problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Example 6 Proof: There are infinitely many primes $p \equiv 1(\bmod 4)$. | Assume there are only finitely many such primes, and let them be \( p_{1}, \cdots, p_{k} \). We consider \(\left(2 p_{1} \cdots p_{k}\right)^{2}+1=P\). By the assumption and \(P \equiv 1(\bmod 4)\), we know that \(P\) is not a prime. Let \(p\) be a prime factor of \(P\), \(p\) is of course odd, so \(-1\) is a quadratic... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,439 |
Example 7 Prove that any odd prime factor $p$ of $x^{4}+1$ satisfies $p \equiv 1(\bmod 8)$, and thus there are infinitely many primes $p \equiv 1(\bmod 8)$. | Let $p$ be an odd prime factor of $x^{4}+1$, i.e.,
$$\left(x^{2}\right)^{2} \equiv x^{4} \equiv -1(\bmod p)$$
Therefore, $\left(\frac{-1}{p}\right)=1$. By Theorem 1, we have $p \equiv 1(\bmod 4)$. On the other hand,
$$x^{4}+1=\left(x^{2}+1\right)^{2}-2 x^{2},$$
so we have
$$\left(x^{2}+1\right)^{2} \equiv 2 x^{2}(\bm... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,440 |
Example 8 Let $p$ be a prime, $p \equiv 3(\bmod 4)$. Prove: $2 p+1$ is a prime if and only if
$$2^{p} \equiv 1(\bmod 2 p+1)$$ | Necessity If $q=2 p+1$ is a prime, then by the condition we know $q \equiv-1(\bmod 8)$, and thus by Theorem 3 we have $\left(\frac{2}{q}\right)=1$. From this and Theorem 1 (ii), we get
$$1 \equiv 2^{(q-1) / 2} \equiv 2^{p}(\bmod 2 p+1)$$
Sufficiency If equation (15) holds. Since $p$ is a prime, by Example 5 in Chapter... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,441 |
1. 计算下列 Legendre 符号:
$$\begin{array}{c}
\left(\frac{13}{47}\right),\left(\frac{30}{53}\right),\left(\frac{71}{73}\right),\left(\frac{-35}{97}\right),\left(\frac{-23}{131}\right),\left(\frac{7}{223}\right), \\
\left(\frac{-105}{223}\right),\left(\frac{91}{563}\right),\left(\frac{-70}{571}\right),\left(\frac{-286}{647}\r... | 1. $-1,-1,1,1,1,1,-1,1,-1,1$. | -1,-1,1,1,1,1,-1,1,-1,1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,442 |
2. Determine whether the following congruence equations have solutions:
(i) $x^{2} \equiv 7(\bmod 227)$;
(ii) $x^{2} \equiv 11(\bmod 511)$;
(iii) $11 x^{2} \equiv -6(\bmod 91)$;
(iv) $5 x^{2} \equiv -14(\bmod 6193)$. | 2. (i) $\left(\frac{7}{227}\right)=1$, has solutions; (ii) $511=7 \cdot 73,\left(\frac{11}{73}\right)=-1$, no solutions;
(iii) $91=7 \cdot 13,\left(\frac{11}{7}\right)=\left(\frac{-6}{7}\right)=1,\left(\frac{11}{13}\right)=\left(\frac{-6}{13}\right)=-1$, has solutions;
(iv) $6193=11 \cdot 563,\left(\frac{5}{11}\right)=... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,443 |
3. (i) Find all prime numbers for which -3 is a quadratic residue;
(ii) Find all prime numbers for which ±3 is a quadratic residue;
(iii) Find all prime numbers for which ±3 is a quadratic non-residue;
(iv) Find all prime numbers for which 3 is a quadratic residue and -3 is a quadratic non-residue;
(v) Find all prime n... | 3. (i) $p \equiv 1(\bmod 6)$. (ii) $p \equiv 1(\bmod 12)$. (iii) $p \equiv 5(\bmod 12)$; (iv) $p \equiv-1$ $(\bmod 12) ;(\mathrm{v}) p \equiv-5(\bmod 12) .(\mathrm{vi})(100)^{2}-3$ 的素因数 $p \equiv \pm 1(\bmod 12)$. $100^{2}-3=13 \cdot 769 ; 150^{2}+3$ 的素因数 $p \equiv 1(\bmod 6)$ 及 $p=3$.
