}$. $\{x\}$ is a periodic function with p... | (i) It follows from $[x] \leqslant x \leqslant y<[y]+1$.
(ii) It follows from $m \leqslant x < m+1$ and the definition. This property is a commonly used technique when proving properties related to $[x]$.
(iii) It follows from $[x]+m \leqslant x+m<([x]+m)+1$ and the definition.
(iv) $x+y=[x]+[y]+\{x\}+\{y\}$ and $0 \le... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,007 |
Example 1 Points with integer coordinates in the plane are called integer points or lattice points. Let $x_{1}<x_{2}$ be real numbers, and $y=f(x)\left(x_{1}<x \leqslant x_{2}\right)$ be a non-negative continuous function. Prove:
(i) The number of integer points in the region $x_{1}<x \leqslant x_{2}, 0<y \leqslant f(x... | To prove (i). All the integer points in the mentioned region lie on such line segments: $x=n, 1 \leqslant y \leqslant f(n)$, where $n$ is an integer satisfying $x_{1}<n \leqslant x_{2}$. The number of integer points on the line segment $x=n$, $1 \leqslant y \leqslant f(n)$ is the number of integers $y$ that satisfy $1 ... | proof | Calculus | proof | Yes | Yes | number_theory | false | 740,008 |
Theorem 2 Let $n$ be a positive integer, and $p$ a prime; let $\alpha=\alpha(p, n)$ satisfy $p^{\alpha} \| n$!. Then
$$\alpha=\alpha(p, n)=\sum_{j=1}^{\infty}\left[\frac{n}{p^{j}}\right] .$$ | The right-hand side of (7) is actually a finite sum, because there must be an integer $k$ such that when $p^{k} \leqslant nk$, $d_{j}=0$, and
$$\alpha=1 \cdot d_{1}+2 \cdot d_{2}+\cdots+k \cdot d_{k}$$
The latter is because we can divide $1,2, \cdots, n$ into $k$ pairwise disjoint sets: the $j$-th set consists of the ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,009 |
Example 2 Find the standard prime factorization of 20!.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | The prime numbers not exceeding 20 are $2,3,5,7,11,13,17,19$. By Theorem 2, we have
$$\begin{aligned}
\alpha(2,20) & =\left[\frac{20}{2}\right]+\left[\frac{20}{4}\right]+\left[\frac{20}{8}\right]+\left[\frac{20}{16}\right] \\
& =10+5+2+1=18 \\
\alpha(3,20) & =\left[\frac{20}{3}\right]+\left[\frac{20}{9}\right]=6+2=8 \\... | 20!=2^{18} \cdot 3^{8} \cdot 5^{4} \cdot 7^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,011 |
Example: How many zeros are at the end of the decimal representation of 320! ?
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve for the positive integer $k$ such that $10^{k} \| 20$!. From equation (12), we know that we only need to find $\alpha(5,20)$, and from the previous example, we know that $k=4$, which is the power of 5. Therefore, there are four zeros at the end. | 4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,012 |
Example 4 Let integers $a_{j}>0(1 \leqslant j \leqslant s)$, and $n=a_{1}+a_{2}+\cdots+a_{s}$. Prove: $n!/\left(a_{1}!a_{2}!\cdots a_{s}!\right)$ is an integer. | To prove using the notation of Theorem 2, it suffices to show that for any prime \( p \),
\[
\alpha(p, n) \geqslant \alpha\left(p, a_{1}\right) + \alpha\left(p, a_{2}\right) + \cdots + \alpha\left(p, a_{s}\right).
\]
By equation (7), this can be deduced from the following inequality: for any \( j \geqslant 1 \),
\[
\l... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 740,013 |
1. Let $a, b$ be integers, $a \geqslant 1, b=q a+r, 0 \leqslant r<a$. Prove:
$$q=[b / a], \quad r=a\{b / a\} .$$ | 1. $b / a=[b / a]+\{b / a\}, b=[b / a] a+\{b / a\} a$, This gives a new proof of the division algorithm (Theorem 1, §3). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,015 |
2. Let $a, b$ be integers, $a \geqslant 1, b=q_{1} a+r_{1}, -a / 2 \leqslant r_{1}<a / 2$. Prove:
$$q_{1}=\left[\frac{2 b}{a}\right]-\left[\frac{b}{a}\right], \quad r_{1}=a\left\{\frac{2 b}{a}\right\}-a\left\{\frac{b}{a}\right\} .$$ | 2. $\begin{aligned} b / a=2 b / a-b / a= & {[2 b / a]-[b / a]+\{2 b / a\}-\{b / a\}, \text { and using } } \\ & -1 / 2 \leqslant\{2 x\}-\{x\}<1 / 2 .\end{aligned}$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,017 |
4. Prove: For any real number $x$,
$$[x]+[x+1 / 2]=[2 x] .$$ | 4. The original expression is equivalent to $[1 / 2+\{x\}]=[2\{x\}]$. Then, we discuss cases for $\{x\}$.
