problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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(2) Commutative Law of Addition For any $a, b \in \boldsymbol{N}$, we have
$$a+b=b+a .$$
| First, we prove that for any $b \in \boldsymbol{N}$, we have
$$e+b=b+e.$$
When $b=e$, equation (10) clearly holds. Assume that equation (10) holds when $b=n$. When $b=n^{+}$, by equations (2), the assumption, and equation (1), we get
$$e+n^{+}=(e+n)^{+}=(n+e)^{+}=\left(n^{+}\right)^{+}=n^{+}+e,$$
which means equation... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,674 |
(3) Additive Cancellation Law Let $a, b, c \in \mathbf{N}$. If $b+a=c+a$, then $b=c$. | First, prove the conclusion when $a=e$. If $b+e=c+e$, by equation (1) we know $b+e=b^{+}, c+e=c^{+}$, so $b^{+}=c^{+}$. By Peano Axiom (iii), it follows that $b=c$, so the conclusion holds when $a=e$. Assume the conclusion holds when $a=n$. When $a=n^{+}$, if $b+n^{+}=c+n^{+}$, then by equation (2) we know
$$(b+n)^{+}=... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,675 |
(5) For any $a, b \in \boldsymbol{N}$, exactly one of the following three cases holds:
(i) $a=b$;
(ii) there exists $x \in \boldsymbol{N}$ such that $a=b+x$;
(iii) there exists $y \in \mathbf{N}$ such that $a+y=b$. | Prove that when $b=e$, the conclusion holds. Because, if $a=e$, then (i) holds; if $a \neq e$, by Theorem 2 of $\S 1$, we know $a=\left(a^{-}\right)^{+}$, and thus by equation (1) and the commutative law, we get $a=a^{-}+e=e+a^{-}$, which means (ii) holds. Moreover, by property (4), it is easy to deduce that for any $a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,677 |
Theorem 3 There exists a unique binary operation $\pi$ on the set of natural numbers $\boldsymbol{N}$, satisfying the conditions:
(i) For any $n \in \boldsymbol{N}$, we have
$$n \pi e=n ;$$
(ii) For any $n, m \in \boldsymbol{N}$, we have
$$n \pi m^{+}=(n \pi m)+n$$ | Theorem 3 Proof of Existence For any $n \in \boldsymbol{N}$, let $h_{n}$ be the mapping determined in Lemma 4. Now define the binary operation $\pi$ as
$$n \pi m=h_{n}(m), \quad n \in \boldsymbol{N}, m \in \boldsymbol{N}$$
We will prove that this binary operation satisfies conditions (11) and (12). From equation (13),... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,679 |
Lemma 4 For any given $n \in \boldsymbol{N}$, there exists a unique mapping $h_{n}$ from $\boldsymbol{N}$ to itself, satisfying
$$h_{n}(e)=n ;$$
and for any $m \in \boldsymbol{N}$, we have
$$h_{n}\left(m^{+}\right)=h_{n}(m)+n .$$ | To prove uniqueness, if there is another such mapping $k_{n}$, then when $m=e$, it is clear that $h_{n}(e)=k_{n}(e)=n$. Assuming for some $m$ that $h_{n}(m)=k_{n}(m)$, then by equation (14) we have
$$k_{n}\left(m^{+}\right)=k_{n}(m)+n=h_{n}(m)+n=h_{n}\left(m^{+}\right)$$
Therefore, by the principle of mathematical ind... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,680 |
(6) Right Distributive Law of Multiplication For any $a, b, c \in N$, we have
$$(a+b) \cdot c=(a \cdot c)+(b \cdot c)$$ | Proof. We prove the statement for $c$ using the principle of mathematical induction. When $c=e$, by the definition of multiplication, we have
$$(a+b) \cdot e=a+b=(a \cdot e)+(b \cdot e),$$
so the conclusion holds. Assume that the conclusion holds for $c=n$. Then, for $c=n^{+}$, by the definition of multiplication, the... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,681 |
(7) For any $a \in \boldsymbol{N}$, we have
$$e \cdot a=a .$$ | Proof. For $a$, use the principle of induction. When $a=e$, by the definition of multiplication, equation (19) holds. Assume that the conclusion holds when $a=n$. When $a=n^{+}$, by the definition of multiplication, the assumption, and the definition of addition (i.e., equation (1)), we get
$$e \cdot n^{+}=(e \cdot n)+... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,682 |
(8) Commutative Law of Multiplication For any $a, b \in N$, we have
$$a \cdot b=b \cdot a .$$ | Proof. We prove the statement for $b$ using the principle of mathematical induction. When $b=e$, by the definition of multiplication, equation (20) is simply equation (19), so the conclusion holds. Assume that the conclusion holds for $b=n$. When $b=n^{+}$, by the definition of addition, property (6), the inductive hyp... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,683 |
(9) Associative Law of Multiplication For any $a, b, c \in \boldsymbol{N}$, we have
$$(a \cdot b) \cdot c=a \cdot(b \cdot c)$$ | Proof. For $c$, we use the principle of induction. When $c=e$, by the definition of multiplication, we have $(a \cdot b) \cdot e=a \cdot b=a \cdot(b \cdot e)$, so the conclusion holds. Assume that the conclusion holds when $c=n$. Then, when $c=n^{+}$, by the definition of multiplication, the assumption, property (8), a... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,684 |
(10) Right Cancellation Law of Multiplication For any $a, b, c \in \boldsymbol{N}$, from $a \cdot c=b \cdot c$ it follows that $a=b$. | Prove by contradiction. If $a \neq b$, by the addition property (5), it must be that $a=b+x$ or $b=a+y$ holds. Without loss of generality, assume $a=b+x$ holds. By the given condition and the multiplication property (6), we have
$$b \cdot c=a \cdot c=(b+x) \cdot c=(b \cdot c)+(x \cdot c)$$ | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,685 |
Theorem 1 For any $a, b \in \boldsymbol{N}$, the following three statements hold and only one of them is true:
$$a=b, a>b \text { or } a<b$$
(1) If $a \neq b$, then either $a>b$ or $a<b$;
(2) For any $a \in \boldsymbol{N}$, there is $a \geqslant e$, i.e., $a=e$ or $a>e$ holds;
(3) From $a>b, b>c$ it follows that $a>c$;... | To prove that $a^{+}=a+e$ implies property (1). When $a \neq e$, property (2) can be derived from Theorem 2 in $\S 1$. Property (3) can be directly derived from the definition and the associative law of addition. Property (4) can be proven as follows: By definition, $a>b$, which means $a=b+y$. From this and the cancell... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,686 |
Theorem 2 (Principle of the Least Natural Number) Any non-empty subset $T$ of the set of natural numbers $\boldsymbol{N}$ must have a least element, i.e., there exists a natural number $t_{0} \in T$, such that for any $t \in T$, we have $t_{0} \leqslant t$. | Consider the set $S$ consisting of all natural numbers $s$ such that for any $t \in T$, we have $s \leqslant t$. By property (2), $e \in S$, so $S$ is non-empty. Moreover, if $t_{1} \in T$ (since $T$ is non-empty, there must be such a $t_{1}$), then $t_{1}+e > t_{1}$. Therefore, $t_{1}+e \notin S$. From these two point... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,687 |
