problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Example 5 Express $\frac{107}{95}$ as a continued fraction. | $$\begin{aligned}
\frac{107}{95} & =1+\frac{12}{95}=1+\frac{1}{\frac{95}{12}}=1+\frac{1}{7+\frac{11}{12}} \\
& =1+\frac{1}{7+\frac{1}{1+\frac{1}{11}}}=[1,7,1,11]
\end{aligned}$$
We have
$$\begin{aligned}
\frac{107}{95} & =1+\frac{12}{95}=1+\frac{1}{\frac{95}{12}}=1+\frac{1}{7+\frac{11}{12}} \\
& =1+\frac{1}{7+\frac{1}... | [1,7,1,11] | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,178 |
Example 6 Express $\frac{225}{43}$ as a continued fraction. | $$\begin{aligned}
\frac{225}{43} & =5+\frac{10}{43}=5+\frac{1}{\frac{43}{10}}=5+\frac{1}{4+\frac{3}{10}}=5+\frac{1}{4+\frac{1}{\frac{10}{3}}} \\
& =5+\frac{1}{4+\frac{1}{3+\frac{1}{3}}}=[5,4,3,3] .
\end{aligned}$$
We have | [5,4,3,3] | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,179 |
Lemma 3 Every rational number can be expressed as a finite continued fraction. | Let $\frac{a}{b}$ be a rational number, where $a$ is an integer and $b$ is a positive integer. Suppose $\frac{a}{b}$ is an integer, i.e., $\frac{a}{b}=c$, where $c$ is an integer, then we have $\frac{a}{b}=[c]$.
Suppose $\frac{a}{b}$ is not an integer, then there exist two integers $q_{1}$ and $r_{1}$, such that $a=$ ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,180 |
Lemma 6 Let $m$ be an integer greater than 1, and let $a_{1}, a_{2}, \cdots, a_{m}$ be a complete residue system modulo $m$, then when $b$ is an integer, $a_{1}+b, a_{2}+b, \cdots, a_{m}+b$ is also a complete residue system modulo $m$. | Proof: Suppose in $a_{1}+b, a_{2}+b, \cdots, a_{m}+b$ there exist two integers $a_{k}+b, a_{\lambda}+b$ (where $1 \leqslant k<\lambda \leqslant m$), such that
$$a_{k}+b \equiv a_{\lambda}+b(\bmod m)$$
We also have
$$b \equiv b(\bmod m)$$
Subtracting (5) from (4), we get
$$a_{k} \equiv a_{\lambda}(\bmod m)$$
By Lemma... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,183 |
Example 7 When $x$ is a real number, we have $0 \leqslant\{x\}<1$.
. | Let $x=n+y$, where $n$ is an integer and $0 \leqslant y<1$. From (II) and (III) we have
$$[x]=[n+y]=n+[y]=n \text {. }$$
From property (I) and equation (33) we have
$$\{x\}=x-[x]=n+y-n=y .$$
Since $0 \leqslant y<1$ and equation (34), we get $0 \leqslant\{x\}<1$. | 0 \leqslant\{x\}<1 | Inequalities | math-word-problem | Yes | Yes | number_theory | false | 741,184 |
Example 8 Prove that equations (3) and (4) hold.
Translate the text above into English, keep the original text's line breaks and format, and output the translation result directly. | Proof In Lemma 5, we take $a_{1}=b, 2 b=a_{2}=a_{3}=a_{4}=$ $a_{5}=\cdots$. By (2), we have
$$\sqrt{b^{2}+1}=[b, 2 b, 2 b, 2 b, 2 b, \cdots]$$
By (2) and Definition 1, we have
$$\begin{array}{l}
\frac{p_{3}}{q_{3}}=b+\frac{1}{2 b+\frac{1}{2 b}} \\
\frac{p_{4}}{q_{4}}=b+\frac{1}{2 b+\frac{1}{2 b+\frac{1}{2 b}}} \\
\fra... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,185 |
Column 11 Let $b$ be a real number $\geqslant 1$, please use Lemma 1 to find $p_{1}$ to $p_{8}$ and $q_{1}$ to $q_{8}$ in $[b, 2 \dot{b}]$ in terms of $b$. | $$\begin{aligned}
p_{1}= & b, p_{2}=2 b^{2}+1 \\
p_{3}= & 2 b\left(2 b^{2}+1\right)+b=4 b^{3}+3 b \\
p_{4}= & 2 b\left(4 b^{3}+3 b\right)+2 b^{2}+1=8 b^{4}+8 b^{2}+1 \\
p_{5}= & 2 b\left(8 b^{4}+8 b^{2}+1\right)+4 b^{3}+3 b=16 b^{5}+20 b^{3}+5 b \\
p_{6}= & 2 b\left(16 b^{5}+20 b^{3}+5 b\right)+8 b^{4}+8 b^{2}+1 \\
= &... | 128 b^{7}+192 b^{5}+80 b^{3}+8 b | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,188 |
Example 13 Prove that $\sqrt{2}=1.414213562373 \cdots$.
| Prove that in (39), taking $b=7$, and by Lemma 5 and in Example 11, taking
$$\frac{p_{7}}{q_{7}}<\sqrt{50}<\frac{p_{6}}{q_{6}}$$
we get
$$\begin{aligned}
\sqrt{50} & \geqslant \frac{64 \times 7^{7}+112 \times 7^{5}+56 \times 7^{3}+7^{2}}{64 \times 7^{6}+80 \times 7^{4}+24 \times 7^{2}+1} \\
& =\frac{54608393}{7722793}... | \sqrt{2} \leqslant 1.4142135623733 | Number Theory | proof | Yes | Yes | number_theory | false | 741,190 |
Example 14 Prove that $\sqrt{26}=5.099019513592 \cdots$.
| Proof: In (39), take $b=5$, and in Lemma 5 and Example 11, take
$$\frac{p_{7}}{q_{7}}<\sqrt{26}<\frac{p_{8}}{q_{8}}$$
we get
$$\begin{aligned}
\sqrt{26} & \leqslant \frac{128 \times 5^{8}+256 \times 5^{6}+160 \times 5^{4}+32 \times 5^{2}+1}{128 \times 5^{7}+192 \times 5^{5}+80 \times 5^{3}+40} \\
& =\frac{54100801}{10... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,191 |
Theorem 2. Let $p$ be a prime and $p \nmid a_{n}$, and
$$f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$$
be an integer-coefficient polynomial. Then the congruence equation
$$f(x) \equiv 0(\bmod p)$$
has at most $n$ solutions (counting multiplicities) $(\bmod p)$. | Proof. If $n=1$, from $p \nmid a_{1}$ we know that when $y$ runs through a complete residue system modulo $p$, $a_{1} y$ also runs through a complete residue system modulo $p$ (see "Elementary Number Theory" $\Pi, p$. 6, Lemma 7). Therefore, there must be a natural number $b$ such that
$$a_{1} b \equiv 1(\bmod p)$$
It... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,201 |
Theorem 1 (Wilson's Theorem) If $p$ is a prime, then $(p-1)!\equiv-1(\bmod p)$ | Prove: If $p=2$, the conclusion is obviously true. Therefore, we can assume $p>3$ is an odd prime. By Fermat's Little Theorem (see Elementary Number Theory II, p. 13, Theorem 2), for $x=1,2, \cdots, p-1$, we have
$$x^{p-1} \equiv 1(\bmod p)$$
holds, i.e., the congruence equation $\square$
$$x^{p-1}-1 \equiv 0(\bmod p)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,204 |
Lemma 1 If $(a, b)=d$, then there must be integers $x_{0}, y_{0}$ such that
$$d=a x_{0}+b y_{0} .$$ | Consider the set $D$ of all integers of the form $a x + b y$ (where $x, y$ are integers). There must be a smallest positive number $c$ in $D$. We will first prove that any integer in $D$ must be a multiple of $c$. Suppose the conclusion is not true, then there is at least one integer $r$ in $D$ such that $c \nmid r$. W... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,205 |
Theorem 3 Let $k>2$ be a natural number, then the number of solutions to the congruence equation
$$x^{k} \equiv 1(\bmod p)$$
is $(k, p-1)$, where $p$ is a prime number. | Let $d=(k, p^{-1})$, by Lemma 1, there must be two integers $s$ and $t$ such that
$$s k+t(p-1)=d$$
If $x_{0}$ is a solution to (3), then it must be true that
$$x_{o}^{d}=x_{o}^{s k} \cdot x_{o}^{\prime(p-1)}=\left(x_{o}^{k}\right)^{s} \cdot\left(x_{o}^{p-1}\right)^{\prime} \equiv 1(\bmod p),$$
Thus, $x_{o}$ must also... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,206 |