---
Please note that the format and line breaks have been preserved as requested. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,019 |
5. Prove: For any integer $n \geqslant 2$ and real number $x$,
$$[x]+[x+1 / n]+\cdots+[x+(n-1) / n]=[n x] .$$ | 5. Let's assume $0 \leqslant x<1$. There must be an integer $k, 0 \leqslant k \leqslant n-1$, such that $k / n \leqslant x<(k+1) / n$. In this way, it is easy to prove that both sides of the equation are equal to $k$, and the 4th problem is a special case of this problem. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,020 |
10. Try to determine for which real numbers $x$ the following equations hold:
(i) $[x+3]=3+x$;
(ii) $[x]+[x]=[2 x]$;
(iii) $[11 x]=11$;
(iv) $[11 x]=10$;
(v) $[x+1 / 2]+[x-1 / 2]=[2 x]$. | 10. (i) All integers $x$.
(ii) Real numbers $x$ satisfying $2\{x\}<1$.
(iii) $1 \leqslant x<12 / 11$.
(iv) $10 / 11 \leqslant x<1$.
(v) The original equation is equivalent to $2[x-1 / 2]=[2(x-1 / 2)]$. From (ii), it is known that this is for real numbers $x$ satisfying $2\{x-1 / 2\}<1$, i.e., real numbers $x$ satisfyin... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,025 |
Theorem 8 (i) $\left(a_{1}, a_{2}\right)=\left(a_{2}, a_{1}\right)=\left(-a_{1}, a_{2}\right)=\left(\left|a_{1}\right|,\left|a_{2}\right|\right)$.
In general, we have
$$\begin{array}{l}
\left(a_{1}, a_{2}, \cdots, a_{i}, \cdots, a_{k}\right)=\left(a_{i}, a_{2}, \cdots, a_{1}, \cdots, a_{k}\right) \\
\quad=\left(-a_{1},... | Prove that according to the definition of common divisors and the properties of divisibility, we have
$$\begin{array}{l}
\mathscr{D}\left(a_{1}, a_{2}\right)=\mathscr{D}\left(a_{2}, a_{1}\right)=\mathscr{D}\left(-a_{1}, a_{2}\right)=\mathscr{D}\left(\left|a_{1}\right|,\left|a_{2}\right|\right), \\
\mathscr{D}\left(a_{1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,027 |
13. Let $m$ be a positive integer. Prove:
(i) $2^{m+1} \|\left[(1+\sqrt{3})^{2 m+1}\right]$;
(ii) $[\sqrt{m}+\sqrt{m+1}]=[\sqrt{m}+\sqrt{m+2}]$. | 13. (i) First prove
$$\begin{array}{l}
{\left[(1+\sqrt{3})^{2 m+1}\right]=}(\sqrt{3}+1)^{2 m+1}-(\sqrt{3}-1)^{2 m+1} \\
\begin{array}{l}
(\sqrt{3}+1)^{2 m+3}-(\sqrt{3}-1)^{2 m+3} \\
= 8\left\{(\sqrt{3}+1)^{2 m+1}-(\sqrt{3}-1)^{2 m+1}\right\} \\
-4\left\{(\sqrt{3}+1)^{2 m-1}-(\sqrt{3}-1)^{2 m-1}\right\}
\end{array}
\en... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,029 |
16. Let $m, n$ be odd positive integers, $(m, n)=1$. Prove
$$\sum_{0<s<m / 2}\left[\frac{n}{m} s\right]+\sum_{0<t<n / 2}\left[\frac{m}{n} t\right]=\frac{m-1}{2} \cdot \frac{n-1}{2}$$ | 16. As the method in the previous question.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | proof | Yes | Yes | number_theory | false | 740,032 |
17. Let $C>0$ be a real number. $M$ is the number of lattice points in the region: $x>0, y>0, xy \leqslant C$. Prove:
(i) $M=\sum_{1 \leq r \leq C}\left[\frac{C}{s}\right]$;
(ii) $M=2 \sum_{1 \leqslant s \leqslant \sqrt{C}}\left[\frac{C}{s}\right]-[\sqrt{C}]^{2}$;
(iii) $M=\sum_{1 \leqslant s \leqslant c} \tau(s)$.