Theorem 3 (Principle of the Greatest Natural Number) Let $M$ be a non-empty subset of the set of natural numbers $\mathbb{N}$. If $M$ has an upper bound, i.e., there exists $a \in \mathbb{N}$, such that for any $m \in M$, we have $m \leqslant a$, then there must exist $m_{0} \in M$, such that for any $m \in M$, we have... | Consider the set $T$ consisting of all natural numbers $t$ such that for any $m \in M$, we have $m \leqslant t$. By the condition, $a \in T$, so $T$ is non-empty. By Theorem 2, there exists a smallest natural number in $T$, denoted as $t_{0}$. We will prove that $t_{0} \in M$. If not, for any $m \in M$, we must have $m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,688 |
5. Under the conditions and notation of Theorem 4 in § 3, let
$$\begin{array}{l}
P_{-1}=1, P_{0}=q_{0}, P_{j}=q_{j} P_{j-1}+P_{j-2}, \quad j=1,2, \cdots, k-1, \\
Q_{-1}=0, Q_{0}=1, Q_{j}=q_{j} Q_{j-1}+Q_{j-2},
\end{array}$$
Then, we have
$$(-1)^{j} u_{j}=Q_{j-2} u_{0}-P_{j-2} u_{1}, \quad j=1,2, \cdots, k+1 .$$ | None
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Theorem 4 The principle of mathematical induction is equivalent to the following principle of non-successor elements: Let $S$ be a non-empty subset of $\boldsymbol{N}$, then there must be $s_{0} \in S$, such that $s_{0}$ is not the successor of any element in $S$, and when $S=\boldsymbol{N}$, $e$ is the only element in... | First, we prove the first part. If $e \in S$, then the conclusion holds. If $e \notin S$, consider the set of all elements in $\boldsymbol{N}$ that do not belong to $S$, denoted as $T$. Clearly, $e \in T$. Since $S$ is non-empty, $T \subset \boldsymbol{N}$, i.e., $T$ is a proper subset of $\boldsymbol{N}$. Therefore, b... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,690 |
5. A set $M$ is called an ordered set if a binary relation, called order, denoted by $\leqslant$, is defined on it, satisfying the following conditions:
(1) Reflexivity: For all $x \in M$, $x \leqslant x$;
(2) Antisymmetry: If $x \leqslant y$ and $y \leqslant x$, then $x=y$;
(3) Transitivity: If $x \preccurlyeq y$ and ... | None
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8. Let $M$ be a totally ordered set. We call the statement "For any non-empty subset $S$ of $M$, if $S$ has an upper bound, i.e., there exists $a \in M$ such that for any $s \in S$, $s \leqslant a$, then there exists $s_{0} \in S$ such that for any $s \in S$, $s \leqslant s_{0}$ holds" the Maximum Element Principle.
(i... | None
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Theorem 1 Let $\pi \in \boldsymbol{Z}[\sqrt{-5}], \pi \neq 0, \pm 1$. If for any $\alpha, \beta \in \boldsymbol{Z}[\sqrt{-5}]$, from $\pi \mid \alpha \beta$ it can always be deduced that $\pi \mid \alpha$ or $\pi \mid \beta$ at least one holds, then $\pi$ must be an irreducible number. | To simplify the derivation, we introduce the following notation. For $\sigma = r + s \sqrt{-5} \in Q(\sqrt{-5})$, denote $\sigma' = r - s \sqrt{-5}$ and
$$N(\sigma) = \sigma \sigma' = r^2 + 5 s^2$$
It is easy to verify (left to the reader):
$$\begin{array}{l}
N(\sigma_1 \sigma_2) = N(\sigma_1) N(\sigma_2), \quad \sigm... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,702 |
8. Let \( M \) be the set of all positive even numbers \( 2, 4, 6, \cdots, 2k, \cdots \). Discuss \( M \) in the same way as \(\boldsymbol{Z}[\sqrt{-5}]\), and point out:
(i) The Fundamental Theorem of Arithmetic does not hold in \( M \);
(ii) A "successor" relation can be introduced in \( M \) such that the Peano axio... | None
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7. Let $u_{0}>u_{1}>1$ as in Theorem 4 of § 3; and let $b_{0}=1, b_{1}=2$ and
$$b_{j+1}=b_{j}+b_{j-1}, \quad j=1,2, \cdots$$
Then, in the notation of Theorem 4 of § 3, we have $u_{1} \geqslant b_{k}$. Furthermore, prove that:
$$k+1 \leqslant 2\left(\ln u_{1}\right) / \ln 2$$
Explain the significance of this result. | 7. Prove by the recursive formula of $b_{j}$: when $k \geqslant 1$, $b_{k} \geqslant 2^{(k+1) / 2}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,711 |
9. Let $f \in \boldsymbol{M}[x], f \neq 0$ (i.e., $f$ is not identically zero, it is a non-zero polynomial). If $1 / f \in \boldsymbol{M}[x]$, then $f$ is called a unit element of $\boldsymbol{M}[x]$. Prove:
(i) The unit elements in $Q[x]$ are and only are all non-zero rational numbers;
(ii) The unit elements in $\bold... | None
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10. Let $f, g \in \boldsymbol{M}[x], g \neq 0$. If there exists $q \in \boldsymbol{M}[x]$, such that $f = q g$, then we say that (in $\boldsymbol{M}[x]$) $g$ divides $f$, denoted by $g \mid f$, and call $g$ a factor of $f$, and $f$ a multiple of $g$. Otherwise, we say that (in $\boldsymbol{M}[x]$) $g$ does not divide $... | None
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11. Let $f \in \boldsymbol{M}[x], f \neq 0$. It is evident that $\varepsilon, \varepsilon f$ are always factors of $f$, as long as $\varepsilon$ is a unit element in $\boldsymbol{M}[x]$. Such factors are called obvious factors of $f$. Other factors of $f$ are called non-obvious factors or proper factors of $f$. If $f \... | None
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14. Let $g_{1}, \cdots, g_{n} \in \boldsymbol{M}[x]$, and not all are zero. If $d \in \boldsymbol{M}[x], d \neq 0$, such that $d \mid g_{j} (1 \leqslant j \leqslant n)$, then $d$ is called a common divisor of $g_{1}, \cdots, g_{n}$. If $g_{1}, \cdots, g_{n}$ have no other common divisors except the unit elements in $\b... | None
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16. Let $g_{1}, \cdots, g_{n} \in \mathbb{Q}[x]$, and not all zero. Consider the set
$$\mathscr{D}=\left\{d=u_{1} g_{1}+\cdots+u_{n} g_{n}: u_{j} \in \mathbb{Q}[x], 1 \leqslant j \leqslant n, d \neq 0\right\} .$$
(i) Prove: All polynomials of the lowest degree in $\mathscr{D}$ are associates in $\mathbb{Q}[x]$. Let $D$... | None
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18. (i) Let $f \in Q[x], f \neq 0$ and a unit element. Prove: $f$ can certainly be expressed as a product of irreducible polynomials in $Q[x]$, and this representation is unique up to order and associates, i.e., if $f=f_{1} f_{2} \cdots f_{s}, f=g_{1} g_{2} \cdots g_{t}$, where $f_{j}, g_{l}$ are all irreducible polyno... | None
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8. In question 2, let $v_{0}>v_{1}>1$; and set $c_{0}=1, c_{1}=2$ and
$$c_{j+1}=2 c_{j}+c_{j-1}, \quad j=1,2, \cdots$$
Then, we have $v_{1} \geqslant c_{h}$. Furthermore, prove:
(i) $h \leqslant\left(\ln v_{1}\right) / \ln 2$;
(ii) When $v_{1} \geqslant 32$, $h+1 \leqslant(\ln v) / \ln 2$. | 8. Estimate the lower bound of $c_{k}$ using the recurrence formula of $c_{j}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | null | Number Theory | proof | Yes | Yes | number_theory | false | 739,722 |
19. (i) Let non-zero polynomials $f_{1}, \cdots, f_{n} \in \mathbb{Q}[x]$. Prove: $f_{1}, \cdots, f_{n}$ are irreducible if and only if they have no common root (roots can be real or complex).