Lemma 2 For any natural number $n$, we have
$$\sum_{d \mid n} \varphi(d)=n .$$ | For any natural number $r \mid n$, we group the natural numbers $1,2, \cdots, n$ that satisfy the condition
$$(m, n)=r$$
into a set denoted as $S_{r}$. Clearly, for any two natural numbers
$$r_{1}\left|n, r_{2}\right| n, \quad r_{1} \neq r_{2}$$
the sets $S_{r_{1}}$ and $S_{r_{2}}$ do not contain the same natural num... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,208 |
Theorem 4 Let $p$ be a prime, $h$ an integer, $p \nmid h$, and the order of $h$ modulo $p$ be $l$. Then for any integer $d$ satisfying
$$(d, l)=1$$
the order of $h^{d}$ modulo $p$ is also equal to $l$. | First, it is easy to see that
$$\left(h^{d}\right)^{\prime}=\left(h^{\prime}\right)^{d} \equiv 1(\bmod p)$$
It remains to prove that for any $m(1<m<l-1)$, we have
$$\left(h^{d}\right)^{m} \not\equiv 1(\bmod p)$$
We use proof by contradiction. If there exists an $m(1 \leqslant m \leqslant l-1)$ such that
$$\left(h^{d}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,209 |
Inference 1 Let $p$ be a prime, if $h$ is not divisible by $p$, and the order of $h$ modulo $p$ is $l$, then in a complete residue system modulo $p$ there are exactly $\varphi(l)$ elements of order $l$. | To prove that among the $l$ numbers $1,2, \cdots, l$, there are exactly $\varphi(l)$ numbers that are coprime with $l$, denoted as $d_{1}, d_{2}, \cdots, d_{\varphi(l)}$. By Theorem 4, the following $\varphi(l)$ numbers
$$h^{d_{1}}, h^{d_{2}}, \cdots, h^{\left.d_{\varphi(}()\right)}$$
all have order $l$ modulo $p$. Ne... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,210 |
Theorem 5 If $p$ is a prime, then there must exist primitive roots modulo $p$, and there are exactly $\varphi(p-1)$ primitive roots. | By Fermat's Little Theorem, if \( p \nmid h \), it must be that
\[ h^{p-1} \equiv 1 \pmod{p} \]
According to Theorem 1, it must be that \( l \mid (p-1) \), where \( l \) is the order of \( h \) modulo \( p \). Conversely, for each natural number \( l \mid (p-1) \), in the reduced residue system modulo \( p \), \( 1, 2... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,211 |
Example 6 Suppose it is known that 2 is a primitive root of $p=13$, try to find all the primitive roots of $p$. | Since $\varphi(p-1)=\varphi(12)=\varphi(4) \varphi(3)=2 \cdot 2=4$, there are exactly 4 primitive roots of 13. Among the 12 natural numbers $1,2, \cdots, p-1=12$, the numbers that are coprime with 12 are exactly the following 4: $1,5,7,11$. Therefore, the following 4 numbers $2^{1} \equiv 2,2^{5} \equiv 6,2^{7} \equiv ... | 2, 6, 7, 11 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,214 |
Theorem 7 Let $g$ be a primitive root of the prime $p$, then the following $p^{-1} 1$ numbers
$$g^{1}, g^{2}, \cdots, g^{p-1}$$
exactly form a reduced residue system modulo $p$. | Obviously, it suffices to prove that any two numbers in (11) are not congruent $(\bmod p)$. By contradiction, if there exist $i, j(1<i<j<p-1)$, such that
$$g^{i} \equiv g^{j}(\bmod p)$$
then we have
$$g^{j-i} \equiv 1(\bmod p)$$
By Theorem 1 and the fact that the order of $g$ modulo $p$ is $p-1$, we have
$$(p-1) \mid... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,215 |
7 Find a primitive root modulo $13^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | From Example 6 of the previous section, we know that 2 is a primitive root of 13. Calculate as follows:
$$\begin{aligned}
2^{12}=(128)(32) & \equiv-(41)(32)=-(164)(8) \\
& \equiv(5)(8)=40\left(\bmod 13^{2}\right)
\end{aligned}$$
Therefore, by Theorem 8, 2 must also be a primitive root modulo $13^{2}$. | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,217 |
Example: Try to calculate the series $\sum_{n=1}^{\infty} \frac{1}{n^{2 k}}$ for $k=1,2,3,4$.
untranslated text remains the same as requested. However, the actual values for these series are well-known and can be provided as follows:
- For $k=1$, the series is $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$.
-... | We have (by (5))
$$\begin{array}{l}
\sum_{n=1}^{\infty} \frac{1}{n^{2}}=(-1)^{2} \frac{(2 \pi)^{2} B_{2}}{2 \cdot 2!}=\pi^{2} B_{2}=\pi^{2} / 6, \\
\sum_{n=1}^{x} \frac{1}{n^{4}}=(-1)^{3} \frac{(2 \pi)^{4} B_{4}}{2 \cdot 4!}=-\frac{\pi^{4}}{3} B_{4}=\pi^{4} / 90, \\
\sum_{n=1}^{\infty} \frac{1}{n^{6}}=(-1)^{4} \frac{(2... | \frac{\pi^2}{6}, \frac{\pi^4}{90}, \frac{\pi^6}{945}, \frac{\pi^8}{9450} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,218 |
Example 8 Given $p=29$, try to find a primitive root of $p^{2}=841$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let's verify that 14 is a primitive root of 29. Since
$$\varphi(p)=28=(4)(7)=(2)(14)$$
To verify that 14 is indeed a primitive root of \( p = 29 \), it is not necessary to check for all \( m (1 \leqslant m < 28) \) that
$$14^{m} \equiv 1 \pmod{29}$$
but only the following conditions need to be verified (see Theorem 1... | 43 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,219 |
Theorem 11 Let $s>3,2 \nmid a$, then
$$a^{2^{s-2}} \equiv 1\left(\bmod 2^{s}\right)$$ | We prove this by induction on $s$.
Let $a=2 b+1$, then
$$a^{2}=(2 b+1)^{2}=4 b(b+1)+1$$
Since one of $b$ and $b+1$ must be even, it follows that $8 \mid 4 b(b+1)$, thus
$$a^{2} \equiv 1\left(\bmod 2^{3}\right)$$
This holds for any odd number $a$, proving the theorem for $s=3$.
Now assume the conclusion holds for $s=u... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,222 |
Theorem 14 If the modulus $n$ has a primitive root, then it must have exactly $\varphi(\varphi(n))$ incongruent primitive roots $(\bmod n)$. If $g$ is one of its primitive roots, then the following sequence
$$1, g, g^{2}, \cdots, g^{\operatorname{lo}(n)-1}$$
exactly forms a complete system of reduced residues modulo $... | Prove that (37) exactly forms the reduced residue system modulo $n$. Each number in (37) is obviously coprime with $n$, and their number is exactly $\varphi(n)$. Therefore, it suffices to prove that they are pairwise incongruent $(\bmod n)$. Use proof by contradiction. Suppose there exist $i, j, 0 < i < j \leq \varphi(... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,225 |
Theorem 15 Let $g$ be a primitive root modulo $n$, $(a, n)=(b, n)=1$, then: (1) ind $1=0$, ind $g=1$,
(2) if $n \geqslant 3$, then ind $(-1)=\varphi(n) / 2$,
(3) ind $(a b) \equiv$ ind $a+\operatorname{ind} b(\bmod \varphi(n))$, in particular, for $m \geqslant 1$ we have
ind $a^{m} \equiv m$ ind $a(\bmod \varphi(n))$,
... | We only prove (2), (3), and (4).
(2) Since $n$ has a primitive root, when $n \geqslant 3$, it can only be one of the following forms: $n=4, p \geqslant 3, p^{c}(p>3, s>2), 2 p^{c}(p>3, s>1)$.