Us... | 17. Using the method and symmetry of the graph in Use Case 1. When deriving the approximate formula for $M$, replace $[C / s]$ with $C / s$. The approximate formula obtained from (i) is
$$C \sum_{1 \leqslant s \leqslant C} 1 / s - [C] < M \leqslant C \sum_{1 \leqslant s \leqslant C} 1 / s$$
The approximate formula obt... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,033 |
19. 求 $2,3,6,12$ 及 70 整除 623 ! 的最高方幂. | 19. $2^{616}$ || $623!, 3^{308}$ || $623!, 6^{308}$ || 623 !, $12^{308}$ || $623!, 70^{102}$ || 623 ! | 2^{616}, 3^{308}, 6^{308}, 12^{308}, 70^{102} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,035 |
20. Find how many zeros are at the end of the decimal expression of 120!.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note above is not part of the translation but is provided to clarify the instruction. The actual translation is above it. | 20. Find the highest power of 10 that divides 120!, which is also the highest power of 5 that divides 120!. Therefore, there are 28 zeros. | 28 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,036 |
22. Find the prime factorization of 32!.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 22. $2^{31} \cdot 3^{14} \cdot 5^{7} \cdot 7^{4} \cdot 11^{2} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$ | 2^{31} \cdot 3^{14} \cdot 5^{7} \cdot 7^{4} \cdot 11^{2} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,039 |
23. Let $p$ be a prime number, and $n$ be a positive integer.
(i) Find the formula for $e$ in $p^{e} \|(2 n)!$, where
$$(2 n)!!=(2 n)(2 n-2) \cdots 2$$
(ii) Find the formula for $f$ in $p^{f} \|(2 n+1)!!$, where
$$(2 n-1)!!=(2 n-1)(2 n-3) \cdots 1$$ | 23. (i) When $p=2$, $e=n+\sum_{j}\left[n / 2^{j}\right] ;$ when $p>2$, $e=\sum_{j}\left[n / p^{j}\right]$.
(ii) When $p=2$, $f=0 ;$ when $p>2$, $f=\sum_{j}\left(\left[2 n / p^{j}\right]-\left[n / p^{j}\right]\right)$. | e=n+\sum_{j}\left[n / 2^{j}\right] \text{ when } p=2, \text{ and } e=\sum_{j}\left[n / p^{j}\right] \text{ when } p>2; \text{ } f=0 \text{ when } p=2, \text{ and } f=\sum_{j}\left(\left[2 n / p^{j}\right]-\left | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,040 |
25. Let $a, b$ be positive integers, $(a, b)=1$; and let $\rho$ be a real number. Prove: If $a \rho, b \rho$ are integers, then $\rho$ is also an integer. | 25. Let's assume $\rho \neq 0$. Set $a \rho=c_{1}, b \rho=c_{2}$, then we have $a c_{2}=b c_{1}$. Using the condition $(a, b)=1$, we get $a\left|c_{1}, b\right| c_{2}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,042 |
28. Prove: $(2 n)!/(n!)^{2}$ is even. | 28. Using Question 23 (i) and Question 27. Or using
$$(2 n)!/(n!)^{2}=2\{(n+1)(n+2) \cdots(n+n-1) /(n-1)!\}$$ | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 740,045 |
29. Let $m, n$ be positive integers. Prove: $n!(m!)^{n} \mid (m n)$!. | 29. To prove that for any prime $p$, we have
$$\sum_{j}\left[n m / p^{j}\right] \geqslant n \sum_{j}\left[m / p^{j}\right]+\sum_{j}\left[n / p^{j}\right]$$
Let $m=p^{l} c, p \nmid c$. When $j \leqslant l$, $\left[n m / p^{j}\right]=n\left[m / p^{j}\right]$; when $j>l$, let $m=q_{j} p^{j}+r_{j}$, $0<l}\left[n m / p^{j}... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 740,046 |
31. Find the greatest common divisor of the binomial coefficients $\binom{n}{1},\binom{n}{2}, \cdots,\binom{n}{n-1}$. | 31. When $n=p^{k}, p$ is a prime, the greatest common divisor is $p$; in other cases, it is $1$. For the case $n=p^{k}$, prove that $p \left\lvert\,\binom{ n}{l}\right., 1 \leqslant l \leqslant n-1$, and $p \|\binom{ n}{p^{k-1}}$. Otherwise, use proof by contradiction. If the greatest common divisor $d>1$, let $p \mid ... | proof | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 740,048 |
Theorem 9 If there exist integers $x_{1}, \cdots, x_{k}$, such that $a_{1} x_{1}+\cdots+a_{k} x_{k}=1$, then $a_{1}, \cdots, a_{k}$ are coprime. | To prove that any common divisor $d$ of $a_{1}, \cdots, a_{k}$ must divide 1, so we must have $d = \pm 1$. This proves the desired conclusion.