$$150^{2}+3=3 \cdot 13 \cdot 577$... | 100^{2}-3=13 \cdot 769 ; 150^{2}+3=3 \cdot 13 \cdot 577 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,444 |
4. Find all prime numbers for which 3 is a quadratic non-residue and 2 is a quadratic residue (i.e., all prime numbers for which 3 is the smallest positive quadratic non-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。 | 4. $p \equiv \pm 7(\bmod 24)$. | p \equiv \pm 7(\bmod 24) | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false | 740,445 |
5. (i) Find all prime numbers $p$ such that $\left(\frac{5}{p}\right)=1$;
(ii) Find all prime numbers $p$ such that $\left(\frac{-5}{p}\right)=1$;
(iii) Find the prime factorization of $121^{2} \pm 5, 82^{2} \pm 5 \cdot 11^{2}, 273^{2} \pm 5 \cdot 11^{2}$;
(iv) Is $x^{4} \equiv 25(\bmod 1013)$ solvable? | 5. (i) $p \equiv \pm 1(\bmod 5)$.
(ii) $p \equiv 1,3,7,9(\bmod 20)$.
(iii) $121^{2}-5=14636=$ $2^{2} \cdot 3659,121^{2}+5=14646=2 \cdot 3 \cdot 2441 ; 82^{2}+5 \cdot 11^{2}=7329=3 \cdot 7 \cdot 349 ; 82^{2}-$ $5 \cdot 11^{2}=6119=29 \cdot 211 ; 273^{2}+5 \cdot 11^{2}=2 \cdot 37567,273^{2}-5 \cdot 11^{2}=2^{2} \cdot 184... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,446 |
6. (i) Find all prime numbers $p$ such that $\left(\frac{-2}{p}\right)=1$; (ii) Find all prime numbers $p$ such that $\left(\frac{10}{p}\right)=1$;
(iii) Find all prime numbers $p$ such that $x^{2} \equiv 13(\bmod p)$ has a solution;
(iv) Prove: The prime factors of $n^{4}-n^{2}+1$ are $\equiv 1(\bmod 12)$. | 6. (i) $p \equiv 1,3(\bmod 8)$; (ii) $p \equiv \pm 1, \pm 3, \pm 9, \pm 13(\bmod 40)$; (iii) $p=2, p=$ 13 and $p \equiv \pm 1, \pm 3, \pm 4(\bmod 13)$; (iv) $n^{4}-n^{2}+1=\left(n^{2}-1\right)^{2}+n^{2}=\left(n^{2}+1\right)^{2}-3 n^{2}$. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,447 |
5. Let $n \geqslant 1$. Prove: $(n!+1,(n+1)!+1)=1$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. | 5. $(n!+1,(n+1)!+1)=(n!+1,-n)=(1,-n)=1$.