When $n=4$, there is only one primitive root $g \equiv 3(\bmod 4), \varphi(4) / 2=1$. In this case, it is clearly true that $-1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,227 |
Example 10 (Solving a Linear Congruence) Let $n$ be a natural number, it has a primitive root, $(a$, $n)=(b, n)=1$, solve the congruence
$$a x \equiv b(\bmod n)$$ | Solution: It is clear that for the solution $x$ of (55), there is also $(x, n)=1$, otherwise there would be $d>1$, $d=(x, n)$, which would necessarily imply $d \mid b$, thus $(b, n) \geq d>1$, leading to a contradiction. Therefore, by Theorem 15 (3), we have
$$\operatorname{ind}_{g} a+\operatorname{ind}_{y} x \equiv \o... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,228 |
Example 11 Solve for x
$$9 x \equiv 13(\bmod 43)$$ | Solving, we look up the table at the end of this chapter and find that a primitive root of 43 can be
$$g \equiv 3(\bmod 43)$$
From the table, we also find
$$\text { ind }_{y} 9=2, \text { ind }{ }_{y} 13=32,$$
Thus, the solution is
$$\operatorname{ind}_{g} x \equiv \operatorname{ind}_{9} 13-\operatorname{ind}_{9} 9=3... | x \equiv 11(\bmod 43) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,230 |
Example 12 Solve the high-order congruence equation
$$x^{6} \equiv 11(\bmod 19)$$ | From Theorem 15 (3), the above equation can be transformed into an equivalent congruence equation in terms of indices:
$$6 \operatorname{ind}_{g} x \equiv \operatorname{ind}_{9} 11(\bmod \varphi(19))$$
By the table, we can take \( g \equiv 2(\bmod 19) \), and from the table we find
$$\text { ind }_{g} 11=12 \text {, }... | x \equiv 4,13,9,15,6,10(\bmod 19) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,231 |
Example 14 Try to convert $-\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$ into decimals. | From the above two lemmas, we know that these six proper fractions can all be converted into pure repeating decimals. Since
$$10^{2} \equiv 2,10^{3} \equiv 20 \equiv-1,10^{6} \equiv 1(\bmod 7)$$
we know that the order of 10 modulo 7 is 6, so the repeating decimals formed by these fractions all have a repeating cycle of... | \begin{array}{ll}
\frac{1}{7}=0 . \dot{1} 4285 \dot{7}, & \frac{2}{7}=0 . \dot{2} 8571 \dot{4} \\
\frac{3}{7}=0 . \dot{4} 2857 \dot{1}, & \frac{4}{7}=0 . \dot{5} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,234 |
Theorem 16 If $b$ is a prime and 10 is a primitive root of $b$, then
$$\frac{1}{b}, \frac{2}{b}, \ldots, \frac{b-1}{b}$$
have cycles consisting of $b-1$ digits, and they differ only by a cyclic permutation. | For each $1 \leqslant j \leqslant b^{-1}$, it is clear that $(j, b)=1$. By Lemma 4, each fraction in (58) must be a pure repeating decimal when converted to a decimal, and the length of the repeating cycle is $b-1$. Let
$$\frac{1}{b}=0 . \dot{a}_{1} a_{2} \cdots \dot{a}_{b-1}$$
If $\{x\}$ denotes the fractional part o... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,237 |
1. Let $l \geqslant 3$, prove that the order of 5 modulo $2^{l}$ is $2^{l-2}$.
保留了源文本的换行和格式，但根据英文书写习惯，这里进行了适当的调整以确保语句的流畅性。如果需要完全按照原文格式翻译，可以如下：
1. Let $l \geqslant 3$, prove that 5 has the order $2^{l-2}$ modulo $2^{l}$. | 1. Proof: First, we use induction to prove that for $a \geqslant 3$,
$$5^{2^{a-3}} \equiv 1+2^{a-1} \quad\left(\bmod 2^{a}\right)$$
For $a=3$, $5^{2^{a-3}}=5, 1+2^{a^{-1}}=5$, the conclusion obviously holds.
Assume the conclusion (1) holds for $a$, $a \geqslant 3$, then we have
$$\begin{aligned}
5^{2^{(a+1)-3}} & =\le... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,242 |
2. Let $l \geqslant 3$, prove: for any odd number $a$, there must be an integer $b\left(0 \leqslant b<2^{l-2}\right)$ such that
$$a \equiv(-1)^{\frac{a^{-1}}{2}} 5^{b}\left(\bmod 2^{l}\right)$$ | 2. Proof: From the previous question, we know that the following $2^{1-2}$ numbers
$$5^{0}, 5^{1}, 5^{2}, \cdots, 5^{2^{l-2}-1}$$
are pairwise incongruent modulo $2^{\prime}$ and are all of the form $4k+1$. Note that in a complete residue system modulo $2^{l}$, the odd numbers are exactly half, i.e., $2^{l-1}$, and am... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,243 |
3. Let $p$ be an odd prime, and $a>1$ be an integer. Prove:
(a) The odd prime factors of $a^{p} - 1$ are factors of $a-1$, or integers of the form $2 p x + 1$, where $x$ is an integer.
(b) The odd prime factors of $a^{p} + 1$ are factors of $a+1$, or integers of the form $2 p x + 1$, where $x$ is an integer. | 3. Proof:
(a) Let $q$ be an odd prime factor of $a^{p-1}$, i.e.,
$$a^{p}-1 \equiv 0(\bmod q)$$
Assume $q \mid a$, then the above equation gives $q \mid 1$, which is impossible. Therefore, $q \nmid a$, and we can set the order of $a$ modulo $q$ to be $d$. By Theorem 1 of this chapter, we have $d \mid p$, so $d=1$ or $d... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,244 |
4. Solve the following congruences
(1) $8 x \equiv 7(\bmod 43)$,
(2) $x^{8} \equiv 17(\bmod 43)$,
(3) $8^{x} \equiv 4(\bmod 43)$ | 4. Solution: From the 13th table attached at the end of this chapter, we know that $g=3$ is a primitive root of $p=43$.
(1) From the table, we know ind $8=39$, ind $7=35$, and let ind $x=y$. Then, from the given congruence, we derive
$$39+y \equiv 35(\bmod \varphi(43))$$
Since ind $x=y=-4 \equiv 38(\bmod 42)$, hence i... | x \equiv 17(\bmod 43), x_{1} \equiv 10, x_{2} \equiv 33(\bmod 43), x \equiv 10,24,38(\bmod 42) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,245 |
5. Prove: $m$ is a prime if and only if there exists an integer $a$ such that the order of $a$ modulo $m$ is $m-1$.
untranslated text remains unchanged. | 5. Proof: First, prove the necessity.
Let $m$ be a prime number, then it must have a primitive root $g$ existing, and by the definition of a primitive root, the order of $g$ modulo $m$ is $m-1$. Taking $a=g$ suffices.
Next, prove the sufficiency. Suppose there exists $a$ such that $a^{m-1} \equiv 1 \pmod{m}$, and for... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,246 |
6. Let $g$ be a primitive root of a prime $p$. Prove: when $p \equiv 1(\bmod 4)$, $-g$ is also a primitive root of $p$, and when $p \equiv 3(\bmod 4)$, the order of $-g$ is $(p-1) / 2$. | 6. Proof: It is clear that
$$(-g)^{p^{-1}} \equiv 1(\bmod p)$$
Case one. Suppose $p \equiv 1(\bmod 4)$, we will prove that the order $h$ of $-g$ must be $p-1$. Proof by contradiction, assume $1 \leqslant h \leqslant p-2$, then $2 \nmid h$, otherwise we would have
$$g^{h}=(-g)^{h} \equiv 1(\bmod p)$$
which implies tha... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,247 |
7. Prove: If $p=2^{n}+1(n \geqslant 2)$ is a prime, then 3 must be a primitive root of $p$. | 7. Proof: By Theorem 5 of this chapter, we know that $p=2^{n}+1$ has exactly
$$\varphi(p-1)=\varphi\left(2^{n}\right)=2^{n-1}$$
primitive roots. Also, $p$ has exactly $(p-1) / 2=2^{n-1}$ quadratic residues and $2^{n-1}$ quadratic non-residues. Let $a$ be any quadratic residue of $p$, then by Euler's criterion we have
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,248 |
8. Let $q$ be an odd prime, $p=4q+1$ also be a prime, prove that $2$ is necessarily a primitive root of $p$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note itself is not part of the translation task and should... | 8. Proof: Still by Theorem 5 of this chapter, if $p=4 q+1$, then
$$\varphi(p-1)=\varphi(4 q)=\varphi(4) \varphi(q)=2(q-1)$$
there are $\varphi(p-1)$ primitive roots, and $p$ has exactly $(p-1) / 2=2 q$ quadratic residues and $2 q$ quadratic non-residues. As in the previous problem, we know that any quadratic residue o... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,249 |
9. (Another solution to problem 8)
Let $q$ be an odd prime, and $p=4q+1$ also be a prime.
(1) Prove that the congruence $x^{2} \equiv -1 \pmod{p}$ has exactly two solutions, and each solution is a quadratic non-residue modulo $p$.
(2) Prove that except for the above solution, all other quadratic non-residues modulo $p... | 9. Proof:
(1) Since $p \equiv 1(\bmod 4)$, we have $\left(\frac{-1}{p}\right)=(-1) \frac{p-1}{2} = 1$, which means -1 is a quadratic residue modulo $p$. Therefore, the congruence $x^{2} \equiv -1(\bmod p)$ has exactly two solutions: $x_{0}$ and $-x_{0}(\bmod p)$. Since
$$\left(\frac{-x_{0}}{p}\right)=\left(\frac{-1}{p}... | 2, 3, 8, 10, 11, 14, 15, 18, 19, 21, 26, 27 | Number Theory | proof | Yes | Yes | number_theory | false | 741,250 |
10. (Generalization of Question 9)
Let $q$ be an odd prime, and $p=2^{n} q+1$ also be a prime. Prove: every quadratic non-residue $a$ of $p$ that satisfies
$$a^{2^{n}} \equiv 1(\bmod p)$$
is a primitive root of $p$. | 10. Proof: It is easy to prove by the same method as the previous two questions that $p$ has $2^{n-1} q$ quadratic residues and $2^{n-1} q$ quadratic non-residues, and $p$ has $\varphi(\varphi(p))=\varphi\left(2^{n} q\right)=\varphi\left(2^{n}\right) \varphi \cdot(q)=2^{n-1}(q-1)$ primitive roots. It is also easy to pr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,252 |
12. Let $m \geqslant 3, (a, m)=1, a$ be a quadratic residue modulo $m$. Prove: The congruence $x^{2} \equiv a(\bmod m)$ has exactly two solutions if and only if: $m$ has a primitive root. | 12. Proof: The sufficiency has been proved in the previous problem. Below, we only prove the necessity.