We will later prove that the condition $a_{1} x_{1}+\cdots+a_{k} x_{k}=1$ is also a necessary condition for $a_{1}, \cdots, a_{k}$ to be coprime. Using Theorem 9, we can also p... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,049 |
32. Let $p$ be a given prime. Prove: there must exist a positive integer $a$, such that for any positive integer $n$, it is impossible to have $p^{a} \| n!$. Propose a method to determine all such $a$. | 32. Let $p^{e} \| n$ !, from the formula
$$e=\sum_{j}\left[n / p^{j}\right]$$
we know, when $n<p$, $e=0$; when $p \leqslant n<p^{2}$, $1 \leqslant e<p$; when $p^{2} \leqslant n<p^{3}$, $p+1 \leqslant$ $e<p^{2}+p-2$; when $p^{3} \leqslant n<p^{4}$, $p^{2}+p+1 \leqslant e<p^{3}+p^{2}+p-2$. From this, we can see that $a=... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,050 |
35. Let $\alpha$ be a positive real number. Suppose
$$a_{n}=[n(1+\alpha)], n=1,2, \cdots ; \quad b_{n}=\left[n\left(1+\alpha^{-1}\right)\right], n=1,2, \cdots .$$
Prove: These numbers are all distinct, and they exactly give all positive integers if and only if $\alpha$ is a positive irrational number. | 35. (i) First, prove that $a_{n}$ are all distinct, and $b_{n}$ are also all distinct. (ii) Then prove that for any $m, n, a_{m} \neq$ $b_{n}$. Since $\alpha$ is a positive irrational number, there must be $0i, nn$ or $m \alpha<n$ holds and only one of them holds. From the above discussion, it follows that $K<m+m \alph... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,053 |
36. Let $\alpha, \beta$ be positive real numbers. Let $a_{n}=[n \alpha], n=1,2, \cdots ; b_{n}=[n \beta]$, $n=1,2, \cdots$. Prove: These numbers are all distinct and exactly give all positive integers if and only if $\alpha, \beta$ are positive irrational numbers and satisfy
$$\frac{1}{\alpha}+\frac{1}{\beta}=1$$ | 36. Proof of sufficiency. It is evident that $\alpha>1, \beta>1$, and one of $\alpha, \beta$ must be less than 2. Without loss of generality, assume $1<\alpha<2, \beta>1$. (ii) Let $N$ be any given positive integer, and let $f(N)$ and $g(N)$ represent the number of positive integers not exceeding $N$ that can be expres... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,054 |
Theorem 1 The necessary and sufficient condition for the indeterminate equation (1) to have a solution is that the greatest common divisor of its coefficients $\left(a_{1}, \cdots, a_{k}\right) \mid c$. Furthermore, when the indeterminate equation (1) has a solution, its solutions are the same as those of the indetermi... | Necessity is obvious. Below we prove sufficiency. If $g \mid c$, let $c=g c_{1}$. By Theorem 8 of Chapter 1, §4, there must be integers $y_{1,0}, \cdots, y_{k, 0}$, such that
$$a_{1} y_{1,0}+\cdots+a_{k} y_{k, 0}=g$$
Therefore, $x_{1}=c_{1} y_{1,0}, \cdots, x_{k}=c_{1} y_{k, 0}$ is a set of solutions to (1), which pro... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,055 |
Theorem 2 Let the binary linear indeterminate equation
$$a_{1} x_{1}+a_{2} x_{2}=c$$
have a solution, and $x_{1,0}, x_{2,0}$ be one of its solutions. Then all solutions are
$$\left\{\begin{array}{l}
\dot{x}_{1}=x_{1,0}+\frac{a_{2}}{\left(a_{1}, a_{2}\right)} t, \\
x_{2}=x_{2,0}-\frac{a_{1}}{\left(a_{1}, a_{2}\right)} ... | To prove that $x_{1}, x_{2}$ given by equation (5) satisfy the indeterminate equation (4) for all integers $t$ is straightforward. Conversely, let $x_{1}, x_{2}$ be a solution to (4). We have
$$a_{1} x_{1}+a_{2} x_{2}=c=a_{1} x_{1,0}+a_{2} x_{2,0}$$
Thus,
$$\begin{aligned}
a_{1}\left(x_{1}-x_{1,0}\right) & =-a_{2}\lef... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,056 |
Example 1 Find all solutions to $10 x_{1}-7 x_{2}=17$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 1 Find all solutions to $10 x_{1}-7 x_{2}=17$.