The translation is as follows:
5. $(n!+1,(n+1)!+1)=(n!+1,-n)=(1,-n)=1$. | 1 | Number Theory | proof | Yes | Yes | number_theory | false | 740,448 |
7. Prove: The congruence equation $x^{2}-x+1 \equiv 0(\bmod m)$, (i) when $m=2^{k}(k \geqslant 1)$, has no solutions; (ii) when $m=3$, the number of solutions is 1, and when $m=3^{k}(k \geqslant 2)$, it has no solutions; (iii) when $m=p^{k}(k \geqslant 1), p$ is an odd prime greater than 3, the number of solutions is $... | 7. For $m=2^{k}, 3^{k}$, verify directly. When $m=p^{k}, p>3$, the congruence equation can be transformed into $(2 x-1)^{2}+3 \equiv 0$ $\left(\bmod p^{k}\right)$ for discussion. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,449 |
8. Prove that the following forms of primes are infinite:
(i) $8 k-1, 8 k+3, 8 k-3$;
(ii) $3 k+1, 6 k+1, 12 k+7, 12 k+1$;
(iii) whose last digit in their decimal representation is 9. | 8. (i) Prove that there are infinitely many primes of the form $8k-1$, using $\left(p_{1} \cdots p_{r}\right)^{2}-2$; prove that there are infinitely many primes of the form $8k+3$, using $\left(p_{1} \cdots p_{r}\right)^{2}+2$; prove that there are infinitely many primes of the form $8k-3$, using $4\left(p_{1} \cdots ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,450 |
10. Let $p>2$ be a prime. Prove:
(i) The congruence equation $\left(x^{2}-a\right)\left(x^{2}-b\right)\left(x^{2}-a b\right) \equiv 0(\bmod p)$ always has a solution, where $a, b$ are any integers;
(ii) $x^{6}-11 x^{4}+36 x^{2}-36 \equiv 0(\bmod p)$ always has a solution. | 10. (i) If $p \mid a b$ holds; if $p \nmid a b$, then $\left(\frac{a}{p}\right),\left(\frac{b}{p}\right),\left(\frac{a b}{p}\right)$ must have at least one equal to 1;
(ii) The original polynomial $=\left(x^{2}-2\right)\left(x^{2}-3\right)\left(x^{2}-6\right)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,452 |
11. Let prime $p>2$. Prove: $x^{4} \equiv-4(\bmod p)$ has a solution if and only if $p \equiv 1(\bmod 4)$. | 11. $x^{4}+4=\left((x-1)^{2}+1\right)\left((x+1)^{2}+1\right)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,453 |
12. Prove: (i) For any prime $p, x^{8} \equiv 16(\bmod p)$ always has a solution;
(ii) When $l \geqslant 3$, $x^{2^{l}} \equiv 2^{2^{l-1}}(\bmod p)$ always has a solution. | 12. (i) $x^{8}-16=\left(x^{4}+4\right)\left(x^{2}-2\right)\left(x^{2}+2\right)$, and use the result from the previous problem; (ii) for $l \geqslant 3$, $x^{8}-16 \mid x^{2^{l}}-2^{2^{l-1}}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,454 |
13. (i) Let $2 \nmid n$, and an odd prime $p \mid a^{n}-1$. Prove: $\left(\frac{a}{p}\right)=1$.
(ii) Let the prime $p>2$. Prove: $2^{p}-1$ has prime factors $\equiv \pm 1(\bmod 8)$. | 13. (i) $\left(a^{(n+1) / 2}\right)^{2} \equiv a(\bmod p)$; (ii) From (i) and Theorem 3, we deduce. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,455 |
14. (i) Without calculation, prove: $23\left|2^{11}-1,47\right| 2^{23}-1,503 \mid 2^{251}-1$;
(ii) If there are infinitely many primes $p=4 n+3$, such that $2 p+1$ is also a prime, then there are infinitely many Mersenne numbers (i.e., of the form $2^{q}-1, q$ being a prime) that are composite. | 14. (i) $2^{22}-1=\left(2^{11}-1\right)\left(2^{11}+1\right) \equiv 0(\bmod 23) \cdot 23 \equiv -1(\bmod 8)$, so $2^{11}+1 \neq 0(\bmod 23)$. The other two can be proved similarly. (ii) $\left(2^{p}-1\right)\left(2^{p}+1\right)=2^{2 p}-1 \equiv 0(\bmod 2 p+1)$, and $2 p+1 \equiv 7(\bmod 8)$. Use the result from problem... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,456 |
15. Let $q=4^{n}+1$. Prove: $q$ is a prime if and only if
$$3^{(q-1) / 2} \equiv-1(\bmod q)$$ | 15. Necessity. Let $q$ be a prime. If the conclusion does not hold, then $\left(\frac{3}{q}\right)=1$, which implies $q \equiv \pm 1$ $(\bmod 12)$, a contradiction. Sufficiency. Let the smallest $h$ such that $3^{h} \equiv 1(\bmod q)$ be $h_{0} . h_{0} \mid q-1=2^{2 n}$, and from $3^{2^{2 n-1}} \not \equiv 1(\bmod q)$ ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,457 |
16. Prove: (i) When the prime $p=4 m+3,\left(\frac{a}{p}\right)=1$, $x_{0}= \pm a^{m+1}$ is a solution to $x^{2} \equiv a(\bmod p)$;
(ii) When $p=8 m+5,\left(\frac{a}{p}\right)=1$, $x_{0}= \pm 2^{3 m+1} a^{m+1}\left(2^{2 m+1}+a^{2 m+1}\right)$ is a solution to $x^{2} \equiv a(\bmod p)$;
(iii) When $p=8 m+1,\left(\frac{... | 16. (i) Use Euler's criterion.
