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Example 9 $3^{8833010}-59049$ can be divided by 24010000. | Given $24010000=245^{4} 7^{4}$ and Lemma 14, we have $\varphi(24010000)=2^{3} \times 5^{3} \times 4 \times 7^{3} \times 6=8232000$. Therefore, by Theorem 1 we have
$$3^{8232000} \equiv 1(\bmod 24010000)$$
By (28) we have
$$3^{8232010}-59049 \equiv 3^{10}-59049(\bmod 24010000)$$
By $3^{10}=59049$ and (29) we know that Example 9 holds. | 3^{10}-59049 \equiv 0(\bmod 24010000) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8. (i) Let $N=9450$, find $\varphi(N)$.
(ii) Find the sum of all positive integers not greater than 9450 and coprime with 9450. | 8. (i) Solution: Given $9450=2 \cdot 3^{3} \cdot 5^{2} \cdot 7$, and by Lemma 14, we have
$$\begin{aligned}
\varphi(N) & =9450\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \\
& =\frac{2 \cdot 3^{3} \cdot 5^{2} \cdot 7 \cdot 2 \cdot 4 \cdot 6}{2 \cdot 3 \cdot 5 \cdot 7}=2160
\end{aligned}$$
(ii) Solution: Let the sum of all positive integers not greater than 9450 and coprime with 9450 be $S$. By Problem 5, we have
$$\begin{aligned}
S & =\frac{1}{2} \cdot 9450 \cdot \varphi(9450) \\
& =\frac{1}{2} \times 9450 \times 2160 \\
& =10206000
\end{aligned}$$ | 10206000 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
9. (i) Determine whether $121^{6}-1$ is divisible by 21.
(ii) Find the remainder when $8^{4965}$ is divided by 13.
(iii) Let $p$ be any prime number other than 2 and 5. Prove that: $p \mid \underbrace{99 \cdots 9}_{(p-1) k \uparrow}, k$ is any positive integer. | 9. (i) Solution: Since $\varphi(21)=(3-1)(7-1)=12$, and $121 = 11^{12}$, and $(11,21)=1$. By Theorem 1, $11^{12} \equiv 1(\bmod 21)$, so $21 \mid\left(121^{6}-1\right)$.
(ii) Solution: Since $\varphi(13)=12$, and $4965=413 \times 12+9$. By Theorem 2, $8^{12} \equiv 1(\bmod 13)$, so $8^{4955} \equiv 8^{9}(\bmod 13)$. Also, $8^{2} \equiv -1(\bmod 13)$, so $8^{9}=8^{8} \cdot 8 \equiv 8(\bmod 13)$. Therefore,
$$8^{4965} \equiv 8(\bmod 13)$$
(iii) Proof: Since $p \neq 2,5$, we have $(10, p)=1$. Thus, $\left(10^{k}, p\right)=1$. By Theorem 2, $\left(10^{k}\right)^{p-1} \equiv 1(\bmod p)$, and
$$\left(10^{k}\right)^{p-1}-1=\underbrace{99 \cdots 9}_{(p-1) k \uparrow},$$
so
$$p \mid \underbrace{99 \cdots 9}_{(p-1) k \uparrow} .$$ | 21 \mid (121^{6}-1), 8^{4965} \equiv 8(\bmod 13), proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
12. Find positive integers $\boldsymbol{n}$ and $m, n>m \geqslant 1$, such that the last three digits of $1978^{n}$ and $1978^{m}$ are equal, and make $n+m$ as small as possible. (20th International Mathematical Olympiad Problem) | 12. Solution: $1978^{n}-1978^{m}=1978^{m}\left(1978^{n-m}-1\right)$
$$=2^{m} \cdot 989^{m}\left(1978^{n-m}-1\right)$$
Since the last three digits of $1978^{n}$ and $1978^{m}$ are the same, the last three digits of $1978^{n}-1978^{m}$ are all 0. Therefore, $1978^{n}-1978^{m}$ is divisible by 1000. And $1000=2^{3} \cdot 5^{3}$. Thus,
$$2^{3} \cdot 5^{3} \mid 2^{m} \cdot 989^{m}\left(1978^{n-m}-1\right)$$
Since $989^{m}$ and $1978^{n-m}-1$ are both odd, $2^{3} \mid 2^{m} . . m$ must be at least 3.
$$\text { Also, }\left(5^{3}, 2^{m} \cdot 989^{m}\right)=1 \text {, so } 5^{3} \mid\left(1978^{n \cdots m}-1\right) \text {, }$$
i.e.,
$$1978^{n-m} \equiv 1(\bmod 125)$$
The problem now is to find the smallest positive integer $n-m$ that satisfies the above congruence. At this point, taking $m=3, n+m=(n-m)+2 m$ will also be the smallest.
Since $\varphi(125)=5^{2} \cdot 4=100,(1978,125)=1$, by Theorem 1 we get
$$1978^{100} \equiv 1(\bmod 125)$$
We can prove that $(n-m) \mid 100$. Because otherwise:
$100=(n-m) q+r, q$ is an integer, $r$ is a positive integer,
and $0<r<n-m$, then
$$1978^{100}=1.978^{(n-m) q} \cdot 1978^{r} \equiv 1978^{r}(\bmod 125)$$
But
$$1978^{100} \equiv 1(\bmod 125)$$
so
$$1978^{r} \equiv 1(\bmod 125)$$
However, $r<n-m$, which contradicts the assumption that $n-m$ is the smallest positive integer that satisfies the congruence. Therefore, $(n-m) \mid 100$.
Since $125 | (1978^{n-m}-1)$, the last digit of $1978^{n-m}$ must be 1 or 6. It is easy to verify that the last digit of $1978^{n-m}$ is 6 only when $4 \mid(n-m)$. So $n-m$ is a multiple of 4, a divisor of 100, and can only be one of 4, 20, 100.
Since
$$\begin{aligned}
1978^{4} & =(125 \times 15+103)^{4} \equiv 103^{4}(\bmod 125) \\
103^{2} & =\left(3+4 \cdot 5^{2}\right)^{2} \equiv 3^{2}+2 \cdot 3 \cdot 4 \cdot 5^{2} \\
& \equiv 609 \equiv-16(\bmod 125) \\
103^{4} & \equiv(-16)^{2} \equiv 6(\bmod 125)
\end{aligned}$$
So
$$1978^{4} \equiv 1(\bmod 125)$$
And
$$1978^{20}=\left(1978^{4}\right)^{5} \equiv 6^{5} \equiv 1(\bmod 125)$$
Therefore, the smallest value of $n-m$ is 100. Now taking $m=3$, we have $n=103, n+m=106$. | 106 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 Please convert $\frac{15}{308}$ into a decimal. | Since $308=4 \times 77=2^{2} \cdot 77, \varphi(77)=(11-1) \times (7-1)=60$.
We also have
$$\begin{array}{ll}
10 \equiv 10(\bmod 77), & 10^{2} \equiv 23(\bmod 77) \\
10^{3} \equiv 76(\bmod 77), & 10^{4} \equiv 67(\bmod 77) \\
10^{5} \equiv 54(\bmod 77), & 10^{6} \equiv 1(\bmod 77)
\end{array}$$
Therefore, in Lemma 5, we can take $a=15, b=308, \alpha=2, \beta=0$, $h=6, b_{1}=77$. By Lemma 5, we have $\frac{15}{308}=0 . a_{1} a_{2} a_{3} a_{4} \cdots a_{8}$, and after calculation, we have
$$\frac{15}{308}=0.04 \dot{8} 7012 \dot{9}$$ | 0.04 \dot{8} 7012 \dot{9} | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Please convert $\frac{1}{17408}$ into a decimal. | Since $17408=1024 \times 17=2^{10} \cdot 17, \varphi(17)=16$. We also have
$$\begin{array}{cc}
10 \equiv 10(\bmod 17), & 10^{2} \equiv 15(\bmod 17) \\
10^{4} \equiv 4(\bmod 17), & 10^{8} \equiv 16(\bmod 17) \\
10^{16} \equiv 1(\bmod 17)
\end{array}$$
Therefore, in Lemma 5, we can take $a=1, b=17408, a=10, \beta=0$, $h=16, b_{1}=17$. By Lemma 5, we have $\frac{1}{17408}=0 . a_{1} a_{2} \cdots a_{10} a_{11}$ $a_{12} \cdots a_{26} g$. After calculation, we have
$$\frac{1}{17408}=0.0000574448 \dot{5} 29411764705882 \dot{3}$$ | 0.0000574448 \dot{5} 29411764705882 \dot{3} | Other | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Find the cube root of 3.652264. | From (31) we have
$$\sqrt[3]{3.652264}=\frac{\sqrt[3]{3652264}}{10^{2}}$$
Since $3652264=2^{3} \times 7^{3} \times 11^{3}$, by (32) we get
$$\sqrt[3]{3.652264}=\frac{2 \times 7 \times 11}{100}=1.54$$ | 1.54 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 8. Find the square root of 7.93. | Solving, from equation (31) we have
$$\sqrt{7.93}=\frac{\sqrt{793}}{10}$$
Since $793=61 \times 13$, where 61 and 13 are both prime numbers, from Table 1 we have
Table of square roots of prime numbers below 150
- (Table 1)
\begin{tabular}{c|c||c|c}
\hline$p$ & $\sqrt{p}$ & $p$ & $\sqrt{p}$ \\
\hline 2 & $1.41421356 \cdots$ & 67 & $8.18535277 \cdots$ \\
3 & $1.73205080 \cdots$ & 71 & $8.42614977 \cdots$ \\
5 & $2.23606797 \cdots$ & 73 & $8.54400374 \cdots$ \\
7 & $2.54575131 \cdots$ & 79 & $8.88819441 \cdots$ \\
11 & $3.3166247 \cdots$ & 83 & $9.11043357 \cdots$ \\
13 & $3.60555127 \cdots$ & 89 & $9.43398113 \cdots$ \\
17 & $4.12310562 \cdots$ & 97 & $9.84885780 \cdots$ \\
19 & $4.35889894 \cdots$ & 101 & $10.0498756 \cdots$ \\
23 & $4.79583152 \cdots$ & 103 & $10.1488915 \cdots$ \\
29 & $5.38516480 \cdots$ & 107 & $10.3440804 \cdots$ \\
31 & $5.56776436 \cdots$ & 109 & $10.4403065 \cdots$ \\
37 & $6.0927625 \cdots$ & 113 & $10.6301458 \cdots$ \\
41 & $6.40312423 \cdots$ & 127 & $11.2694276 \cdots$ \\
43 & $6.55743852 \cdots$ & 131 & $11.4455231 \cdots$ \\
47 & $6.85565460 \cdots$ & 137 & $11.7046999 \cdots$ \\
53 & $7.28010988 \cdots$ & 139 & $11.7898261 \cdots$ \\
59 & $7.68114574 \cdots$ & 149 & $12.2065556 \cdots$ \\
61 & $7.81024967 \cdots$ & & \\
\hline
\end{tabular}
Table of cube roots and fifth roots of prime numbers below 60
- (Table 2)
\begin{tabular}{r|l|l|l}
\hline$p$ & $\sqrt[3]{p}$ & $p$ & $\sqrt[5]{p}$ \\
\hline 2 & $1.2599210 \cdots$ & 2 & $1.1486983 \cdots$ \\
3 & $1.4422495 \cdots$ & 3 & $1.2457309 \cdots$ \\
5 & $1.7099759 \cdots$ & 5 & $1.37972966 \cdots$ \\
7 & $1.9129311 \cdots$ & 7 & $1.4757731 \cdots$ \\
11 & $2.22398009 \cdots$ & 11 & $1.6153942 \cdots$ \\
13 & $2.35133468 \cdots$ & 13 & $1.6702776 \cdots$ \\
17 & $2.57128159 \cdots$ & 17 & $1.7623403 \cdots$ \\
19 & $2.66840164 \cdots$ & 19 & $1.8019831 \cdots$ \\
23 & $2.8438669 \cdots$ & 23 & $1.8721712 \cdots$ \\
29 & $3.07231682 \cdots$ & 29 & $1.9610090 \cdots$ \\
31 & $3.14138065 \cdots$ & 31 & $1.9873407 \cdots$ \\
37 & $3.3322218 \cdots$ & 37 & $2.0589241 \cdots$ \\
41 & $3.44821724 \cdots$ & 41 & $2.10163247 \cdots$ \\
43 & $3.50339806 \cdots$ & 43 & $2.1217474 \cdots$ \\
47 & $3.6088260 \cdots$ & 47 & $2.1598300 \cdots$ \\
53 & $3.7562857 \cdots$ & 53 & $2.2123568 \cdots$ \\
59 & $3.8929964 \cdots$ & 59 & $2.2603224 \cdots$ \\
\hline
\end{tabular}
$$\sqrt{61}=7.81024967 \cdots, \quad \sqrt{13}=3.60555127 \cdots$$
From equations (33) and (34) we have
$$\begin{aligned}
\sqrt{7.93} & =\frac{(7.81024967 \cdots) \times(3.60555127 \cdots)}{10} \\
& =2.81602556 \cdots
\end{aligned}$$ | 2.81602556 \cdots | Other | math-word-problem | Yes | Yes | number_theory | false |
Example 10 Find the tenth root of 194400. | Since $194400=2^{2} \times 3^{5} \times 5^{2}$, we have
$$\sqrt[10]{194400}=\sqrt{2} \times \sqrt{3} \times \sqrt[5]{5}$$
From Table 1 we have
$$\sqrt{2}=1.41421356 \cdots, \quad \sqrt{3}=1.73205080 \cdots$$
From Table 2 we have
$$\sqrt[5]{5}=1.37972966 \cdots$$
From equations (37) to (39) we have
$$\begin{array}{l}
\sqrt[10]{194400}=(1.41421356 \cdots) \times(1.7320508 \cdots) \\
\times(1.37972966 \cdots)=3.3796336 \cdots
\end{array}$$ | 3.3796336 \cdots | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 12 Find $\sum_{k=1}^{8} \frac{1}{k!}$ equals what. | Solve $\begin{aligned} \sum_{k=1}^{8} \frac{1}{k!}= & 1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!} \\ = & 1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\frac{1}{720}+\frac{1}{5040} \\ & +\frac{1}{40320}=1.71827877 \cdots\end{aligned}$ | 1.71827877 \cdots | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 13 Find $\sum_{k=1}^{15} \frac{1}{k^{2}}$ equals what.
| Solve $\begin{aligned} \sum_{k=1}^{15} \frac{1}{k^{2}}= & 1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\frac{1}{5^{2}}+\frac{1}{6^{2}}+\frac{1}{7^{2}} \\ & +\frac{1}{8^{2}}+\frac{1}{9^{2}}+\frac{1}{10^{2}}+\frac{1}{11^{2}}+\frac{1}{12^{2}}+\frac{1}{13^{2}} \\ & +\frac{1}{14^{2}}+\frac{1}{15^{2}} \\ = & 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\frac{1}{49} \\ & +\frac{1}{64}+\frac{1}{81}+\frac{1}{100}+\frac{1}{121}+\frac{1}{144}+\frac{1}{169} \\ & +\frac{1}{196}+\frac{1}{225} \\ & =1.58044028 \cdots .\end{aligned}$ | 1.58044028 \cdots | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Column 14 Find $\sum_{k=1}^{11} \frac{1}{k!}$ equals what. | Solve $\begin{aligned} \sum_{k=1}^{11} \frac{1}{k!}= & \sum_{k=1}^{8} \frac{1}{k!}+\frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!} \\ = & 1.71827877 \cdots+\frac{1}{362880}+\frac{1}{3628800} \\ & +\frac{1}{39916800}=1.71828182 \cdots\end{aligned}$ | 1.71828182 \cdots | Algebra | math-word-problem | Yes | Yes | number_theory | false |
1. Convert the following fractions to decimals:
(i) $\frac{371}{6250}$,
(ii) $\frac{190}{37}$,
(iii) $\frac{13}{28}$,
(iv) $\frac{a}{875}, a=4,29,139,361$. | 1. (i) Solution: Since $6250=2 \times 5^{5}$, $\frac{371}{6250}$ is a finite decimal. After calculation, we get
$$\frac{371}{6250}=0.05936$$
(ii) Solution: $\frac{190}{37}=5+\frac{5}{37}$. Since $(10,37)=1$, $\frac{5}{37}$ is a pure repeating decimal. Also, $p(37)=36$, and $10^{2} \neq 1(\bmod 37), 10^{3} \equiv 1(\bmod 37)$, so the length of the repeating cycle is 3. After calculation, we get
$$\frac{190}{37}=5 . \dot{1} 3 \dot{5}$$
(iii) Solution: Since $28=2^{2} \times 7$, $\frac{13}{28}$ is a mixed repeating decimal. Also, $\varphi(7)=6$, and $10^{2} \equiv 1(\bmod 7), 10^{3} \equiv 1(\bmod 7), 10^{6} \equiv 1(\bmod 7)$, so the length of the repeating cycle is 6. After calculation, we get
$$\frac{13}{28}=0.46 \dot{4} 2857 \dot{1}$$
(iv) Solution: Since $875=5^{3} \times 7$, $\frac{a}{875}$ is a mixed repeating decimal. From the calculation in (iii), we know the length of the repeating cycle is 6. After calculation, we get
$$\begin{array}{l}
\frac{4}{875}=0.004 \dot{5} 7142 \dot{8} \\
\frac{29}{875}=0.033 \dot{1} 4285 \dot{7} \\
\frac{139}{875}=0.158 \dot{8} 5714 \dot{2} \\
\frac{361}{875}=0.41257142 \dot{8}
\end{array}$$ | \begin{array}{l}
\text{(i) } 0.05936 \\
\text{(ii) } 5 . \dot{1} 3 \dot{5} \\
\text{(iii) } 0.46 \dot{4} 2857 \dot{1} \\
\text{(iv) } \begin{array}{l}
\frac{4}{875}=0.00 | Algebra | math-word-problem | Yes | Yes | number_theory | false |
11. If $3^{k} \mid 1000!$ and $3^{k+1} \nmid 1000!$, find $k$. | 11. Solution: From the result of the previous problem, we have.
$$\begin{aligned}
k= & {\left[\frac{1000}{3}\right]+\left[\frac{1000}{9}\right]+\left[\frac{1000}{27}\right]+\left[\frac{1000}{81}\right] } \\
& +\left[\frac{1000}{243}\right]+\left[\frac{1000}{729}\right] \\
= & 333+111+37+12+4+1=498
\end{aligned}$$ | 498 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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
$$\begin{aligned}
\frac{107}{95} & =1+\frac{12}{95}=1+\frac{1}{\frac{95}{12}}=1+\frac{1}{7+\frac{11}{12}} \\
& =1+\frac{1}{7+\frac{1}{1+\frac{1}{11}}}=[1,7,1,11]
\end{aligned}$$ | [1,7,1,11] | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 |
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 |
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 $b$. | $$\begin{aligned}
p_{1}= & b, p_{2}=2 b^{2}+1 \\
p_{3}= & 2 b\left(2 b^{2}+1\right)+b=4 b^{3}+3 b \\
p_{4}= & 2 b\left(4 b^{3}+3 b\right)+2 b^{2}+1=8 b^{4}+8 b^{2}+1 \\
p_{5}= & 2 b\left(8 b^{4}+8 b^{2}+1\right)+4 b^{3}+3 b=16 b^{5}+20 b^{3}+5 b \\
p_{6}= & 2 b\left(16 b^{5}+20 b^{3}+5 b\right)+8 b^{4}+8 b^{2}+1 \\
= & 32 b^{6}+48 b^{4}+18 b^{2}+1 \\
p_{7}= & 2 b\left(32 b^{6}+48 b^{4}+18 b^{2}+1\right)+16 b^{5}+20 b^{3}+5 b \\
= & 64 b^{7}+112 b^{5}+56 b^{3}+7 b \\
p_{8}= & 2 b\left(64 b^{7}+112 b^{5}+56 b^{3}+7 b\right)+32 b^{6}+48 b^{4} \\
& +18 b^{2}+1 \\
= & 128 b^{8}+256 b^{6}+160 b^{4}+32 b^{2}+1 \\
q_{1}= & 1, q_{2}=2 b, q_{3}=4 b^{2}+1 \\
q_{4}= & 2 b\left(4 b^{2}+1\right)+2 b=8 b^{3}+4 b, \\
q_{5}= & 2 b\left(8 b^{3}+4 b\right)+4 b^{2}+1=16 b^{4}+12 b^{2}+1 \\
q_{6}= & 2 b\left(16 b^{4}+12 b^{2}+1\right)+8 b^{3}+4 b \\
= & 32 b^{5}+32 b^{3}+6 b, \\
q_{7}= & 2 b\left(32 b^{5}+32 b^{3}+6 b\right)+16 b^{4}+12 b^{2}+1 \\
= & 64 b^{6}+80 b^{4}+24 b^{2}+1, \\
q_{8}= & 2 b\left(64 b^{6}+80 b^{4}+24 b^{2}+1\right)+32 b^{5}+32 b^{3}+6 b \\
= & 128 b^{7}+192 b^{5}+80 b^{3}+8 b
\end{aligned}$$ | 128 b^{7}+192 b^{5}+80 b^{3}+8 b | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 $2^{1} \equiv 2,2^{5} \equiv 6,2^{7} \equiv 11,2^{11} \equiv 7(\bmod 13)$ are precisely all the primitive roots of 13.
This method is also fully applicable to finding all $l$-th power residues of $p$ $(l \mid(p-1))$. Since $p-1=12$ has the divisors $1,2,3,4,6,12$, we know that modulo 13 there are $\varphi(1)=1$ first-order elements, $\varphi(2)=1$ second-order elements, $\varphi(3)=2$ third-order elements, $\varphi(4)=2$ fourth-order elements, $\varphi(6)=2$ sixth-order elements, and $\varphi(12)=4$ primitive roots. Clearly, 1 is a first-order element; $-1 \equiv 12$ is a second-order element; since $3^{3} \equiv 1$ and $3^{2} \equiv 1(\bmod 13)$, 3 is a third-order element. Noting that among $1,2,3$ the numbers coprime with 3 are 1 and 2, the complete set of third-order elements modulo 13 is $3^{1}$ and $3^{2} \equiv 9(\bmod 13)$. Since $5^{2} \equiv-1,5^{3} \equiv-5$, and $5^{4} \equiv 1(\bmod 13)$, 5 is a fourth-order element of 13. Noting that among $1,2,3,4$ the numbers coprime with 4 are 1 and 3, the complete set of fourth-order elements modulo 13 is the following two: $5^{1}, 5^{3}=8(\bmod 13)$. From the above calculations, it is clear that the remaining 4 and 10 must be precisely all the sixth-order elements modulo 13. | 2, 6, 7, 11 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$.
- For $k=2$, the series is $\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$.
- For $k=3$, the series is $\sum_{n=1}^{\infty} \frac{1}{n^6} = \frac{\pi^6}{945}$.
- For $k=4$, the series is $\sum_{n=1}^{\infty} \frac{1}{n^8} = \frac{\pi^8}{9450}$.