Keep the line breaks and format of the original text, and output the translation result directly as above. | It is easy to see that $(10,7)=1$, so the equation has a solution. By inspection, we can find that $x_{1,0}=$ $1, x_{2,0}=-1$ is a particular solution. Therefore, all solutions are
$$x_{1}=1-7 t, \quad x_{2}=-1-10 t, \quad t=0, \pm 1, \pm 2, \cdots$$ | x_{1}=1-7 t, \quad x_{2}=-1-10 t, \quad t=0, \pm 1, \pm 2, \cdots | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,057 |
Example 2 Find the solution to $18 x_{1}+24 x_{2}=9$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | From $(18,24)=6 \nmid 9$ we know there is no solution.
To solve the indeterminate equation (4), it is necessary (i) to find the greatest common divisor $g=\left(a_{1}, a_{2}\right)$, and determine whether $g \mid c$; (ii) if $g \mid c$, i.e., there is a solution, then try to find a particular solution $x_{1,0}, x_{2,0}... | not found | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,058 |
Example 3 Find the solution to $907 x_{1}+731 x_{2}=2107$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solving, if there is a solution, then 731 must divide $-907 x_{1}+2107$, so from the original equation we get
$$\begin{aligned}
x_{2}= & \frac{1}{731}\left(-907 x_{1}+2107\right) \\
= & -x_{1}+3+\frac{1}{731} \\
& \cdot\left(-176 x_{1}-86\right)
\end{aligned}$$
Thus, the original equation is equivalent to the indeterm... | x_{1}=-258+731 x_{6}, \quad x_{2}=323-907 x_{6}, \quad x_{6}=0, \pm 1, \pm 2, \cdots | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,059 |
Theorem 10 Let $m \mid\left(a_{1}, \cdots, a_{k}\right)$. We have
$$m\left(a_{1} / m, \cdots, a_{k} / m\right)=\left(a_{1}, \cdots, a_{k}\right)$$
In particular, we have
$$\left(\frac{a_{1}}{\left(a_{1}, \cdots, a_{k}\right)}, \cdots, \frac{a_{k}}{\left(a_{1}, \cdots, a_{k}\right)}\right)=1$$ | Let $D=\left(a_{1}, \cdots, a_{k}\right)$. From $m|D, D| a_{j}(1 \leqslant j \leqslant k)$ we know
$$m \mid a_{j} \quad(1 \leqslant j \leqslant k),$$
thus we have
$$(D / m) \mid\left(a_{j} / m\right), \quad j=1, \cdots, k,$$
which means $D / m$ is a common divisor of $a_{1} / m, \cdots, a_{k} / m$, and it is positive... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,060 |
Example 5 Find all solutions to $15 x_{1}+10 x_{2}+6 x_{3}=61$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | To solve for the coefficient of $x_{3}$ with the smallest absolute value, we transform the original equation into
$$\begin{aligned}
x_{3} & =\frac{1}{6}\left(-15 x_{1}-10 x_{2}+61\right) \\
& =-2 x_{1}-2 x_{2}+10+\frac{1}{6}\left(-3 x_{1}+2 x_{2}+1\right)
\end{aligned}$$
The original indeterminate equation is equivale... | x_{1}=1+2 x_{5}, \quad x_{2}=1+3 x_{4}+3 x_{5}, \quad x_{3}=6-5 x_{4}-10 x_{5}, \quad x_{4}, x_{5}=0, \pm 1, \pm 2, \cdots | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,062 |
Theorem 3 Let $g_{1}=a_{1}, g_{2}=\left(g_{1}, a_{2}\right)=\left(a_{1}, a_{2}\right), g_{3}=\left(g_{2}, a_{3}\right)=$ $\left(a_{1}, a_{2}, a_{3}\right), \cdots, g_{j}=\left(a_{1}, \cdots, a_{j}\right), \cdots, g_{k}=\left(g_{k-1}, a_{k}\right)=\left(a_{1}, \cdots, a_{k}\right)$, then the indeterminate equation (1) i... | To prove the equivalence. If $x_{1}, \cdots, x_{k}, y_{2}, \cdots, y_{k-1}$ are solutions to the system of equations (6), then it is evident that $x_{1}, \cdots, x_{k}$ are solutions to (1). Conversely, if $x_{1}, \cdots, x_{k}$ are solutions to (1), then take
$$y_{j}=\frac{1}{g_{j}}\left(a_{1} x_{1}+\cdots+a_{j} x_{j}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,063 |