(ii) Let non-zero polynomials $f_{1}, \cdots, f_{n} \in \mathbb{Z}[x]$, and at least one of them is primitive. Prove: $f_{1}, \... | None
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20. A complex number $\alpha$ is called an algebraic number if there exists a non-zero $f(x) \in \mathbb{Q}[x]$ such that $f(\alpha)=0$; $\alpha$ is called an algebraic integer if there exists a monic polynomial $g(x) \in \mathbb{Z}[x]$ such that $g(\alpha)=0$. Prove:
(i) If $\alpha$ is an algebraic number, then there ... | None
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21. Let $\alpha$ be an algebraic number. Define the set of polynomials
$$P(\alpha)=\{f(x): f(x) \in \mathbb{Q}[x], f(\alpha)=0\}.$$
(i) Prove: For non-zero polynomials in $P(\alpha)$, the following three properties of $h(x)$ are equivalent: (a) $h(x)$ is the polynomial of lowest degree in $P(\alpha)$; (b) $f(x) \in P(\... | None
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22. (i) Prove: A complex number $\alpha$ is a quadratic algebraic number if and only if
$$\alpha=r+s \sqrt{d},$$
where $r, s$ are rational numbers, $s \neq 0, d \in \mathbb{Z}, d \neq 0,1$ and $d$ is square-free.
(ii) Prove: $\alpha$ is a quadratic algebraic integer if and only if, in addition to the conditions in (i)... | None
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26. Let $\alpha_{0}, \alpha_{1} \in \boldsymbol{Z}[\sqrt{-1}]$, and not both zero. If $\Delta \in \boldsymbol{Z}[\sqrt{-1}]$, satisfies:
(1) $\Delta$ is a common divisor of $\alpha_{0}, \alpha_{1}$, i.e., $\Delta\left|\alpha_{0}, \Delta\right| \alpha_{1}$;
(2) For any common divisor $\delta$, i.e., $\delta\left|\alpha_... | None
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9. Let $a>b>1, k$ be obtained when taking $u_{0}=a, u_{1}=b$ in Theorem 4 of $\S 3$, and $h$ be obtained when taking $v_{0}=a, v_{1}=b$ in Problem 2. Prove: $h \leqslant k$. | 9. Use induction on $k$, and by sequentially comparing the relationships of $u_{j}, v_{j}$, find the pattern. | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,733 |
6. Let $a \geqslant 2$ be a given positive integer. Prove:
(i) For any positive integer $n$, there must be $n < a^{n}$;
(ii) For any positive integer $n$, there must be a unique integer $k \geqslant 0$, such that $a^{k} \leqslant n < a^{k+1}$. | 6. (i) Prove by mathematical induction; (ii) Consider the set $T=\left\{k: a^{k} \leqslant n, k \in \boldsymbol{Z}\right\}$, and apply the conclusion from question 5 to $T$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,744 |
10. . Let $p$ be an odd prime, and $q$ be a prime factor of $2^{p}-1$. Prove: $q=2 k p+1$. | 10. $q\left|2^{q-1}-1 . q\right| 2^{(p, q-q-1)}-1$. Therefore $p \mid q-1$. This, together with $q$ being odd, implies $q=2 k p+1$. | q=2kp+1 | Number Theory | proof | Yes | Yes | number_theory | false | 739,745 |
[8.1] In a competition, three problems $A, B, C$ were given. It is known:
(i) Among all the participating students, 25 people each solved at least one problem;
(ii) Among the students who did not solve problem $A$, the number of students who solved problem $B$ is twice the number of students who solved problem $C$; $\s... | None
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Theorem $1 a_{j} \mid c(1 \leqslant j \leqslant k)$ if and only if $\left[a_{1}, \cdots, a_{k}\right] \mid c$. This means that a common multiple must be a multiple of the least common multiple. This is the essential property of the least common multiple. | Theorem 1 Proof: The sufficiency is obvious. Next, we prove the necessity. Let $L=\left[a_{1}, \cdots, a_{k}\right]$. By Theorem 1 in §3, there exist $q, r$ such that
$$c=q L+r, \quad 0 \leqslant r<L$$
From this and $a_{j} \mid c$, it follows that $a_{j} \mid r(1 \leqslant j \leqslant k)$, so $r$ is a common multiple.... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,778 |
[20.3] Let $\boldsymbol{N}$ be the set of natural numbers, and $f, g$ be strictly increasing functions from $\boldsymbol{N} \rightarrow \boldsymbol{N}$. It is known that the union $f(\boldsymbol{N}) \cup g(\boldsymbol{N})=\boldsymbol{N}$, the intersection $f(\boldsymbol{N}) \cap g(\boldsymbol{N})$ is the empty set, and... | None
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[21.6] Let $A$ and $E$ be two opposite vertices of a regular octagon. A frog starts jumping from point $A$. If the frog is at any vertex other than $E$, it can jump to either of the two adjacent vertices. When it reaches vertex $E$, it stops there. Let $a_{n}$ be the number of different paths that the frog can take to ... | None
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[22.4] (i) For which positive integers $n>2$ is it possible to have $n$ consecutive positive integers such that the largest positive integer is a divisor of the least common multiple of the other $n-1$ numbers?
(ii) For which $n$ is the above sequence of $n$ consecutive positive integers unique? | None
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[23.1] Let $f(n)$ be a function defined on the set of positive integers, with values in the set of non-negative integers, and satisfying:
(i) $f(2)=0, f(3)>0, f(9999)=3333$;
(ii) For any $m, n$, $f(m+n)-f(m)-f(n)=0$ or 1. Find $f(1982)$. | None
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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 | math-word-problem | Yes | Yes | number_theory | false | 739,786 |
[31.5] Given an initial integer value $n_{0}>1$, two players $A, B$ take turns to choose integers $n_{1}, n_{2}, n_{3}, \cdots$ according to the following rules: When $n_{2 k}$ is known, $A$ can choose any integer $n_{2 k+1}$ such that $n_{2 k} \leqslant n_{2 k+1} \leqslant n_{2 k}^{2}$; When $n_{2 k+1}$ is known, $B$ ... | None
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Theorem 4 (i) $\left(a_{1}, a_{2}, a_{3}, \cdots, a_{k}\right)=\left(\left(a_{1}, a_{2}\right), a_{3}, \cdots, a_{k}\right)$;
(ii) $\left(a_{1}, \cdots, a_{k+r}\right)=\left(\left(a_{1}, \cdots, a_{k}\right),\left(a_{k+1}, \cdots, a_{k+r}\right)\right)$.
Conclusion (i) means that finding the greatest common divisor of... | Theorem 4 Proof We prove (i). If $d \mid a_{j}(1 \leqslant j \leqslant k)$, then by Theorem $2(k=2)$, we know that $d\left|\left(a_{1}, a_{2}\right), d\right| a_{j}(3 \leqslant j \leqslant k)$; conversely, if $d\left|\left(a_{1}, a_{2}\right), d\right| a_{j}$ $(3 \leqslant j \leqslant k)$, then by definition, $d \mid a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,811 |
[33.6] For each positive integer $n$, let $S(n)$ denote the largest integer such that for every positive integer $k \leqslant S(n)$, $n^2$ can be expressed as the sum of $k$ positive integer squares.