Let $m \geqslant 3, (a, m)=1$, and $x^{2} \equiv a(\bmod m)$ has exactly two solutions. We need to prove that $\boldsymbol{m}$ must have a primitive root. We use proof by contradiction. Assume $m$ does not have a pr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,254 |
13. Let $S_{n}(p)=\sum_{k=1}^{p-1} k^{n}$, where $p$ is an odd prime, $n \geqslant 2$, prove $S_{n}(p) \equiv\left\{\begin{array}{l}0(\bmod p), \text { if } n \equiv 0(\bmod p-1), \\ -1(\bmod p), \text { if } n \not\equiv 0(\bmod p-1) .\end{array}\right.$ | 13. Proof: Since $(k, p)=1$, we have
$$k^{p-1} \equiv 1(\bmod p)$$
When $n \equiv 0(\bmod p-1)$, we have $n=(p-1) r$, where $r$ is an integer, thus
$$k^{n}=\left(k^{p-1}\right)^{r} \equiv 1(\bmod p) \quad(1 \leqslant k \leqslant p-1)$$
Therefore, in this case,
$$\sum_{k=1}^{p-1} k^{n} \equiv p-1 \equiv-1(\bmod p)$$
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,255 |
17. Let $p=2^{2^{k}}+1$ be a prime number, try to prove that 7 is a primitive root of $p$ under the condition. | 17. Proof: From the previous question, we only need to prove that 7 is a quadratic non-residue modulo \( p \).
For \( k=0 \), we have \( p=3,7 \equiv 1 \pmod{3} \), in which case 7 is clearly a quadratic residue modulo \( p \), so 7 is not a primitive root of \( \boldsymbol{p} \) when \( k=0 \).
For \( k \geqslant 1 ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,259 |
19. Given the prime $p=71$ with 7 as one of its primitive roots, try to find all primitive roots of 71, and find a primitive root for $p^{2}$ and $2 p^{2}$. | 19. Solution: From $\varphi(p-1)=\varphi(70)=24$, we know that $p=71$ has 24 primitive roots. The 24 natural numbers between 1 and 70 that are coprime with 70 are: 1, 3, 9, 11, 13, 17, 19, 23, 27, 29, 31, 33, 37, 39, 41, 43, 47, 51, 53, 57, 59, 61, 67, 69.
$$\begin{array}{l}
7^{1}=7,7^{3}=343 \equiv 59,7^{9} \equiv 59^... | 7,59,47,31,28,62,56,53,21,35,11,42,22,13,69,44,67,52,63,33,55,68,65,61 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,261 |
20. Find the necessary and sufficient condition for $x^{n} \equiv a(\bmod p), p \nmid a, n<p$, to have $n$ solutions modulo $p$. | 20. Solution: Let $g$ be a primitive root of $p$. Since $p \nmid a$, if there is a solution, it must also be that $p \nmid x$. Therefore, we can set $b=$ ind $_{g} a, y=\operatorname{ind}_{g} x$, and the original congruence equation becomes
$$g^{n y} \equiv g^{b}(\bmod p)$$
i.e.,
$$g^{n y - b} \equiv 1(\bmod p)$$
Sin... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,263 |
21. Let $a+b=p, p$ be an odd prime, prove ind $a-$ ind $b \equiv \frac{p^{-1}}{2}(\bmod p-1)$ | 21. Proof: From $a+b=p$ we have $a=p-b \equiv -b$, thus ind $a \equiv$ ind $(-b) \equiv$ ind $(-1) +$ ind $b$ $(\bmod p-1)$,
Furthermore, when $g$ is a primitive root, it must hold that
$$g^{\frac{n-1}{2}} \neq 1(\bmod p)$$
Therefore, from
$$\left(g^{\frac{p-1}{2}}+1\right)\left(g^{\frac{p-1}{2}}-1\right)=g^{p-1}-1=0... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,264 |
Theorem 1 Any prime of the form $4 k+3$ cannot be expressed as the sum of squares of two integers. | Proof: Let $x$ be an integer. Then, when $x=2k$ is even, we have $x^{2}=4k^{2} \equiv 0(\bmod 4)$. When $x=2k+1$ is odd, we have $x^{2}=4k(k+1)+1 \equiv 1(\bmod 4)$. Therefore, for any two integers $x$, $y$, we always have:
$$x^{2}+y^{2}=\left\{\begin{array}{ll}
0(\bmod 4), & \text { if } 2 \mid x, 2 \mid y, \\
1(\bmod... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,265 |
Lemma 1 Let the prime $p \equiv 1(\bmod 4)$, denote $q=(p-1) / 2$ and $a=q!$, then $a^{2} \equiv-1(\bmod p)$. | By Wilson's theorem, we have
$$(p-1)!\equiv-1(\bmod p)$$
On the other hand, we have
$$\begin{array}{l}
(p-1)!=1 \cdot 2 \cdots\left(\frac{p-1}{2}\right)\left(\frac{p+1}{2}\right) \cdots(p-2)(p-1) \\
\equiv 1 \cdot 2 \cdots\left(\frac{p-1}{2}\right)(-1)\left(\frac{p-1}{2}\right) \cdots(-1)(2) \\
\cdot(-1)(1) \equiv(-1)... | a^{2} \equiv -1(\bmod p) | Number Theory | proof | Yes | Yes | number_theory | false | 741,266 |
Lemma 2 Let $p$ be a prime, $m$ an integer, $p \nmid m$, prove that there must exist integers $x, y$, such that
$$\begin{aligned}
m x \equiv y(\bmod p), \\
1<x<\sqrt{p} \quad 1<y<\sqrt{p},
\end{aligned}$$ | Consider the set formed by $mt + u$ when $t, u$ take all the $[\sqrt{p}] + 1$ integers $0, 1, \cdots, [\sqrt{p}]$. Since $t$ and $u$ each take $[\sqrt{p}] + 1$ values, this results in $([\sqrt{p}] + 1)^2$ values of $mt + u$. Note that
$$([\sqrt{p}] + 1)^2 > (\sqrt{p})^2 = p$$
Thus, there are at least two pairs of numbe... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,267 |
Theorem 2 If $p \equiv 1(\bmod 4)$ is a prime, then it can necessarily be expressed as the sum of two squares of integers. | By Lemma 2, we know that there must be integers $x, y, 1 \leqslant x<\sqrt{p}$, $1 \leqslant|y|<\sqrt{p}$, such that for a given $m, p \nmid m$, we have
$$m x \equiv y(\bmod p)$$
In particular, taking $m=a$, where $a$ is defined as in Lemma 1, we have
$$p \mid(a x-y)$$
Thus, we also have
$$p \mid(a x-y)(a x+y)$$
whic... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,268 |
Lemma 3 Let $x_{1}, x_{2}, y_{1}, y_{2}$ be any integers, then
$$\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)=\left(x_{1} x_{2}+y_{1} y_{2}\right)^{2}+\left(x_{1} y_{2}-x_{2} y_{1}\right)^{2} .$$ | Prove that expanding both sides immediately yields the proof.