Example 6 Find the solution to $15 x_{1}+10 x_{2}+6 x_{3}=61$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve using the method of Theorem 3. $a_{1}=15, a_{2}=10, a_{3}=6$, so $g_{1}=$ $a_{1}=15, g_{2}=\left(a_{1}, a_{2}\right)=(15,10)=5, g_{3}=\left(g_{2}, a_{3}\right)=(5,6)=1$. Therefore, this indeterminate equation is equivalent to a system of indeterminate equations with 4 variables and two equations:
$$\left\{\begin{... | \begin{array}{c}
x_{1}=5+2 t_{1}+6 t_{2}, \quad x_{2}=-5-3 t_{1}-6 t_{2}, \quad x_{3}=6-5 t_{2}, \\
t_{1}, t_{2}=0, \pm 1, \cdots .
\end{array} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,064 |
Theorem 4 Let $a_{1}, a_{2}$, and $c$ be positive integers, $\left(a_{1}, a_{2}\right)=1$, then when $c>a_{1} a_{2}-a_{1}-a_{2}$, the indeterminate equation (4) has non-negative solutions, the number of solutions is $\left[c /\left(a_{1} a_{2}\right)\right]$ or $\left[c /\left(a_{1} a_{2}\right)\right]+1$; when $c=a_{1... | Prove that since $\left(a_{1}, a_{2}\right)=1$, equation (4) must have a solution. Let $x_{1,0}, x_{2,0}$ be a particular solution of equation (4). By the general solution formula (5), all non-negative solutions $x_{1}, x_{2}$ are given by parameters $t$ satisfying the following conditions: $\square$
$$\begin{array}{l}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,065 |
Theorem 5 Let $a_{1}, a_{2}$, and $c$ be positive integers, $\left(a_{1}, a_{2}\right)=1$, then when $c>a_{1} a_{2}$, equation (4) has positive solutions, the number of solutions is equal to $-\left[-c /\left(a_{1} a_{2}\right)\right]-1$ or $-\left[-c /\left(a_{1} a_{2}\right)\right]$; when $c=a_{1} a_{2}$, equation (4... | Given that $\left(a_{1}, a_{2}\right)=1$, equation (4) must have a solution. Let $x_{1,0}, x_{2,0}$ be a particular solution of equation (4). By the general solution formula (5), all positive solutions $x_{1}, x_{2}$ are given by parameters $t$ satisfying the following conditions:
$$\begin{array}{l}
-\left[x_{1,0} / a_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,066 |
Example 7 Find all positive solutions of $5 x_{1}+3 x_{2}=52$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given $x_{1}=8, x_{2}=4$ is a particular solution, from equations (5) and (14) we know all positive solutions are:
$$\begin{array}{c}
x_{1}=8+3 t, \quad x_{2}=4-5 t \\
-2=[-8 / 3]+1 \leqslant t \leqslant-[-4 / 5]-1 \leqslant 0
\end{array}$$
Therefore, there are three sets of positive solutions: 8, 4; 5, 9; 2, 14. It i... | 8, 4; 5, 9; 2, 14 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,067 |
Example 8 Proof: $101 x_{1}+37 x_{2}=3189$ has positive integer solutions. | Here, $c=3189<a_{1} a_{2}=101 \cdot 37$, so from the conclusion of Theorem 5, we cannot determine whether the equation has a positive solution (of course, it can be deduced that there is at most one). Therefore, we need to use formula (15) (or (14)). It can be found that $x_{1}=11 \cdot 3189, x_{2}=-30 \cdot 3189$ is a... | 1 | Number Theory | proof | Yes | Yes | number_theory | false | 740,068 |
Example 9 Rooster one, worth five coins; hen one, worth three coins; three chicks, worth one coin. A hundred coins buy a hundred chickens. Question: How many roosters and hens are there? | Let $x_{1}, x_{2}, x_{3}$ represent the number of roosters, hens, and chicks, respectively. From the conditions, we can derive the following system of indeterminate equations:
$$\left\{\begin{array}{l}
5 x_{1}+3 x_{2}+x_{3} / 3=100, \\
x_{1}+x_{2}+x_{3}=100 .