(i) Prove that for each $n \geqslant 4$, $S(n) \leqslant n^2 - 14$;
(ii) Find a positive integer $n$ such that $S(n) = n... | None
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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 | 739,814 |
[34.6] Let $n>1$ be an integer, and let $n$ lamps $L_{0}, L_{1}, \cdots, L_{n-1}$ be arranged in a circle. Each lamp can be in one of two states: "on" or "off". A series of steps $S_{0}$, $S_{1}, \cdots, S_{j}, \cdots$ are performed, with each step $S_{j}$ affecting the state of lamp $L_{j}$ according to the following ... | None
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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 | Combinatorics | proof | Yes | Yes | number_theory | false | 739,817 |
$[35.3]$ For any positive integer $k$, let $A_{k}$ denote the subset of elements in the set $\{k+1, k+2, \cdots, 2 k\}$ that satisfy the following condition: their binary representation contains exactly three 1s. Let the number of elements in $A_{k}$ be denoted by $f(k)$.
(i) Prove: For any positive integer $m, f(k)=m$... | None
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Theorem 5 If $(m, a)=1$, then $(m, a b)=(m, b)$. This means that when finding the greatest common divisor of $m$ and another number, the factor in the other number that is coprime with $m$ can be removed. | Theorem 5 Proof: When $m=0$, $a= \pm 1$, the conclusion is obviously true. When $m \neq 0$, by Theorem 8, Theorem 3, and Theorem 4 in §2, we have
$$(m, b)=(m, b(m, a))=(m,(m b, a b))=(m, m b, a b)=(m, a b).$$
This completes the proof of the desired conclusion. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,822 |
[37.1] Let $A B C D$ be a rectangle with side lengths $|A B|=20,|B C|=12$. Divide this rectangle into $20 \times 12$ unit squares. Let $r$ be a given positive integer. Place a coin in one of the unit squares, and the coin can move from one square to another if and only if the distance between the centers of these two s... | None
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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 | 739,824 |
Theorem 6 If $(m, a)=1$, then if $m \mid a b$, it follows that $m \mid b$. This means that if a number is divisible by $m$, then removing a factor that is coprime with $m$ from this number still results in a number divisible by $m$. | Theorem 6 Proof: By Theorem 8 of § 2 and Theorem 5, we have \( |m| = (m, ab) = (m, b) \), which implies \( m \mid b \).
Theorems 5 and 6 are frequently used. For example, when \( m \) is odd, Theorem 5 implies \( \left(m, 2^{k} b\right) = (m, b) \), and Theorem 6 implies: if \( m \mid 2^{k} b \), then \( m \mid b \).
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,833 |
[43.4] Let $n>1$ be a positive integer, and let its positive divisors be $d_{1}, d_{2}, \cdots, d_{k-1}, d_{k}$, satisfying $1=d_{1}<d_{2}<\cdots<d_{k-1}<d_{k}=n$. Define
$$D=d_{1} d_{2}+d_{2} d_{3}+\cdots+d_{k-1} d_{k}$$
(i) Prove: $D<n^{2}$;
(ii) Determine all $n$ such that $D$ divides $n^{2}$. | None
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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 | 739,839 |
Theorem $7\left[a_{1}, a_{2}\right]\left(a_{1}, a_{2}\right)=\left|a_{1} a_{2}\right|$. This means, the least common multiple of two numbers multiplied by their greatest common divisor is equal to the absolute value of the product of these two numbers. | Theorem 7 Proof: First, assume $\left(a_{1}, a_{2}\right)=1$. Let $L=\left[a_{1}, a_{2}\right]$. By Theorem 1, we know $L \mid a_{1} a_{2}$. On the other hand, since $a_{1} \mid L$, we have $L=a_{1} L^{\prime}$. Furthermore, by $a_{2} \mid L=a_{1} L^{\prime}$ and $\left(a_{2}, a_{1}\right)=1$, by Theorem 6, we know $a_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,844 |
Theorem 1 (i) $a|b \Longleftrightarrow -a| b \Longleftrightarrow a|-b \Longleftrightarrow| a|||b|$.
(ii) $a \mid b$ and $b|c \Rightarrow a| c$.
(iii) $a \mid b$ and $a \mid c \Longleftrightarrow$ for any $x, y \in \boldsymbol{Z}$, $a \mid b x+c y$.
In general, $a\left|b_{1}, \cdots, a\right| b_{k}$ hold simultaneously... | To prove (i): $b=a q \Leftrightarrow b=(-a)(-q) \Leftrightarrow-b=a(-q) \Longleftrightarrow|b|=|a||q|$.
To prove (ii): Since $b=a q_{1}$ and $c=b q_{2}$ can lead to $c=a\left(q_{1} q_{2}\right)$, this proves (ii).
To prove (iii): From $b=a q_{1}, c=a q_{2}$, we can derive $b x+c y=a\left(q_{1} x+q_{2} y\right)$, whic... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,855 |
Example 1 ([9.6]) A sports competition lasted for $n$ days, and a total of $m$ medals were awarded. It is known that on the first day, 1 medal was awarded plus $1 / 7$ of the remaining $m-1$ medals; on the second day, 2 medals were awarded plus $1 / 7$ of the remaining medals; each day the medals were awarded in this m... | Solve: The problem requires finding $n$ and $m$. The conditions that $n$ and $m$ must satisfy are given through the conditions that the number of medals awarded on the $k$-th day must meet. Additionally, it is known that $1 < n < m$ (why).
(i) Formulate the system of equations. Let the number of medals awarded on the $... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,864 |
Example 2 ([18.4]) Find the maximum value of the product of positive integers whose sum is 1976. | Solve: Represent 1976 as the sum of several positive integers:
$1976=x_{1}+\cdots+x_{k}$, where the number of terms $k$ and each term $x_{1}, \cdots, x_{k}$ are positive integers.
Since $1 \leqslant k \leqslant 1976,1 \leqslant x_{j} \leqslant 1976,1 \leqslant j \leqslant k$, there are only a finite number of such rep... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,865 |
Example 1 Let $p$ be a prime. Prove:
(i) $p \left\lvert\,\binom{ p}{j}\right., 1 \leqslant j \leqslant p-1$, where $\binom{p}{j}$ denotes the binomial coefficient;
(ii) For any positive integer $a, p \mid a^{p}-a$;
(iii) If $(a, p)=1$, then $p \mid a^{p-1}-1$. | Given the combination number
$$\binom{p}{j}=\frac{p!}{j!(p-j)!}$$
is an integer, i.e., $j!(p-j)!\mid p!$ (this result is derived from combinatorial theory, and will be directly proven in Theorem 4 of § 7). Since $p$ is a prime number, for any $1 \leqslant i \leqslant p-1$, we have $(p, i)=1$. Therefore, by Theorem 5, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,867 |
Example $4([34.1])$ Let $n>1$ be an integer, $f(x)=x^{n}+5 x^{n-1}+3$. Prove: $f(x)$ cannot be expressed as the product of two integer-coefficient polynomials, both of which have degrees no less than one. | Solve by contradiction. Assume that $f(x)$ can be expressed in this way, i.e., let
$$f(x)=x^{n}+5 x^{n-1}+3=g(x) h(x),$$
where the integer-coefficient polynomials $g(x), h(x)$ are respectively
$$\begin{array}{c}
g(x)=x^{l}+a_{l-1} x^{l-1}+\cdots+a_{1} x+a_{0} \\
h(x)=x^{m}+b_{m-1} x^{m-1}+\cdots+b_{1} x+b_{0} \\
1 \le... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,868 |
Example 5([25.6]) Let $a, b, c, d$ be positive odd integers, and satisfy $a<b<c<d, ad=bc$. Prove: If there are positive integers $k, m$, such that $a+d=2^{k}, b+c=2^{m}$, then $a=1$.