These results are specific cases of the Riemann zeta function, $\zeta(s)$, where $s = 2k$. | We have (by (5))
$$\begin{array}{l}
\sum_{n=1}^{\infty} \frac{1}{n^{2}}=(-1)^{2} \frac{(2 \pi)^{2} B_{2}}{2 \cdot 2!}=\pi^{2} B_{2}=\pi^{2} / 6, \\
\sum_{n=1}^{x} \frac{1}{n^{4}}=(-1)^{3} \frac{(2 \pi)^{4} B_{4}}{2 \cdot 4!}=-\frac{\pi^{4}}{3} B_{4}=\pi^{4} / 90, \\
\sum_{n=1}^{\infty} \frac{1}{n^{6}}=(-1)^{4} \frac{(2 \pi)^{6} B_{6}}{2 \cdot 6!}=\frac{2 \pi^{6}}{45} B_{6}=\pi^{6} / 945, \\
\sum_{n=1}^{\infty} \frac{1}{n^{8}}=(-1)^{5} \frac{(2 \pi)^{8} B_{8}}{2 \cdot 8!}=-\frac{\pi^{8}}{315} B_{8}=\pi^{8} / 9450 .
\end{array}$$ | \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 |
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 in the previous section):
$$14^{2} \equiv 1, 14^{4} \equiv 1, 14^{7} \equiv 1, 14^{14} \equiv 1 \pmod{29}$$
Calculations yield:
$$\begin{array}{c}
14^{2} \equiv 22, \quad 14^{4} \equiv 22^{2} \equiv (-7)^{2} \equiv 22 \\
14^{7} \equiv (14)^{4}(14)^{2}(14) \equiv (20)(22)(14) \equiv 12 \\
14^{14} \equiv (12)^{2} \equiv -1 \pmod{29}
\end{array}$$
Thus, the conditions in (21) are indeed satisfied, and 14 must be a primitive root modulo 29.
Further calculations yield:
$$\begin{aligned}
(14)^{28} & = (196)^{14} = (38416)^{7} \equiv (571)^{7} = (571)(326041)^{3} \\
& \equiv (571)(574)^{3} = (327754)(329476) \equiv (605)(645) \\
& = 390225 \equiv 1 \pmod{29^2}
\end{aligned}$$
Therefore, 14 is not a primitive root modulo \( 29^2 \). To find a primitive root of \( 29^2 \) from 14, we consider \( 14 + 29 = 43 \), since
$$43 \equiv 14 \pmod{29}$$
43 is still a primitive root modulo 29. However,
$$\begin{aligned}
(43)^{28} & = (3418801)^{7} \equiv (136)^{7} = (136)(18496)^{3} \\
& \equiv (136)(-6)^{3} = -29376 \equiv 59 \equiv 1 \pmod{29^2}
\end{aligned}$$
Thus, by Theorem 8, 43 is a primitive root modulo \( 29^2 \).
The method used in the above example is generally applicable. That is, if \( g \) is a primitive root modulo \( p \geqslant 3 \) and
$$g^{p-1} \equiv 1 \pmod{p^2}$$
then \( g + p \) must be a primitive root modulo \( p^2 \).
Since it is known that a primitive root modulo a prime \( p \) always exists, the above result implies that a primitive root modulo \( p^2 \) also always exists for an odd prime \( p \geqslant 3 \).
Surprisingly, for any odd prime power \( p^s, s \geqslant 2 \), condition (17) is also a necessary and sufficient condition for a primitive root \( g \) of \( p \) to remain a primitive root of \( p^s \). When \( g \) does not satisfy (17), taking \( g + p \) will yield a primitive root of \( p^s \). This is the result stated below. | 43 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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
$$\operatorname{ind}_{g} x \equiv \operatorname{ind}_{9} 13-\operatorname{ind}_{9} 9=30(\bmod \varphi(43))$$
Therefore, we have
$$\operatorname{ind}_{g} x=30,$$
Looking up the table again, we get
$$x \equiv 11(\bmod 43)$$ | x \equiv 11(\bmod 43) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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
$$\text { ind }_{g} 11=12 \text {, }$$
Substituting this into the congruence equation, we get
$$6 \text { ind }_{g} x \equiv 12(\bmod 18)$$
Eliminating the common factor 6, we obtain
$$\operatorname{ind}_{g} x \equiv 2(\bmod 3)$$
Thus, there are six solutions:
$$\text { ind }_{g} x=2,5,8,11,14,17$$
Referring to the index table again, the corresponding six solutions are
$$x \equiv 4,13,9,15,6,10(\bmod 19)$$ | x \equiv 4,13,9,15,6,10(\bmod 19) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 exactly 6 digits. The calculations yield
$$\begin{array}{ll}
\frac{1}{7}=0 . \dot{1} 4285 \dot{ }= & \frac{2}{7}=0 . \dot{2} 8571 \dot{4} \\
\frac{3}{7}=0 . \dot{4} 2857 \dot{i}, & \frac{4}{7}=0 . \dot{5} 7142 \dot{8} \\
\frac{5}{7}=0 . \dot{7} 1428 \dot{5}, & \frac{6}{7}=0 . \dot{8} 5714 \dot{2}
\end{array}$$
This set of decimals exhibits an interesting phenomenon: the repeating cycle of each decimal is composed of the same six digits $1,4,2,8,5,7$. If we follow the order of the repeating cycle of $\frac{1}{7}$
Figure 1 shows
$$1 \rightarrow 4 \rightarrow 2 \rightarrow 8 \rightarrow 5 \rightarrow 7$$
and add an arrow from 7 to 1, we get a circle (see left Figure 1). Comparing the order of the digits in the repeating cycles of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$, we find that they all have the same order as in Figure 1, just starting from different digits among the six, for example, $\frac{2}{7}$ starts from the digit 2, $\frac{3}{7}$ starts from the digit 4, etc. | \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 |
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 ind $x=38$. Checking the 13th table again, we get $x \equiv 17(\bmod 43)$.
(2) Checking the table, we find ind $17=38$. Let ind $x=y$, then we have $8 y \equiv 38(\bmod 42)$,
which simplifies to
$$4 y \equiv 19 \equiv 40(\bmod 21)$$
Solving this, we get
$$y \equiv 10(\bmod 21)$$
Thus,
$$y_{1}=10, y_{2}=31$$
Checking the table, we get the two solutions $x_{1} \equiv 10, x_{2} \equiv 33(\bmod 43)$.
(3) Checking the table, we find ind $8=39$, ind $4=12$, hence we get
$$39 x \equiv 12(\bmod 42)$$
Thus,
$$13 x \equiv 4(\bmod 14)$$
Since $2|4,2| 14$, it must be that $2 \mid 13 x$, i.e., $x=2 y$, so
$$13 y \equiv 2(\bmod 7)$$
Hence,
$$y \equiv-2(\bmod 7)$$
Thus,
$$x=2 y \equiv-4 \equiv 3(\bmod 7)$$
To ensure $2 \mid x$, we get $x \equiv 10,24,38(\bmod 42)$. | 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 |
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.
$$\begin{array}{l}
7^{1}=7,7^{3}=343 \equiv 59,7^{9} \equiv 59^{3}=205379 \equiv 47,7^{11} \\
\equiv 7^{2} \cdot 47 \equiv 31,7^{13} \equiv 31 \cdot 7^{2} \equiv 28,7^{17} \equiv 7^{4} \cdot 28 \equiv \\
7 \cdot 59 \cdot 28 \equiv 62,7^{19} \equiv 7^{2} \cdot 62 \equiv 56,7^{23} \equiv 7^{3} \cdot 7^{3} \\
\cdot 56 \equiv 7 \cdot 59 \cdot 56 \equiv 53,7^{27} \equiv 7 \cdot 59 \cdot 53 \equiv 21, \\
7^{29} \equiv 7^{2} \cdot 21 \equiv 35,7^{31} \equiv 7^{2} \cdot 35 \equiv 11,7^{33} \equiv 7^{2} \\
\cdot 11 \equiv 42,7^{37} \equiv 7 \cdot 59 \cdot 42 \equiv 22,7^{39} \\
\equiv 7^{2} \cdot 22=1078 \equiv 13,7^{41} \equiv 7^{2} \cdot 13=637 \equiv 69,7^{43} \\
\equiv 7^{2} \cdot 69=3381 \equiv 44,7^{47} \equiv 7 \cdot 59 \cdot 44=18172 \\
\equiv 67,7^{51} \equiv 7 \cdot 59 \cdot 67=27671 \equiv 52,7^{53} \equiv 7^{2} \cdot 52 \\
=2548 \equiv 63,7^{57} \equiv 7^{2} \cdot 59 \cdot 63=26019 \equiv 33,7^{59} \equiv 7^{2} \\
\cdot 33=1617 \equiv 55,7^{61} \equiv 7^{2} \cdot 55=2695 \equiv 68,7^{67} \\
\equiv\left(7^{3}\right)^{2} \cdot 7^{61} \equiv(59)^{2} \cdot 68=236708 \equiv 65,7^{69} \equiv 7^{2} \\
\cdot 65=3185 \equiv 61 \neq(\bmod 71) .
\end{array}$$
Thus, the 24 primitive roots modulo 71 are
$61(\bmod 71)$.
Further calculations yield
$$\begin{array}{c}
7^{5}=16807 \equiv 1684\left(\bmod 71^{2}\right) \\
7^{10} \equiv(1684)^{2}=2835856 \equiv 2814\left(\bmod 71^{2}\right) \\
7^{20} \equiv(2814)^{2}=7918596 \equiv-815\left(\bmod 71^{2}\right) \\
7^{40} \equiv(-815)^{2}=664225 \equiv-1187\left(\bmod 71^{2}\right) \\
7^{70}=7^{10} \cdot 7^{20} \cdot 7^{40} \equiv(2814)(-815)(-1187)= \\
(2814)(967405) \equiv(2814)(-467)=-1314138 \\
\equiv 1563\left(\bmod 71^{2}\right)
\end{array}$$
Therefore, $7^{p-1} \not \equiv 1\left(\bmod p^{2}\right)$, and by Theorem 8 of this chapter, 7 is also a primitive root modulo $71^{2}=5041$. Furthermore, by Theorem 12 of this chapter, 7 is also a primitive root modulo $2 p^{2}=10082$.
$$\begin{array}{l}
7,59,47,31,28,62,56,53,21,35,11,42,22, \\
13,69,44,67,52,63,33,55,68,65,
\end{array}$$ | 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 |
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 $\square$
$$(a-2d)(a^{2}+5ad+7d^{2})=0$$
Since $25-4 \times 7=-30$
From (1), we get $a=2d$, so the solution is
$$x=3d, y=4d, z=5d, w=6d$$
where $d$ is any natural number. | x=3d, y=4d, z=5d, w=6d | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 following four cases:
( 1$)\left\{\begin{array}{l}x-y=2, \\ x+y=30\end{array}\right.$
(2) $\left\{\begin{array}{l}x-y=6, \\ x+y=10,\end{array}\right.$
(3) $\left\{\begin{array}{l}x-y=10, \\ x+y=6,\end{array}\right.$
(4) $\left\{\begin{array}{l}x-y=30, \\ x+y=2\end{array}\right.$
The solutions are
$$\left\{\begin{array}{l}
x_{1}=16 \\
y_{1}=14
\end{array},\left\{\begin{array} { l }
{ x _ { 2 } = 8 , } \\
{ y _ { 2 } = 2 , }
\end{array} \left\{\begin{array} { l }
{ x _ { 3 } = 8 , } \\
{ y _ { 3 } = - 2 , }
\end{array} \left\{\begin{array}{l}
x_{4}=16 \\
y_{4}=-14
\end{array}\right.\right.\right.\right.$$
Thus, the integers $x$ that make $x^{2}-60$ a perfect square are 16 or 8. | 16 \text{ or } 8 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 by 6", and property $P_{3}$ is "an integer is divisible by 8". $A_{i}(i=1,2,3)$ is the subset of $S$ consisting of integers with property $P_{i}$. Note that
$$\begin{array}{l}
\left|A_{1}\right|=\left[\frac{1000}{5}\right]=200, \\
\left|A_{2}\right|=\left[\frac{1000}{6}\right]=166, \\
\left|A_{3}\right|=\left[\frac{1000}{8}\right]=125,
\end{array}$$
Since $\operatorname{LCM}\{5,6\}=30$, we have
$$\left|A_{1} \cap A_{2}\right|=\left[\frac{1000}{30}\right]=33 \text {, }$$
Since $\operatorname{LCM}\{5,8\}=40$, we have
$$\left|A_{1} \cap A_{3}\right|=\left[\frac{1000}{40}\right]=25$$
Since $\operatorname{LCM}\{6,8\}=24$, we have
$$\left|A_{2} \cap A_{3}\right|=\left[\frac{1000}{24}\right]=41$$
Since $\operatorname{LCM}\{5,6,8\}=120$, we have
$$\left|A_{1} \cap A_{2} \cap A_{3}\right|=\left[\frac{1000}{120}\right]=8$$
Thus, by Theorem 1, the number of integers in $S$ that are not divisible by 5, 6, or 8 is
$$\begin{aligned}
\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3}\right|= & |S|-\left|A_{1}\right|-\left|A_{2}\right|-\left|A_{3}\right|+\left|A_{1} \cap A_{2}\right|+\left|A_{1} \cap A_{3}\right| \\
& +\left|A_{2} \cap A_{3}\right|-\left|A_{2} \cap A_{2} \cap A_{3}\right| \\
= & 1000-200-166-125+33+25+41-8 \\
= & 600
\end{aligned}$$ | 600 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 $p_{1}, \cdots, p_{s}$ are distinct prime numbers, $\alpha_{j} \geqslant 1,1 \leqslant j \leqslant s, s \geqslant 1$.
Let $P_{i}$ represent the property that a natural number in the set $S=\{1,2, \cdots, n\}$ is divisible by $P_{i}$ $(i=1, \cdots s)$. The subset of $S$ with property $P_{i}$ is denoted as $A_{i}$. Thus, we have
$$\begin{aligned}
\varphi(n)= & \left|\bar{A}_{1} \cap \cdots \cap \bar{A}_{s}\right|=|S|-\sum_{i}\left|A_{i}\right|+\sum_{i, j}\left|A_{i} \cap A_{j}\right|-+\cdots \\
& +(-1)^{s}\left|A_{1} \cap \cdots \cap A_{s}\right| \\
= & n-\sum_{i} \frac{n}{p_{i}}+\sum_{i<j} \frac{n}{p_{i} p_{j}}-+\cdots+(-1)^{s} \frac{n}{p_{1} \cdots p_{s}} \\
= & n\left(1-\frac{1}{p_{1}}\right) \cdots\left(1-\frac{1}{p_{s}}\right)=n \prod_{p \mid n}\left(1-\frac{1}{p}\right)
\end{aligned}$$
Example: From $60=2^{2} \cdot 3 \cdot 5$, we get
$$\varphi(60)=60\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)=16 .$$ | 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". Here is the formatted output as requested:
None
| 16 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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
$$\begin{array}{c}
10^{5}-\left[\frac{10^{5}}{7}\right]-\left[\frac{10^{5}}{11}\right]-\left[\frac{10^{5}}{13}\right]+\left[\frac{10^{5}}{7 \times 11}\right] \\
+\left[\frac{10^{5}}{7 \times 13}\right]+\left[\frac{10^{5}}{11 \times 13}\right]-\left[\frac{10^{5}}{7 \times 11 \times 13}\right]=10^{5} \\
-14285-9090-7692+1298+1098+699-99=71929
\end{array}$$ | 71929 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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 groups, with 550 students joining the mathematics group, 460 students joining the Chinese group, and 350 students joining the foreign language group. Among them, 100 students participate in both the mathematics and foreign language groups, 120 students participate in both the mathematics and Chinese groups, and 140 students participate in all three groups. How many students participate in both the Chinese and foreign language groups? | 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, we have
$$\left|A_{1}\right|=550,\left|A_{2}\right|=460,\left|A_{3}\right|=350$$
We also use $A_{12}$ to represent the set of students who joined both the math and Chinese groups, $A_{13}$ to represent the set of students who joined both the math and English groups, and $A_{23}$ to represent the set of students who joined both the Chinese and English groups, then we have
$$\left|A_{13}\right|=100, \quad\left|A_{12}\right|=120$$
We use $A_{123}$ to represent the set of students who joined all three groups, then
$$\left|A_{123}\right|=140$$
Noting the initial explanation, by Theorem 1 of this chapter, we get
$$\begin{array}{c}
0=1200-\left|A_{1}\right|-\left|A_{2}\right|-\left|A_{3}\right|+\left|A_{12}\right| \\
+\left|A_{13}\right|+\left|A_{23}\right|-\left|A_{123}\right|,
\end{array}$$
Thus, we have
$$\begin{aligned}
\left|A_{23}\right| & =-1200+550+460+350-100-120+140 \\
& =80
\end{aligned}$$
That is, the number of students who joined both the Chinese and English groups is 80. | 80 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
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 $i$-th chair? | 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 problem is reduced to finding the number of permutations $a_{1}, a_{2}, \cdots, a_{n}$ that satisfy condition (1). This special type of permutation without repetition is called a derangement of $1,2, \cdots, n$. Generally, $D_{n}$ denotes the number of all possible derangements of the set $\{1,2, \cdots, n\}$. The problem is to find the value of $D_{n}$.
Let $S$ be the set of all permutations of the set $\{1,2, \cdots, n\}$, then $|S|=n!$. For $j=1,2, \cdots, n$, let $P_{j}$ be the property that "in a permutation, the number $j$ is still in the $j$-th position". Let $A_{j}(j=1,2, \cdots, n)$ be the subset of $S$ consisting of all permutations that have property $P_{j}$. By definition, a derangement of $\{1,2, \cdots, n\}$ is a permutation in $S$ that does not have any of the properties $P_{1}, P_{2}, \ldots, P_{n}$. Thus, by Theorem 1 of this chapter, we have
$$\begin{aligned}
D_{n}= & |S|-\sum\left|A_{j}\right|+\sum\left|A_{i} \cap A_{j}\right|-+\cdots+(-1)^{n} \\
& \left|A_{1} \cap A_{2} \cap \cdots \cap A_{n}\right|
\end{aligned}$$
Since the set $A_{j}(j=1,2, \cdots, n)$ consists of all permutations that keep the number $j$ in the $j$-th position, $\left|A_{j}\right|$ is the number of permutations of the remaining $n-1$ elements, i.e.,
$$\left|A_{j}\right|=(n-1)!\quad(j=1,2, \cdots, n)$$
Similarly, we have
$$\begin{array}{l}
\left|A_{i} \cap A_{j}\right|=(n-2)!\quad(1<i<j<n) \\
\cdots \cdots \cdots \\
\left|A_{1} \cap A_{2} \cap \cdots \cap A_{n}\right|=0!=1
\end{array}$$
Thus, from (2), we have
$$\begin{aligned}
D_{n}= & n!-\binom{n}{1}(n-1)!+\binom{n}{2}(n-2)!-+\cdots \\
& +(-1)^{n}\binom{n}{n} 0!=n!\left(1-\frac{1}{1!}+\frac{1}{2!}-+\cdots\right. \\
& \left.+(-1)^{n} \frac{1}{n!}\right) .
\end{aligned}$$ | 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 |
*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 $\{1,2, \cdots, n\}$. The number of all $k$ - derangements of the set $\{1,2, \cdots, n\}$ is denoted by $D_{n}(k)$, try to find the value of $D_{n}(k)$. | 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 of $S$ consisting of all permutations with property $P_{i}$. Thus, the problem requires the number of permutations in $S$ that have exactly $n-k$ properties. From the result of the previous problem (taking $r=n-k$ there),
$$\begin{aligned}
D_{n}(k) & =\binom{n-k}{n-k}\binom{n}{n-k} k!-\binom{n-k+1}{n-k} \\
& \cdot\binom{n}{n-k+1}(k-1)!+\cdots+(-1)^{k}\binom{n}{n-k}\binom{n}{n} 0! \\
& =\sum_{s=n-k}^{n}(-1)^{s-(n-k)}\binom{s}{n-k}\binom{n}{s}(n-s)! \\
& =\sum_{s=n-k}^{n}(-1)^{s-n+k} \frac{s!}{(n-k)!(s-n+k)!} \frac{n!}{s!(n-s)!}(n-s)! \\
& =\frac{n!}{(n-k)!} \sum_{s=n-k}^{n}(-1)^{s-n+k} \frac{1}{(s-n+k)!}
\end{aligned}$$
Note: In formula (8), by setting $k=n$, we get
$$D_{n}=D_{n}(n)=n!\sum_{s=0}^{n}(-1)^{s} \frac{1}{s!}$$
which is precisely the result from Problem 3. | 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 |
Example 2 Find an odd prime $p$ such that
$$x^{2} \equiv 5(\bmod p)$$
List the solution | From $5 \equiv 1(\bmod 4)$ and Theorem 2, we know
$$\left(\frac{5}{p}\right)=\left(\frac{p}{5}\right)$$
Thus, when $p \equiv 1$ or $-1(\bmod 5)$, $\left(\frac{p}{5}\right)=1$. In this case, equation (44) has a solution. When $p \equiv 2$ or $-2(\bmod 5)$, $\left(\frac{p}{5}\right)=-1$, in which case equation (44) has no solution. | p \equiv 1 \text{ or } -1(\bmod 5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find an odd prime $p$ such that
$$x^{2}+3 \equiv 0(\bmod p)$$
has a solution. | From $p$ being a prime number and Lemma 8 and Theorem 2, we have
$$\begin{array}{l}
\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) \\
\quad=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p-1}{2}}\left(\frac{p}{3}\right)=\left(\frac{p}{3}\right) .
\end{array}$$
Thus, when $p \equiv 1(\bmod 3)$, we have $\left(\frac{-3}{p}\right)=\left(\frac{1}{3}\right)=1$, meaning that equation (45) has a solution. When $p \equiv -1(\bmod 3)$, we have $\left(\frac{-3}{p}\right)=\left(\frac{-1}{3}\right)=(-1)^{\frac{3-1}{2}}=-1$, meaning that equation (45) has no solution. Additionally, for $p=3$, equation (45) clearly also has a solution. | p \equiv 1(\bmod 3) \text{ or } p = 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Discuss the congruence equation
$$x^{2} \equiv -286 \pmod{4272943}$$
whether it has a solution, where 4272943 is a prime number. | Let $p=4272943$, by Lemma 7 we have
$$\left(\frac{-286}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)\left(\frac{143}{p}\right)$$
Since $4272943 \equiv 7(\bmod 8)$, we have
$$\left(\frac{-1}{p}\right)=-1,\left(\frac{2}{p}\right)=1$$
Thus,
$$\left(\frac{-286}{p}\right)=-\left(\frac{143}{p}\right)$$
Since $143=4 \times 35+3, p=3(\bmod 4)$, by Theorem 3 we have
$$\left(\frac{143}{p}\right)=-\left(\frac{p}{143}\right)$$
From $p=143 \times 29880+103$, we get
$$\left(\frac{p}{143}\right)=\left(\frac{103}{143}\right)$$
By Theorem 3 and $103=3(\bmod 4), 143 \equiv 3(\bmod 4)$, we have
$$\begin{array}{l}
\left(\frac{103}{143}\right)=-\left(\frac{143}{103}\right)=-\left(\frac{40}{103}\right)=-\left(\frac{2^{2} \times 2 \times 5}{103}\right) \\
\quad=-\left(\frac{2 \times 5}{103}\right)=-\left(\frac{2}{103}\right)\left(\frac{5}{103}\right)=-\left(\frac{5}{103}\right) \\
=-\left(\frac{103}{5}\right)=-\left(\frac{3}{5}\right)=1
\end{array}$$
Therefore,
$$\left(\frac{-286}{p}\right)=1$$
This means that equation (51) has a solution. | 1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
4. Find the prime $p \geqslant 5$ for which 3 is a quadratic residue.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note above should not be included in the final translation, as it is a meta-instruction. Here is the final translation:
4. Find the prime $p \geqslant 5$ for which 3 is a quadratic residue. | 4 . Solution: We have
$$\left(\frac{3}{p}\right)=(-1)^{\frac{p-1}{2}}\left(\frac{p}{3}\right)$$
It is easy to see that
$$(-1)^{\frac{p-1}{2}}=\left\{\begin{aligned}
1, & \text { when } p \equiv 1(\bmod 4), \\
-1, & \text { when } p \equiv 3(\bmod 4),
\end{aligned}\right.$$
and on the other hand,
$$\left(\frac{p}{3}\right)=\left\{\begin{aligned}
1, & \text { when } p \equiv 1(\bmod 3), \\
-1, & \text { when } p \equiv 2(\bmod 3).