(ii) Use Euler's criterion and $\left(\frac{-2}{p}\right)=-1$. The solutions can be written separately as: when $a^{2 m+1} \equiv 1$ $(\bmod p)$, $x_{0}= \pm a^{m+1}$; when $a^{2 m+1} \equiv-1(\bmod p)$, $x_{0}= \pm 2^{2 m+1} a^{m+1}$.
(iii) Using the same method as (ii), the role of $b$ ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,458 |
6. Find the greatest common divisor
(i) $(2 t+1,2 t-1)$;
(ii) $(2 n, 2(n+1))$;
(iii) $(k n, k(n+2))$;
(iv) $\left(n-1, n^{2}+n+1\right)$. | 6. (i) $(2 t+1,2 t-1)=(2 t+1,-2)=(1,-2)=1$.
(ii) $(2 n, 2(n+1))=(2 n, 2)=2$.
(iii) $(k n, k(n+2))=(k n, 2 k)$. When $n=2 a$, $(k n, 2 k)=(2 a k, 2 k)=2 k$; when $n=2 a+1$, $(k n, 2 k)=(2 a k+k, 2 k)=(k, 2 k)=k$.
(iv) $\left(n-1, n^{2}+n+1\right)=(n-1,2 n+1)=(n-1,3)=3$, when $n=3 k+1 ;=1$, when $n=3 k-1,3 k$. | 2k \text{ when } n=2a, k \text{ when } n=2a+1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,459 |
17. (i) For $p=11,17,19,29, d=2,3,5,7,13$, specifically calculate $n$ in Lemma 2 of §6, and verify whether the results match those in Example 1 of §5.
(ii) Directly use Lemma 2 of §6 to prove: For a prime $p>5$,
$$\left(\frac{5}{p}\right)=(-1)^{[p / s]-[p / 10]+[2 p / 5]-[3 p / 10]}$$ | 17. (i)
\begin{tabular}{|l|l|lllll|}
\hline & \multicolumn{1}{|r|}{$d$} & \multirow{2}{*}{2} & 3 & 5 & 7 & 13 \\
\hline$p$ & $n$ & & & & & \\
\hline 11 & 3 & 2 & 2 & 3 & \\
17 & & 4 & 3 & 3 & 3 & 4 \\
19 & & 5 & 3 & 4 & 4 & 3 \\
29 & & 7 & 5 & 6 & 6 & 6 \\
\hline
\end{tabular}
(ii) Following the method in Example 3 to ... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,460 |