(10) tells us that if $m_{1}, m_{2}$ are two positive integers, both of which can be expressed as the sum of two squares of integers, then $m_{1}, m_{2}$ must also be expressible as the sum of two squares of integers. Let $n$ be any given positive integer wi... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,269 |
Theorem 9 We have $g(8) \leqslant 42273$. | First, by applying the method of comparing coefficients or direct expansion, it is easy to verify that the following identity holds:
$$\begin{array}{l}
5040\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{4} \\
\quad=6 \sum(2 a)^{8}+60 \sum(a \pm b)^{8} \\
\quad+\sum(2 a \pm b \pm c)^{8}+6 \sum(a \pm b \pm c+d)^{8}
\end{array}$$
... | 42273 | Number Theory | proof | Yes | Yes | number_theory | false | 741,278 |
Lemma 5 Let $\Delta f(x)=f(x+1)-f(x)$, when $m \geqslant 1$, define $\Delta^{m+1} f(x)=\Delta\left(\Delta^{m} f(x)\right)$, where $f(x)$ is a $k$-degree polynomial with integer coefficients, then when $k \geqslant 1$, we have
$$\Delta^{k-1} x^{k}=(k!) x+d,$$
where $d$ is an integer. | Prove that any polynomial $f(x)$ of degree $k \geqslant 2$ with leading coefficient $a$ can obviously be expressed in the following form:
$$f(x)=a x^{k}+b x^{k-1}+f_{1}(x),$$
where $f_{1}(x)$ is a polynomial of degree $k-2$. By definition, we have
$$\begin{aligned}
\Delta f(x)= & f(x+1)-f(x) \\
= & a\left[(x+1)^{k}-x^... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,279 |
Lemma 6 When $k \geqslant 2$, we have
$$v(k) \leqslant 2^{k-1}+\frac{1}{2} k!$$ | Prove that $\Delta x^{k}=(x+1)^{k}-x^{k}$, we get
$$\begin{aligned}
\Delta^{2} x^{k} & =\Delta\left((x+1)^{k}-x^{k}\right)=\Delta(x+1)^{k}-\Delta x^{k} \\
& =(x+2)^{k}-(x+1)^{k}-(x+1)^{k}+x^{k}
\end{aligned}$$
Continuing this method, we can easily see that $\Delta^{k-1} x^{k}$ is the algebraic sum of $2^{k-1}$ integer... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,280 |
Lemma 8 We have
$$\begin{array}{l}
720 x-360=x^{5}+(x-1)^{5} \\
\quad+(x-4)^{5}+(x+3)^{5}-2(x+2)^{5}-2(x-3)^{5}
\end{array}$$ | The right side of (48) is a polynomial in $x$, where the coefficient of the $x^{5}$ term is $1+1+1+1-2-2=0$, the coefficient of the $x^{4}$ term is $(5)(-1-4+3-(2) + (2) (3)) =0$, the coefficient of the $x^{3}$ term is
$$\text { (10) }\left(1+4^{2}+3^{2}-(2)(2)^{2}-(2)(3)^{2}\right)=0 \text {, }$$
the coefficient of t... | 720 x-360=x^{5}+(x-1)^{5}+(x-4)^{5}+(x+3)^{5}-2(x+2)^{5}-2(x-3)^{5} | Algebra | proof | Yes | Yes | number_theory | false | 741,286 |
Lemma 9 Let $a_{1}$ and $a_{2}$ be integers, and $m_{1}$ and $m_{2}$ be positive integers satisfying the condition $\left(m_{1}, m_{2}\right)=1$, then there must exist an integer $a$ such that $a \equiv a_{1}\left(\bmod m_{1}\right), a \equiv a_{2}\left(\bmod m_{2}\right)$. | Given that $\left(m_{1}, m_{2}\right)=1$, there exist integers $x_{1}$ and $y_{1}$ such that
$$x_{1} m_{1}+y_{1} m_{2}=1$$
Thus, we have
$$\begin{aligned}
a_{2}-a_{1} & =\left(a_{2}-a_{1}\right)\left(x_{1} m_{1}+y_{1} m_{2}\right) \\
& =x_{1}\left(a_{2}-a_{1}\right) \dot{m}_{1}-y_{1}\left(a_{1}-a_{2}\right) m_{2}
\end... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,287 |
Lemma 10 For any integer $n$, there exist two integers $a, b$ such that
$$n \equiv a^{5}+b^{5}(\bmod 720)$$ | From the obvious congruence relations $1 \equiv 0^{5}+1^{5}(\bmod 16), 3 \equiv$ $0^{5}+3^{5}(\bmod 16), 5 \equiv 0^{5}+5^{5}(\bmod 16), 7 \equiv 0^{5}+$ $7^{5}(\bmod 16)$, we know that for any integer $r, -8 \leqslant r<8$, there must exist two integers $c, d$ such that $r \equiv c^{5}+d^{5}(\bmod 16)$. Here, we also ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,288 |
1. Prove: If $8 \mid\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$, then $a, b, c, d$ must all be even. | 1. Proof: Let $x$ be an integer. If $2 \mid x$, then when $4 \nmid x$, we have $x=2(2 y+1)$, thus
$$x^{2}=4(2 y+1)^{2}=4\left(4 y^{2}+4 y+1\right) \equiv 4(\bmod 8)$$
While when $4 \mid x$, it is obvious that
$$x^{2}=(4 y)^{2}=16 y^{2} \equiv 0(\bmod 8)$$
If $2 \nmid x$, we can set $x=2 y+1$,
$$x^{2}=(2 y+1)^{2}=8 \c... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,290 |
2. Prove: A positive integer $m$ can be expressed as the difference of two squares
$$m=a^{2}-b^{2}$$
if and only if $m$ can be expressed as the product of two numbers, both of which are either odd or even. | 2. Proof: We first prove the necessity.
Assume $m$ can be expressed as the difference of squares of two integers, then we have
$$m=a^{2}-b^{2}=(a+b)(a-b)$$
If $2 \mid (a-b)$, then $a-b \equiv 0(\bmod 2)$, and thus
$$a+b=a-b+2 b \equiv a-b \equiv 0 \quad(\bmod 2),$$
hence both $a+b$ and $a-b$ must be even.
If $2 \nmi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,291 |
3. Prove: The cube of any integer is the difference of two squares. | 3. Proof: If $n$ is odd, then from the previous problem we have
$$\begin{aligned}
n^{3} & =n^{2} \cdot(n \times 1)=n^{2}\left\{\left(\frac{n+1}{2}\right)^{2}-\left(\frac{n-1}{2}\right)^{2}\right\} \\
& =\left(\frac{n(n+1)}{2}\right)^{2}-\left(\frac{n(n-1)}{2}\right)^{2}
\end{aligned}$$
Notice that when $n$ is odd, $\f... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,292 |
4. Prove that there are infinitely many triples of integers with the following property: these three numbers are three consecutive integers, two of which can be expressed as the sum of two squares. | 4. Proof: Note that for $x \geqslant 1, y \geqslant 1$ we have
$$\begin{array}{l}
(x+1)^{2}-x^{2}=2 x+1 \geqslant 3 \\
y^{2}-(y-1)^{2}=2 y-1 \geqslant 1
\end{array}$$
It follows that $(2 x+1)-(2 y-1)=2(x-y)+2 \geqslant 2$, to make $(2 x+1)-(2 y-1)=2$ exactly, it is clear that $x=y$. Therefore, we can consider taking
$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,293 |
5. Find all positive integers $x, y, z, w$ such that:
(1) $x, y, z, w$ are four consecutive terms of an arithmetic sequence,
(2) $x^{3}+y^{3}+z^{3}=w^{3}$. | 5. Solution: Let $x=a+d, y=a+2d, z=a+3d, w=a+4d$. Then from
$$x^{3}+y^{3}+z^{3}=w^{3}$$
we get
$$(a+d)^{3}+(a+2d)^{3}+(a+3d)^{3}=(a+4d)^{3}$$
Expanding and combining like terms, we obtain
$$a^{3}+3a^{2}d-3ad^{2}-14d^{3}=0$$
This is $\square$
$$a^{3}-2a^{2}d+5a^{2}d-10ad^{2}+7ad^{2}-14d^{3}=0$$
Thus, we have $\squar... | x=3d, y=4d, z=5d, w=6d | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,294 |
*6. Prove that there do not exist positive integers $x, y, z, w, t$ with the following properties:
(1) $x, y, z, w, t$ are five consecutive terms of an arithmetic sequence,
(2) $x^{3}+y^{3}+z^{3}+w^{3}=t^{3}$. | 6. Let
$$x=a, y=a+d, z=a+2 d, w=a+3 d, t=a+4 d \text {, }$$
then from
$$x^{3}+y^{3}+z^{3}+w^{3}=t^{3}$$
we get
$$a^{3}+(a+d)^{3}+(a+2 d)^{3}+(a+3 d)^{3}=(a+4 d)^{3},$$
Expanding and combining like terms, we obtain
$$3 a^{3}+6 a^{2} d-6 a d^{2}-28 d^{3}=0 .$$
Since the first to third terms are all divisible by 3, we... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,296 |
7. Find the positive integer $x$ that makes $x^{2}-60$ a perfect square
Find the positive integer $x$ that makes $x^{2}-60$ a perfect square | 7. Solution: Let $x^{2}-60=y^{2}$, then
$$(x-y)(x+y)=60$$
If $x \neq y(\bmod 2)$, then $x+y=x-y+2 y \equiv x-y \neq 0(\bmod 2)$, so both $x-y$ and $x+y$ are odd, which contradicts equation (6). Therefore, $x$ and $y$ must both be odd or both be even, thus $2|(x-y), 2|(x+y)$. Since $60=4 \times 15$, there are only the ... | 16 \text{ or } 8 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,297 |