\end{aligned}\right.$$
From $\left\{\begin{array}{l}p \equiv 1(\bmod 4) \\ p \equiv 1(\bmod 3)\end{array} \quad\right.$ we get $p \equiv 1(\bmod 12)$,
$$\left\{\begin{array}{l}
p \equiv-1(\bmod 4) \\
p \equiv-1(\bmod 3)
\end{array} \quad \text { we get } p \equiv-1(\bmod 12),\right. \text { thus }$$
we obtain
$$\left(\frac{3}{p}\right)=\left\{\begin{array}{r}
1, \text { when } p \equiv \pm 1(\bmod 12), \\
-1, \text { when } p \equiv \pm 5(\bmod 12).
\end{array}\right.$$ | p \equiv \pm 1(\bmod 12) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5. Find the prime $p$ for which 10 is a quadratic residue. | 5. Solution: We have
$$\left(\frac{10}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{5}{p}\right)$$
By Example 2 of this chapter, we have
And by Theorem 9, we have
$$\begin{array}{l}
\left(\frac{5}{p}\right)=\left\{\begin{array}{ll}
1, & p \equiv \pm 1(\bmod 5), \\
-1, & p \equiv \pm 2(\bmod 5),
\end{array}\right. \\
\text { and } \\
\left(\frac{2}{p}\right)=\left\{\begin{array}{ll}
1, & p \equiv \pm 1(\bmod 8), \\
-1, & p \equiv \pm 3(\bmod 8) .
\end{array}\right.
\end{array}$$
Solve the following systems of congruences:
$$\begin{array}{l}
\left\{\begin{array}{l}
p \equiv 1(\bmod 5), \\
p \equiv 1(\bmod 8),
\end{array}, \begin{array}{l}
p \equiv 1(\bmod 5), \\
p \equiv-1(\bmod 8)
\end{array},\left\{\begin{array}{l}
p \equiv-1(\bmod 5) \\
p \equiv 1(\bmod 8),
\end{array}\right.\right. \\
\left\{\begin{array}{l}
p \equiv-1(\bmod 5), \\
p=-1(\bmod 8) .
\end{array}\right.
\end{array}$$
We get
$$\begin{array}{l}
p \equiv 1(\bmod 40), p \equiv 31(\bmod 40), p \equiv 9(\bmod 40) \\
p \equiv-1(\bmod 40),
\end{array}$$
Thus, for $p \equiv \pm 1, \pm 9(\bmod 40)$, we have $\left(\frac{10}{p}\right)=1$.
Solve the following systems of congruences:
$$\begin{array}{l}
\left\{\begin{array}{l}
p \equiv 2(\bmod 5), \\
p \equiv 3(\bmod 8),
\end{array},\left\{\begin{array}{l}
p \equiv 2(\bmod 5), \\
p \equiv-3(\bmod 8),
\end{array},\left\{\begin{array}{l}
p \equiv-2(\bmod 5) \\
p \equiv 3(\bmod 8),
\end{array}\right.\right.\right. \\
\left\{\begin{array}{l}
p \equiv=-2(\bmod 5), \\
p \equiv-3(\bmod 8) .
\end{array}\right.
\end{array}$$
We get $p \equiv 27,-3,3,-27(\bmod 40)$, and thus we also know that when $p= \pm 27, \pm 3$, we have $\left(\frac{10}{p}\right)=1$.
Therefore, we conclude that if and only if
$$p \equiv \pm 1, \pm 3, \pm 9, \pm 13,(\bmod 40)$$
$p$ has 10 as a quadratic residue. Similarly, we can prove that if and only if
$$p \equiv \pm 7, \pm 11, \pm 17, \pm 19(\bmod 40)$$
$p$ has 10 as a quadratic non-residue. | p \equiv \pm 1, \pm 3, \pm 9, \pm 13,(\bmod 40) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
6. Find the prime $p$ for which 6 is a quadratic residue.
The text has been translated while preserving the original line breaks and format. | 6. Solution: We have
$$\left(\frac{6}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{3}{p}\right)$$
By Lemma 9 and Problem 4, we have respectively
$$\begin{array}{l}
\left(\frac{2}{p}\right)=\left\{\begin{array}{l}
1, p \equiv \pm 1(\bmod 8) \\
-1, p \equiv \pm 3(\bmod 8)
\end{array}\right. \\
\left(\frac{3}{p}\right)=\left\{\begin{array}{ll}
1, & p \equiv \pm 1(\bmod 12) \\
-1, & p \equiv \pm 5(\bmod 12)
\end{array}\right.
\end{array}$$
Therefore, for $\left(\frac{6}{p}\right)=1$, it is necessary that $\left(\frac{2}{p}\right)$ and $\left(\frac{3}{p}\right)$ are both 1 or both -1. This leads to the following eight systems of congruences:
$$\begin{array}{c}
\left\{\begin{array}{l}
p \equiv 1(\bmod 8) \\
p \equiv 1(\bmod 12)
\end{array},\left\{\begin{array}{l}
p \equiv 1(\bmod 8) \\
p \equiv-1(\bmod 12)
\end{array},\left\{\begin{array}{l}
p \equiv-1(\bmod 8) \\
p \equiv 1(\bmod 12)
\end{array}\right.\right.\right. \\
\left\{\begin{array}{l}
p \equiv-1(\bmod 8), \\
p \equiv-1(\bmod 12),
\end{array}, \begin{array}{l}
p \equiv 3(\bmod 8), \\
p \equiv 5(\bmod 12),
\end{array},\left\{\begin{array}{l}
p \equiv 3(\bmod 8) \\
p \equiv-5(\bmod 12)
\end{array}\right.\right. \\
\left\{\begin{array}{l}
p \equiv-3(\bmod 8), \\
p \equiv 5(\bmod 12),
\end{array},\left\{\begin{array}{c}
p \equiv-3(\bmod 8) \\
p \equiv-5(\bmod 12)
\end{array}\right.\right.
\end{array}$$
The second system, when simplified modulo 4, leads to a contradictory system of congruences $p \equiv 1(\bmod 4), p \equiv-1(\bmod 4)$, hence it has no solution; similarly, the 3rd, 5th, and 8th systems of congruences also have no solutions.
From the 1st, 4th, 6th, and 7th systems of congruences, we obtain the following four equivalent systems of congruences:
Solving these, we find that $p$ is a quadratic residue of 6 if and only if $p \equiv \pm 1, \pm 5(\bmod 24)$. Similarly, by solving the remaining systems of congruences, we can prove that $p$ is a quadratic non-residue of 6 if and only if $p \equiv \pm 7, \pm 11(\bmod 24)$.
$$\begin{array}{l}
\left\{\begin{array} { l }
{ p \equiv 1 ( \operatorname { m o d } 8 ) , } \\
{ p \equiv 1 ( \operatorname { m o d } 3 ) , }
\end{array} \quad \left\{\begin{array}{l}
p \equiv-1(\bmod 8), \\
p \equiv-1(\bmod 3),
\end{array}\right.\right. \\
\left\{\begin{array} { l }
{ p \equiv 3 ( \operatorname { m o d } 8 ) , } \\
{ p \equiv - 5 ( \operatorname { m o d } 3 ) , }
\end{array} \quad \left\{\begin{array}{l}
p \equiv-3(\bmod 8), \\
p \equiv 5(\bmod 3),
\end{array}\right.\right.
\end{array}$$ | p \equiv \pm 1, \pm 5(\bmod 24) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
9. Solve the congruence equation $x^{2} \equiv 59(\bmod 125)$ | 9. Solution: First, it is easy to see that $59 \equiv 9\left(\bmod 5^{2}\right)$, hence
$$x^{2} \equiv 59\left(\bmod 5^{2}\right)$$
has solutions $x \equiv \pm 3(\bmod 25)$. Next, we solve
$$x^{2} \equiv 59\left(\bmod 5^{3}\right)$$
(1) Let $x=25 t+3$, substituting we get
$$(25 t+3)^{2} \equiv 59\left(\bmod 5^{3}\right)$$
Thus,
$$(6)(25) t \equiv 50\left(\bmod 5^{3}\right)$$
Dividing both sides by 5, we get
$$6 t \equiv 2(\bmod 5)$$
This simplifies to
$$t \equiv 2(\bmod 5)$$
Substituting $x=25 t+3$ gives one solution as $x=25(5 k+2)+3=53\left(\bmod 5^{3}\right)$.
(2) Let $x=25 t-3$, substituting we get
$$(25 t-3)^{2} \equiv 59\left(\bmod 5^{3}\right)$$
Expanding, we get
$$(-6)(25) t \equiv 50\left(\bmod 5^{3}\right)$$
Dividing by 25, we get
$$-6 t \equiv 2(\bmod 5)$$
This simplifies to
$$t \equiv -2(\bmod 5)$$
Thus, the second solution is $x_{2}=25(5 k-2)-3 \equiv -53\left(\bmod 5^{3}\right)$.
In summary, the solutions are $x \equiv \pm 53\left(\bmod 5^{3}\right)$.
Note: Regarding the solution of higher-order congruence equations of the form
$$f(x) \equiv 0\left(\bmod p^{\alpha}\right)$$
where $f(x)$ is an $n$-degree polynomial with integer coefficients, $p$ is a prime, and $x \geqslant 1$, it is closely related to the corresponding congruence equation
$$f(x) \equiv 0(\bmod p)$$
Due to the depth of the involved knowledge, it will not be elaborated here. Readers interested in this topic can refer to Professor Hua Luogeng's "Introduction to Number Theory" and other specialized works. | x \equiv \pm 53\left(\bmod 5^{3}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
15. Let $p \geqslant 3$, try to calculate the value of the following expression:
$$\left(\frac{1 \cdot 2}{p}\right)+\left(\frac{2 \cdot 3}{p}\right)+\cdots+\left(\frac{(p-2)(p-1)}{p}\right)$$ | 15. Solution: To solve this problem, we need to study the properties of the Legendre symbol with the general term $\left(\frac{n(n+1)}{p}\right)$, where $(n, p)=1$.
By $(n, p)$ being coprime, we know there must exist an integer $r_{n}$ such that $p \nmid r_{n}$ and $n r_{n} \equiv 1(\bmod p)$. This $r_{n}$ is called the modular inverse of $n$ modulo $p$. From the properties of the Legendre symbol, it is easy to see that
$$\begin{aligned}
\left(\frac{n(n+1)}{p}\right) & =\left(\frac{n\left(n+n r_{n}\right)}{p}\right)=\left(\frac{n^{2}\left(1+r_{n}\right)}{p}\right) \\
& =\left(\frac{n}{p}\right)^{2}\left(\frac{1+r_{n}}{p}\right)=\left(\frac{1+r_{n}}{p}\right)
\end{aligned}$$
We will prove that for $n \neq m(\bmod p), p \nmid n m$, it must also be true that
$$r_{n} \not\equiv r_{m}(\bmod p)$$
That is, different numbers in the reduced residue system modulo $p$ must correspond to different inverses. We use proof by contradiction. If $r_{n} \equiv r_{m}(\bmod p)$, multiplying both sides by $n m$ gives
$$n\left(m r_{m}\right) \equiv m\left(n r_{n}\right)(\bmod p)$$
Then, by the definition of the inverse, we get
$$n \equiv m(\bmod p)$$
This leads to a contradiction. This proves that when $n$ runs through $1,2, \ldots, p-1$, the corresponding inverse $r_{n}$ also runs through $1,2, \ldots, p-1(\bmod p)$, just in a different order. Also, note that from $p-1 \equiv-1(\bmod p)$, we immediately get
$$(p-1)^{2} \equiv(-1)^{2} \equiv 1(\bmod p)$$
So $r_{p-1}=p-1$. Therefore, when $n$ takes $1,2, \ldots, p-2$, the inverse $r_{n}$ of $n$ also takes $1,2, \ldots, p-2(\bmod p)$, just in a different order. Thus, we get
$$\begin{array}{l}
\left(\frac{1.2}{p}\right)+\left(\frac{2.3}{p}\right)+\ldots+\left(\frac{(p-2)(p-1)}{p}\right) \\
= \sum_{n=1}^{p-2}\left(\frac{1+r_{n}}{p}\right)=\sum_{r=1}^{p-2}\left(\frac{1+r}{p}\right) \\
=\sum_{r=1}^{p-1}\left(\frac{r}{p}\right)-\left(\frac{1}{p}\right)
\end{array}$$
Since in a reduced residue system modulo $p$, there are exactly $\frac{p-1}{2}$ quadratic residues and $\frac{p-1}{2}$ quadratic non-residues, we have $\sum_{r=1}^{p-1}\left(\frac{r}{p}\right)=0$. Therefore, the required sum is $-\left(\frac{1}{p}\right)=-1$. | -1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
17. Find all values of $m$ for which the congruence
$$x^{2} \equiv 6(\bmod m)$$
may have a solution. | 17. Solution: If $m$ is odd, then the necessary condition for $x^{2} \equiv 6(\bmod m)$ to have a solution is $\left(\frac{6}{m}\right)=1$. By the properties of the Legendre symbol, we have
$$\left(\frac{6}{m}\right)=\left(\frac{2}{m}\right)\left(\frac{3}{m}\right)=(-1)^{\frac{m^{2}-1}{8}} \cdot(-1)^{\frac{m-1}{2}}\left(\frac{m}{3}\right)$$
Since
Thus, the $m$ that makes $\left(\frac{6}{m}\right)=1$ must be the solutions to the following system of congruences:
Among these, the second, third, fifth, and eighth systems of congruences have no solutions. From the first, fourth, sixth, and seventh systems, we get
$$m \equiv 1,-1,-5,5(\bmod 24)$$
- That is, when $m$ is odd, $\left(\frac{6}{m}\right)=1$ if and only if $m \equiv \pm 1, \pm 5(\bmod 24)$.
If $m$ is even, we can set $m=2^{k} n(k \geqslant 1,2 \times n)$. The congruence $x^{2} \equiv 6(\bmod m)$ can be decomposed into
$$\begin{array}{l}
x^{2} \equiv 6\left(\bmod 2^{k}\right) \\
x^{2} \equiv 6(\bmod n)
\end{array}$$
For (26), as shown above, the necessary condition for (26) to have a solution is
$$n \equiv \pm 1, \pm 5(\bmod 24)$$
Now consider (25). When $k=1$, (25) clearly has a solution $x \equiv 0 (\bmod 2)$. When $k=2$, (25) clearly has no solution, so (25) has no solution for $k \geqslant 2$. Combining these, we get that the given congruence equation may have a solution when $m \equiv \pm 1, \pm 2, \pm 5, \pm 10(\bmod 24)$.
$$\begin{array}{l}
(-1)^{\frac{m^{2}-1}{8}}=\left\{\begin{array}{c}
1, \quad m \equiv \pm 1(\bmod 8), \\
-1, m \equiv \pm 3(\bmod 8),
\end{array}\right. \\
(-1)^{\frac{m-1}{2}}=\left\{\begin{array}{ll}
1, & m \equiv 1(\bmod 4), \\
-1, & m \equiv-1(\bmod 4),
\end{array}\right. \\
\left(\frac{m}{3}\right)=\left\{\begin{array}{ll}
1, & m \equiv 1 \quad(\bmod 3), \\
-1 & m \equiv-1 \quad(\bmod 3) .
\end{array}\right.
\end{array}$$
$$\begin{array}{l}
\left\{\begin{array}{l}
m \equiv 1(\bmod 8) \\
m \equiv 1(\bmod 4) \\
m \equiv 1(\bmod 3)
\end{array},\left\{\begin{array}{l}
m \equiv-1(\bmod 8) \\
m \equiv 1(\bmod 4) \\
m \equiv 1(\bmod 3)
\end{array}, \quad\left\{\begin{array}{l}
m \equiv 1(\bmod 8) \\
m \equiv-1(\bmod 4) \\
m \equiv-1(\bmod 3)
\end{array}\right.\right.\right. \\
\left\{\begin{array}{l}
m \equiv-1(\bmod 8) \\
m \equiv-1(\bmod 4), \\
m \equiv-1(\bmod 3),
\end{array},\left\{\begin{array}{l}
m \equiv 3(\bmod 8), \\
m \equiv 1(\bmod 4) \\
m \equiv-1(\bmod 3)
\end{array},\left\{\begin{array}{l}
m \equiv 3(\bmod 8) \\
m \equiv-1(\bmod 4) \\
m \equiv 1(\bmod 3)
\end{array}\right.\right.\right. \\
\left\{\begin{array}{l}
m \equiv-3(\bmod 8), \\
m \equiv 1(\bmod 4), \\
m \equiv-1(\bmod 3), \\
m \equiv-3(\bmod 8), \\
m \equiv-1(\bmod 4), \\
m \equiv 1(\bmod 3) .
\end{array}\right.
\end{array}$$ | m \equiv \pm 1, \pm 2, \pm 5, \pm 10(\bmod 24) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Try to solve the congruence equation
$$x^{2} \equiv 73(\bmod 127)$$ | First, it is easy to see that $p=127$ is a prime number. Moreover, we have $p \equiv 3(\bmod 4)$, $73 \equiv 1(\bmod 8)$. By the properties of the Legendre symbol and the Jacobi symbol, we have
$$\begin{aligned}
\left(\frac{73}{127}\right)= & \left(\frac{127}{73}\right)=\left(\frac{54}{73}\right)=\left(\frac{2}{73}\right)\left(\frac{3}{73}\right)^{3}=\left(\frac{3}{73}\right) \\
& =\left(\frac{73}{3}\right)=\left(\frac{1}{3}\right)=1
\end{aligned}$$
Therefore, equation (3) has a solution. From the previous discussion about case $I$, we immediately obtain the solution to the original congruence equation as
$$\begin{array}{l}
x_{0} \equiv \pm 73^{\frac{127+1}{4}}= \pm 73^{32} \equiv \pm(5329)^{16} \equiv \pm(-5)^{16} \equiv \\
\equiv \pm(390625)^{2} \equiv \pm(-27)^{2} \equiv \pm 729 \equiv \mp 33(\bmod 127)
\end{array}$$ | \pm 33 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 Try to solve the congruence equation
$$x^{2} \equiv 22(\bmod 29)$$ | Solving: Since 29 is a prime number,
$$\begin{aligned}
\left(\frac{22}{29}\right) & =\left(\frac{2}{29}\right)\left(\frac{11}{29}\right)=-\left(\frac{11}{29}\right)=-\left(\frac{29}{11}\right) \\
& =-\left(\frac{7}{11}\right)=\left(\frac{11}{7}\right)=\left(\frac{4}{7}\right)=1
\end{aligned}$$
Therefore, the original congruence equation has a solution.
Since $29=4 \times 7+1$, we have $\frac{u+1}{2}=\frac{7+1}{2}=4$, and
$$22^{4} \equiv(-7)^{4}=49^{2} \equiv 20^{2} \equiv(-9)^{2}=81 \equiv 23(\bmod 29)$$
Because
$$(22)^{7} \equiv(23)(22)(-9) \equiv-1(\bmod 29)$$
and
$$29=3 \times 8+5$$
2 is a quadratic non-residue modulo 29. Since $n=7$, we get
$$b^{n}=2^{7}=2^{5} \times 2^{2}=32 \times 4 \equiv 3 \times 4=12(\bmod 29)$$
From equation (9), we have
$$x \equiv 23 \times 12 \equiv-6 \times 12=-72 \equiv-14(\bmod 29)$$
Therefore, the solution to the original congruence equation is
$$x \equiv \pm 14(\bmod 29)$$ | x \equiv \pm 14(\bmod 29) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Try to solve the congruence equation
$$x^{2} \equiv -4(\bmod 41)$$ | Since 41 is a prime number, it is easy to see that -1 and 4 are quadratic residues modulo 41, so -4 is also a quadratic residue modulo 41, which means the original congruence equation must have a solution.
Because $41=8 \times 5+1$, we have $u=5, \frac{u+1}{2}=3$, and
$$(-4)^{3}=-64 \equiv-23 \equiv 18(\bmod 41)$$
Since
$$(-4)^{5} \equiv(18)(16) \equiv 1(\bmod 41)$$
By (8), the roots we are looking for are
$$x \equiv \pm 18(\bmod 41)$$ | x \equiv \pm 18(\bmod 41) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 Try to solve the congruence equation
$$x^{2} \equiv 34(\bmod 257)$$ | Since 257 is a prime number, we have
$$\begin{aligned}
\left(\frac{34}{257}\right) & =\left(\frac{2}{257}\right)\left(\frac{17}{257}\right)=\left(\frac{17}{257}\right) \\
& =\left(\frac{257}{17}\right)=\left(\frac{2}{17}\right)=1
\end{aligned}$$
Therefore, the original congruence equation must have a solution.
Since $257=4 \times 2^{6}+1$, we have $n=2^{6}, \lambda=6, \mu=1$, $\frac{u+1}{2}=1$, so $a^{u}=34$. We also have
$$\begin{array}{c}
34^{2}=1156 \equiv 128(\bmod 257) \\
34^{2^{2}} \equiv 128^{2}=16384 \equiv 193 \equiv-64(\bmod 257) \\
34^{2^{3}} \equiv(-64)^{2}=4096 \equiv 241 \equiv-16(\bmod 257) \\
34^{2^{4}} \equiv(-16)^{2}=256 \equiv-1 \quad(\bmod 257)
\end{array}$$
Thus,
$$\mu=4, \lambda-\mu=6-4=2,$$
Hence,
$$h=2^{2} \cdot t=4 t, \quad t \text { is odd }$$
And we have $\square$
$$1 \leqslant t \leqslant 2^{4}-1=15$$
Since $257=12 \times 21+5$, 3 is a quadratic non-residue modulo 257, so we have
$$\left(b^{2 u}\right)^{h}=\left(3^{2}\right)^{4 t}=\left(3^{8}\right)^{t}$$
However, we have
$$\begin{array}{c}
3^{8}=9^{4}=81^{2}=6561 \equiv 136 \equiv-121(\bmod 257) \\
\left(3^{8}\right)^{3} \equiv(-121)^{3}=(14641)(-121) \equiv(-8)(-121) \\
\equiv 968 \equiv 197 \equiv-60(\bmod 257) \\
\left(3^{8}\right)^{5} \equiv(-8)(-60)=480 \equiv-34(\bmod 257)
\end{array}$$
Therefore, we must have $t=5$, so $h=4 t=4 \times 5=20$,
$$n-h=2^{6}-20=44,$$
Since
$$\left(3^{8}\right)^{5} \equiv-34(\bmod 257)$$
We have
$$\begin{aligned}
3^{44} & \equiv 3^{4} \times 3^{40} \equiv 81 \times(-34) \equiv-2754 \\
& \equiv-184 \equiv 73(\bmod 257)
\end{aligned}$$
And
$$34 \times 73=2482 \equiv 169 \equiv-88(\bmod 257)$$
Thus, by (10), the solution to the original equation is
$$x \equiv \pm 88(\bmod 257)$$
For the case of $p^{\alpha}$ modulo, where $p>2, \alpha>1$.