---
The translation maintains the original text's format and line breaks as requested. | Given the positive odd numbers $a, b, c, d$ such that $ac - b > 0$ and $(d - a)^2 > (c - b)^2$. From these and the condition $ad = bc$, it follows that $(d + a)^2 > (c + b)^2$, which implies
$$a + d > b + c.$$
From this and equation (2), condition (1) leads to the important numerical relationship
$$k > m \geqslant 3.$... | a = 1 | Number Theory | proof | Yes | Yes | number_theory | false | 739,869 |
Example $6([47.4])$ Find all integer pairs $\{x, y\}$ such that $1+2^{x}+2^{2 x+1}=y^{2}$. | Solve: First, exclude simple cases. $x$ cannot be a negative integer (why). When $x=0$, $y=\pm 1$. Below, we discuss the case where $x \geqslant 1$.
(i) $y$ must be an odd number, so $y^{2}$ is a positive integer of the form $8k+1$. This implies $x \geqslant 3$ (why). We will assume $y$ is positive for now.
(ii) Rewrit... | x=0, y=\pm 2; x=4, y=\pm 23 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,870 |
Example 7([43.4]) Let $n$ be a positive integer greater than 1, and let its all positive divisors be $d_{1}, d_{2}, \cdots, d_{k}$, satisfying $1=d_{1}<d_{2}<\cdots<d_{k}=n$. Let $D=d_{1} d_{2}+d_{2} d_{3}+\cdots+d_{k-1} d_{k}$.
(i) Prove: $D<n^{2}$;
(ii) Determine all $n$ such that $D$ divides $n^{2}$. | Using the knowledge from sections $\S 1$ and $\S 2$ of Chapter 1, we can prove (i). The proof of (i) relies on the fundamental properties of the divisors of a positive integer $n$: If $d_{1}, d_{2}, \cdots, d_{k}$ are all the (positive) divisors of $n$, then $n / d_{1}, n / d_{2}, \cdots, n / d_{k}$ are also all the (p... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,871 |
Example $\mathbf{8}([35.4])$ Find all pairs of positive integers $\{m, n\}$ such that $m n-1 \mid n^{3}+1$ is an integer.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | In the given form, $m, n$ are asymmetric, which brings difficulty to solving the problem. Noting that both $m, n$ are coprime with $m n-1$, we can use the analysis method from Example 1 ([9.6]) to derive an equivalent divisibility relation to $m n-1 \mid n^{3}+1$ through algebraic operations, thus proving that $m, n$ a... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,872 |
Example 9([47.5]) Let $P(x)$ be an integer-coefficient polynomial of degree $n(n>1)$, and $k$ be a positive integer. Consider the iterated polynomial $Q(x)=P(P(\cdots P(P(x)) \cdots))$, where $P$ appears $k$ times. Prove: There are at most $n$ integers $t$ such that $Q(t)=t$.
---
The translation maintains the origina... | A point $a$ that satisfies $f(a)=a$ is called a fixed point of $f$, i.e., $a$ is a root of $f(x)-x$. Typically, $Q(x)$ is called the $k$-th iteration of $P(x)$, denoted as $P_{(k)}(x)$. For any positive integer $k$, this problem requires proving that the $k$-th iteration $P_{(k)}(x)$ of an $n$-degree integer-coefficien... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,873 |
Example 10 ([37.3]) Let $S$ be the set of all non-negative integers. Find all functions $f(m)$ defined on $S$ that take values in $S$ and satisfy the condition $f(m+f(n))=f(f(m))+f(n)$ for all $m, n \in S$. | (i)
$$\begin{array}{l}
f(f(0))=f(0+f(0))=f(f(0))+f(0) \Rightarrow f(0)=0 \\
\Rightarrow f(f(m))=f(m) \Rightarrow f(m+f(n))=f(m)+f(n), \\
f(m+f(n))=f(f(m))+f(n) \\
\Leftrightarrow f(m+f(n))=f(m)+f(n), f(0)=0 .
\end{array}$$
(ii) $f(2 f(n))=2 f(n)$. By induction, we get $f(k f(n))=k f(n)$.
(iii) From $f(f(m))=f(m)$, we ... | f(n)=([n / a]+g(r)) a | Algebra | math-word-problem | Yes | Yes | number_theory | false | 739,874 |
Example 12([28.6]) Let $n \geqslant 2$ be a positive integer. Prove: If $k^{2}+k+n$ is prime for $0 \leqslant k \leqslant (n / 3)^{1 / 2}$, then $k^{2}+k+n$ is also prime for $0 \leqslant k \leqslant n-2$. | Let $f(k)=k^{2}+k+n$. The problem is to prove its contrapositive: if there exists some $k(0 \leqslant k \leqslant n-2)$ such that $f(k)$ is not a prime, then there must be a $k_{0}\left(0 \leqslant k_{0} \leqslant\right.$ $\left.(n / 3)^{1 / 2}\right)$ such that $f\left(k_{0}\right)$ is not a prime. That is, if the set... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,876 |
Example 2 Prove: (i) $(a, u v)=(a,(a, u) v)$;
(ii) $(a, u v) \mid(a, u)(a, v)$;
(iii) if $(u, v)=1$, then $(a, u v)=(a, u)(a, v)$. | Proof: By Theorem 8(i) and (iii) of § 2, Theorem 4, and Theorem 3, we have
$$(a, u v)=(a, u v, a v)=(a,(u v, a v))=(a,(a, u) v)$$
By Theorem 2 and Theorem 3, we get
$$(a,(a, u) v) \mid((a, u) a,(a, u) v)=(a, u)(a, v) .$$
From this and (i), we obtain (ii). It is evident that (i) is a generalization of Theorem 5. The p... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,877 |
Example 14([13.3]) Proof: In the sequence $2^{n}-3(n=2,3, \cdots)$, there must exist an infinite subsequence, in which any two numbers are coprime. | Prove (i) For any odd number $a$, there must be a $d \geqslant 2$ such that $a \mid 2^{d}-1$, and $a$ and $2^{d}-3$ are coprime (Chapter 1, §3, Example 5).
(ii) Based on (i), use induction to construct the sequence. Take $a_{1}=2^{2}-3$. If $a_{1}, \cdots, a_{k}$ have been chosen, then take
$a_{k+1}=2^{d(k+1)}-3$, wher... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,879 |
Example $\mathbf{1 5}$ ([46.2]) Let $a_{1}, a_{2}, \cdots$ be a sequence of integers, where there are infinitely many positive integers and infinitely many negative integers. Prove: If for every positive integer $n$, the integers $a_{1}, a_{2}, \cdots, a_{n}$ yield distinct remainders when divided by $n$, then each int... | Prove that the condition given in this problem, "for every positive integer $n$, the integers $a_{1}, a_{2}, \cdots, a_{n}$ yield distinct remainders when divided by $n$", is very strong and strictly limits the values of each $a_{j}$. From this, we can deduce the following properties:
(i) No two numbers in the sequenc... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,880 |
Example 16([49.3]) Prove: There exist infinitely many positive integers $n$, such that $n^{2}+1$ has a prime factor greater than $2 n+(2 n)^{1 / 2}$. | Consider $n$ is an even number greater than 2. If $n^{2}+1$ is a prime $p$, then $p>2 n+(2 n)^{1 / 2}$. If $n^{2}+1$ is not a prime, then it has an odd prime factor $p$. By the division algorithm, we have
$$n=q p+r, \quad|r| \leqslant p / 2$$
Let $n_{1}=|r|$, from the above equation, $p$ divides $n_{1}^{2}+1$. Here we... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,881 |
Example 17 ([12.4]) Find all positive integers $n$ such that the six numbers $n$, $n+1, n+2, n+3, n+4, n+5$ can be divided into two groups, with the product of the numbers in one group equal to the product of the numbers in the other group. | Solving this problem, dividing these 6 numbers into two groups has the following three scenarios:
(i) One group has 1 number, and the other group has 5 numbers. This is obviously impossible.