8. Find the positive integer $x$ that makes both $x^{2}-5$ and $x^{2}+5$ perfect squares. | 8. Solution: Let $x^{2}-5=y^{2}$, $x^{2}+5=z^{2}$, then we have
$$\left\{\begin{array}{l}
z^{2}-y^{2}=10 \\
2 x^{2}=y^{2}+z^{2}
\end{array}\right.$$
Since when $t$ is even, $4 \mid t^{2}$, and when $t$ is odd,
$$t^{2}=\left(2 t_{1}+1\right)^{2}=4\left(t_{1}^{2}+t_{1}\right)+1 \equiv 1(\bmod 4),$$
we have
$$y^{2}+z^{2... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,298 |
9. Prove that $x^{n}+1=y^{n+1}$ has no positive integer solutions, where
$$n \geqslant 2, \quad(x, n+1)=1$$ | 9. Proof: From $x^{n}+1=y^{n+1}$, we have
$$x^{n}=y^{n+1}-1=(y-1)\left(y^{n}+y^{n-1}+\cdots+1\right),$$
Let $p \mid(y-1)$, then it must be that $p \mid x$. And by $(x, n+1)=1$, we have $p \nmid(n+1)$. Thus, $(y-1, n+1)=1$. We also have
$$y^{n}+y^{n-1}+\cdots+1 \equiv n+1(\bmod y-1),$$
Therefore, $y^{n}+y^{n-1}+\cdots... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,299 |
11. Prove: Every positive integer $n$ can be expressed in the form
$$n=x^{2}+y^{2}-z^{2}$$
where $x, y, z$ are non-negative integers. | 11. Proof: From the method used in problem 2, it is easy to see that for any odd number \( r \),
$$r = r \cdot 1 = \left(\frac{r+1}{2}\right)^{2} - \left(\frac{r-1}{2}\right)^{2},$$
Given \( n \), if \( n \) is even, then \( n-1 \) is odd, so the above method gives
$$n = 1^{2} + \left(\frac{n}{2}\right)^{2} - \left(\f... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,301 |
12. Prove: For any positive integer $n$, the indeterminate equation
$$x^{2}+y^{2}=z^{n}$$
always has integer solutions. | 12. Proof: By Theorem 2 on p. 65 of the first penalty of this book,
$$X^{2}+Y^{2}=Z^{2}$$
must have positive integer solutions. In fact, the solutions satisfying the conditions
$$(X, Y)=1, \quad 2 \mid X$$
can be expressed by the formulas
$$X=2 a b, \quad Y=a^{2}-b^{2}, Z=a^{2}+b^{2},$$
where $a > b$ are positive int... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,302 |
*13. Prove: The indeterminate equation
$$3^{x}+4^{y}=5^{2}$$
has only the positive integer solution $x=y=2$. | 13. Proof: From
$$3^{x}+4^{y} \neq 5^{z},$$
we have
$$(-1)^{x} \equiv 1(\bmod 4)$$
and
$$1 \equiv(-1)(\bmod 3)$$
Thus, $x$ and $y$ must both be even. Let $x=2 x_{1}, z=2 z_{1}$, and substitute into (14) to get
$$5^{2 z_{1}} - 3^{2 x_{1}}=4^{y},$$
which is $\square$
$$\left(5^{z_{1}}+3^{x_{1}}\right)\left(5^{z_{1}}-... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,303 |
14. Let the positive integer $m$ have the standard factorization
$$m=2^{a_{0}} p_{1}^{\alpha_{1}} \cdots p_{s}^{\alpha_{s}}$$
where $p_{1}, \cdots, p_{s}$ are distinct odd primes, $\alpha_{0} \geqslant 0, \alpha_{1} \geqslant 1, \cdots, \alpha_{s} \geqslant 1$, $s \geqslant 1$, or $m=2^{\alpha}, \alpha_{0} \geqslant 0... | 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 | 741,304 |
15. Prove: For a right-angled triangle with integer side lengths, when the difference between the hypotenuse and one of the legs is 1, its three side lengths can be expressed as
$$2 b+1, 2 b^{2}+2 b, 2 b^{2}+2 b+1$$
where $b$ is any positive integer. | 15. Proof: Let the three sides be $x, y, x+1$, then
$$x^{2}+y^{2}=(x+1)^{2}=x^{2}+2 x+1$$
Thus, we have
$$y^{2}=2 x+1$$
which means $y$ must be an odd number. Let $y=2 b+1$ and substitute it into the above equation to get
$$2 x+1=(2 b+1)^{2}=2\left(2 b^{2}+2 b\right)+1$$
Thus, we solve for
$$x=2 b^{2}+2 b,$$
so the... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,305 |
Example 1 Prove that $n^{3}+5 n$ is a multiple of 6 (here $n$ is a positive integer). | We prove this by applying mathematical induction.
(1) When $n=1$, we have $n^{3}+5 n=6$, so the proposition holds for $n=1$.
(2) Assume the proposition holds for $n=k-1$, where $k \geqslant 2$ is a natural number, i.e., there exists an integer $m$ such that
$$(k-1)^{3}+5(k-1)=6 \, m,$$
We now prove that the propositio... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,306 |
Theorem 2 Proof: For $n \geqslant 1$,
$$\left(F_{n}, F_{n+1}\right)=\left(L_{n}, L_{n+1}\right)=1$$ | Proof: Let $a, b$ be two given integers. We will first prove a relation concerning the greatest common divisor of two numbers:
$$(a+b, a)=(b, a)$$
Obviously, if we let
$$d_{1}=(a+b, a), d_{2}=(b, a)$$
then we only need to prove that $d_{1}\left|d_{2}, d_{2}\right| d_{1}$. By definition, $d_{1}\left|a, d_{1}\right|(a+... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,307 |
16. Prove: The indeterminate equation
$$x^{2}+(x+1)^{2}=k y^{2} \quad(k \geqslant 2)$$
has positive integer solutions if and only if -1 is a quadratic residue modulo $k$, but this condition is not sufficient. | 16. Proof: From the given equation, we have
$$x^{2}+(x+1)^{2} \equiv 0(\bmod k)$$
Thus,
$$2 x^{2}+2 x+1 \equiv 0(\bmod k)$$
Therefore, we also have
$$4 x^{2}+4 x+2 \equiv 0(\bmod k)$$
This is equivalent to
$$(2 x+1)^{2} \equiv -1(\bmod k)$$
Thus, -1 is a quadratic residue modulo $k$.
For $k=2$, -1 is clearly a quad... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,308 |
17. Let $n$ be a positive integer, prove that the indeterminate equation
$$x^{n}+y^{n}=z^{n} \quad \because \quad(n \geqslant 2)$$
has no positive integer solutions satisfying $0<x<n, 0<y<n$. | 17. Proof: Consider the indeterminate equation
$$x^{n}+y^{n}=z^{n}$$
has positive integer solutions $x_{0}, y_{0}, z_{0}$, we need to prove that $x_{0} \geqslant n$ or $y_{0} \geqslant n$.
First, it is easy to show that $x_{0} \neq y_{0}$. If $x_{0}=y_{0}$, then we have
$$2 x_{0}^{n}=z_{0}^{n},$$
which implies
$$\fra... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,309 |
18. Prove:
(1) A positive integer can be expressed as the sum of two squares if and only if: twice this number also has this property.
(2) Let \( p \) be an odd prime, then
\[
\frac{2}{p} = \frac{1}{x} + \frac{1}{y}
\]
always has positive integer solutions satisfying \( x \neq y \), and there is only one such represen... | 18. Proof:
(1) First prove the necessity: Let $m=x^{2}+y^{2}, m$ be a positive integer, then we have
$$2 m=2 x^{2}+2 y^{2}=(x+y)^{2}+(x-y)^{2},$$
Thus, $2 m$ can also be expressed as the sum of two squares.
Next, prove the sufficiency: Let $2 m=x^{2}+y^{2}$, note that
$$2 m \equiv\left\{\begin{array}{ll}
2(2 k) \equiv... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,310 |
19. (1) Prove: If $7 \mid n, 7^{2} \nmid n$, then $n$ cannot be expressed as the sum of two squares.
(2) Suppose $m, n$ are integers that can be expressed as the sum of two squares, and $m \mid n$, prove that $\frac{n}{m}$ can also be expressed as the sum of two squares.
(3) Suppose $m, n$ are integers that can be expr... | 19. Proof:
(1) By contradiction. Suppose $n \equiv 0(\bmod 7), n \geqslant 0(\bmod 7^{2})$, but
$$n=x^{2}+y^{2}$$
Since
$$0^{2} \equiv 0,1^{2} \equiv 1,2^{2} \equiv 4,3^{2} \equiv 2,4^{2} \equiv 2,5^{2} \equiv 4,6^{2} \equiv$$
$1(\bmod 7)$,
we have $x^{2} \equiv 0,1,2,4$ and $y^{2} \equiv 0,1,2,4(\bmod 7)$. From $x^{2... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,311 |
20. (1) Prove: If $n=a^{2}+b^{2}=c^{2}+d^{2}$, then
$$n=\frac{\left[(a-c)^{2}+(b-d)^{2}\right]\left[(a+c)^{2}+(b-d)^{2}\right]}{4(b-d)^{2}},$$
Therefore, if $n$ can be expressed as the sum of two squares in two different ways, then $n$ must be a composite number.
(2) Given that
$$533=23^{2}+2^{2}=22^{2}+7^{2}, 1073=32... | 20. Solution:
(1) Let's assume $a>0, b>0, c>0, d>0$. Since the given expression represents two different representations of $n$, the following two conditions must be satisfied simultaneously:
1) $a \neq c, b \neq d$,
2) $a \neq d, b \neq c$.