We already know how to determine whether
$$x^{2} \equiv a(\bmod p),(a, p)=1$$
has a solution, and if it does, how to find it. In this section, we will discuss, if $p$ is an odd prime, $\alpha$ is an integer greater than 1, the congruence equation
$$x^{2} \equiv a\left(\bmod p^{\alpha}\right),(\alpha, p)=1$$ | x \equiv \pm 88(\bmod 257) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Try to solve the congruence equation
$$x^{2} \equiv 33(\bmod 128)$$ | We have $1^{2}-33=-32$. So 1 is a root modulo 32, therefore, $1+16=17$ is a root modulo 64, because
$$17^{2}-33=289-33=256,$$
Hence 17 is also a root modulo 128, and the other root modulo 128 is
$$17+64=81 \equiv-47(\bmod 128)$$
Thus, the roots of the original congruence equation are
$$x \equiv \pm 17, \pm 47(\bmod 128)$$ | x \equiv \pm 17, \pm 47(\bmod 128) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 9 Try to solve the congruence equation
$$x^{2} \equiv 105(\bmod 256)$$ | From $1^{2}-105=-104=-8 \times 13$ we know that 1 is not a root modulo 16, so 5 is a root modulo 16. Then, by
$$5^{2}-105=-80=-16 \times 5$$
we know that 5 is not a root modulo 32, hence $5+8=13$ is a root modulo 32. And by
$$13^{2}-105=169-105=64$$
we know that 13 is not only a root modulo 32 but also a root modulo 64, thus $13+32=45$ is a root modulo 128, because
$$45^{2}-105=2025-105=1920=128 \times 15,$$
so 45 is not a root modulo 256, hence $45+64=109$ is a root modulo 256, and the other root modulo 256 is
$$109+128=237 \equiv-19(\bmod 256)$$
Therefore, the roots of the original congruence equation are
$$x \equiv \pm 19, \pm 109(\bmod 256)$$ | x \equiv \pm 19, \pm 109(\bmod 256) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 10 Try to find the solution of the congruence equation
$$x^{2} \equiv 19(\bmod 45)$$ | It is obvious that
$x^{2} \equiv 19 \equiv 1(\bmod 9)$ has two roots $x \equiv \pm 1(\bmod 9)$, and $x^{2} \equiv 19 \equiv 4(\bmod 5)$ has two roots $x \equiv \pm 2(\bmod 5)$. From
$$x \equiv a(\bmod 9), x \equiv b(\bmod 5)$$
and the Chinese Remainder Theorem, we get
$$x \equiv 5 \times 2 a+9 \times 4 b=10 a+36 b(\bmod 45)$$
Thus, from
$$x \equiv 1(\bmod 9), x \equiv 2(\bmod 5)$$
we get
$$x \equiv 10+72=82 \equiv-8(\bmod 45)$$
And from
$$x \equiv 1(\bmod 9), x \equiv-2(\bmod 5)$$
we get
$$x \equiv 10-72=-62 \equiv-17(\bmod 45)$$
Therefore, the solutions to the original congruence equation are
$$x \equiv \pm 8, \pm 17(\bmod 45)$$ | x \equiv \pm 8, \pm 17(\bmod 45) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. Find the smallest positive integer $a$, such that there exists a positive odd integer $n$, satisfying
$$2001 \mid\left(55^{n}+a \cdot 32^{n}\right)$$ | 2. From $2001=3 \times 23 \times 29$ and the conditions, we have
$$\left\{\begin{array}{ll}
a \equiv 1 & (\bmod 3) \\
a \equiv 1 & (\bmod 29), \\
a \equiv-1 & (\bmod 23)
\end{array}\right.$$
From the first two equations, we can set $a=3 \times 29 \times k+1$, substituting into the last equation gives
$$k \equiv 5(\bmod 23),$$
Thus, we get $a \geqslant 3 \times 29 \times 5+1=436$. When $a=436$, $2001 \mid(55+436 \times 32)$. Therefore, the smallest value is 436. | 436 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3. Find the smallest prime $p$ such that there do not exist $a, b \in \mathbf{N}$, satisfying
$$\left|3^{a}-2^{b}\right|=p$$ | 3. Notice that, $2=3^{1}-2^{0}, 3=2^{2}-3^{0}, 5=2^{3}-3^{1}, 7=2^{3}-3^{0}, 11=3^{3}-$ $2^{4}, 13=2^{4}-3^{1}, 17=3^{4}-2^{6}, 19=3^{3}-2^{3}, 23=3^{3}-2^{2}, 29=2^{5}-3^{1}, 31=$ $2^{5}-3^{0}, 37=2^{6}-3^{3}$. Therefore, the required $p \geqslant 41$.
On the other hand, if $\left|3^{a}-2^{b}\right|=41$, there are two cases.
Case one: $3^{a}-2^{b}=41$, taking modulo 3 on both sides, we know $b$ is even; taking modulo 4 on both sides, we know $a$ is even. Let $a=2 m, b=2 n$, then $\left(3^{m}-2^{n}\right)\left(3^{m}+2^{n}\right)=41$, which gives $\left(3^{m}-2^{n}, 3^{m}+\right.$ $\left.2^{n}\right)=(1,41)$, leading to
$$2 \times 3^{m}=\left(3^{m}-2^{n}\right)+\left(3^{m}+2^{n}\right)=42$$
i.e., $3^{m}=21$, a contradiction.
Case two: $2^{b}-3^{a}=41$, in this case $b \geqslant 3$, taking modulo 8 on both sides leads to a contradiction.
In summary, the smallest prime $p=41$. | 41 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8. Let $n \in \mathbf{N}^{\cdot}, n \geqslant 2$. The array of positive integers $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ satisfies:
$$a_{1}+a_{2}+\cdots+a_{n}=2 n$$
If it is not possible to divide $a_{1}, a_{2}, \cdots, a_{n}$ into two groups with equal sums, then $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ is called "good". Find all "good" arrays. | 8. Let $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ be a "good" array. For $1 \leqslant i \leqslant n$, consider the following $n+1$ numbers:
$$a_{i}, a_{i+1}, a_{i}+a_{i+1}, \cdots, a_{i}+a_{i+1}+\cdots+a_{i+n-1},$$
where $a_{n+j}=a_{j}$.
Since $a_{1}, a_{2}, \cdots, a_{n}$ cannot be divided into two groups with equal sums, i.e., there are no several numbers among them whose sum equals $n$, therefore, in (6), except for $a_{i}, a_{i+1}$, any two of the remaining terms are not congruent modulo $n$ (otherwise, the difference between the larger sum and the smaller sum would equal $n$). Since (6) contains $n+1$ numbers, there must be two numbers that are congruent modulo $n$, so, $a_{i} \equiv a_{i+1}(\bmod n)$. This holds for $i=1,2, \cdots, n$, hence
$$a_{1} \equiv a_{2} \equiv \cdots \equiv a_{n}(\bmod n)$$
Combining this with $a_{1}+a_{2}+\cdots+a_{n}=2 n$, we have
$$a_{1} \equiv a_{2} \equiv \cdots \equiv a_{n} \equiv 1 \text { or } 2(\bmod n),$$
Thus, $\left(a_{1}, a_{2}, \cdots, a_{n}\right)=(1,1, \cdots, 1, n+1)$ or $\left(a_{1}, a_{2}, \cdots, a_{n}\right)=(2,2, \cdots, 2)$ (this is only valid when $n$ is odd), and these arrays clearly meet the requirements.
In summary, when $n$ is odd, $a_{1}, a_{2}, \cdots, a_{n}$ are either one $n+1$ and the rest are 1, or $a_{1}=a_{2}=\cdots=a_{n}=2$; when $n$ is even, $a_{1}, a_{2}, \cdots, a_{n}$ have exactly one $n+1$ and the rest are 1. | (1,1, \cdots, 1, n+1) \text{ or } (2,2, \cdots, 2) \text{ when } n \text{ is odd}; (1,1, \cdots, 1, n+1) \text{ when } n \text{ is even} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
25. For $n \in \mathbf{N}^{*}$, let $f(n)$ denote the smallest positive integer such that: $n \mid \sum_{k=1}^{f(n)} k$. Find all $n \in \mathbf{N}^{*}$ such that $f(n)=2 n-1$. | 25. First, we prove: If \( n=2^{m}, m \in \mathbf{N} \), then \( f(n)=2 n-1 \).
In fact, on one hand,
\[
\sum_{k=1}^{2 n-1} k=(2 n-1) n=\left(2^{m+1}-1\right) 2^{m}
\]
is divisible by \( n \). On the other hand, if \( l \leqslant 2 n-2 \), then
\[
\sum_{k=1}^{l} k=\frac{1}{2} l(l+1)
\]
Since one of \( l \) and \( l+1 \) is odd, and
\[
l+1 \leqslant 2 n-1=2^{m+1}-1,
\]
the above sum cannot be divisible by \( 2^{m} \) (because \( 2^{m+1} \times l(l+1) \)).
Next, we prove: When \( n \) is not a power of 2, \( f(n) < 2 n-1 \). Let \( n=2^{m} p \), where \( p \) is an odd number. We show that there exists \( l < 2 n-1 \) such that \( 2^{m+1} \mid l \) and \( p \mid (l+1) \) (in this case, of course, \( 2^{m} p \left\lvert\, \frac{l(l+1)}{2}\right. \), thus \( f(n) < 2 n-1 \)).
Since \( \left(2^{m+1}, p\right)=1 \), by the Chinese Remainder Theorem,
\[
l \equiv 0\left(\bmod 2^{m+1}\right), \quad l \equiv p-1(\bmod p),
\]
has a solution \( l \equiv x_{0}\left(\bmod 2^{m+1} p\right) \). Therefore, there exists \( l_{0}, 0 < l_{0} \leqslant 2^{m+1} p \) satisfying the above system of congruences. Note that \( 2 n-1 \neq 0\left(\bmod 2^{m+1}\right) \), and \( 2 n+1 \not \equiv 0(\bmod p) \), so \( 2 n-1 \) and \( 2 n \) are not solutions to this system of congruences, hence \( 0 < l_{0} < 2 n-1 \), i.e., \( f(n) < 2 n-1 \).
In summary, \( f(n) = 2 n-1 \) if and only if \( n \) is a power of 2. | n = 2^m, m \in \mathbf{N} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 uses $P^{*}$ to denote the set of all odd prime numbers less than 10000. Let $p$ be a number in $P^{*}$, satisfying: for any subset $S=\left\{p_{1}, p_{2}, \cdots, p_{k}\right\}$ of $P^{*}$ that does not contain $p$ and where $k \geqslant 2$, there exists $q \in \mathbf{P}^{*} \backslash S$ such that
$$(q+1) \mid\left(\left(p_{1}+1\right) \cdot\left(p_{2}+1\right) \cdot \cdots \cdot\left(p_{k}+1\right)\right)$$
Find all possible values of $p$. | A basic idea is to minimize the number of distinct prime factors in the numerator of (1), the simplest case being to make each $p_{i}+1$ a power of 2. Therefore, we examine the Mersenne primes in $P^{*}$.
Let $T=\left\{M_{2}, M_{3}, M_{5}, M_{7}, M_{13}\right\}=\{3,7,31,127,8191\}$ (note that $M_{11}=$ $23 \times 89$ is not a prime), then $T$ is the set of all Mersenne primes in $P^{*}$.
Now we discuss all possible values of $p$.
On one hand, if $p \notin T$, then by taking $S=T$ in the condition, there exists $q \in P^{*} \backslash S$ such that $(q+1) \mid\left(M_{2}+1\right)\left(M_{3}+1\right)\left(M_{5}+1\right)\left(M_{7}+1\right)\left(M_{13}+1\right)$, hence $(q+1) \mid 2^{30}$, which indicates that $q+1$ is a power of 2, thus $q \in T$, a contradiction. Therefore, $p \in T$.
On the other hand, for any $p \in T$, we prove that $p$ meets the requirement.
If $p$ does not meet the requirement, there should exist a set $S=\left\{p_{1}, p_{2}, \cdots, p_{k}\right\} \subseteq P^{*}, k \geqslant 2$, not containing $p$, such that any prime $q$ satisfying (1) belongs to $S$. By this property, since $4 \mid\left(p_{1}+1\right)\left(p_{2}+1\right)$, we have $M_{2} \in S$, then by $8 \mid\left(M_{2}+1\right)\left(p_{2}+1\right)$, we have $M_{3} \in S$, by $2^{5} \mid\left(M_{2}+1\right)\left(M_{3}+1\right)$, we know $M_{5} \in S$, by $2^{7} \mid\left(M_{2}+1\right)\left(M_{3}+1\right)\left(M_{5}+1\right)$, we know $M_{7} \in S$, and by $2^{13} \mid\left(M_{2}+1\right)\left(M_{3}+1\right)\left(M_{5}+1\right)\left(M_{7}+1\right)$, we know $M_{13} \in S$. This leads to $T \subseteq S$, contradicting $p \notin S$. Therefore, when $p \in T$, $p$ meets the requirement.
In summary, the $p$ that satisfies the condition is $p \in\left\{M_{2}, M_{3}, M_{5}, M_{7}, M_{13}\right\}$. | \{3,7,31,127,8191\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
28. Let $a$ be an integer, and $n, r$ be integers greater than 1. $p$ is an odd prime, and $(n, p-1)=1$. Find the number of solutions to the following congruence equation:
$$x_{1}^{n}+x_{2}^{n}+\cdots+x_{r}^{n} \equiv a(\bmod p)$$
Here, the solutions $\left(x_{1}, x_{2}, \cdots, x_{r}\right)$ and $\left(x_{1}^{\prime}, x_{2}^{\prime}, \cdots, x_{r}^{\prime}\right)$ are considered the same if and only if: for $1 \leqslant j \leqslant r$, we have
$$x_{j} \equiv x_{j}^{\prime}(\bmod p)$$ | 28. First, prove: For any $b \in \mathbf{Z}$, the congruence equation $x^{n} \equiv b(\bmod p)$ has a unique solution.
In fact, by $(n, p-1)=1$ and Bézout's theorem, there exist $u, v \in \mathbf{N}^{*}$ such that $n u-(p-1) v=1$. If $b \equiv 0(\bmod p)$, then the equation $x^{n} \equiv b(\bmod p)$ has only the unique solution $x \equiv 0(\bmod p)$; if $b \not\equiv 0(\bmod p)$, then the solution to the equation $x^{n} \equiv b(\bmod p)$ satisfies $(x, p)=1$, and $x^{m m} \equiv b^{u}(\bmod p)$. By Fermat's little theorem, we know that $x^{p-1} \equiv 1(\bmod p)$, thus
$$x^{m}=x^{(p-1) v+1} \equiv x(\bmod p),$$
which means $x \equiv b^{u}(\bmod p)$. This shows that the solution to $x^{n} \equiv b(\bmod p)$ is unique.
Using the above conclusion and the structure of the congruence equation, we can get: For any $a \in \mathbf{Z}$, the original congruence equation has exactly $p^{r-1}$ distinct solutions (arbitrarily choose the remainders of $x_{1}^{n}, x_{2}^{n}, \cdots, x_{-1}^{n}$ modulo $p$, and adjust $x_{r}^{n}$ to make $x_{1}^{n}+x_{2}^{n}+\cdots+x_{r}^{n} \equiv a(\bmod p)$, thus obtaining all $p^{r-1}$ solutions). | p^{r-1} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
29. Let $p$ be a prime, $a, b \in \mathbf{N}^{*}$, satisfying: $p>a>b>1$. Find the largest integer $c$, such that for all $(p, a, b)$ satisfying the conditions, we have
$$p^{c} \mid\left(\mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}\right)$$ | 29. When taking $p=5, a=3, b=2$, we should have $5^{c} \mid 3000$, so $c \leqslant 3$. Below, we prove that for any $p, a, b$ satisfying the conditions, we have $p^{3} \mid \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}$.
In fact, notice that
$$\begin{aligned}
& \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}=\frac{(a p)(a p-1) \cdot \cdots \cdot((a-b) p+1)}{(b p)!}-\frac{a!}{b!(a-b)!} \\
= & \frac{a(a-1) \cdot \cdots \cdot(a-b+1) \prod_{k=a-b}^{a-1}(k p+1)(k p+2) \cdots \cdots \cdot(k p+(p-1))}{b!\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdots \cdots(k p+(p-1))} \\
& -\frac{a!}{b!(a-b)!} \\
= & \frac{a!}{b!(a-b)!} \cdot \frac{1}{\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdot \cdots \cdot(k p+(p-1))}\left\{\prod_{k=a-b}^{a-1}(k p+1)\right. \\
& \left.\cdot(k p+2) \cdot \cdots \cdot(k p+(p-1))-\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdot \cdots \cdot(k p+(p-1))\right\},
\end{aligned}$$
Thus, we only need to prove that $p^{3} \mid A$, where
$$\begin{aligned}
A= & \prod_{k=a-b}^{a-1}(k p+1) \cdot(k p+2) \cdot \cdots \cdot(k p+(p-1)) \\
& -\prod_{k=0}^{b-1}(k p+1) \cdot(k p+2) \cdot \cdots \cdot(k p+(p-1))
\end{aligned}$$
For this, we set
$$\begin{aligned}
f(x) & =(x+1)(x+2) \cdot \cdots \cdot(x+(p-1)) \\
& =x^{p-1}+\alpha_{p-2} x^{p-2}+\cdots+\alpha_{1} x+(p-1)!
\end{aligned}$$
Then, by the conclusion of Example 1 in Section 2.4, we know that $p^{2} \mid \alpha_{1}$. Therefore,
$$\begin{aligned}
A= & \prod_{k=a-b}^{a-1} f(k p)-\prod_{k=0}^{b-1} f(k p) \\
\equiv & ((p-1)!)^{b-1} \sum_{k=a-b}^{a-1} \alpha_{1} k p+((p-1)!)^{b} \\
& -((p-1)!)^{b-1} \sum_{k=0}^{b-1} \alpha_{1} k p-((p-1)!)^{b} \\
\equiv & 0\left(\bmod p^{3}\right) .
\end{aligned}$$
Therefore, the maximum integer $c=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
31. Find all positive integers $n$, such that
$$\frac{2^{n}-1}{3} \in \mathbf{N}^{*}$$
and there exists $m \in \mathbf{N}^{*}$, satisfying ${ }^{*}$ :
$$\left.\frac{2^{n}-1}{3} \right\rvert\,\left(4 m^{2}+1\right)$$ | 31. From $\frac{2^{n}-1}{3} \in \mathbf{N}^{\cdot}$, we know that $n$ is even. If there exists an odd number $q \geqslant 3$ such that $q \mid n$, then using factorization we know that $\left(2^{q}-1\right) \mid\left(2^{n}-1\right)$. Combining $2^{q}-1 \equiv-2(\bmod 3)$, we know that $\left(2^{q}-1\right) \left\lvert\, \frac{2^{n}-1}{3}\right.$, which indicates $\left(2^{q}-1\right) \mid\left(4 m^{2}+1\right)$. However, when $q \geqslant 3$, $2^{q}-1 \equiv-1$ $(\bmod 4)$, so $2^{q}-1$ has a prime factor $p$ satisfying $p \equiv 3(\bmod 4)$. For this $p$, we have $p \mid\left(4 m^{2}+1\right)$, i.e.,
$$(2 m)^{2} \equiv-1(\bmod p)$$
By Euler's criterion, we know that -1 is not a quadratic residue modulo $p$, leading to a contradiction. Therefore, $n$ can only be a power of 2, i.e., there exists $k \in \mathbf{N}^{*}$ such that $n=2^{k}$.
On the other hand, if $n=2^{k}, k \in \mathbf{N}^{*}$, then
$$\frac{2^{n}-1}{3}=\left(2^{2}+1\right)\left(2^{2^{2}}+1\right) \cdot \cdots \cdot\left(2^{2^{k-1}}+1\right)$$
Using the fact that when $r \in \mathbf{N}^{*}$, $\left(2^{2^{r}}-1,2^{2^{r}}+1\right)=1$, and
$$2^{2^{r}}-1=(2+1)\left(2^{2}+1\right) \cdot \cdots \cdot\left(2^{2^{r-1}}+1\right)$$
we know that $2^{2}+1,2^{2^{2}}+1, \cdots, 2^{2^{k-1}}+1$ are pairwise coprime. Therefore, by the Chinese Remainder Theorem, there exists $m \in \mathbf{N}^{*}$ such that for $1 \leqslant i \leqslant k-1$, we have
$$m \equiv 2^{2^{i-1}}-1\left(\bmod 2^{2^{i}}+1\right)$$
Thus, $4 m^{2} \equiv\left(2^{2^{i-1}}\right)^{2}=2^{2^{i}} \equiv-1\left(\bmod 2^{2^{i}}+1\right)$, which implies $\left.\frac{2^{n}-1}{3} \right\rvert\,$ $\left(4 m^{2}+1\right)$.
In summary, $n$ is all numbers of the form a power of 2. | n = 2^k, k \in \mathbf{N}^{*} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find all positive integers $n$, such that $2^{n} \equiv 1(\bmod n)$.
untranslated text remains unchanged. | When $n=1$, it is obviously true.
Below is the proof: When $n>1$, we always have
$$2^{n} \not \equiv 1(\bmod n)$$
In fact, if there exists $n>1$, such that
$$2^{n} \equiv 1(\bmod n),$$
take the smallest prime factor $p$ of $n$, then $2^{n} \equiv 1(\bmod p)$. By Fermat's Little Theorem, we know $2^{p-1} \equiv$ $1(\bmod p)$, so,
$$2^{(n, p-1)} \equiv 1(\bmod p)$$
Using the fact that $p$ is the smallest prime factor of $n$, we know $(n, p-1)=1$, leading to $2 \equiv 1(\bmod$ $p)$, i.e., $p \mid 1$. This is a contradiction. $\square$
In summary, only $n=1$ meets the requirement. | n=1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 If $p$ and $p+2$ are both prime numbers, then these two prime numbers are called "twin primes". Consider the following two sequences.
Fibonacci sequence: $1,1,2,3,5,8, \cdots$ (the sequence satisfying $F_{1}=1, F_{2}=1$, $F_{n+2}=F_{n+1}+F_{n}, n=1,2, \cdots$).