7. Let $a, b$ be positive integers. Prove: if $[a, b]=(a, b)$, then $a=b$.
The translation is complete as requested, maintaining the original format and line breaks. | 7. $[a, b] \geqslant a \geqslant(a, b),[a, b] \geqslant b \geqslant(a, b)$, so, $a=b=(a, b)=[a, b]$. | a=b | Number Theory | proof | Yes | Yes | number_theory | false | 740,470 |
27. Let the prime $p \equiv 3(\bmod 4)$, and the number of even quadratic residues among $1,2, \cdots, p-1$ be denoted as $R^{(2)}$. Prove:
$$R^{(2)}=\left\{\begin{array}{ll}
N_{1}, & p \equiv 3(\bmod 8), \\
(p-1) / 2-N_{1}, & p \equiv 7(\bmod 8),
\end{array}\right.$$
where $N_{1}$ is given by Exercise 12 (iii) of §5 ... | 27. When $p \equiv 3(\bmod 8)$, $\left(\frac{2}{p}\right)=-1.2,4,6, \cdots, p-1$ contains the same number of quadratic residues as the number of quadratic non-residues in $1,2, \cdots,(p-1) / 2$. From this and Exercise 5, Question 12 (iii), we deduce that $R^{(2)}=N_{1}$. When $p \equiv 7(\bmod 8)$, $\left(\frac{2}{p}\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,471 |
28. Let the prime $p \equiv 3(\bmod 4), R_{1}$ be the number of quadratic residues modulo $p$ in the set $1,2, \cdots,(p-1) / 2$. Prove:
(i) $2 \cdot 4 \cdots(p-1) \equiv(-1)^{R_{1}+(p-3) / 4}(\bmod p)$;
(ii) $1 \cdot 3 \cdots(p-2) \equiv(-1)^{R_{1}+(p+1) / 4}(\bmod p)$;
(iii) $((p-1) / 2)!\equiv(-1)^{R_{1}+1}(\bmod p)... | 28. First prove (iii). Let $N_{1}$ be the same as in problem 27, $N_{1}+R_{1}=(p-1) / 2$. It is easy to prove: $((p-1) / 2)!\equiv$ $(-1)^{N_{1}}(((p-1) / 2)!)^{2}(\bmod p)$. Then use problem 4 (ii) from Chapter 3, Exercise 4 to get (iii). From (iii) and $2 \cdot 4 \cdots(p-1)=2^{(p-1) / 2}((p-1) / 2)$!, we can deduce ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,472 |
30. Let $a, b$ be integers, $b^{2}>1$. Prove:
(i) $b^{2}+2 \nmid 4 a^{2}+1$;
(ii) $b^{2}-2 \nmid 4 a^{2}+1$;
(iii) $2 b^{2}+3 \nmid a^{2}-2$;
(iv) $3 b^{2}+4 \nmid a^{2}+2$. | 30. (i) By contradiction. If $b^{2}+2 \mid 4 a^{2}+1$, then $2 \nmid b, b^{2}+2 \equiv 3(\bmod 4)$. Therefore, $b^{2}+2$ must have a prime factor $p \equiv 3(\bmod 4)$, but $p \mid 4 a^{2}+1$ must have $p \equiv 1(\bmod 4)$, which is a contradiction.
(ii) The proof is the same as (i).
(iii) If $2 b^{2}+3 \mid a^{2}-2$,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,474 |
31. Using Problem 19, prove Theorem 5 of §6 by the following method. Let $p, q$ be distinct odd primes. For the pair $\{j, k\}$, define
$$\begin{array}{c}
L(j, k)=(2 j-1) q-(2 k-1) p \\
1 \leqslant j \leqslant(p-1) / 2,1 \leqslant k \leqslant(q-1) / 2
\end{array}$$
Let $n_{1}$ be the $n_{1}$ defined in Problem 19 when... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 740,475 |
32. Let prime $p \geqslant 3$, $p \nmid a$. Prove: $\sum_{x=1}^{p}\left(\frac{a x+b}{p}\right)=0$. | 32. $x$ and $a x+b$ simultaneously traverse the complete residue modulo $p$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,476 |
33. Let prime $p \geqslant 3, p \nmid a$. Prove:
$$\sum_{x=1}^{p}\left(\frac{x^{2}+a x}{p}\right)=\sum_{x=1}^{p}\left(\frac{x^{2}+x}{p}\right)=-1 .$$ | 33. Let $x^{-1}=y$ denote the inverse of $x$ modulo $p$, where $x, y$ simultaneously traverse the reduced residue system modulo $p$.