Thus, we have
$$\begin{aligned}
n= & \frac{1}{4}\left(2 a^{2}+2 b^{2}+2 c^{2}... | 533 = 13 \times 41, \quad 1073 = 29 \times 37 | Number Theory | proof | Yes | Yes | number_theory | false | 741,312 |
22. Prove:
$$\begin{aligned}
{[0,3]_{1} } & =[1,2]_{1} \\
{[1,2,6]_{2} } & =[0,4,5]_{2}, \\
{[0,4,7,11]_{3} } & =[1,2,9,10]_{3} .
\end{aligned}$$ | 22. Proof: Direct verification gives
$$0+3=1+2,0^{2}+3^{2} \neq 1^{2}+2^{2}$$
Thus,
$$[0,3]_{1}=[1,2]_{1}.$$
Taking $d=3$ and applying the result from the previous problem to equation (43), we get
$$[3,6,1,2]_{2}=[0,3,4,5]_{2},$$
removing the common number 3 from both sides, we obtain
$$[1,2,6]_{2}=[0,4,5]_{2}.$$
T... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,314 |
Theorem 3 Proof: When $n \geqslant 1$
(1) If $F_{n}$ is odd, then $L_{n}$ is also odd, and
$$\left(F_{n}, L_{n}\right)=1$$
(2) If $F_{n}$ is even, then $L_{n}$ is also even, and
$$\left(F_{n}, L_{n}\right)=2$$ | (1) From (20) and (8), we have
$$L_{n}=F_{n-1}+F_{n+1}=2 F_{n-1}+F_{n} \text {, }$$
Thus, when $F_{n}$ is odd (even), $L_{n}$ is also odd (even). Moreover,
$$\begin{aligned}
\left(F_{n}, L_{n}\right) & =\left(F_{n}, 2 F_{n-1}+F_{n}\right) \\
& =\left(F_{n}, 2 F_{n-1}\right) \\
& =\left(F_{n}, F_{n-1}\right) \\
& =1
\e... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,318 |
Theorem 1
$$\begin{aligned}
\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \cdots \cap \bar{A}_{m}\right|= & |S|-\Sigma\left|A_{i}\right|+\Sigma\left|A_{i} \cap A_{j}\right|-\Sigma\left|A_{i} \cap A_{j} \cap A_{k}\right| \\
& +\cdots+(-1)^{m}\left|A_{1} \cap A_{2} \cap \cdots \cap A_{m}\right|
\end{aligned}$$
where the first... | Let $Q$ be an element with $k_{0}$ properties. Then it appears once in $|S|$, $k_{0}$ times in $\Sigma\left|A_{i}\right|$, $\left({ }_{2}^{k 0}\right)$ times in $\Sigma\left|A_{i} \cap A_{j}\right|$, $\cdots \cdots$, and $\left.{ }_{k 0}^{k 0} k_{0}\right)$ times in $\Sigma\left|A_{i_{1}} \cap A_{i_{2}} \cap \cdots \ca... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 741,319 |
Inference The number of elements in set $S$ that have at least one of the properties $P_{1}, \cdots, P_{m}$ is
$$\begin{aligned}
\left|A_{1} \cup A_{2} \cup \cdots \cup A_{m}\right| & =\Sigma\left|A_{i}\right| - \Sigma\left|A_{i} \cap A_{j}\right|+\Sigma\left|A_{i} \cap A_{j} \cap A_{k}\right| \\
& -+\cdots+(-1)^{m+1}\... | Prove that we have
$$\left|A_{1} \cup A_{2} \cup \cdots \cup A_{m}\right|=|S|-|B|$$
Here
$$B=\overline{A_{1} \cup A_{2} \cup \cdots \cup A_{m}}$$
By applying the definition of set equality, it is easy to directly prove that
$$B=\bar{A}_{1} \cap \bar{A}_{2} \cap \cdots \cap \bar{A}_{m},$$
Thus, applying the above the... | proof | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,320 |
Example 1 Find the number of integers from 1 to 1000 that are not divisible by 5, nor by 6 and 8. | We use the notation $\operatorname{LCM}\left\{a_{1}, \cdots, a_{n}\right\}$ to represent the least common multiple of $n$ integers $a_{1}, \cdots, a_{n}$. Let $S$ be the set consisting of the natural numbers from 1 to 1000. Property $P_{1}$ is "an integer is divisible by 5", property $P_{2}$ is "an integer is divisible... | 600 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,321 |
Example 2 Let $a_{1}, \cdots, a_{n}$ be $n$ non-negative integers, then
$$\begin{array}{l}
\max \left\{a_{1}, \cdots, a_{n}\right\}=\sum_{i} a_{i}-\sum_{i, j} \min \left\{a_{i}, a_{j}\right\}+\cdots \\
+(-1)^{n+1} \min \left\{a_{1} \cdots, a_{n}\right\}
\end{array}$$ | Let $\max \left\{a_{1}, \cdots, a_{n}\right\}=M_{n}$. Take any natural number $N>M_{n}$, and let $S$ be the set composed of $1,2, \cdots N$. Let the property $P_{i}(1 \leqslant i \leqslant n)$ be "a natural number is not greater than $a_{i}$". The set of elements with property $P_{i}$ is denoted as $A_{i}(1 \leqslant i... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,322 |
Example 3 Let $b_{1}, b_{2}, \cdots, b_{m}$ be $m$ non-negative integers, and let $\left(b_{i_{1}}, \cdots, b_{i_{s}}\right)$ denote the greatest common divisor of $b_{i 1}, \cdots, b_{i s}$, then
$$\begin{array}{c}
\operatorname{LCM}\left\{b_{1}, b_{2}, \cdots, b_{m}\right\}=b_{1} \cdots b_{m}\left(b_{1}, b_{2}\right)... | By definition, $\operatorname{LCM}\left\{b_{1}, b_{2}, \cdots b_{m}\right\}$ is the smallest positive integer that can be divided by $b_{1}, \cdots, b_{m}$. If we set their factorization as
$$\left\{\begin{array}{l}
b_{1}=p_{1}^{a_{1}^{(1)}} \cdots \cdots p_{n}^{a_{n}^{(1)}}, a_{1}^{(1)} \geqslant 0, \cdots, a_{n}^{(1)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,323 |
Example 4 (Euler's $\varphi$ function calculation formula)
Euler's $\varphi$ function value at $n$, $\varphi(n)$, is defined as the number of natural numbers that are coprime to $n$ in the set $\{1,2, \cdots, n\}$. Suppose $n$ has the standard factorization
$$n=p_{1}^{z_{1}} \cdots p_{s}^{\alpha_{s}},$$
where $p_{1}, ... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | 16 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,324 |
1. Find the number of positive integers among the first $10^{5}$ that are not divisible by $7, 11, 13$. | 1. Solution: According to Theorem 1 of this chapter, the number of integers sought is
$$\begin{array}{c}
10^{5}-\left[\frac{10^{5}}{7}\right]-\left[\frac{10^{5}}{11}\right]-\left[\frac{10^{5}}{13}\right]+\left[\frac{10^{5}}{7 \times 11}\right] \\
+\left[\frac{10^{5}}{7 \times 13}\right]+\left[\frac{10^{5}}{11 \times 13... | 71929 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,327 |
2. A school organized three extracurricular activity groups in mathematics, Chinese, and foreign language. Each group meets twice a week, with no overlapping schedules. Each student can freely join one group, or two groups, or all three groups simultaneously. A total of 1200 students participate in the extracurricular ... | 2. Solution: Since all 1200 students have joined at least one extracurricular group, the number of students who did not join any group is 0. We use $A_{1}, A_{2}, A_{3}$ to represent the sets of students who joined the math group, the Chinese group, and the English group, respectively. Thus, by the problem statement, w... | 80 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,328 |
Theorem 4 For $n \geqslant 1$ we have
$$\begin{array}{c}
F_{n}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n}, \\
L_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\left(\frac{1-\sqrt{5}}{2}\right)^{n} .