Twin prime sequence: $3,5,7,11,13,17,19, \cdots$ (the sequence formed by writing all twin prime pairs in ascending order).
Question: Which positive integers appear in both of the sequences above? | This is a problem related to the properties of the Fibonacci sequence. A certain term $F_{n}$ appears in the twin prime sequence if and only if $F_{n}-2$ and $F_{n}$ are both prime, or $F_{n}$ and $F_{n}+2$ are both prime. Therefore, to negate that $F_{n}$ appears in the twin prime sequence, we need to prove that $F_{n}$ is composite, or that $F_{n}-2$ and $F_{n}+2$ are both composite.
Notice that, $F_{4}=3, F_{5}=5, F_{7}=13$ all appear in the twin prime sequence. Below, we prove that when $n \geqslant 8$, $F_{n}$ does not appear in the twin prime sequence.
Using the recursive formula of the Fibonacci sequence, we have:
$$\begin{aligned}
F_{n+m+1} & =F_{1} F_{n+m-1}+F_{2} F_{n+m} \\
& =F_{1} F_{n+m-1}+F_{2}\left(F_{n-m-1}+F_{n+m-2}\right) \\
& =F_{2} F_{n+m-2}+F_{3} F_{n+m-1} \\
& =\cdots \\
& =F_{n-1} F_{m+1}+F_{n} F_{m+2}
\end{aligned}$$
In (2), let $m=n-1$, then we have $F_{2 n}=F_{n}\left(F_{n+1}+F_{n-1}\right)$. Since $F_{2}=1$, when $n \geqslant 3$, $F_{2 n}$ is composite.
On the other hand, it is well-known that: $F_{n}^{2}=F_{n-1} F_{n+1}+(-1)^{n}, n=1,2, \cdots$ (this conclusion is easily proven by mathematical induction). Combining with (2), we have
$$\begin{aligned}
F_{4 n+1} & =F_{2 n}^{2}+F_{2 n+1}^{2} \\
& =F_{2 n}^{2}+\left(F_{2 n-1}+F_{2 n}\right)^{2} \\
& =2 F_{2 n}^{2}+\left(2 F_{2 n}+F_{2 n-1}\right) F_{2 n-1} \\
& =2 F_{2 n+1} F_{2 n-1}+\left(2 F_{2 n}+F_{2 n-1}\right) F_{2 n-1}+2
\end{aligned}$$
Therefore,
$$\begin{aligned}
F_{4 n+1}-2 & =F_{2 n-1}\left(2 F_{2 n+1}+2 F_{2 n}+F_{2 n-1}\right) \\
& =F_{2 n-1}\left(F_{2 n+1}+F_{2 n+2}+F_{2 n+1}\right) \\
& =F_{2 n-1}\left(F_{2 n+3}+F_{2 n+1}\right)
\end{aligned}$$
Thus, when $n \geqslant 2$, $F_{4 n+1}-2$ is composite.
Furthermore, we also have
$$\begin{aligned}
F_{4 n+1} & =F_{2 n}^{2}+F_{2 n+1}^{2}=\left(F_{2 n+2}-F_{2 n+1}\right)^{2}+F_{2 n+1}^{2} \\
& =2 F_{2 n+1}^{2}+F_{2 n+2}\left(F_{2 n+2}-2 F_{2 n+1}\right) \\
& =2 F_{2 n+1}^{2}-F_{2 n+2} F_{2 n-1} \\
& =2 F_{2 n} F_{2 n+2}-F_{2 n+2} F_{2 n-1}-2 \\
& =F_{2 n+2}\left(2 F_{2 n}-F_{2 n-1}\right)-2 \\
& =F_{2 n+2}\left(F_{2 n}+F_{2 n-2}\right)-2
\end{aligned}$$
Therefore, when $n \geqslant 2$, $F_{4 n+1}+2$ is also composite.
Similarly, it can be proven that when $n \geqslant 2$, the numbers $F_{4 n+3}-2$ and $F_{4 n+3}+2$ are also composite. In summary, only the numbers $3,5,13$ appear in both of the above sequences. | 3,5,13 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 Let $p$ be a given odd prime, and call a positive integer $m$ a "good number" if it satisfies the following conditions:
(1) $m \in\{1,2, \cdots, p-1\}$;
(2) there exists $n \in \mathbf{N}^{*}$, such that $m^{n} \equiv-1(\bmod p)$.
Find the number of "good numbers". | Let $g$ be a primitive root modulo $p$, then $g^{\frac{p-1}{2}} \equiv -1 \pmod{p}$. (Because $\delta_{p}(g)=p-1$, and $g^{p-1} \equiv 1 \pmod{p}$, and $g, g^{2}, \cdots, g^{p-1}$ form a reduced residue system modulo $p$.)
Now for $m \in \{g, g^{2}, \cdots, g^{p-1}\}$, let $m=g^{k}$. According to the definition of "good numbers", $m$ is a "good number" if and only if there exists $n \in \mathbf{N}^{*}$ such that
$$k n = \frac{p-1}{2} \cdot q$$
where $q$ is a positive odd number. This is equivalent to the power of 2 in the prime factorization of $k$ being no greater than the power of 2 in $\frac{p-1}{2}$.
Let $\frac{p-1}{2} = 2^{a} \cdot u$, where $a \in \mathbf{N}^{*}$ and $u$ is an odd number. Then $m = g^{k}$ meets the requirement if and only if $k \notin \{2^{a+1}, 2^{a+1} \cdot 2, \cdots, 2^{a+1} \cdot u\}$.
In summary, there are $p-1-u$ "good numbers", where $u$ is the largest odd divisor of $p-1$. | p-1-u | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. Let $n$ be a given positive integer, find the smallest positive integer $m$, such that
$$2^{m} \equiv 1\left(\bmod 5^{n}\right)$$ | 2. This is equivalent to finding $\delta_{5^{n}}(2)$, the answer is: the smallest positive integer $m=4 \times 5^{n-1}$.
We prove the above conclusion by induction on $n$. Direct calculation shows that $\delta_{5}(2)=4$. Now assume $\delta_{5^{n}}(2)=4 \times 5^{n-1}$, then we can set $2^{4 \times 5^{n-1}}=t \cdot 5^{n}+1, t \in \mathbf{N}^{*}$, thus
$$2^{4 \times 5^{n}}=\left(t \cdot 5^{n}+1\right)^{5} \equiv \mathrm{C}_{5}^{1} \cdot t \cdot 5^{n}+1 \equiv 1\left(\bmod 5^{n+1}\right),$$
Therefore, $\delta_{5^{n+1}}(2) \leqslant 4 \times 5^{n}$. Any $m$ that satisfies $2^{m} \equiv 1\left(\bmod 5^{n+1}\right)$ also satisfies $2^{m} \equiv 1(\bmod 5^{n})$, hence $\delta_{5^{n}}(2) \mid \delta_{5^{n+1}}(2)$. Thus, $\delta_{5^{n+1}}(2)=4 \times 5^{n-1}$ or $4 \times 5^{n}$. Since $2^{4 \times 5^{n-1}}=16^{5^{n-1}}=(15+1)^{5^{n-1}}$, using the binomial theorem expansion, we get $5^{n} \|\left(2^{4 \times 5^{n-1}}-1\right)$, so it must be that $\delta_{5^{n+1}}(2)=4 \times 5^{n}$. | 4 \times 5^{n-1} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
4. Let $n, b_{0} \in \mathbf{N}^{*}, n \geqslant 2, 2 \leqslant b_{0} \leqslant 2 n-1$. The sequence $\left\{b_{i}\right\}$ is defined as follows:
$$b_{i+1}=\left\{\begin{array}{ll}
2 b_{i}-1, & b_{i} \leqslant n, \\
2 b_{i}-2 n, & b_{i}>n,
\end{array} \quad i=0,1,2, \cdots\right.$$
Let $p\left(b_{0}, n\right)$ denote the smallest index $p$ such that $b_{p}=b_{0}$.
(1) For $k \in \mathbf{N}^{*}$, find the values of $p\left(2,2^{k}\right)$ and $p\left(2,2^{k}+1\right)$;
(2) Prove that for any $n$ and $b_{0}$, $p\left(b_{0}, n\right) \mid p(2, n)$. | 4. Let $m=n-1, a_{i}=b_{i}-1$, then $1 \leqslant a_{0} \leqslant 2 m$, and
$$a_{i+1}=\left\{\begin{array}{ll}
2 a_{i}, & a_{i} \leqslant m, \\
2 a_{i}-(2 m+1), & a_{i}>m .
\end{array}\right.$$
This indicates: $a_{i+1} \equiv 2 a_{i}(\bmod 2 m+1)$, and for $i \in \mathbf{N}$, we have $1 \leqslant a_{i} \leqslant 2 m$.
(1) The required values are equivalent to finding $p\left(1,2^{k}-1\right)$ and $p\left(1,2^{k}\right)$ for $\left\{a_{i}\right\}$. The former is equivalent to finding the smallest $l \in \mathbf{N}^{*}$ such that
$$2^{l} \equiv 1\left(\bmod 2\left(2^{k}-1\right)+1\right) ;$$
The latter is equivalent to finding the smallest $t \in \mathbf{N}^{*}$ such that
$$2^{t} \equiv 1\left(\bmod 2^{k+1}+1\right)$$
Since $2\left(2^{k}-1\right)+1=2^{k+1}-1$, and for $1 \leqslant l \leqslant k$, it is clear that $2^{l} \neq 1\left(\bmod 2^{k+1}-1\right)$, hence $p\left(1,2^{k}-1\right)=k+1$. Also, $2^{2(k+1)} \equiv 1\left(\bmod 2^{k+1}+1\right)$, so $\delta_{2^{k+1}+1}(2) \mid 2(k+1)$. But for $1 \leqslant t \leqslant k+1$, we have $2^{t} \neq 1\left(\bmod 2^{k+1}+1\right)$, thus $p\left(1,2^{k}\right)=\delta_{2^{k+1}+1}(2)=2(k+1)$.
Therefore, for $\left\{b_{i}\right\}$, we have $p\left(2,2^{k}\right)=k+1, p\left(2,2^{k}+1\right)=2(k+1)$.
(2) We still discuss in terms of $\left\{a_{i}\right\}$, and need to prove: $p\left(a_{0}, m\right) \mid p(1, m)$.
First, let $p(1, m)=t$, then $2^{t} \equiv 1(\bmod 2 m+1)$, and thus $2^{t} a_{0} \equiv a_{0}(\bmod 2 m+1)$, so $p\left(a_{0}, m\right) \leqslant p(1, m)$.
Second, if $p\left(a_{0}, m\right) \times p(1, m)$, then we can set
$$p(1, m)=p\left(a_{0}, m\right) q+r, 0<r<p\left(a_{0}, m\right)$$
Let $s=p\left(a_{0}, m\right), t=p(1, m)$, then combining $2^{s} a_{0} \equiv a_{0}(\bmod 2 m+1)$, we know
$$a_{0} \equiv 2^{t} a_{0}=2^{n+r} a_{0} \equiv 2^{s(q-1)+r} a_{0} \equiv \cdots \equiv 2^{r} a_{0}(\bmod 2 m+1),$$
which contradicts the minimality of $s$.
Therefore, $p\left(a_{0}, m\right) \mid p(1, m)$, and the proposition is proved. | p\left(2,2^{k}\right)=k+1, p\left(2,2^{k}+1\right)=2(k+1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5. Find all two-digit numbers $n=\overline{a b}$ (where $a \geqslant 1, a, b \in\{0,1,2, \cdots, 9\}$), such that for any $x \in \mathbf{Z}$, we have $n \mid\left(x^{a}-x^{b}\right)$. | 5. First, $n=11,22, \cdots, 99$ satisfy the condition. When $a \neq b$, let $p$ be a prime factor of $n$. By the condition, for any $x \in \mathbf{Z}$, we have $p \mid (x^a - x^b)$. Therefore, for any $x \in \{1, 2, \cdots, p-1\}$, we have $x^{|a-b|} \equiv 1 \pmod{p}$. Let $g$ be a primitive root modulo $p$, then we also have $g^{|a-b|} \equiv 1 \pmod{p}$, thus $(p-1) \mid |a-b|$. Noting that $|a-b| \leq 9$, we have $p \leq 10$, hence $p \in \{2,3,5,7\}$.
We discuss the cases separately.
(1) If $7 \mid n$, then similarly, we should have $6 \mid |a-b|$, in this case, $n=28$. Using Fermat's Little Theorem and the property of squares modulo 4, we can see that $n=28$ satisfies the condition.
(2) If $7 \nmid n$, but $5 \mid n$. Similarly, we have $4 \mid |a-b|$, so $n \in \{15,40\}$. Direct verification shows that $n=15$ satisfies the condition.
(3) If $7 \nmid n$, $5 \nmid n$, but $3 \mid n$, we have $2 \mid |a-b|$, so $n \in \{24,48\}$. Direct verification shows that $n=48$ satisfies the condition.
(4) If $7 \nmid n$, $5 \nmid n$, $3 \nmid n$, but $2 \mid n$, then $n \in \{16,32,64\}$, none of which meet the requirements.
In summary, the $n$ that satisfy the condition are $n=11,22, \cdots, 99,15,28,48$. | n=11,22, \cdots, 99,15,28,48 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
7. Find all prime triples $(p, q, r)$ such that
$$p \mid \left(q^{r}+1\right), q \mid \left(r^{p}+1\right), r \mid \left(p^{q}+1\right) .$$ | 7. Let $(p, q, r)$ be an array of prime numbers, and by cyclic symmetry, assume $p = \max \{p, q, r\}$. From $p \mid (q^r + 1)$, we know $p \neq q$, similarly $q \neq r$, $r \neq p$, i.e., $p, q, r$ are pairwise distinct. We discuss the following two cases:
(1) $p > q > r$. From $q \mid (r^p + 1)$, i.e., $r^p \equiv -1 \pmod{q}$, we know $r^{2p} \equiv 1 \pmod{q}$, hence $\delta_q(r) \times p$ but $\delta_q(r) \mid 2p$. By Fermat's Little Theorem, we know $r^{q-1} \equiv 1 \pmod{q}$, hence $\delta_q(r) \mid (q-1)$. Since $q-1 < r$, we have $\delta_q(r) \mid 2$.
By $p \mid (q^r + 1)$, similarly, we get $\delta_p(q) \mid 2r$. Since $p \geq r + 2 > q + 2$, we have $\delta_p(q) > 2$, thus $\delta_p(q) = 2r$. Combining Fermat's Little Theorem, we get $2r \mid (p-1)$, hence $p \equiv 1 \pmod{r}$, leading to
$$0 \equiv p^q + 1 \equiv 2 \pmod{r},$$
i.e., $r \mid 2$, a contradiction. $\square$
In summary, $(p, q, r) = (5, 3, 2), (3, 2, 5)$, or $(2, 5, 3)$. | (5, 3, 2), (3, 2, 5), (2, 5, 3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
11. Find all prime pairs $(p, q)$ such that $p q \mid\left(5^{p}-2^{p}\right)\left(5^{q}-2^{q}\right)$. | 11. If $p \mid\left(5^{p}-2^{p}\right)$, then $p$ is coprime with 2 and 5. Using Fermat's Little Theorem, we know $5^{p}-2^{p} \equiv 5-2(\bmod p)$, hence $p \mid 3$. Therefore, $p=3$. In this case, $5^{p}-2^{p}=3^{2} \times 13$, so $q \mid\left(5^{q}-2^{q}\right)$ or $q \mid\left(3^{2} \times 13\right)$, which gives $q=3$ or 13. Similarly, discussing the case $q \mid\left(5^{q}-2^{q}\right)$, we get the solutions $(p, q)=(3,3),(3,13)$ or $(13,3)$.
Finally, suppose $p\left|\left(5^{q}-2^{q}\right), q\right|\left(5^{p}-2^{q}\right)$, and $p \neq 3, q \neq 3$. In this case, without loss of generality, assume $q < p$, then $(p, q-1)=1$. By Bézout's Theorem, there exist $a, b \in \mathbf{N}^{*}$ such that $a p-b(q-1)=1$. Combining $(5, q)=1,(2, q)=1$ and Fermat's Little Theorem, we know $5^{q^{-1}} \equiv 2^{q-1} \equiv 1(\bmod q)$. From $5^{p} \equiv 2^{p}(\bmod q)$, we also have $5^{a p} \equiv 2^{a p}(\bmod q)$, i.e.,
$$5^{1+b q-1)} \equiv 2^{1+b q-1)}(\bmod q)$$
This leads to $5 \equiv 2(\bmod q)$, i.e., $q \mid 3$, which is a contradiction.
Therefore, the prime pairs $(p, q)=(3,3),(3,13)$ or $(13,3)$ satisfy the conditions. | (3,3),(3,13),(13,3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
12. Find all prime pairs $(p, q)$ such that
$$p q \mid\left(2^{p}+2^{q}\right) \text {. }$$ | 12. Let $p \leqslant q$.
If $p=2$, then $q=2$ satisfies the condition. When $q>2$, $q$ is odd. From $2^{p}+2^{q}=2^{2}(1+2^{q-2})$, we know that $q \mid (2^{q-2}+1)$. Therefore,
$$2^{q-2} \equiv -1 \pmod{q}$$
By Fermat's Little Theorem, $2^{q-1} \equiv 1 \pmod{q}$, so
$$1 \equiv 2^{q-1} = 2^{q-2} \cdot 2 \equiv -2 \pmod{q}$$
Thus, $q \mid 3$. Therefore, in this case, there are two solutions: $(p, q) = (2, 2)$ and $(2, 3)$.
If $p>2$, by Fermat's Little Theorem, $2^{q-1} \equiv 1 \pmod{q}$. Combining this with the condition $q \mid (2^{p} + 2^{q})$, we have
$$2^{p} + 2^{q} \equiv 2^{p} + 2 \pmod{q}$$
Thus, $q \mid (2^{p-1} + 1)$, and hence $2^{2(p-1)} \equiv 1 \pmod{q}$, so
$$\delta_{q}(2) \mid 2(p-1)$$
But $\delta_{q}(2) \times (p-1)$. Therefore, let $p-1 = 2^{s} \cdot t$, where $t$ is odd, $s, t \in \mathbf{N}^{*}$, then $\delta_{q}(2) = 2^{s+1} \cdot k$, where $k$ is a factor of $t$. By $2^{q-1} \equiv 1 \pmod{q}$, we know
$$\delta_{q}(2)^{4} \mid (q-1)$$
This indicates that in the prime factorization of $q-1$ and $p-1$, the power of 2 in $q-1$ is higher than that in $p-1$. However, if we replace $q$ with $p$ and repeat the above discussion, we should have the power of 2 in $p-1$ higher than that in $q-1$, which is a contradiction. Therefore, when $q \geqslant p > 2$, there is no solution.
In summary, the only solutions for $(p, q)$ are $(2, 2)$, $(2, 3)$, or $(3, 2)$. | (2, 2), (2, 3), (3, 2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
18. Let $p$ be a prime, and $p \equiv 1(\bmod 12)$. Find the number of tuples $(a, b, c, d)$ satisfying:
(1) $a, b, c, d \in\{0,1,2, \cdots, p-1\}$;
(2) $a^{2}+b^{2} \equiv c^{3}+d^{3}(\bmod p)$. | 18. Under the condition $p \equiv 1(\bmod 12)$, first prove the following lemma.
Lemma 1: The number of integer pairs $(x, y)$ that satisfy $x^{2}+y^{2} \equiv 0(\bmod p), x, y \in\{0,1,2, \cdots, p-1\}$ is $2 p-1$.
Since $p \equiv 1(\bmod 4)$, we know that -1 is a quadratic residue modulo $p$, so there exists $u$ such that $u^{2} \equiv -1(\bmod p)$. Therefore, if $x^{2} \equiv -y^{2}(\bmod p)$, then $x^{2} \equiv (u y)^{2}(\bmod p)$, so $p \mid (x - u y)$ or $p \mid (x + u y)$. This indicates that for each $y \in \{1,2, \cdots, p-1\}$, there are exactly two different $x (x \equiv u y$ or $-u y(\bmod p)) \in \{1,2, \cdots, p-1\}$ such that $x^{2} + y^{2} \equiv 0(\bmod p)$, and when $y \equiv 0(\bmod p)$, only $x \equiv 0(\bmod p)$ satisfies $x^{2} + y^{2} \equiv 0(\bmod p)$. Lemma 1 is proved.
Lemma 2: For a given $k \in \{1,2, \cdots, p-1\}$, the number of integer pairs $(x, y)$ that satisfy $x^{2} + y^{2} \equiv k(\bmod p), x, y \in \{0,1,2, \cdots, p-1\}$ is $p-1$.
Notice that the number of integer pairs $(r, t)$ that satisfy $r t \equiv k(\bmod p)$ is $p-1$, where $r, t \in \{0,1,2, \cdots, p-1\}$ (since $k \neq 0(\bmod p)$, when $t$ takes $1,2, \cdots, p-1$, $r \equiv t^{-1} k(\bmod p)$ is determined). Therefore, for $x^{2} + y^{2} \equiv k(\bmod p)$, using $u^{2} \equiv -1(\bmod p)$ from Lemma 1, we have
$$(x - u y)(x + u y) \equiv k(\bmod p)$$
Thus, $(x - u y, x + u y) = (r, t)$ has $p-1$ pairs. When $r, t$ are determined,
$$x \equiv 2^{-1}(r + t)(\bmod p), y \equiv (2 u)^{-1}(t - r)(\bmod p)$$
are also determined, so Lemma 2 is proved.
Lemma 3: The number of integer pairs $(x, y)$ that satisfy $x^{3} + y^{3} \equiv 0(\bmod p), x, y \in \{0,1,2, \cdots, p-1\}$ is $3 p - 2$.
When $y = 0$, we must have $x = 0$; when $y \neq 0$, for a given $y$, by Lagrange's theorem, $x^{3} \equiv -y^{3}(\bmod p)$ has at most three solutions. On the other hand, taking a primitive root $g$ modulo $p$, since $p \equiv 1(\bmod 3)$, $1, g^{\frac{p-1}{3}}, g^{\frac{2(p-1)}{3}}$ are the three distinct solutions of the equation $t^{3} \equiv 1(\bmod p)$, so
$$x \equiv -y, -g^{\frac{p-1}{3}} y, -g^{\frac{2(p-1)}{3}} y(\bmod p)$$
are the three distinct solutions of $x^{3} \equiv -y^{3}(\bmod p)$. Therefore, Lemma 3 is proved.
Returning to the original problem. By Lemma 3, there are $3 p - 2$ pairs $(c, d)$ that satisfy $c^{3} + d^{3} \equiv 0(\bmod p)$, and for each such pair $(a, b)$, there are $2 p - 1$ pairs (Lemma 1), i.e., the number of $(a, b, c, d)$ that satisfy $a^{2} + b^{2} \equiv c^{3} + d^{3} \equiv 0(\bmod p)$ is $(3 p - 2)(2 p - 1)$.
By Lemma 3, there are $p^{2} - (3 p - 2)$ pairs $(c, d)$ such that $c^{3} + d^{3} \not\equiv 0(\bmod p)$, and for each such pair $(c, d)$, by Lemma 2, the corresponding $(a, b)$ has $p - 1$ pairs.