(ii) One group has 2 numbers, and the other group has 4 numbers. When \( n=1 \), these 6 numbers are 1, 2, 3, 4, 5, and 6, so it ... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,882 |
Example 18([27.1]) Let $d$ be a positive integer such that $d \neq 2,5,13$. Prove: In the set $\{2,5,13, d\}$, there must exist two distinct elements $a, b$ such that $a b-1$ is not a perfect square. | To prove that it is easy to verify, take any two numbers from $2, 5, 13$ as $a, b$, then $ab-1$ must be a perfect square. Therefore, the selection of $a, b$ must be one as $d$, and the other chosen from $2, 5, 13$. We will use proof by contradiction to show that there must be a selection such that $ab-1$ is not a perfe... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,883 |
Example 19([23.4(ii)]) Prove: When $n=2891$, the equation $x^{3}-3 x y^{2}+y^{3}=n$ has no integer solutions. | To be concise, use the congruence symbol. If there is a solution to \(2891 = x^3 - 3xy^2 + y^3\), taking both sides modulo 3, we have
$$2 \equiv 2891 = x^3 - 3xy^2 + y^3 \equiv x + y \pmod{3}$$
Thus, \(x = 3k + 2 - y\). Taking both sides modulo 9, we get
$$\begin{array}{c}
2 \equiv 2891 = x^3 - 3xy^2 + y^3 = (3k + 2 -... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,884 |
Example 20([37.4]) Let positive integers $a, b$ be such that $15a + 16b$ and $16a - 15b$ are both squares of positive integers. Find the smallest possible values of these two squares. | Let $15a + 16b = x^2$, $16a - 15b = y^2$ ($x, y \in \mathbb{N}$). Solving for $a, b$ gives:
$$15x^2 + 16y^2 = (13 \cdot 37)a, \quad 16x^2 - 15y^2 = (13 \cdot 37)b$$
We need to prove that $x, y$ are both multiples of $13 \cdot 37$. 13 and 37 are both prime numbers. The solution uses the concept of modular inverses (but... | (13 \cdot 37)^2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,885 |
Example 21([31.2]) Let $n \geqslant 3$ be a positive integer, and let $E$ be a set of $2 n-1$ distinct points on the same circle. Some of the points in $E$ are colored black, and the rest are not colored. If there is at least one pair of black points such that one of the two arcs with these points as endpoints has exac... | Let the $2n-1$ points be numbered clockwise as $1, 2, 3, \cdots, 2n-1$.
(i). Take the modulus $m = 2n-1$, and under modular arithmetic, assign each point infinitely many labels. Label the $2n-1$ points cyclically starting from point 1 in the clockwise direction: 1, 2, 3, ..., $2n-2, 2n-1; 2n, 2n+1, \cdots$. Labels in t... | k = n-1 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 739,886 |
Example 22([39.4]) Find all positive integer pairs $\{a, b\}$, such that $a^{2} b+a+b$ is divisible by $a b^{2}+b+7$.
| (i) First, analyze the size relationship between $a$ and $b$. Since it must be that $a^{2} b + a + b \geqslant a b^{2} + b + 7$, we have
$$a - 7 \geqslant a b (b - a).$$
This implies that (why)
$$a \geqslant b \geqslant 1.$$
We need to subtract a multiple of $a b^{2} + b + 7$ from $a^{2} b + a + b$ so that the absolu... | \{a, b\} = \{7 t^{2}, 7 t\}, \quad t \text{ is any positive integer.} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,887 |
Example 3 Let $k$ be a positive integer. If the $k$-th power of a rational number is an integer, then this rational number must be an integer. | Proof: Without loss of generality, let this rational number be $b / a, a \geqslant 1,(a, b)=1$. If $(b / a)^{k}=c$ is an integer, then $c a^{k}=b^{k}$, so $a \mid b^{k}$. Since $(a, b)=1$, it follows from Theorem 6 that $a \mid b$, hence $1=(a, b)=a$. This proves the desired result. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,888 |
Example 24([43.3]) Find all pairs of positive integers $\{m, n\}(m \geqslant 3, n \geqslant 3)$ such that there exist infinitely many positive integers $a$ for which the value of $\frac{a^{m}+a-1}{a^{n}+a^{2}-1}$ is an integer. | First, if the conclusion holds, then it must be that $m>n$ (why). By the division algorithm for polynomials, there exist integer-coefficient polynomials $q(x)$ and $r(x)$ (with degree less than $n$), such that
$$x^{m}+x-1=q(x)\left(x^{n}+x^{2}-1\right)+r(x) \text {. }$$
If there are infinitely many positive integers $... | m=5, n=3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,890 |
Example 25([46.4]) Let the sequence $a_{1}, a_{2}, \cdots$ be defined as follows: $a_{n}=2^{n}+3^{n}+6^{n}-1$ $(n=1,2, \cdots)$. Find all integers that are coprime with every term of this sequence. | It is evident that this problem can be reduced to finding all prime numbers that are coprime with each term of the sequence, i.e., primes that do not divide any term of the sequence. From \(a_{1}=10, a_{2}=48\), we know that we can consider primes \(p \geqslant 5\). For the integer sequence \(a^{n} (n=1,2, \cdots)\) an... | 1, -1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,891 |
Example 26([22.4]) (i) For which positive integers $n>2$ is it possible to have $n$ consecutive positive integers such that the largest positive integer is a divisor of the least common multiple of the other $n-1$ numbers?
(ii) For which $n$ is the sequence of $n$ consecutive positive integers in (i) unique? | First, we need to choose a convenient form for the $n$ consecutive positive integers to facilitate discussion. According to requirement (i), we should take them as:
$$K, K-1, \cdots, K-(n-1), K \geqslant n>2 \text{.}$$
Thus, the problem is to find a positive integer $K$ that satisfies:
$$K \mid [K-1, \cdots, K-(n-1)]$... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,892 |
Example 27 ([38.5]) Find all positive integer pairs $\{a, b\}$, satisfying the equation
$$a^{b^{2}}=b^{a}.$$ | (i) Discuss the simplest case. If $a=b$, then $a=1$; if $a=1$, then $b=1$; if $b=1$, then $a=1$. Therefore, we only need to consider
$$a \neq b, \quad a \geqslant 2, \quad b \geqslant 2$$
(ii) Discuss the relationship between $a$ and $b$. First, prove that it is impossible to have $b > a \geqslant 2$. It is easy to pro... | \{1,1\}, \{16,2\}, \{27,3\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,893 |
Example 28([39.3]) For any positive integer $n$, let $d(n)$ denote the number of all positive divisors (including 1 and $n$) of $n$. Determine all possible positive integers $k$ such that there exists a positive integer $n$ satisfying: $d\left(n^{2}\right) / d(n)=k$.