\end{array}\right.$$
We need to find the non-negative solut... | \begin{array}{|c|cccc|}
\hline
x_{1} & 0 & 4 & 8 & 12 \\
\hline
x_{2} & 25 & 18 & 11 & 4 \\
\hline
x_{3} & 75 & 78 & 81 & 84 \\
\hline
\end{array} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,069 |
Example 10 Find all non-negative solutions of $15 x_{1}+10 x_{2}+6 x_{3}=61$.
untranslated text remains the same as requested. However, the instruction was to translate the text, which I have done while keeping the format and structure intact. | From Example 6, we know the general solution formula is
$$x_{1}=5+2 t_{1}+6 t_{2}, \quad x_{2}=-5-3 t_{1}-6 t_{2}, \quad x_{3}=6-5 t_{2}.$$
Therefore, the non-negative solutions for $t_{1}, t_{2}$ are
$$5+2 t_{1}+6 t_{2} \geqslant 0, \quad-5-3 t_{1}-6 t_{2} \geqslant 0, \quad 6-5 t_{2} \geqslant 0$$
From this, we get... | 1,1,6 ; 3,1,1 ; 1,4,1 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,070 |
1. Solve the following equations:
(i) $3 x_{1}+5 x_{2}=11$;
(ii) $60 x_{1}+123 x_{2}=25$;
(iii) $903 x_{1}+731 x_{2}=1106$;
(iv) $21 x_{1}+35 x_{2}=98$;
(v) $1402 x_{1}-1969 x_{2}=2$ | 1. (i) $x_{1}=2+5 t, x_{2}=1-3 t$; (ii) no solution, $(60,123)=3 \nmid 25$; (iii) no solution,
$$\begin{array}{l}
43=(903,731), 43 \nmid 1106 ; \text { (iv) } x_{1}=3+5 t, x_{2}=1-3 t ; \text { (v) } x_{1}=1778+1969 t, \\
x_{2}=1266+1402 t .
\end{array}$$ | x_{1}=2+5 t, x_{2}=1-3 t; \text{ no solution}; \text{ no solution}; x_{1}=3+5 t, x_{2}=1-3 t; x_{1}=1778+1969 t, x_{2}=1266+1402 t | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,072 |
2. Solve the following equations:
(i) $x_{1}-2 x_{2}-3 x_{3}=7$;
(ii) $3 x_{1}+6 x_{2}-4 x_{3}=7$;
(iii) $6 x_{1}+10 x_{2}-21 x_{3}+14 x_{4}=1$. | 2. (i) $x_{1}=7+s, x_{2}=-s+3 t, x_{3}=s-2 t$;
(ii) $x_{1}=1+4 s+2 t, x_{2}=-t, x_{3}=-1+3 s$;
(iii) Let $3 x_{1}+5 x_{2}=y_{1}, 3 x_{3}-2 x_{4}=y_{2}$, the original equation becomes $2 y_{1}-7 y_{2}=1 . y_{1}=4+$ $7 s, y_{2}=1+2 s$, then we get
$$\begin{array}{ll}
x_{1}=2 y_{1}+5 t=8+14 s+5 t, & x_{2}=-y_{1}-3 t=-4-7 ... | not found | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,073 |
4. Let $(a, b)=1, c$ be an integer. Prove: In the Cartesian coordinate system, on the line with the equation $a x+b y=c$, any segment of length $\geqslant\left(a^{2}+b^{2}\right)^{1 / 2}$ (including endpoints) must contain a point whose coordinates are integers. | 4. Let the solution be $x=x_{0}+b t, y=y_{0}-a t$. The distance between the integer points given by two consecutive solutions (i.e., corresponding to $t, t+1$) is equal to $\left(a^{2}+b^{2}\right)^{1 / 2}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,075 |
5. Prove: The general solution of $a_{1} x_{1}+a_{2} x_{2}=c$ is $x_{1}=e+f t, x_{2}=g+h t, t=0$, $\pm 1, \pm 2, \cdots$ (where $e, f, g, h$ are integers) if and only if $x_{1}=e, x_{2}=g$ is a solution and
or $\square$
$$\begin{array}{ll}
f=a_{2} /\left(a_{1}, a_{2}\right), & h=-a_{1} /\left(a_{1}, a_{2}\right) \\
f=-... | 5. The sufficiency is obvious. Taking $t=0$ gives $x_{1}=e, x_{2}=g$ as a solution. Furthermore, for any integer $t$ we have $a_{1} f t + a_{2} h t = 0$, thus $a_{1} f + a_{2} h = 0$. Therefore, $a_{1} / (a_{1}, a_{2}) | h, a_{2} / (a_{1}, a_{2}) | f$. On the other hand, $x_{1}=e + a_{2} / (a_{1}, a_{2}), x_{2}=f - a_{... | proof | Algebra | proof | Yes | Yes | number_theory | false | 740,076 |