In summary, the number of tuples $(a, b, c, d)$ that satisfy the conditions is $(3 p - 2)(2 p - 1) + (p^{2} - 3 p + 2)(p - 1) = p^{3} + 2 p^{2} - 2 p$. | p^3 + 2p^2 - 2p | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Arrange the simplest fractions with denominators not exceeding 99 between $[0,1]$ in ascending order, find the two numbers adjacent to $\frac{17}{76}$. | Let $x, y \in \mathbf{N}^{*}, (x, y)=1$, and $\frac{x}{y}$ is the number to the left of $\frac{17}{76}$ in the above sequence, then
$$\frac{17}{76}-\frac{x}{y}=\frac{17 y-76 x}{76 y}>0$$
Notice that $17 y-16 x$ is an integer, so $17 y-76 x \geqslant 1$.
We first solve the indeterminate equation
$$17 y-76 x=1$$
for positive integer solutions $(x, y)$ satisfying $1 \leqslant y \leqslant 99$.
We can use the following method to find a particular solution to (3):
$$y=4 x+\frac{8 x+1}{17} \in \mathbf{Z}$$
By trial, we find that $(x, y)=(2,9)$ is a particular solution. Therefore, all integer solutions to (3) are
$$\left\{\begin{array}{l}
x=2+17 t \\
y=9+76 t
\end{array} t \in \mathbf{Z}\right.$$
Among the positive integer solutions to (3), $(x, y)=(19,85)$ is the solution that satisfies $1 \leqslant y \leqslant 99$ and has the largest $y$, and at this point $y=85>\frac{99}{2}$, so the number to the left of $\frac{17}{76}$ is $\frac{19}{85}$.
Similarly, the number to the right of $\frac{17}{76}$ is $\frac{15}{67}$. | \frac{19}{85}, \frac{15}{67} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Find all integer solutions to the equation $3 x+7 y+16 z=40$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Since $(3,7)=1$, the equation has a solution.
Let $3 x+7 y=t$, this can be regarded as a linear Diophantine equation in two variables, its general solution is
$$x=-2 t+7 u, y=t-3 u$$
On the other hand, the general solution of $t+16 z=40$ is
$$t=40-16 v, z=v$$
Therefore, all solutions of the original equation are
$$x=-80+32 v+7 u, y=40-16 v-3 u, z=v$$
where $u, v$ are arbitrary integers. | x=-80+32 v+7 u, y=40-16 v-3 u, z=v | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 For which $n \in \mathbf{N}^{\cdot}, n \geqslant 5$, can the vertices of a regular $n$-gon be colored using no more than 6 colors, such that any 5 consecutive vertices have distinct colors? | Let the colors be $a, b, c, d, e, f$. Define the sequence $A: a, b, c, d, e$ and the sequence $B: a, b, c, d, e, f$.
If there exist non-negative integers $x, y$, such that $n=5 x+6 y$, then for a regular $n$-gon, the vertices can be colored by first coloring $y$ sequences of $B$, followed by $x$ sequences of $A$, ensuring that any 5 consecutive vertices are of different colors.
Using the conclusion from Example 3, we know that when $n \geqslant 5 \times 6-(5+6)+1=20$, the equation $n=5 x+6 y$ always has non-negative integer solutions. For $5 \leqslant n \leqslant 19$, direct calculation shows that the equation $n=5 x+6 y$ has no non-negative integer solutions only when $n \in \{7,8,9,13,14,19\}$.
On the other hand, for $n \in \{7,8,9,13,14,19\}$, there exists $k \in \mathbf{N}^{*}$ such that $6 k<n<6(k+1)$. Therefore, there must be a color that appears $k+1$ times. Since these $k+1$ points of the same color are at least 4 points apart, we have
$$n \geqslant 5(k+1)$$
Now,
when $n \in \{7,8,9\}$, $k=1$, requiring $n \geqslant 10$, which is a contradiction.
when $n \in \{13,14\}$, $k=2$, requiring $n \geqslant 15$, which is a contradiction.
when $n=19$, $k=3$, requiring $n \geqslant 20$, which is also a contradiction.
In summary, when $n \geqslant 5$, except for the numbers in the set $\{7,8,9,13,14,19\}$, all other positive integers meet the requirement. | n \geqslant 5, \text{ except } n \in \{7,8,9,13,14,19\} | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Let $p_{1}, p_{2}, \cdots, p_{n}$ be $n(\geqslant 2)$ pairwise coprime positive integers, and let
$$\pi_{i}=\frac{p_{1} p_{2} \cdots p_{n}}{p_{i}}, i=1,2, \cdots, n$$
Find the largest positive integer $m$ such that the indeterminate equation
$$\pi_{1} x_{1}+\pi_{2} x_{2}+\cdots+\pi_{n} x_{n}=m$$
has no non-negative integer solutions. | Let $M=(n-1) p_{1} p_{2} \cdots p_{n}-\sum_{i=1}^{n} \pi_{i}$, we prove that the required maximum positive integer is
$M$.
In fact, if there exist non-negative integers $x_{1}, x_{2}, \cdots, x_{n}$, such that
$$\pi_{1} x_{1}+\pi_{2} x_{2}+\cdots+\pi_{n} x_{n}=M$$
Taking modulo $p_{i}$ on both sides of (6), we get
$$\pi_{i} x_{i} \equiv M \equiv-\pi_{i} \quad\left(\bmod p_{i}\right)$$
That is, $p_{i} \mid\left(x_{i}+1\right) \pi_{i}, 1 \leqslant i \leqslant n$.
Using the fact that $p_{1}, p_{2}, \cdots, p_{n}$ are pairwise coprime and the definition of $\pi_{i}$, we know $\left(p_{i}, \pi_{i}\right)=1$, so $p_{i} \mid\left(x_{i}+1\right)$. Since $x_{i}+1 \in \mathbf{N}^{*}$, it follows that $x_{i}+1 \geqslant p_{i}$, i.e., $x_{i} \geqslant p_{i}-1$. This implies (by (6))
$$M=\sum_{i=1}^{n} x_{i} \pi_{i} \geqslant \sum_{i=1}^{n}\left(p_{i}-1\right) \pi_{i}=n p_{1} p_{2} \cdots p_{n}-\sum_{i=1}^{n} \pi_{i}$$
which contradicts the definition of $M$.
On the other hand, if $m>M$, we prove that (5) has a non-negative integer solution.
Note that $\left(\pi_{1}, \pi_{2}, \cdots, \pi_{n}\right)=1$, so (5) has an integer solution $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$. Similar to Example 3, since $\left(x_{1} \pm p_{1}, \cdots, x_{i} \pm p_{i}, \cdots, x_{n}+y p_{n}\right)$, where $y$ is some integer, is also a solution to (5), we can assume $0 \leqslant x_{i} \leqslant p_{i}-1, i=1,2, \cdots, n-1$. At this time,
$$M-\pi_{n}$$
i.e., $\left(x_{n}+1\right) \pi_{n}>0$. Therefore, $x_{n}>-1$, so $x_{n} \geqslant 0$. Hence, when $m>M$, (5) has a non-negative integer solution.
In summary, the required maximum positive integer is
$$M=p_{1} p_{2} \cdots p_{n}\left((n-1)-\sum_{i=1}^{n} \frac{1}{p_{i}}\right)$$ | M=p_{1} p_{2} \cdots p_{n}\left((n-1)-\sum_{i=1}^{n} \frac{1}{p_{i}}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 3 For every positive integer $n$, how many primitive right triangles are there such that their area (numerically) equals $n$ times their perimeter? | Let the sides of a primitive right-angled triangle be $x, y, z$. Then $x, y$ are one odd and one even. Without loss of generality, assume $2 \mid y$. By the theorem, there exist $u, v \in \mathbf{N}^{*}, (u, v)=1$, and $u, v$ are one odd and one even, such that
$$x=u^{2}-v^{2}, y=2 u v, z=u^{2}+v^{2}.$$
By the given condition, we have $\frac{1}{2} x y=n(x+y+z)$, which means
$$u v\left(u^{2}-v^{2}\right)=n\left(2 u^{2}+2 u v\right),$$
Thus,
$$v(u-v)=2 n$$
Let the standard factorization of $n$ be $n=2^{r} p_{1}^{\varepsilon_{1}} p_{2}^{\sigma_{2}} \cdots p_{k}^{\sigma_{k}}$, where $p_{1}, p_{2}, \cdots, p_{k}$ are the odd prime factors of $n$.
Since $u, v$ are one odd and one even, $u-v$ is odd. From (4), we know $v$ is even. Additionally,
$$(u, v)=1,$$
So $(u-v, v)=1$. From (4), we can conclude
$$2^{r+1} \mid v$$
Let $v=2^{r+1} A, u-v=B$. Then from (4) and the above discussion, we know that for each $i=1,2, \cdots, k$, either $p_{i}^{s_{i}}\left|A, p_{i}\right| A$ or $p_{i}^{s_{i}^{*}}\left|B, p_{i}\right| B$ must hold, but not both. Therefore, there are exactly $2^{k}$ possible values for $v$ and $u-v$.
In conclusion, there are exactly $2^{k}$ primitive right-angled triangles that satisfy the condition, where $k$ is the number of odd prime factors of $n$. | 2^{k} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Let positive integers $a, b$ satisfy $a b>1$. Find all positive integer values that the algebraic expression
$$f(a, b)=\frac{a^{2}+a b+b^{2}}{a b-1}$$
can take. | Let $a, b \in \mathbf{N}^{*}$ satisfy $a b>1$, and such that
$$f(a, b)=\frac{a^{2}+a b+b^{2}}{a b-1}=k \in \mathbf{N}^{*},$$
with the conditions: $a \geqslant b$ and $b$ is the smallest positive integer.
From this, we know that the quadratic equation in $x$
i.e., $\square$
$$\begin{aligned}
x^{2}+b x+b^{2}-k(b x-1) & =0 \\
x^{2}+(1-k) b x+b^{2}+k & =0
\end{aligned}$$
has one solution $a$. Let the other solution be $\bar{a}$, then by Vieta's formulas we have
$$\bar{a}=(k-1) b-a \in \mathbf{Z}$$
Since $a \cdot \bar{a}=b^{2}+k>0$, it follows that $\bar{a} \in \mathbf{N}^{\top}$.
Using the fact that $\bar{a}$ is a solution to (5), we know $f(\bar{a}, b)=k$. Since $f$ is symmetric in $a$ and $b$, we have $f(b, \bar{a})=k$. Given $\bar{a}, b \in \mathbf{N}^{*}$ and the minimality of $b$, we have
$$\bar{a}=\frac{b^{2}+k}{a} \geqslant b$$
This implies
$$\frac{a+b+b^{3}}{a b-1} \geqslant b$$
If $a=b$, then $k=\frac{3 a^{2}}{a^{2}-1}=3+\frac{3}{a^{2}-1}$, hence
$$\left(a^{2}-1\right) \mid 3,$$
which means $a^{2}-1=1$ or 3, solving for $a=2$, thus $b=2, k=4$.
If $a>b$, we classify the values of $b$.
(1) If $b \geqslant 4$, then
$$\begin{aligned}
& (a b-1) b-\left(a+b+b^{3}\right)=a\left(b^{2}-1\right)-\left(b^{3}+2 b\right) \\
\geqslant & (b+1)\left(b^{2}-1\right)-\left(b^{3}+2 b\right)=b^{2}-3 b-1 \\
= & (b-2)(b-1)-3>0
\end{aligned}$$
This contradicts (6), so it does not meet the requirements.
(2) If $b=3$, then
$$\begin{aligned}
(a b-1) b-\left(a+b+b^{3}\right) & =3(3 a-1)-(a+3+27) \\
& =8 a-33
\end{aligned}$$
In this case, if $a \geqslant 5$, then $8 a-33>0$, contradicting (6). Therefore, $a=4$, leading to $k=\frac{37}{11}$, which does not meet the conditions.
(3) If $b=2$, then $k=\frac{a^{2}+2 a+4}{2 a-1} \in \mathbf{N}^{*}$, hence
$$\frac{4 a^{2}+8 a+16}{2 a-1} \in \mathbf{N}^{*}$$
i.e., $\square$
$$2 a+5+\frac{21}{2 a-1} \in \mathbf{N}^{\cdot}$$
Thus, $(2 a-1) \mid 21$, so $a \in\{1,2,4,11\}$. Combining $a>b$, we have $a \in\{4, 11\}$, yielding $k \in\{4,7\}$.
(4) If $b=1$, then
$$k=\frac{a^{2}+a+1}{a-1}=a+2+\frac{3}{a-1} \in \mathbf{N}^{*},$$
hence $(a-1) \mid 3$, so $a \in\{2,4\}$, giving $k=7$.
In summary, the positive integer values that $f(a, b)$ can take are 4 or 7. | 4, 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find all positive integers $m(>1)$, such that the number $m^{3}$ can be expressed as the sum of squares of $m$ consecutive positive integers. | Let $m$ be a positive integer that meets the requirement, then there exists $k \in \mathbf{N}$, such that
$$m^{3}=(k+1)^{2}+(k+2)^{2}+\cdots+(k+m)^{2} .$$
Equation (11) is equivalent to
$$k^{2}+(m+1) k+\frac{1}{6}(m+1)(2 m+1)-m^{2}=0$$
Solving the above quadratic equation in $k$, we get
$$k=-\frac{1}{2}(m+1) \pm \frac{1}{6} t$$
where $t=\sqrt{33 m^{2}+3}$. It is easy to prove that when and only when $t \in \mathbf{N}^{*}$,
$$k=-\frac{1}{2}(m+1)+\frac{1}{6} t \in \mathbf{N} .$$
Therefore, it is converted to finding all positive integer solutions of the indeterminate equation
$$t^{2}-33 m^{2}=3$$
For this purpose, we first provide all solutions of
$$x^{2}-33 y^{2}=1$$
Direct calculation shows that the fundamental solution of (14) is $\left(x_{0}, y_{0}\right)=(23,4)$, so all positive integer solutions $\left(x_{n}, y_{n}\right)$ of (14) are determined by:
$$x_{n}+y_{n} \sqrt{33}=(23+4 \sqrt{33})^{n} .$$
Notice that, $\left(t_{0}, m_{0}\right)=(6,1)$ is a positive integer solution of (13), therefore, for any positive integer solution $(x, y)$ of (14), the positive integer pair $(t, m)$ satisfying
$$t+m \sqrt{33}=(6+\sqrt{33}) \cdot(x+y \sqrt{33})$$
is a positive integer solution of (13).
In fact, by (15) we know
$$t-m \sqrt{33}=(6-\sqrt{33})(x-y \sqrt{33}),$$
thus
$$\begin{aligned}
t^{2}-33 m^{2} & =(6-\sqrt{33})(6+\sqrt{33})(x-y \sqrt{33})(x+y \sqrt{33}) \\
& =3\left(x^{2}-33 y^{2}\right)=3
\end{aligned}$$
On the other hand, we prove: the positive integer pair $(t, m)$ determined by (15) is all positive integer solutions of (13).
Suppose $(t, m)$ is a positive integer solution of (13), then $3 \mid t^{2}$, hence $3 \mid t$. Combining this with (13), we know $(x, y)=\left(2 t-11 m, \frac{6 m-t}{3}\right)$ (this relationship is obtained by solving (15) in reverse)
is a pair of positive integers, and
$$\begin{aligned}
x^{2}-33 y^{2} & =(2 t-11 m)^{2}-33\left(\frac{6 m-t}{3}\right)^{2} \\
& =\frac{1}{3}\left(3(2 t-11 m)^{2}-11(6 m-t)^{2}\right) \\
& =\frac{1}{3}\left(t^{2}-33 m^{2}\right)=1
\end{aligned}$$
Thus $x, y$ is a positive integer solution of (13), and
$$t=6 x+33 y, m=x+6 y$$
which is consistent with the relationship between $(t, m)$ and $(x, y)$ corresponding to (15).
In summary, the set of all positive integers $m$ that meet the requirement is $\left\{m_{n} \mid m_{n}=x_{n}+6 y_{n}\right.$, where $x_{n}, y_{n} \in \mathbf{N}^{*}$, determined by $x_{n}+y_{n} \sqrt{33}=(23+4 \sqrt{33})^{n}, n \in \mathbf{N}^{*} \}$. | m_{n}=x_{n}+6 y_{n} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Find all positive integer pairs $(m, n)$ that satisfy the equation
$$3^{m}=2 n^{2}+1$$ | It is easy to verify that $(m, n)=(1,1),(2,2)$ and $(5,11)$ satisfy (19). Below, we prove that these pairs are all the solutions that meet the conditions.
Case 1: $m$ is even. In this case, the problem turns into finding the positive integer solutions of the Pell equation
$$x^{2}-2 y^{2}=1$$
that make $x$ a power of 3.
Since the fundamental solution of (20) is $\left(x_{0}, y_{0}\right)=(3,2)$, all positive integer solutions $\left(x_{k}, y_{k}\right)$ are given by
$$x_{k}+y_{k} \sqrt{2}=(3+2 \sqrt{2})^{k}, k \in \mathbf{N}^{*}$$
Thus,
$$x_{k}-y_{k} \sqrt{2}=(3-2 \sqrt{2})^{k},$$
solving for
$$x_{k}=\frac{1}{2}\left((3+2 \sqrt{2})^{k}+(3-2 \sqrt{2})^{k}\right)$$
Using the binomial theorem, for any $t \in \mathbf{N}^{*}$, we have
$$x_{2 t}=3^{2 t}+\mathrm{C}_{2 t}^{2} \cdot 3^{2 t-2} \cdot 8+\cdots+\mathrm{C}_{2 t}^{2 t-2} \cdot 3^{2} \cdot 8^{t-1}+8^{t}$$
which is not a multiple of 3, and thus cannot be a power of 3.
$$\begin{aligned}
x_{2 t+1}= & 3^{2 t+1}+\mathrm{C}_{2 t+1}^{2} \cdot 3^{2 t-1} \cdot 8+\cdots+\mathrm{C}_{2 t+1}^{2 t-2} \cdot 3^{3} \cdot 8^{t-1} \\
& +\mathrm{C}_{2 t+1}^{2 t} \cdot 3 \cdot 8^{t} \\
= & \mathrm{C}_{2 t+1}^{1} \cdot 3 \cdot 8^{t}+\mathrm{C}_{2 t+1}^{3} \cdot 3^{3} \cdot 8^{t-1}+\cdots+\mathrm{C}_{2 t+1}^{2 t+1} \cdot 3^{2 t+1}
\end{aligned}$$
Now, consider the term in (21)
$$T_{l}=\mathrm{C}_{2 t+1}^{2 l+1} \cdot 3^{2 l+1} \cdot 8^{t-t}$$
in the prime factorization of 3. Let $v_{3}(2 t+1)=\alpha$. For $1 \leqslant l \leqslant t$, we have
$$v_{3}((2 l+1)!)=\sum_{i=1}^{+\infty}\left[\frac{2 l+1}{3^{i}}\right] \leqslant \sum_{i=1}^{+\infty} \frac{2 l+1}{3^{i}}=\frac{1}{2}(2 l+1)<l$$
Thus,
$$v_{3}\left(T_{l}\right) \geqslant \alpha+(2 l+1)-l=\alpha+l+1 \geqslant \alpha+2,$$
while $v_{3}\left(T_{0}\right)=\alpha+1$. This indicates that in the right-hand side of (21), the prime factorization of 3 in all terms except the first is at least $\alpha+2$. Hence, $3^{a+1} \mid x_{2 t+1}$, but $3^{a+2} \nmid x_{2 t+1}$.
Therefore, when $m$ is even, there is only one solution $(m, n)=(2,2)$.
Case 2: $m$ is odd. In this case, the problem turns into finding the solutions of the Pell equation
$$3 x^{2}-2 y^{2}=1$$
that make $x$ a power of 3.
Multiplying both sides of (22) by 3, we can transform the problem into finding the solutions of the Pell equation
$$x^{2}-6 y^{2}=3$$
that make $x$ a power of 3.
Using a similar approach as in Example 4, we can find all positive integer solutions $\left(x_{k}, y_{k}\right)$ of (23), which are determined by
$$x_{k}+y_{k} \sqrt{6}=(3+\sqrt{6})(5+2 \sqrt{6})^{k-1}, k \in \mathbf{N}^{*}$$
Thus, we have
$$x_{k}=\frac{1}{2}\left((3+\sqrt{6})(5+2 \sqrt{6})^{k-1}+(3-\sqrt{6})(5-2 \sqrt{6})^{k-1}\right) .$$
Using $5 \pm 2 \sqrt{6}=\frac{1}{3}(3 \pm \sqrt{6})^{2}$, we get
$$x_{k}=\frac{1}{2 \cdot 3^{2(k-1)}}\left((3+\sqrt{6})^{2 k-1}+(3-\sqrt{6})^{2 k-1}\right)$$
Now, we need to find all positive integers $k$ such that
$$A_{k}=\frac{1}{2}\left((3+\sqrt{6})^{2 k-1}+(3-\sqrt{6})^{2 k-1}\right)$$
is a power of 3.
Notice that,
$$\begin{aligned}
A_{k} & =3^{2 k-1}+\mathrm{C}_{2 k-1}^{2} \cdot 3^{2 k-3} \cdot 6+\cdots+\mathrm{C}_{2 k-1}^{2 k-2} \cdot 3 \cdot 6^{k-1} \\
& =3^{2 k-1}+\mathrm{C}_{2 k-1}^{2} \cdot 3^{2 k-2} \cdot 2+\cdots+\mathrm{C}_{2 k-1}^{2 k-1} \cdot 3^{k} \cdot 2^{k-1} \\
& =\mathrm{C}_{2 k-1}^{1} \cdot 3^{k} \cdot 2^{k-1}+\mathrm{C}_{2 k-1}^{3} \cdot 3^{k-1} \cdot 2^{k-2}+\cdots+\mathrm{C}_{2 k-1}^{2 k-1} \cdot 3^{2 k-1}
\end{aligned}$$
Using a similar analysis method for the right-hand side of (24) as for (21), and classifying based on $3 \mid(2 k-1)$ and $3 \nmid (2 k-1)$, we can show that for $k \geqslant 3$, $A_{k}$ is not a power of 3 (the detailed process is left to the reader). Therefore, $k=1$ or 2, corresponding to $(m, n)=(1,1)$ and $(5,11)$. | (m, n)=(1,1),(2,2),(5,11) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the integer solutions of the indeterminate equation $\frac{1}{2}(x+y)(y+z)(z+x)+(x+y+z)^{3}=1-$ $x y z$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Let's make the substitution, set $x+y=u, y+z=v, z+x=w$, then the original equation transforms to $4 u v w+(u+v+w)^{3}=8-(u+v-w)(u-v+w)(-u+v+w)$.
After expanding and combining like terms, we get
$$4\left(u^{2} v+v^{2} w+w^{2} u+u v^{2}+v w^{2}+w u^{2}\right)+8 u v w=8,$$
which simplifies to
$$u^{2} v+v^{2} w+w^{2} u+u v^{2}+v w^{2}+w u^{2}+2 u v w=2 \text {. }$$
Factoring the left side of the equation, we get
$$(u+v)(v+w)(w+u)=2,$$
Thus, $(u+v, v+w, w+u)=(1,1,2),(-1,-1,2),(-2,-1,1)$ and their symmetric cases. Solving each case, we obtain
$$(u, v, w)=(1,0,1),(1,-2,1),(-1,0,2)$$
Consequently, $(x, y, z)=(1,0,0),(2,-1,-1)$.