---
The translation maintains the original text's ... | (i) Let $n=p_{1}^{a(1)} \cdots p_{r}^{a(r)}$, we have
$$\begin{aligned}
k & =d\left(n^{2}\right) / d(n) \\
& =(1+2 a(1)) \cdots(1+2 a(r)) /[(1+a(1)) \cdots(1+a(r))]
\end{aligned}$$
Thus, $k$ must be an odd number. Therefore, $n$ must be a perfect square.
(ii) Now we prove that any odd number $k$ can be expressed in th... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,894 |
Example 29([42.6]) Let $a, b, c, d$ be integers, $a>b>c>d>0$, and satisfy
$$a c+b d=(b+d+a-c)(b+d-a+c) .$$
Prove: $a b+c d$ is not a prime. | The given second condition can be transformed into
$$a^{2}-a c+c^{2}=b^{2}+b d+d^{2} .$$
To relate the condition to $(a b+c d)$, multiplying both sides of equation (1) by $a c+b d$ yields
$$\begin{aligned}
(a c+b d)\left(a^{2}-a c+c^{2}\right) & =a c\left(b^{2}+b d+d^{2}\right)+b d\left(a^{2}-a c+c^{2}\right) \\
& =(a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,895 |
Example 30([32.2]) Let $n>6$ be an integer, and $a_{1}, a_{2}, \cdots, a_{k}$ be all positive integers less than $n$ and coprime with $n$. Prove: If $a_{2}-a_{1}=a_{3}-a_{2}=\cdots=a_{k}-a_{k-1}=d>0$, then $n$ is either a prime or equal to $2^{s}$, where the integer $s \geqslant 3$. | The condition of this problem states that the smallest positive reduced residue system modulo $n$ forms an arithmetic sequence. First, we prove that under the given condition, the following three properties hold:
(i) It is evident that $a_{1}=1, a_{k}=n-1$, hence we have
$$a_{j}=1+(j-1) d \quad \text { and } \quad a_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,896 |
Example 4 Let $k$ be a positive integer. Prove:
(i) $\left(a^{k}, b^{k}\right)=(a, b)^{k}$;
(ii) Let $a, b$ be positive integers. If $(a, b)=1, a b=c^{k}$, then $a=(a, c)^{k}, \quad b=(b, c)^{k}$. | Proof: By Theorem 3, we have
$$\left(a^{k}, b^{k}\right)=(a, b)^{k}\left(\left(\frac{a}{(a, b)}\right)^{k},\left(\frac{b}{(a, b)}\right)^{k}\right)$$
By Theorem 10 in §2, we know
$$\left(\frac{a}{(a, b)}, \frac{b}{(a, b)}\right)=1$$
From the above and Theorem 5, we get
$$\left(\left(\frac{a}{(a, b)}\right)^{k},\left(... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,899 |
(i) There must exist a positive integer $d \leqslant m-1$, such that
$$m \mid a^{d}-1 \text {, i.e., } a^{d}-1 \equiv 0(\bmod m) \text {. }$$
(ii) Let $d_{0}$ be the smallest positive integer $d$ that satisfies (i), then a positive integer $h$ satisfies $m \mid a^{h}-1$, i.e., $a^{h}-1 \equiv 0(\bmod m)$ if and only if... | Consider $m$ numbers $a^{0}, a^{1}, \cdots, a^{m-1}$. They are all not divisible by $m$, so the $m$ non-negative remainders obtained when they are divided by $m$ must take values in the $m-1$ numbers $1,2, \cdots, m-1$. Therefore, there must be two numbers with the same remainder, let them be $a^{i}, a^{j}(0 \leqslant ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,900 |
Property II Suppose $(a, m)=1$ (or there exist $x_{0}$ and $y_{0}$, such that $a x_{0}+m y_{0}=1$), $m \geqslant 2$. If there is a positive integer $d$, such that $m \mid a^{d}+1$, let $\delta_{m}^{-}(a)$ be the smallest such $d$, then when $m \mid a^{h}-1$ or $m \mid a^{h}+1$, we have $\delta_{m}^{-}(a) \mid h$. Speci... | To prove (i) is obviously true. Now let's prove (ii). Let $\delta_{m}^{+}(a)=\delta(+), \delta_{m}^{-}(a)=\delta(-)$. First, we prove $\delta(+)>\delta(-)$. If not, then $\delta(+) \leqslant \delta(-)$. By $a^{\delta(-)}+1=$ $a^{\delta(+)}\left(a^{\delta(-)-\delta(+)}+1\right)-\left(a^{\delta(+)}-1\right)$, and $m \mid... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,901 |
Example 32([41.5]) Does there exist a positive integer $n$ satisfying the conditions: (i) $n$ divides $2^{n}+1$; (ii) $n$ has exactly 2000 distinct prime factors? | (a) From the problem, the number 2000 has no special significance, it is only because the competition was held in the year 2000. The 2000 in condition (ii) can generally be changed to any given positive integer $k$.
(b) First, the positive integer $n$ that satisfies condition (i) must be odd, and $n=1,3$ satisfy this c... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,902 |
Example 33([40.4]) Find the set of positive integer pairs $\{p, n\}$, satisfying the conditions:
(i) $p$ is a prime; $\square$
(ii) $n \leqslant 2 p$;
(iii) $(p-1)^{n}+1$ is divisible by $n^{p-1}$. | For any prime $p$, the pair $\{p, 1\}$ is a solution; when $p=2$, the only solutions are the pairs $\{2,1\},\{2,2\}$. Therefore, we only need to discuss the case $p \geqslant 3, n>1$.
If the pair $\{p, n\}$ is a solution, $p \geqslant 3, n>1$, then $n$ must be an odd number. We need to further discuss the conditions t... | n=p=3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,903 |
Example 34([44.6]) Let $p$ be any given prime. Prove: there must exist a prime $q$, such that for any integer $n$, the number $n^{p}-p$ cannot be divisible by $q$.
保留源文本的换行和格式，直接输出翻译结果。 | From the problem, we know that the prime number \( q \) is only related to \( p \) and not to \( n \). Therefore, we need to analyze the conclusion "for any integer \( n \), the number \( n^p - p \) cannot be divisible by \( q \)" to directly link \( q \) and \( p \).
Since \( n^p - p = (n^p - 1) - (p - 1) \) and we h... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,904 |
Example 37([31.6]) Prove: There exists a convex 1990-gon that simultaneously has the following properties:
(i) All interior angles are equal;
(ii) The lengths of the 1990 sides are a permutation of the numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 1989^{2}, 1990^{2}$. | Suppose this polygon exists, with vertices $A_{1}, A_{2}, \cdots, A_{1990}$. Choose a Cartesian coordinate system, placing vertex $A_{1990}$ at the origin and edge $A_{1990} A_{1}$ along the $x$-axis. From condition (i), we know that each interior angle of this convex polygon is $1-\theta$, where $\theta=2 \pi / 1990$ ... | proof | Geometry | proof | Yes | Yes | number_theory | false | 739,907 |
Example 38([24.5]) Can we find 1983 distinct positive integers, each not greater than $10^{5}$, such that no three of them are consecutive terms of some arithmetic progression? | It is hard to think that this problem is related to numeral systems. Taking a ternary number with 11 digits is determined by the conditions "$10^{5}$" and "1983". Consider the set
$$M=\left\{n=\left(a_{1} \cdots a_{11}\right)_{3}, a_{j}=0,1\right\}$$
Thus, when $n \in M$, it must be true that
$$2 n=\left(b_{1} \cdots ... | proof | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 739,908 |
Example 39([31.4]) Let the set of all positive rational numbers be denoted as $Q^{+}$. Construct a function $f: Q^{+} \rightarrow Q^{+}$, such that for any $x, y \in Q^{+}$, it satisfies $y f(x f(y))=f(x)$. | Assume the function $f$ satisfies the given condition $y f(x f(y))=f(x)$. Let's first discuss the basic properties such a function $f$ should satisfy. Taking $x=1$, from the condition we get
$$f(f(y))=f(1) / y.$$
If $f(x)=f(y)$, then from this and equation (1) we derive
$$f(1) / y=f(f(y))=f(f(x))=f(1) / x, \text{ i.e.... | f\left(p_{2 j-1}\right)=1 / p_{2 j}, \quad f\left(p_{2 j}\right)=p_{2 j-1} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 739,909 |
Example 5 Let $m \geqslant 2, (m, a)=1$. Prove:
(i) There exists a positive integer $d \leqslant m-1$, such that $m \mid a^{d}-1$.