6. Let $k>h$. We write the system of indeterminate equations $a_{1 j} x_{1}+\cdots+a_{k j} x_{k}=c_{j}(1 \leqslant j \leqslant h)$ in matrix form:
$$\boldsymbol{A}\left(\begin{array}{c}
x_{1} \\
\vdots \\
x_{k}
\end{array}\right)=\left(\begin{array}{c}
c_{1} \\
\vdots \\
c_{h}
\end{array}\right)$$
where the matrix $\b... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | proof | Algebra | proof | Yes | Yes | number_theory | false | 740,077 |
7. Under the notation of Theorem 3 in §1, prove:
(i) The indeterminate equation (1) in §1 is equivalent to the system of indeterminate equations
$$a_{1} x_{1}+a_{2} x_{2}=g_{2} y_{2}, \quad g_{2} y_{2}+a_{3} x_{3}+\cdots+a_{k} x_{k}=c ;$$
(ii) For any fixed $2 \leqslant h<k$, the indeterminate equation (1) in §1 is als... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 740,078 |
8. Find all non-negative solutions and all positive solutions of the following equations:
(i) $5 x_{1}+7 x_{2}=41$;
(ii) $96 x_{1}+97 x_{2}=1000$;
(iii) $7 x_{1}+3 x_{2}=123$;
(iv) $15 x_{1}+12 x_{2}+20 x_{3}=59$. | 8. (i) $x_{1}=4+7 t, x_{2}=3-5 t$. $x_{1}=4, x_{2}=3$.
(ii) $x_{1}=-1000+97 t, x_{2}=1000-96 t$. No non-negative or positive solutions.
(iii) $x_{1}=3 t, x_{2}=41-7 t$. All positive solutions are given by $t=1,2,3,4,5$; non-negative solutions also include $t=0$
(iv) $x_{1}=1, x_{2}=2, x_{3}=1$. | t=1,2,3,4,5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,079 |
10. There are 50 RMB notes with denominations of 1 yuan, 2 yuan, and 5 yuan. To make their total value 100 yuan, how can the number of notes of these denominations be chosen? | 10. $x, y, z$ represent the number of 1 yuan, 2 yuan, and 5 yuan notes respectively. $x+y+z=50, x+2 y+5 z=100$.
$$x=36+3 t, y=2-4 t, z=12+t, t=0,-1,-2, \cdots,-12 \text {. }$$ | x=36+3t, y=2-4t, z=12+t, t=0,-1,-2, \cdots,-12 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,081 |
Theorem 1 (Mathematical Induction) Let $P(n)$ be a property or proposition concerning the natural number $n$. If
(i) $P(1)$ is true when $n=1$;
(ii) the truth of $P(n)$ implies the truth of $P(n+1)$, then $P(n)$ is true for all natural numbers $n$. | Proof: Let the set of all natural numbers $n$ for which $P(n)$ holds be $S$. $S$ is a subset of $\boldsymbol{N}$. From condition (i), we know $1 \in S$; from condition (ii), we know that if $n \in S$, then $n+1 \in S$. Therefore, by the principle of induction, $S=N$. Proof completed.
The theory of divisibility and the... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,082 |
Theorem 12 Let $m>0$. We have
$$\left[m a_{1}, \cdots, m a_{k}\right]=m\left[a_{1}, \cdots, a_{k}\right] .$$ | Proof: Let $L=\left[m a_{1}, \cdots, m a_{k}\right], L^{\prime}=\left[a_{1}, \cdots, a_{k}\right]$. From $m a_{j} \mid L(1 \leqslant j \leqslant k)$, we deduce that $a_{j} \mid L / m(1 \leqslant j \leqslant k)$, and by the definition of the least common multiple, we have $L^{\prime} \leqslant L / m$. On the other hand,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,083 |
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