In summary, considering symmetry, the integer solutions to the original equation are
$$\begin{aligned}
(x, y, z)= & (1,0,0),(0,1,0),(0,0,1),(2,-1,-1), \\
& (-1,2,-1),(-1,-1,2)
\end{aligned}$$
There are 6 solutions in total. | (x, y, z)= (1,0,0),(0,1,0),(0,0,1),(2,-1,-1),(-1,2,-1),(-1,-1,2) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Find the integer solutions of the equation $x^{2}+x=y^{4}+y^{3}+y^{2}+y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solving as in the previous example, multiply both sides of the equation by 4 and complete the square on the left side:
$$(2 x+1)^{2}=4\left(y^{4}+y^{3}+y^{2}+y\right)+1$$
Next, we estimate the right side of equation (1). Since
$$\begin{aligned}
4\left(y^{4}+y^{3}+y^{2}+y\right)+1 & =\left(2 y^{2}+y+1\right)^{2}-y^{2}+2 y \\
& =\left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1
\end{aligned}$$
Thus, when $y>2$ or $y<-1$, we have
$$\left(2 y^{2}+y\right)^{2}<(2 x+1)^{2}<\left(2 y^{2}+y+1\right)^{2} .$$
Since $2 y^{2}+y$ and $2 y^{2}+y+1$ are two consecutive integers, there cannot be a perfect square between their squares, so the above inequality does not hold.
Therefore, we only need to consider the solutions of the equation when $-1 \leqslant y \leqslant 2$, which is trivial. It is easy to find that all integer solutions to the original equation are
$$(x, y)=(0,-1),(-1,-1),(0,0),(-1,0),(-6,2),(5,2) .$$ | (x, y)=(0,-1),(-1,-1),(0,0),(-1,0),(-6,2),(5,2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Find all positive integer tuples $(a, b, c, d)$ such that
$$\left\{\begin{array}{l}
b d > a d + b c, \\
(9 a c + b d)(a d + b c) = a^{2} d^{2} + 10 a b c d + b^{2} c^{2}
\end{array}\right.$$ | Let $(a, b, c, d)$ be a set of positive integers that satisfy the conditions. Then, from (3), we have $b > a$, $d > \frac{bc}{b-a}$.
Consider the following quadratic function:
\[
\begin{aligned}
f(x) & = (9ac + bx)(ax + bc) - a^2x^2 - 10abcx - b^2c^2 \\
& = a(b-a)x^2 + c(9a-b)(a-b)x + bc^2(9a-b)
\end{aligned}
\]
The coefficient of the quadratic term is $a(b-a) > 0$, and $f(d) = 0$.
Notice that the discriminant of the function $f(x)$ is
\[
\Delta = -(9a-b)(b-3a)^2c^2
\]
Therefore, if $b < 3a$, then $\Delta > 0$, which contradicts $f(d) = 0$.
If $b \geqslant 9a$, then the axis of symmetry of the quadratic function $f(x)$ is $x = \frac{c(9a-b)}{2a} < \frac{bc}{b-a} > 0$, so
\[
f(d) > f\left(\frac{bc}{b-a}\right) = \frac{ab^2c^2}{b-a} > 0
\]
which also contradicts $f(d) = 0$.
Thus, it can only be that $b = 3a$. In this case, $\Delta = 0$, so $f(d) = 0$ is the minimum value of the function $f(x)$, hence $d = \frac{c(9a-b)}{2a}$, i.e., $d = 3c$.
Direct verification shows that when $a, c \in \mathbf{N}^{*}$, the array $(a, 3a, c, 3c)$ is a set of positive integers that meet the requirements, and these are the answers we seek. | (a, 3a, c, 3c) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 5 Find all integer arrays $(a, b, c, x, y, z)$, such that
$$\left\{\begin{array}{l}
a+b+c=x y z, \\
x+y+z=a b c,
\end{array}\right.$$
where $a \geqslant b \geqslant c \geqslant 1, x \geqslant y \geqslant z \geqslant 1$. | By symmetry, we only need to consider the case $x \geqslant a$. At this time, $x y z=a+b+c \leqslant 3 a \leqslant 3 x$, so $y z \leqslant 3$. Thus, $(y, z)=(1,1),(2,1),(3,1)$.
When $(y, z)=(1,1)$, $a+b+c=x$ and $x+2=a b c$, thus $a b c=a+b+c+2$. If $c \geqslant 2$, then
$$a+b+c+2 \leqslant 3 a+2 \leqslant 4 a \leqslant a b c$$
Equality holds if and only if $a=b=c=2$.
If $c=1$, then $a b=a+b+3$, which means
$$(a-1)(b-1)=4$$
Thus, $(a, b)=(5,2)$ or $(3,3)$.
When $(y, z)=(2,1)$,
$$2 a b c=2 x+6=a+b+c+6,$$
Similar discussion shows that $c=1$, then $(2 a-1)(2 b-1)=15$, yielding $(a, b)=(3, 2)$.
When $(y, z)=(3,1)$,
$$3 a b c=3 x+12=a+b+c+12,$$
Discussion shows that there is no solution for $x \geqslant a$.
In summary, we have $(a, b, c, x, y, z)=(2,2,2,6,1,1),(5,2,1,8,1,1)$, $(3,3,1,7,1,1),(3,2,1,3,2,1),(6,1,1,2,2,2),(8,1,1,5,2,1)$ and $(7, 1,1,3,3,1)$. | (2,2,2,6,1,1),(5,2,1,8,1,1),(3,3,1,7,1,1),(3,2,1,3,2,1),(6,1,1,2,2,2),(8,1,1,5,2,1),(7,1,1,3,3,1) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Find all integer pairs $(m, n)$ such that
$$\left\{\begin{array}{l}
2 m \equiv-1(\bmod n), \\
n^{2} \equiv-2(\bmod m) .
\end{array}\right.$$ | This problem is formally about divisibility, but in essence, it is about solving indeterminate equations.
Using the conditions $n|(2 m+1), m|\left(n^{2}+2\right)$, we know that $n, m$ are both odd numbers. If $(m, n)$ meets the requirements, then $(m, -n)$ also meets them, so we can assume $n$ is a positive odd number.
Now, let $2 m+1=a n, n^{2}+2=b m$, then $a, b$ are also odd numbers. Eliminating $n$ from the two equations, we get
$$a^{2} b m=a^{2}\left(n^{2}+2\right)=(a n)^{2}+2 a^{2}=(2 m+1)^{2}+2 a^{2},$$
which is
$$2 a^{2}+1=m\left(a^{2} b-4 m-4\right)$$
If $a= \pm 1$, then $m(b-4 m-4)=3$, so $(m, b-4 m-4)=$ $( \pm 1, \pm 3)$ or $( \pm 3, \pm 1)$, which respectively yield all solutions $(m, n, a, b)=(3, \pm 7, \pm 1,17),(1, \pm 3, \pm 1,11),(-1, \mp 1, \pm 1,-3)$ or $(-3, \mp 5, \pm 1,-9)$.
Similarly, for $a= \pm 3$, the solutions to (5) are $(m, n, a, b)=(19, \pm 13, \pm 3,9)$ or $(1, \pm 1, \pm 3,3)$.
On the other hand, eliminating $m$, we get
$$2 n^{2}+4=2 b m=a b n-b$$
Let $k=a b-2 n$, then $k$ is an odd number. From (6), we know $n=\frac{1}{k}(b+4)$, and substituting back into (6), we get
$$\begin{aligned}
& \frac{2(b+4)^{2}}{k^{2}}+4=\frac{a b(b+4)}{k}-b \\
\Rightarrow & k a=\frac{1}{b(b+4)}\left(2(b+4)^{2}+4 k^{2}+b k^{2}\right)=2+\frac{k^{2}+8}{b}
\end{aligned}$$
If $k= \pm 1$, then from (7), $\pm a=2+\frac{9}{b}$. There are three solutions, which are $(m, n, a, b)=(17, \pm 7, \pm 5,3),(27, \pm 5, \pm 1,1)$ or $(-11, \pm 3, \mp 7,-1)$. Similarly, for $k= \pm 3, \pm 5$, the new solutions are $(m, n, a, b)=(-3, \pm 1, \mp 5,-1)$ and $(3, \pm 1, \pm 7,-1)$.
Now, consider the case where $|a| \geqslant 5$ and $|k| \geqslant 7$. From (7), we know $b=\frac{k^{2}+8}{a k-2}$, and combining with $n=\frac{1}{k}(b+4)$, we get $n=\frac{k+4 a}{a k-2}$. Since $n$ is odd, we should have
$$|a k-2| \leqslant|k+4 a|$$
If $a, k$ have different signs, then
$$\begin{array}{l}
|a k-2|=|a k|+2>\max \{|k|,|4 a|\} \\
>\max \{|k|-4|a|, 4|a|-k\}=|k+4 a|
\end{array}$$
which contradicts (8).
If $a, k$ have the same sign, then
$$\begin{aligned}
|a k-2| & =|a k|-2=4|a|+|k|+(|a|-1)(|k|-1)-6 \\
& \geqslant 4|a|+|k|+4 \times 6-6>4|a|+|k|=|k+4 a|
\end{aligned}$$
which also contradicts (8).
In summary, the integer pairs $(m, n)$ that satisfy the conditions are $(27, \pm 5),(19, \pm 13),(17$,
$$\begin{array}{l}
\pm 7),(3, \pm 1),(3, \pm 7),(1, \pm 1),(1, \pm 3),(-1, \pm 1),(-3, \pm 1) \\
(-3, \pm 5) \text { or }(-11, \pm 3) .
\end{array}$$ | (27, \pm 5),(19, \pm 13),(17, \pm 7),(3, \pm 1),(3, \pm 7),(1, \pm 1),(1, \pm 3),(-1, \pm 1),(-3, \pm 1),(-3, \pm 5),(-11, \pm 3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 7 Find the smallest positive integer $n$, such that the indeterminate equation
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}=1599$$
has integer solutions $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$. | Note that for any $x \in \mathbf{Z}$, if $x$ is even, then $x^{4} \equiv 0(\bmod 16)$; if $x$ is odd, then $x^{2} \equiv 1(\bmod 8)$, and in this case, $x^{4} \equiv 1(\bmod 16)$.
The above discussion shows that $x_{i}^{4} \equiv 0$ or $1(\bmod 16)$, thus the remainder of $x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}$ modulo 16 lies between 0 and $n$. Therefore, when $n \leqslant 14$, equation (9) cannot hold (since $1599 \equiv 15(\bmod 16)$).
Notice that $5^{4}+12 \times 3^{4}+1^{4}+1^{4}=1599$, so when $n=15$, equation (9) has an integer solution $(5, \underbrace{3, \cdots, 3}_{12 \uparrow}, 1,1)$.
In summary, the smallest positive integer $n=15$. | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Find all non-negative integer arrays $(m, n, p, q)$, such that $0<p<q,(p, q)=1$, and
$$\left(1+3^{m}\right)^{p}=\left(1+3^{n}\right)^{q} .$$ | If $p>1$, by $(p, q)=1$, we can set $q=p s+r, 0<p$. Then, $\left(c^{p}-1\right) \mid\left(c^{q}-1\right)$, and
$$c^{g}-1=c^{\infty+r}-1=\left(c^{p}\right)^{s} \cdot c^{r}-1 \equiv 1^{s} \cdot c^{r}-1\left(\bmod c^{\rho}-1\right),$$
Therefore,
$$\left(c^{p}-1\right) \mid\left(c^{r}-1\right)$$
This cannot hold when $r<p$, so $p=1$, and (10) becomes
$$1+3^{m}=\left(1+3^{n}\right)^{q}$$
Noting that $q>p=1$, hence $q \geqslant 2$, at this time the right side of (11) $\equiv 0(\bmod 4)$, so
$$3^{m} \equiv-1(\bmod 4)$$
Thus, $m$ is odd, at this moment
$$1+3^{m}=(1+3)\left(1-3+3^{2}-\cdots+3^{m-1}\right)=4 \cdot A,$$
Here $A=1-3+3^{2}-\cdots+3^{m-1}$ is the sum of $m$ odd numbers, so $A$ is odd, thus $(4, A)=1$. This requires 4 and $A$ to both be $q$-th power numbers, from which we can deduce $q=2$, and then we can find
$$(m, n, p, q)=(1,0,1,2),$$
It is easy to verify that it meets the requirements. | (m, n, p, q)=(1,0,1,2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 9 Find the positive integer solutions $(x, y, z, t)$ of the indeterminate equation $1+5^{x}=2^{y}+2^{x} \cdot 5^{t}$. | Let $(x, y, z, t)$ be a positive integer solution to the original equation. Taking both sides of the equation modulo 5, we have $2^{y} \equiv 1(\bmod 5)$. Since 2 is a primitive root modulo 5, it follows that $4 \mid y$. At this point, taking both sides of the equation modulo 4, we get $2^{z} \equiv 2(\bmod 4)$, so $z=1$.
We set $y=4r$, yielding
$$5^{x}-2 \cdot 5^{t}=16^{r}-1$$
Taking both sides of (12) modulo 3, we have $(-1)^{x}-(-1)^{t+1} \equiv 0(\bmod 3)$, hence
$$x \equiv t+1(\bmod 2)$$
Further, taking both sides modulo 8, we get $5^{x} \equiv 2 \cdot 5^{t}-1 \equiv 1(\bmod 8)$ (using $5 \equiv 5$ or $1(\bmod 8)$). Thus, $x$ is even, and $t$ is odd.
Notice that if $t=1$, then $5^{x}=16^{r}+9$. Setting $x=2m$, we have
$$\left(5^{m}-3\right)\left(5^{m}+3\right)=16^{r}$$
Since $\left(5^{m}-3,5^{m}+3\right)=\left(5^{m}-3,6\right)=2$, it follows that
$$5^{m}-3=2, \quad 5^{m}+3=2^{4r-1},$$
thus $m=1, r=1$. If $t>1$, then $5^{3} \mid\left(5^{x}-2 \cdot 5^{t}\right)$, and hence $5^{3} \mid\left(16^{r}-1\right)$. By the binomial theorem, expanding $16^{r}-1=(15+1)^{r}-1$, we see that $5 \mid r$. Thus, we can set $r=5k$, yielding
$$16^{r}-1=16^{5k}-1 \equiv\left(5^{5}\right)^{k}-1 \equiv\left(5 \times 3^{2}\right)^{k}-1 \equiv 0 \quad(\bmod 11)$$
This requires $11 \mid\left(5^{x}-2 \cdot 5^{t}\right)$, hence $11 \mid\left(5^{x-t}-2\right)$. However, for any $m \in \mathbf{N}^{*}$, we have $5^{n}=5,3,4,9$ or $1(\bmod 11)$, and it is impossible for
$$5^{x-t} \equiv 2(\bmod 11)$$
to occur, leading to a contradiction.
In summary, the equation has a unique solution
$$(x, y, z, t)=(2,4,1,1)$$ | (x, y, z, t)=(2,4,1,1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
3. Find all integer pairs $(a, b)$ such that the system of equations
$$\left\{\begin{array}{l}
x^{2}+2 a x-3 a-1=0 \\
y^{2}-2 b y+x=0
\end{array}\right.$$
has exactly three distinct real solutions $(x, y)$. | 3. Let $(a, b)$ be a pair of integers that satisfy the conditions, then the system of equations
$$\left\{\begin{array}{l}
x^{2}+2 a x-3 a-1=0, \\
y^{2}-2 b y+x=0
\end{array}\right.$$
has exactly 3 different real solutions, so equation (1) should have two different real roots, with the discriminant $\Delta=4 a^{2}+12 a+4=4\left(a^{2}+3 a+1\right)>0$.
Let the two roots be $x_{1}<0$ and $x_{2}>0$. For the equation $y^{2}-2 b y+x_{1}=0$ to have no real roots, the discriminant must be less than 0, i.e., $4 b^{2}-4 x_{1}<0$, which implies $b^{2}<x_{1}$. For the equation $y^{2}-2 b y+x_{2}=0$ to have two different real roots, the discriminant must be greater than 0, i.e., $4 b^{2}-4 x_{2}>0$, which implies $b^{2}>x_{2}$. Therefore, we have $b^{2}<x_{1}<0$ and $b^{2}>x_{2}>0$, leading to $b^{2}=-a-\sqrt{a^{2}+3 a+1}$ and $b^{2}=-a+\sqrt{a^{2}+3 a+1}$.
This requires $a^{2}+3 a+1$ to be a perfect square. Let $a^{2}+3 a+1=c^{2}(c \geqslant 0, c \in \mathbf{Z})$. Multiplying both sides by 4 and completing the square, we get: $(2 a+3)^{2}-(2 c)^{2}=5$, i.e., $(2 a-2 c+3)(2 a+2 c+3)=5$, which gives $(2 a-2 c+3,2 a+2 c+3)=(-5,-1)$ or $(1,5)$. Solving these, we get $a=0$ or -3. Further, we have $(a, b)=(0,1)$, $(0,-1)$, $(-3,2)$, or $(-3,-2)$. | (0,1), (0,-1), (-3,2), (-3,-2) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
4. Find all integer solutions to the indeterminate equation $5 x^{2}-14 y^{2}=11 z^{2}$. | 4. Taking both sides modulo 7, we know $5 x^{2} \equiv 11 z^{2} \equiv 4 z^{2}(\bmod 7)$. If $7 \times x$, then $\left(2 z x^{-1}\right)^{2} \equiv 5(\bmod 7)$, where $x^{-1}$ is the modular inverse of $x$ modulo 7. However, a square number $\equiv 0,1$ or $4(\bmod 7)$, which is a contradiction. Therefore, $7 \mid x$. Consequently, we can deduce that $7|z, 7| y$. Thus, $\left(\frac{x}{7}, \frac{y}{7}, \frac{z}{7}\right)$ is also an integer solution to the equation. By induction, the only solution to the equation is $(x, y, z)=(0, 0,0)$. | (x, y, z)=(0, 0,0) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5. Find all integer pairs $(x, y)$ such that $x^{3}=y^{3}+2 y^{2}+1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. Notice that, when $y>0$ or $y<-3$, we always have $y^{3}<y^{3}+2 y^{2}+1<(y+$ $1)^{3}$, at this time $y^{3}+2 y^{2}+1$ is not a cube number, so the original equation has no solution. Therefore, we only need to consider the cases $y=$ $-3,-2,-1,0$. Substituting them respectively, we get the integer solutions of the equation as $(x, y)=(-2$, $-3),(1,-2)$ or $(1,0)$. | (x, y)=(-2, -3), (1, -2), (1, 0) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
6. Find the integer solutions of the indeterminate equation $x^{2}(y-1)+y^{2}(x-1)=1$.
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6. Find the integer solutions of the indeterminate equation $x^{2}(y-1)+y^{2}(x-1)=1$. | 6. Let $(x, y)$ be an integer solution of the equation, and assume $x \leqslant y$, then $y \geqslant 2$. In this case, consider the original equation as a quadratic equation in $x$:
$$(y-1) x^{2}+y^{2} x-\left(y^{2}+1\right)=0$$
Since the equation has integer solutions, $\Delta=y^{4}+4(y-1)\left(y^{2}+1\right)$ must be a perfect square, i.e., $\left(y^{2}+2 y\right)^{2}-8 y^{2}+4 y-4$ is a perfect square.
Notice that when $y \geqslant 8$, $\left(y^{2}+2 y-3\right)^{2}>\Delta>\left(y^{2}+2 y-4\right)^{2}$, and when $2\Delta>\left(y^{2}+2 y-3\right)^{2}$, thus $y>2$, $\Delta$ is not a perfect square. Therefore, $y=2$, and then $x=1$ or -5.
In summary, the integer solutions of the equation are $(x, y)=(1,2),(2,1),(-5,2)$ or $(2,-5)$. | (x, y)=(1,2),(2,1),(-5,2),(2,-5) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
7. Let $a, b, c, d$ all be prime numbers, satisfying: $a>3 b>6 c>12 d$, and
$$a^{2}-b^{2}+c^{2}-d^{2}=1749$$
Find the value of $a^{2}+b^{2}+c^{2}+d^{2}$. | 7. From the conditions, we know that $a, b, c$ are all odd numbers. If $d$ is odd, then $a^{2}-b^{2}+c^{2}-d^{2}$ is even, which is a contradiction. Therefore, $d$ is even, and thus $d=2$. Consequently, $a^{2}-b^{2}+c^{2}=1753$. From the conditions, we also know that $a \geqslant 3 b+2, b \geqslant 2 c+1, c \geqslant 5$, hence
$$\begin{aligned}
1753 & \geqslant(3 b+2)^{2}-b^{2}+c^{2}=8 b^{2}+12 b+4-c^{2} \\
& \geqslant 8(2 c+1)^{2}+12 b+4=33 c^{2}+32 c+12 b+12 \\
& \geqslant 33 c^{2}+160+132+12
\end{aligned}$$
Therefore, $c^{2}3 b$, we know $a-b>2 b \geqslant 22$. Also, $a-b$ and $a+b$ are both even, and $a, b$ are both odd, so $a-b=32$, and $a+b=54$. Thus, $a=43, b=11$, and consequently, $a^{2}+b^{2}+c^{2}+d^{2}=1999$. | 1999 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
10. Find all positive integer triples $(x, y, z)$ such that $y$ is a prime number, 3 and $y$ are not divisors of $z$, and $x^{3}-y^{3}=z^{2}$. | 10. From the original equation, we get $z^{2}=(x-y)\left(x^{2}+x y+y^{2}\right)$. Let
$$\begin{aligned}
\left(x-y, x^{2}+x y+y^{2}\right) & =\left(x-y,(x-y)^{2}+3 x y\right)=(x-y, 3 x y) \\
& =\left(x-y, 3 y^{2}\right)=d
\end{aligned}$$
Then $d \mid 3 y^{2}$. Since $y$ is a prime and 3 and $y$ are not factors of $z$, we have $d=1$. Thus, $x-y$ and $x^{2}+x y+y^{2}$ are both perfect squares. We set $x-y=u^{2}, x^{2}+x y+y^{2}=v^{2}$, where $u, v \in \mathbf{N}^{*}$. Multiplying both sides of the equation $x^{2}+x y+y^{2}=v^{2}$ by 4, completing the square, rearranging, and factoring, we get
$$(2 v-2 x-y)(2 v+2 x+y)=3 y^{2}$$
Noting that $y$ is a prime, $2 v+2 x+y \in \mathbf{N}^{*}$, and $2 v-2 x-y<2 v+2 x+y$, we can only have
$$\left\{\begin{array}{l}
2 v-2 x-y=1, y \text { or } 3, \\
2 v+2 x+y=3 y^{2}, 3 y \text { or } y^{2} .
\end{array}\right.$$
For the first case, we have
$$3 y^{2}-1=2(2 x+y)=4 u^{2}+6 y$$
Thus, $u^{2}+1 \equiv 0(\bmod 3)$. But $u^{2} \equiv 0$ or $1(\bmod 3)$, so there is no solution in this case.