(ii) Let $d_{0}$ be the smallest positive integer $d$ satisfying (i), then $m \mid a^{h}-1(h \geqslant 1)$ if and only if $d_{0} \mid h$. We denote $d_{0}$ as $\delta_{m}(a)$, called the in... | Given $m \geqslant 2, (m, a)=1$, we know $m \nmid a$. From this and $(m, a)=1$, Theorem 6 implies $m \nmid a^{j}, j \geqslant 1$. Furthermore, by Theorem 1 in § 3, we have
$$a^{j}=q_{j} m+r_{j}, \quad 0<r_{j}<m$$
Thus, the $m$ remainders $r_{0}, r_{1}, \cdots, r_{m-1}$ can only take $m-1$ values, so there must be two ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,910 |
Example 40 ([35.6]) Find a set of positive integers $A$ such that: for any set $P$ consisting of infinitely many primes, there exist positive integers $m \in A$ and $n \notin A$ such that they are the products of the same number of different elements from $P$. | From the problem, we know that the numbers in $A$ can be composed of the product of different primes (i.e., square-free numbers). The arbitrariness of $P$ implies no restriction on the prime factors of the numbers in $A$. Since the condition requires these two positive integers to be "products of the same number of dif... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,911 |
Example 6 If $(a, b)=1$, then any integer $n$ can be expressed as $n=a x+b y, \quad x, y$ are integers. | From $(a, b)=1$ and Theorem 8, we know that there exist $x_{0}, y_{0}$ such that $a x_{0}+b y_{0}=1$. Therefore, taking $x=$ $n x_{0}, y=n y_{0}$ satisfies the requirement. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,912 |
Example 8 Let $n, k$ be positive integers, $(k, n)=1, 0<k<n$. Let the set $M=\{1, 2, \cdots, n-1\}$. Now, each number $i$ in the set $M$ is painted blue or white, satisfying the following conditions: (i) $i$ and $n-i$ must be painted the same color; (ii) when $i \neq k$, $i$ and $|k-i|$ must be painted the same color. ... | To prove: All $i \in M$ must have the same color as $k$. By Example 6, there exist $x$, $y$, such that
$$i=x k+y n$$
By $1 \leqslant i \leqslant n-1$, we know that the values of $x, y$ can fall into three cases:
(a) $x>0, y=0$;
(b) $x>0, y<0$.
In case (a), by condition (ii), $i$ and $k$ have the same color.
In case (... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,914 |
Example 9 Find the greatest common divisor of $198,252,924$, and express it as an integer linear combination of 198,252 and 924. | From Theorem 4(i) of §4 and Example 6 of §2, we know
From this and Example 6 of §2, we get: $(198,252,924)=6$,
$$\begin{aligned}
6 & =924-51 \cdot 18=924-51(4 \cdot 252-5 \cdot 198) \\
& =924-204 \cdot 252+255 \cdot 198 .
\end{aligned}$$
$$\begin{array}{l}
(198,252,924)=((198,252), 924)=(18,924) \text {. } \\
\begin{a... | 6 = 924 - 204 \cdot 252 + 255 \cdot 198 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,915 |
Example 1 Proof: If $3 \mid n$ and $7 \mid n$, then $21 \mid n$.
| Given $3 \mid n$, we know $n=3m$, so $7 \mid 3m$. From this and $7 \mid 7m$, we get $7 \mid (7m - 2 \cdot 3m) = m$, hence $21 \mid n$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,916 |
1. Let $p$ be a prime, $\left(a, p^{2}\right)=p,\left(b, p^{3}\right)=p^{2}$. Find $\left(a b, p^{4}\right), \quad\left(a+b, p^{4}\right)$. | 1. $\begin{aligned} a=p a_{1}, b=p^{2} b_{1},\left(a_{1}, p\right) & =\left(b_{1}, p\right)=1 .\left(a b, p^{4}\right)=p^{3}\left(a_{1} \cdot b_{1}, p\right)=p^{3} . \\ \left(a+b, p^{4}\right) & =p\left(a_{1}+p b_{1}, p^{3}\right)=p .\end{aligned}$ | p^3, p | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,917 |
2. Let $p$ be a prime, $(a, b)=p$. Find $\left(a^{2}, b\right),\left(a^{3}, b\right),\left(a^{2}, b^{3}\right)$ and the possible values they can take. | 2. $\begin{array}{l}a=p a_{1}, b=p b_{1},\left(a_{1}, b_{1}\right)=1 .\left(a^{2}, b\right)=p\left(p a_{1}^{2}, b_{1}\right)=p\left(p, b_{1}\right)=p, p^{2} . \\ \left(a^{3}, b\right)=p\left(a_{1}^{3} p^{2}, b_{1}\right)=p\left(p^{2}, b_{1}\right)=p, p^{2}, p^{3} . \\ \left(a^{2}, b^{3}\right)=p^{2}\left(a_{1}^{2}, p b... | p, p^2, p^3, p^2, p^3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,918 |
3. Determine whether the following statements are true. If true, provide a proof; if false, provide a counterexample:
(i) If $(a, b)=(a, c)$, then $[a, b]=[a, c]$;
(ii) If $(a, b)=(a, c)$, then $(a, b, c)=(a, b)$;
(iii) If $d|a, d| a^{2}+b^{2}$, then $d \mid b$;
(iv) If $a^{4} \mid b^{3}$, then $a \mid b$;
(v) If $a^{2... | 3. (i) Not true. For example, $a=b=1, c=2$. (ii) True. $(a, b, c)=((a, b), c)=((a, c), c)=(a, c)=(a, b)$. (iii) Not true. $d=4, a=4, b=2$. (iv) True. Since $a^{4} \mid b^{4}$, use Example 3 or Example 4. (v) Not true. $a=8, b=4$. (vi) True. See Example 3 or Example 4. (vii) True. $\left[a^{2}, b^{2}\right]=a^{2} b^{2} ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,919 |
6. Let $\theta=r \pi, r$ be a rational number. Prove: except for $\cos \theta=0, \pm 1 / 2, \pm 1$, $\cos \theta$ must be an irrational number, i.e., except for $\theta=k \pi, k \pi \pm \pi / 3, k \pi+\pi / 2$, $\cos \theta$ must be an irrational number, where $k$ is any integer (Hint: For any given positive integer $n... | 6. Prove the conclusion in the prompt by induction, utilizing
$$\cos (2 m+1) \alpha=\cos (2 m-1) \alpha \cos 2 \alpha-2 \cos \alpha(\cos (2 m-2) \alpha-\cos 2 m \alpha)$$
and when $2 \nmid n$, the constant term of $f_{n}(x)$ is zero; when $2 \mid n$, the constant term of $f_{n}(x)$ is $\pm 2$. Then use the result from... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,922 |
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