\end{array}$$ | Let $F_{n}=q^{n}$ in (8), we get
$$q^{n}=q^{n^{-1}}+q^{n^{-2}},$$
thus
$$q^{2}-q-1=0$$
Equation (35) has two roots
$$q_{1}=\frac{1+\sqrt{5}}{2}, \quad q_{2}=\frac{1-\sqrt{5}}{2}$$
Therefore, it is easy to see that for any real numbers $c_{1}, c_{2}$,
$$@_{n}=c_{1}\left(\frac{1+\sqrt{5}}{2}\right)^{n}+c_{2}\left(\fra... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,329 |
3. (One of the more listed problems) There are $n$ people, each labeled with numbers from 1 to $n$, and there are $n$ chairs, also labeled with numbers from 1 to $n$. Question: How many different ways can these $n$ people sit on these $n$ chairs such that the $i$-th person ($i=1,2, \cdots, n$) does not sit on the $i$-t... | 3. Solution: Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of the $n$ natural numbers $1,2, \cdots, n$, and satisfy
$$a_{i} \neq i \quad(i=1,2, \cdots, n)$$
Then, the permutation $a_{1}, a_{2}, \cdots, a_{n}$ clearly gives a seating arrangement for these $n$ people that meets the problem's requirements. Thus, the... | D_{n} = n!\left(1-\frac{1}{1!}+\frac{1}{2!}-+\cdots+(-1)^{n} \frac{1}{n!}\right) | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,330 |
*6. ( $k-$ derangement problem)
If a permutation of the set $\{1,2, \cdots, n\}$ without repetition satisfies the following conditions:
$$a_{1} a_{2} \cdots a_{n}$$
(1) For $k$ indices $i$, $a_{i} \neq i$,
(2) For the remaining $n-k$ indices $j$, $a_{j}=j$, then this permutation is called a $k$ - derangement of $\{1,2,... | 6. Solution: Let $S$ denote the set of all permutations of $\{1,2, \cdots, n\}$ without repetition, then $|S|=n!$. Let
$$a_{1} a_{2} \cdots a_{n}$$
be a permutation in $S$. If for some $i(1 \leqslant i \leqslant n)$, $a_{i}=i$, then the permutation (7) is said to have property $p_{i}$, and $A_{i}$ denotes the subset o... | D_{n}(k) = \frac{n!}{(n-k)!} \sum_{s=n-k}^{n}(-1)^{s-n+k} \frac{1}{(s-n+k)!} | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,333 |
Lemma 1 Let $n$ and $k$ be positive integers, then we have
$$\begin{array}{c}
2 \sum_{i=1}^{k}(2 k-1) S_{2 i-1}(n)=(n+1)^{2 k}+n^{2 k} \\
-1-2 k n(n+1) \quad(k \geqslant 2), \\
2 \sum_{i=1}^{k}(2 k+1) S_{2 i}(n)=(n+1)^{2 k+1}+n^{2 k+1}-2 n-1 .
\end{array}$$ | We prove (81) and (82) by mathematical induction on $n$. When $n=1$, by $\mathrm{f}^{*} S_{2 i-1}(1)=1$, thus both sides of (81) are
$$\begin{aligned}
2^{2 k}-4 k & =(1+1)^{2 k}-(1-1)^{2 k}-2\binom{2 k}{1} \\
& =\sum_{i=0}^{2 k}\left(\frac{2 k}{i}\right)-\sum_{i=0}^{2 k}(-1)^{i(2 k}\right)-2\left(\frac{2 k}{1}\right) \... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,339 |
1. Use mathematical induction to prove the following general term formula for the Fibonacci sequence: for $n \geqslant 1$
$$F_{n}=\frac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{2^{n} \sqrt{5}}$$
and prove that, for $n \geqslant 1$,
$$F_{n}=\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n} / 2\right]+C_{n}$$
where $[a]$ denotes t... | 1. Proof: Since $F_{1}=F_{2}=1$, (91) holds for $n=1$. Now assume (91) holds for $1 \leqslant n \leqslant k$, we consider the case $n=k+1$. By equation (8) in this chapter, we have
$$\begin{array}{c}
F_{k+1}=F_{k}+F_{k-1} \\
=\frac{(1+\sqrt{5})^{k}-(1-\sqrt{5})^{k}+2(1+\sqrt{5})^{k-1}-2(1-\sqrt{5})^{k-1}}{2^{k} \sqrt{5... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,341 |
3. Try to prove the following results:
(1) $F_{n+1}^{2}-F_{n} F_{n+2}=(-1)^{n}(n \geqslant 1)$,
(2) $F_{1} F_{2}+F_{2} F_{3}+\cdots+F_{2 n-1} F_{2 n}=F_{2 n}^{2}(n \geqslant 1)$,
(3)
$$\begin{array}{l}
F_{1} F_{2}+F_{2} F_{3}+\cdots+F_{2 n} F_{2 n+1} \\
=F_{2 n+1}^{2}-1 \quad(n \geqslant 1)
\end{array}$$
(4)
$$\begin{a... | This proves that (95) holds for $n=k+1$. Therefore, (95) holds for any natural number.
(2) We still use mathematical induction to prove (96). Since
$$\begin{aligned}
& F_{1}=F_{2}=1, F_{3}=2, F_{4}=3, \\
\text { therefore } \quad F_{1} F_{2} & =1=F_{2}^{2}, F_{1} F_{2}+F_{2} F_{3}+F_{3} F_{4} \\
& =1+2+6=9=F_{4}^{2},
\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,343 |
Example 2 Let $n$ be a positive integer, and $x_{1} \cdots, x_{n}, y_{1} \cdots, y_{n}$ be real numbers, then the inequality holds
$$\left(x_{1} y_{1}+\cdots+x_{n} y_{n}\right)^{2} \leqslant\left(x_{1}^{2}+\cdots+x_{n}^{2}\right)\left(y_{1}^{2}+\cdots+y_{n}^{2}\right)$$ | To prove that the proposition here is the inequality (1).
(1) When $n=1$, we have $\left(x_{1} y_{1}\right)^{2}=x_{1}^{2} y_{1}^{2}$, hence (1) holds.
(2) Now assume that inequality (1) holds for the natural number $n=k-1$, i.e., assume
$$\begin{array}{l}
\left(x_{1} y_{1}+\cdots+x_{k-1} y_{k-1}\right)^{2} \leqslant\l... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 741,344 |
5. Let $k$ be a positive integer, prove
$$\left(F_{4 k}, F_{4 k+2}\right)=1$$ | $$\begin{array}{l}
\text { 5. Proof: From the formula in Theorem } 4 \text {, we have } \\
F_{4 k}^{2}-F_{4 k-2} F_{4 k+2} \\
=\left(\frac{1}{5 \cdot 2^{8 k}}\right)\left(\left((1+\sqrt{5})^{4 k}-(1-\sqrt{5})^{4 k}\right)^{2}-\left((1+\sqrt{5})^{4 k-2}\right.\right. \\
\left.\left.-(1-\sqrt{5})^{4 k-2}\right) \cdot\le... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,346 |
6. Let $n$ be a positive integer, prove that
(1) $2 \mid F_{n}$ if and only if $3 \mid n$;
(2) $3 \mid F_{n}$ if and only if $4 \mid n$. | 6. Proof: If $3 \mid n$, we can set $n=3m$. In formula (102) of this chapter, take $m=3$ and take $n$ as the $m$ here, we get $F_{3} \mid F_{3m}$, i.e., $F_{3} \mid F_{n}$. Since $F_{3}=2$, it follows that $2 \mid F_{n}$. Conversely, suppose $2 \mid F_{n}$, we need to prove that $3 \mid n$.
Without loss of generality,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,347 |
8. There are two piles of chess pieces, with unequal numbers. Two people play a game, each can take any number of pieces from any one pile, but cannot take from both piles at the same time. The player who takes the last piece wins. Prove that the first player can always win. | 8. Proof: Let the number of stones in the two piles be $n_{1}, n_{2}$, and denote
$$n=n_{1}+n_{2} \text {, }$$
Since $n_{1} \neq n_{2}$, we can assume without loss of generality that $1 \leqslant n_{1}<n_{2}$. Thus, $n \geqslant 3$.
If $n=3$, then it must be that $n_{1}=1, n_{2}=2$. In this case, the first player can ... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 741,349 |
9. Proof
(1) Let $p>2$ be a prime number, and $r$ be a natural number, then $n=p \cdot 2^{r}$ is a perfect number when $p=2^{r+1}-1$, a deficient number when $p>2^{r+1}-1$, and a deficient number when $p<2^{r+1}-1$.
(2) Let $p>2$ be a prime number, and $q$ be a prime number, then when $p \neq q$ and $\frac{1}{q}+2\left... | 9. Proof:
(1) It is easy to see that all positive divisors of $n=p \cdot 2^{r}$ are the following $2(r+1)$ numbers:
$$\begin{array}{ll}
1,2, \cdots \cdots \cdots & , 2^{r} \\
p, 2 p, \cdots \cdots & , 2^{r} p
\end{array}$$
Thus, we have
$$\sigma(n)=(p+1)\left(1+2+\cdots+2^{r}\right)=(p+1)\left(2^{r+1}-1\right)$$
When... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,350 |
Lemma 2 Let $f(x)=a_{n} x^{\prime \prime}+\cdots+a_{1} x+a_{0}$ be an $n \geqslant 1$ degree polynomial with integer coefficients, $a_{n} \not \equiv 0(\bmod p)$, where $p$ is an odd prime. Then the number of solutions to the congruence equation
$$f(x) \equiv 0(\bmod p)$$
does not exceed its degree $n$. | We use proof by contradiction. Assume that the number of solutions to equation (15) exceeds $n$. Let
$$x \equiv x_{i}(\bmod p) \quad(i=1,2, \cdots, n, n+1)$$
be $n+1$ solutions to (15) that are mutually incongruent modulo $p$. By polynomial division, we have
$$f(x)=\left(x-\alpha_{1}\right) f_{1}(x)+r$$
where $f_{1}(... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,352 |
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