For the second case, we have
$$2 y=4 x+2 y$$
Leading to $x=0$, which also has no solution.
For the third case, we have
$$y^{2}-3=4 u^{2}+6 y$$
That is, $(y-2 u-3)(y+2 u-3)=12$, solving which gives $(y, u)=(7,1)$, and thus $(x, z)=(8,13)$.
The solution to the original equation that satisfies the conditions is $(x, y, z)=(8,7,13)$. | (8,7,13) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
11. Find the integer solutions of the indeterminate equation $x^{5}+y^{5}=(x+y)^{3}$. | 11. Clearly, the integer pairs $(x, y)$ that satisfy $x+y=0$ are all solutions to the equation. Now consider the solutions to the equation when $x+y \neq 0$. In this case, dividing both sides by $x+y$ gives
$$x^{4}-x^{3} y+x^{2} y^{2}-x y^{3}+y^{4}=(x+y)^{2},$$
Rearranging and simplifying, we get
$$\left(x^{2}+y^{2}\right)^{2}+x^{2} y^{2}=(x+y)^{2}(x y+1)$$
This indicates that $x y+1>0$, i.e., $x y \geqslant 0$. If $x, y \geqslant 0$, then by the power mean inequality, we have
$$(x+y)^{3}=x^{5}+y^{5} \geqslant 2\left(\frac{x+y}{2}\right)^{5},$$
Thus, $x+y \leqslant 4$. If $x, y \leqslant 0$, similarly, we have $x+y \geqslant-4$. Therefore, we always have $x y \geqslant 0$ and $|x+y| \leqslant 4$. Discussing separately for $|x+y|=1,2,3,4$, we find that the solutions are $(x, y)=$ $(0, \pm 1),( \pm 1,0),(2,2),(-2,-2)$.
In summary, the integer solutions to the equation are $\{(t,-t) \mid t \in \mathbf{Z}\} \cup\{(0, \pm 1),( \pm 1,0)$, $(2,2),(-2,-2)\}$. | (t,-t) \mid t \in \mathbf{Z}\ | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Let $n$ be a positive integer of the form $a^{2}+b^{2}$, where $a, b$ are two coprime positive integers. It satisfies: if $p$ is a prime and $p \leqslant \sqrt{n}$, then $p \mid a b$. Find all positive integers $n$ that meet the requirements. | Let $n=a^{2}+b^{2}$ be a number that meets the requirement.
If $a=b$, then by $(a, b)=1$, we know $a=b=1$, in this case $n=2$ meets the requirement.
If $a \neq b$, without loss of generality, assume $a < b$. Take a prime factor $p$ of $b-a$. By $(a, b)=1$, we know $(b-a, b)=(b-a, a)=1$, thus $(b-a, ab)=1$, hence $p \nmid ab$. On the other hand, $p \leqslant b-a < \sqrt{a^{2}+b^{2}} = \sqrt{n}$, which contradicts the requirement that $n$ meets. Therefore, $b-a=1$.
At this point, $n=a^{2}+(a+1)^{2}=2a^{2}+2a+1$. Direct verification shows that when $a=1,2$, $n=5,13$ meet the requirement. When $a \geqslant 3$, if $a$ is odd, then $(a+2, a(a+1))=1$, at this time a prime factor $p$ of $a+2$ satisfies $p \leqslant a+2 \leqslant \sqrt{2a^{2}+2a+1} = \sqrt{n}$, and $p \nmid a(a+1)$, a contradiction; if $a$ is even, then $(a-1, a(a+1))=1$, at this time a prime factor $p$ of $a-1$ satisfies $p \leqslant a-1 < \sqrt{2a^{2}+2a+1} = \sqrt{n}$, and $p \nmid a(a+1)$, a contradiction.
In summary, the only $n$ that meet the condition are 2, 5, and 13. | 2, 5, 13 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
12. Find the integer solutions of the indeterminate equation $x^{6}+x^{3} y=y^{3}+2 y^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 12. Let $(x, y)$ be the integer solution of the equation, then when $x=0$, $y^{2}(y+2)=0$, we get $y=$ 0 or -2. When $y=0$, $x^{6}=0$, we get $x=0$. It is known that when $x, y$ have one equal to zero, the solutions are $(x, y)=(0,0)$ or $(0,-2)$.
Below, we discuss the case where $x y \neq 0$. For any prime factor $p$ of $y$, let $p^{m} \| y$, then $p \mid x^{6}$, hence $p \mid x$. Suppose $p^{n} \| x$, there are two possibilities.
(1) $p$ is an odd prime, then $p^{2 m} \|\left(y^{3}+2 y^{2}\right)$. And $x^{6}+x^{3} y=x^{3}\left(x^{3}+y\right)$. If $3 n \leq m$, then we get $3 n+m=2 m$, which implies $m=3 n$. Therefore, we always have $m=3 n$.
(2) $p=2$, if $m \geqslant 2$, then $2^{2 m+1} \|\left(y^{3}+2 y^{2}\right)$, similarly to (1) we get $m=3 n$ or $3 n-1$.
Using the conclusions from (1) and (2), we can set $(x, y)=\left(a b, 2 b^{3}\right),\left(a b, b^{3}\right)$ or $\left(a b, \frac{b^{3}}{2}\right)$, where $a, b \in \mathbf{Z}$. Substituting into the original equation, we get
$1^{\circ} a^{6}+a^{3}=b^{3}+2$,
$2^{\circ} a^{6}+2 a^{3}=8 b^{3}+8$,
$3^{\circ} \quad 8 a^{6}+4 a^{3}=b^{3}+4$.
For equation $1^{\circ}$, if $a>1$, then
$$\left(a^{2}+1\right)^{3}>b^{3}=a^{6}+a^{3}-2>\left(a^{2}\right)^{3},$$
no solution. If $a b^{3}>\left(a^{2}-1\right)^{3},$
also no solution. Hence $a=0, x=0$ or $a=1, b=0, y=0$, both contradict $x y \neq 0$.
For equation $2^{\circ}$, if $a>0$, then
$$\left(a^{2}+1\right)^{3}>(2 b)^{3}=a^{6}+2 a^{3}-8>\left(a^{2}\right)^{3},$$
no solution. If $a(2 b)^{3}>\left(a^{2}-1\right)^{3},$
also no solution. For $a=-2,-1,0$, all lead to $x y=0$.
For equation $3^{\circ}$, if $a>1$, then
$$\left(2 a^{2}+1\right)^{3}>b^{3}=8 a^{6}+4 a^{3}-4>\left(2 a^{2}\right)^{3} .$$
If $a b^{3}>\left(2 a^{2}-1\right)^{3}$
Hence, only $a \in\{-1,0,1\}$. Only when $a=1, b=2$, there is a solution $(x, y)=(2,4)$ such that $x y \neq 0$.
In summary, the integer solutions of the equation are $(x, y)=(0,0),(0,-2)$ or $(2,4)$. | (x, y)=(0,0),(0,-2) \text{ or } (2,4) | Algebra | math-word-problem | Yes | Yes | number_theory | false |
13. Find all pairs of positive integers $(x, y)$ such that $x^{x+y}=y^{y-x}$. | 13. Let the pair of positive integers $(x, y)$ satisfy the condition, then $x^{y}(x y)^{x}=y^{y}$, hence $x^{y} \mid y^{y}$, so $x \mid y$. Let $y=k x, k \in \mathbf{N}^{*}$, then $k^{y}=\left(k x^{2}\right)^{x}$, thus $k^{k}=k x^{2}$, which means $x^{2}=k^{k-1}$. Therefore, all pairs of positive integers $(x, y)$ that satisfy the condition are $\left((2 n+1)^{n},(2 n+1)^{n+1}\right)$ or $\left((2 m)^{4 m^{2}-1},(2 m)^{4 m^{2}+1}\right)$, where $m, n$ are any positive integers. | (2 n+1)^{n},(2 n+1)^{n+1} \text{ or } (2 m)^{4 m^{2}-1},(2 m)^{4 m^{2}+1} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
25. Find the integer solutions of the indeterminate equation $8 x^{4}+1=y^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 25. Notice that $(x, y)=(0, \pm 1),( \pm 1, \pm 3)$ are solutions to the equation. If $(x, y)$ is an integer solution to the equation and $|x|>1$, we may assume $y>1$.
From $8 x^{4}=(y-1)(y+1)$, and since $y$ is odd, we can set
$$\left\{\begin{array} { l }
{ y - 1 = 2 u ^ { 4 } , } \\
{ y + 1 = 4 v ^ { 4 } , }
\end{array} \text { or } \left\{\begin{array}{l}
y-1=4 v^{4}, \\
y+1=2 u^{4},
\end{array}\right.\right.$$
where $u, v \in \mathbf{N}^{*}$. Thus, we always have $u^{4}-2 v^{4}= \pm 1$, and therefore
$$\left(u^{4}-v^{4}\right)^{2}=v^{8} \pm 2 v^{4}+1=v^{8} \pm u^{4}$$
Using the result from Example 5 in Section 4.2 and the conclusion from the previous problem, we know that $u v^{2}\left(u^{4}-v^{4}\right)=0$, hence $u=v$, leading to $u^{4}-2 v^{4}=-1$, i.e., $u=v=1$. This implies $y=3$, leading to $|x|=1$, a contradiction. Therefore, the only solutions to the equation are $(x, y)=(0, \pm 1),( \pm 1, \pm 3)$. | (x, y)=(0, \pm 1),( \pm 1, \pm 3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
26. Find all integer solutions $(x, y, z)$ that satisfy the following system of equations:
$$\left\{\begin{array}{l}
x+1=8 y^{2}, \\
x^{2}+1=2 z^{2} .
\end{array}\right.$$ | 26. First, $(x, y, z)=(-1,0, \pm 1)$ is a solution to the system of equations. Next, consider the solutions to the system of equations when $y \neq 0$. By eliminating $x$ from the system, we get
$$1+\left(2(2 y)^{2}-1\right)^{2}=2 z^{2},$$
which simplifies to $(2 y)^{4}+\left((2 y)^{2}-1\right)^{2}=z^{2}$. Therefore, there exist $a, b \in \mathbf{N}^{*}$ such that
$$4 y^{2}=2 a b, 4 y^{2}-1=a^{2}-b^{2},(a, b)=1, a>b .$$
From the above equations, it is easy to see that $a$ is even and $b$ is odd. Thus, we can set
$$a=2 s^{2}, b=t^{2},(s, t)=1, s, t \in \mathbf{N}^{*} .$$
This leads to $4 s^{2} t^{2}-1=4 s^{4}-t^{4}$, hence $\left(2 s^{2}+t^{2}\right)^{2}=8 s^{4}+1$. Using the result from the previous problem, we know $s=1, 2 s^{2}+t^{2}=3$, which implies $s=t=1, a=2, b=1$, so $y= \pm 1, x=7, z=$ $\pm 5$.
Therefore, the integer solutions to the system of equations are $(x, y, z)=(-1,0, \pm 1),( \pm 1,7, \pm 5)$. | (x, y, z)=(-1,0, \pm 1),(7, \pm 1, \pm 5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
30. Find all $n \in \mathbf{N}^{*}$, such that there exist $x, y, z \in \mathbf{N}^{*}$, satisfying:
$$n=\frac{(x+y+z)^{2}}{x y z}$$ | 30. Let the set of $n$ that meets the requirements be $S$. By taking $(x, y, z)=(9,9,9)$, $(4,4,8)$, $(3,3,3)$, $(2,2,4)$, $(1,4,5)$, $(1,2,3)$, $(1,1,2)$, $(1,1,1)$, we know $\{1,2,3,4,5,6,8,9\} \subseteq S$.
Now, let $n \in S$, and $x, y, z$ be the positive integers that satisfy $n=\frac{(x+y+z)^{2}}{x y z}$, such that $x+y+z$ is minimized.
We prove: $x \leqslant y+z$, $y \leqslant z+x$, $z \leqslant x+y$.
To prove (10) holds, by symmetry, we only need to prove: $x \leqslant y+z$.
If $x>y+z$, then by the condition, $x \mid (x+y+z)^{2}$, so $x \mid (y+z)^{2}$. Now let $x^{\prime}=\frac{(y+z)^{2}}{x}$, then $x^{\prime} \geq 1$. If $x^{\prime} \geq 1$, then $y z - (y+z) = (y-1)(z-1) - 1 \geq 0$, i.e., $y z \geq y + z \geq x$. At this point, combining $x \geq y \geq z$, we have
$$\begin{aligned}
n & =\frac{x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x}{x y z}=2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y} \\
& \leqslant 2\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)+1+1+1=6
\end{aligned}$$
Therefore, $n \in \{1,2,3,4,5,6,8,9\}$.
In summary, $S=\{1,2,3,4,5,6,8,9\}$, which is the set of all $n$ that meet the conditions. | S=\{1,2,3,4,5,6,8,9\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
35. Find the positive integer tuples $(a, b, x, y)$ that satisfy the following condition:
$$(a+b)^{x}=a^{y}+b^{y}$$ | 35. When $a=b=1$, it is easy to see that the solution is $(a, b, x, y)=(1,1,1, y)$, where $y$ is any positive integer.
When $y=1$, the solution is $(a, b, 1,1)$, where $a, b$ are any positive integers.
When $y=2$, it is known that $x=1$, at this time $a+b=a^{2}+b^{2}$, so $a(a-1)=b(1-b) \geqslant 0$, which gives $a=b=1$.
When $y=3$, it is known that $x2^{\frac{3}{2}}$, we can get $2^{y^{-1}}>2^{\frac{3}{2}(t+k)\left(y^{-x}\right)}$, i.e.,
$$y-1>\frac{3}{2}(t+k)(y-x)$$
Now compare the exponents of the prime factorization of $p$ on both sides of (19), we should have
$$\begin{aligned}
v_{p}\left(a_{1}^{y}+b_{1}^{\gamma}\right) & =k x-t(y-x)>k x-\frac{2}{3}\left((y-1)-\frac{3}{2} k(y-x)\right) \\
& =k y-\frac{2}{3}(y-1)
\end{aligned}$$
Here we use (21).
On the other hand, note that $\left(a_{1}, b_{1}\right)=1$, so $\left(a_{1}, a_{1}+b_{1}\right)=1$. If $y$ is even, then $a_{1}^{Y}+b_{1}^{Y} \equiv 2 a_{1}^{Y} \neq 0(\bmod p)$, i.e., $p \nmid\left(a_{1}^{Y}+b_{1}^{Y}\right)$, which contradicts (22), so $y$ is odd. At this time, set $b_{1}=p^{k} \cdot u-a_{1}, u \in \mathbf{N}^{*}, p \nmid u$, then using the binomial theorem we know
$$\begin{aligned}
a_{1}^{y}+b_{1}^{y} & =a_{1}^{y}+\left(p^{k} \cdot u-a_{1}\right)^{y} \\
& =\sum_{n=2}^{y}(-1)^{n-1} C_{y}^{n} p^{n k} u^{n} a_{1}^{y-n}+y \cdot p^{k} \cdot u \cdot a_{1}^{y-1} .
\end{aligned}$$
Set $p^{s} \| y, s \in \mathbf{N}$. It is known that when $n \geqslant 2$, we have
$$\begin{aligned}
v_{p}\left(\mathrm{C}_{y}^{n} \cdot p^{n k} \cdot u^{n} \cdot a_{1}^{y^{-n}}\right) & =v_{p}\left(\frac{y}{n} \mathrm{C}_{y}^{n-1} \cdot p^{*}\right) \\
& \geqslant s+n k-v_{p}(n) \geqslant s+k+1
\end{aligned}$$
Here we use $p \geqslant 3, n \geqslant 3$ when, $p^{v} p^{(n)} \leqslant n \leqslant p^{n-2}$, so $v_{p}(n) \leqslant n-2$ (note that when $n=2$, $v_{p}(n)=0$, this inequality also holds).
And $v_{p}\left(y \cdot p^{k} \cdot u \cdot a_{1}^{Y-1}\right)=s+k$. Therefore, $v_{p}\left(a_{1}^{Y}+b_{1}^{y}\right)=s+k$. Thus, combining (22) we have
$$s+k>k y-\frac{2}{3}(y-1)$$
Thus, $s>k(y-1)-\frac{2}{3}(y-1) \geqslant \frac{1}{3}(y-1)$, so $s \geqslant\left[\frac{y-1}{3}\right]+1=$ $\left[\frac{y+2}{3}\right]$. Therefore, $p^{\left[\frac{2+2}{3}\right]} \leqslant p^{s} \leqslant y$, but this inequality does not hold when $p \geqslant 3, y \geqslant 4$.
The above discussion shows that $a_{1}+b_{1}$ can only have one prime factor 2. Set $a_{1}+b_{1}=2^{k}$, combining $\left(a_{1}, b_{1}\right)=1$, we know that $a_{1}, b_{1}$ are both odd. Using (19) we can set $d=2^{2}$, then
$$a_{1}^{y}+b_{1}^{y}=2^{k-x(y-x)} .$$
If $k=1$, then $a_{1}=b_{1}=1$, we get $a=b=2^{t}$, and require $x-t(y-x)=$ 1, solving this linear Diophantine equation in $x, y$ (for fixed $t$), we get the solution $(a, b, x, y)=$ $\left(2^{t}, 2^{t}, t s+1,(t+1) s+1\right), t, s \in \mathbf{N}$.
If $k \geqslant 2$, similarly analyzing the modulus 4 of both sides of (23), we know that $y$ is odd, at this time we have
$$a_{1}^{y}+b_{1}^{y}=a_{1}^{y}+\left(2^{k}-a_{1}\right)^{y}=\sum_{n=2}^{y}(-1)^{n-1} C_{y}^{n} 2^{2^{k}} a_{1}^{y-n}+y \cdot 2^{k} \cdot a_{1}^{y-1} .$$
Similarly analyzing the power of 2 on both sides of the above equation, we get $2^{k} \|\left(a_{1}^{y}+b_{1}^{*}\right)$, by (23) we have $a_{1}^{\gamma}+b_{1}^{y}=2^{k}$, thus $y=1$, which is a contradiction.
In summary, all positive integer solutions to the equation are $(a, b, x, y)=(1,1,1, y),(a, b$, $1,1),(2,1,2,3),(1,2,2,3),(2,2,2,3)$ or $\left(2^{t}, 2^{t}, t s+1,(t+1) s+1\right)$, where $a, b, y \in \mathbf{N}^{*}, t, s \in \mathbf{N}$. | (a, b, x, y)=(1,1,1, y),(a, b, 1,1),(2,1,2,3),(1,2,2,3),(2,2,2,3),\left(2^{t}, 2^{t}, t s+1,(t+1) s+1\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
36. Find all positive integer triples $(a, m, n)$ such that
$$\left(a^{m}+1\right) \mid(a+1)^{n}$$ | 36. Notice that when $a, m$ is equal to 1, it is easy to get the positive integer solutions $(a, m, n)=(1, m, n)$ or $(a, 1, n)$, where $a, m, n \in \mathbf{N}^{*}$.
Below, we discuss the case where $a, m$ are both greater than 1.
First, we prove a lemma: Let $u, v, l \in \mathbf{N}^{*}$, and $u \mid v^{l}$, then $u \mid(u, v)^{l}$.
In fact, let $u=p_{1}^{\sigma_{1}} p_{2}^{\sigma_{2}} \cdots p_{k}^{\sigma_{k}}$, $v=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}}$, here $p_{1}1$, thus $\left(a^{m}+1, a+1\right)$ $>1$, so $\left(a^{m}+1, a+1\right)=2$. This way, by the lemma, $\left(a^{m}+1\right) \mid 2^{n}$, hence $a^{m}+1=$ $2^{s}$, where $s$ is some positive integer. Given $a>1$, we know $s \geqslant 2$, requiring $a^{m}+1 \equiv 0(\bmod 4)$, but when $m$ is even, $a^{m} \equiv 0$ or $1(\bmod 4)$, which is a contradiction.
If $m$ is odd, by $m>1$, we know $n>1$. Let $p$ be a prime factor of $m$, and set $m=p r$, $b=a^{r}$, $r$ is a positive odd number. Then, by the condition, $\left(b^{p}+1\right) \mid(a+1)^{n}$. And $(a+1) \mid\left(a^{r}+\right.$ $1)$, i.e., $(a+1) \mid(b+1)$, so, $\left(b^{p}+1\right) \mid(b+1)^{n}$, i.e., $\left.\frac{b^{p}+1}{b+1} \right\rvert\,(b+1)^{n-1}$. Note that,
$$\begin{aligned}
B & =\frac{b^{p}+1}{b+1}=b^{p-1}-b^{p-2}+\cdots-b+1 \\
& \equiv(-1)^{p-1}-(-1)^{p-2}+\cdots-(-1)+1 \\
& =p(\bmod b+1)
\end{aligned}$$
Thus, $(B, b+1) \mid p$. Combining $B>1$ and $B \mid(b+1)^{n-1}$, we know $(B, b+1)>1$, hence $(B, b+1)=p$. Using the lemma, we get $B \mid p^{n-1}$, thus $B$ is a power of $p$.
Now, let $b=p k-1, k \in \mathbf{N}^{*}$, then by the binomial theorem, we know
$$\begin{aligned}
b^{p}+1 & =(p k-1)^{p}+1 \\
& =(k p)^{p}-\mathrm{C}_{p}^{p-1}(k p)^{p-1}-\cdots-\mathrm{C}_{p}^{2}(k p)^{2}+k p^{2} \\
& \equiv k p^{2}\left(\bmod k p^{3}\right)
\end{aligned}$$
Therefore, $B=\frac{b^{p}+1}{b+1}=\frac{b^{p}+1}{k p} \equiv p\left(\bmod p^{2}\right)$, hence $p^{2} \times B$, so $B=p$.
When $p \geqslant 5$,
$$\begin{aligned}
B & =\frac{b^{p}+1}{b+1}=b^{p-1}-b^{p-2}+\cdots-b+1>b^{p-1}-(b-1) b^{p-2} \\
& =b^{p-2} \geqslant 2^{p-2}>p
\end{aligned}$$
Thus, by $B=p$, we know, it can only be $p=3$. At this time, $b^{2}-b+1=3$, solving gives $b=2$, then $a=$ $2, r=1$. At this moment, $m=p r=3$. And obviously, $(a, m, n)=(2,3, n)$ satisfies the condition when $n \geqslant 2$.
Therefore, the positive integer solutions $(a, m, n)=(1, m, n)(a, 1, n)$ or $(2,3, r)$, where $a, m, n, r \in \mathbf{N}^{*}, r \geqslant 2$. | (a, m, n)=(1, m, n), (a, 1, n), (2,3, r) | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find all positive integers $n$ such that $2^{n}-1$ is divisible by 7. | Since $2^{n}=(\underbrace{100 \cdots 0}_{n \uparrow})_{2}$, then $2^{n}-1=(\underbrace{11 \cdots 1}_{n \uparrow})_{2}$. And $7=(111)_{2}$, therefore, for $7 \mid\left(2^{n}-1\right)$, i.e., $(111)_{2} \mid(\underbrace{11 \cdots 1}_{n \uparrow})_{2}$, $n$ must be a multiple of 3. | n \text{ must be a multiple of 3} | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
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