$$\begin{aligned}
\sum_{x=1}^{p}\left(\frac{x^{2}+a x}{p}\right) & =\sum_{x=1}^{p}\left(\frac{(a x)^{2}+a(a x)}{p}\right) \\
& =\sum_{x=1}^{p-1}\left(\frac{x^{2}+x}{p}\right)=\sum_{x=1}^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,477 |
34. Let prime $p \geqslant 3, p \nmid a$ and $f(x)=a x^{2}+b x+c, \Delta=b^{2}-4 a c$. Prove:
(i) If $p \nmid \Delta$, then $\sum_{x=1}^{p}\left(\frac{f(x)}{p}\right)=-\left(\frac{a}{p}\right)$;
(ii) If $p \mid \Delta$, then $\sum_{x=1}^{p}\left(\frac{f(x)}{p}\right)=(p-1)\left(\frac{a}{p}\right)$. | 34. $4 a f(x)=(2 a x+b)^{2}-\Delta . \sum_{x=1}^{p}\left(\frac{f(x)}{p}\right)=\left(\frac{a}{p}\right) \sum_{x=1}^{p}\left(\frac{x^{2}-\Delta}{p}\right)$. When $p \mid \Delta$, it follows that (ii) holds. When $p \nmid \Delta$, we consider two cases: $\left(\frac{\Delta}{p}\right)=1$ and $\left(\frac{\Delta}{p}\right)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,478 |
35. Prove: For any prime $p$, there must be integers $a, b, c, d$, such that
$$x^{4}+1 \equiv\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)(\bmod p) .$$ | 35. $x^{4}+1 \equiv\left(x^{2}+1\right)^{2}(\bmod 2)$, so we can assume $p \geqslant 3$. If $\left(\frac{-1}{p}\right)=1$, then take $b$ such that $b^{2} \equiv-1(\bmod p)$, and we have $x^{4}+1 \equiv\left(x^{2}-b\right)\left(x^{2}+b\right)(\bmod p)$. If $\left(\frac{-1}{p}\right)=-1$, let $a, b, c, d$ be undetermined... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,479 |
8. Prove: If $(a, 4)=(b, 4)=2$, then $(a+b, 4)=4$. | 8. $a=4 k_{1}+2, b=4 k_{2}+2$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
8. $a=4 k_{1}+2, b=4 k_{2}+2$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,481 |
Lemma 2 Let $a_{j} \equiv 1(\bmod m)(1 \leqslant j \leqslant s), a=a_{1} \cdots a_{s}$, we have
$$\frac{a-1}{m} \equiv \frac{a_{1}-1}{m}+\cdots+\frac{a_{s}-1}{m}(\bmod m) .$$ | It is obvious that we only need to prove the case where $s=2$. We have
$$a-1=a_{1} a_{2}-1=\left(a_{1}-1\right)+\left(a_{2}-1\right)+\left(a_{1}-1\right)\left(a_{2}-1\right) .$$
From $a_{j} \equiv 1(\bmod m)$, we know $a \equiv 1(\bmod m)$, so
$$\begin{aligned}
\frac{a-1}{m} & =\frac{a_{1}-1}{m}+\frac{a_{2}-1}{m}+\fra... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,482 |
Theorem 3 We have
$$\begin{array}{l}
\left(\frac{-1}{P}\right)=(-1)^{(P-1) / 2} \\
\left(\frac{2}{P}\right)=(-1)^{\left(P^{2}-1\right) / 8}
\end{array}$$ | Let $P=p_{1} \cdots p_{s}, p_{j}$ be odd primes. By definition and Theorem 1(v) of §6, we know
$$\left(\frac{-1}{P}\right)=\left(\frac{-1}{p_{1}}\right) \cdots\left(\frac{-1}{p_{s}}\right)=(-1)^{\left(p_{1}-1\right) / 2+\cdots+\left(p_{s}-1\right) / 2} .$$
Taking $m=2, a_{j}=p_{j}(1 \leqslant j \leqslant s)$ in Lemma ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,483 |
Theorem 4 Let $P>1$ be an odd number, $Q>1$ be an odd number, and $(P, Q)=1$. We have
$$\left(\frac{Q}{P}\right) \cdot\left(\frac{P}{Q}\right)=(-1)^{(P-1) / 2 \cdot(Q-1) / 2}$$ | Let $P=p_{1} \cdots p_{s}, Q=q_{1} \cdots q_{r}, p_{j}, q_{i}$ all be odd primes. By definition, Theorem 1, and Theorem 5 in §6 (note $q_{i} \neq p_{j}$), we have
$$\begin{aligned}
\left(\frac{Q}{P}\right) & =\prod_{j=1}^{s}\left(\frac{Q}{p_{j}}\right)=\prod_{j=1}^{s} \prod_{i=1}^{r}\left(\frac{q_{i}}{p_{j}}\right) \\
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,484 |
2. Let $a, b$ be positive integers, $2 \nmid b$. Prove that for the Jacobi symbol, the formula is:
$$\left(\frac{a}{2 a+b}\right)=\left\{\begin{array}{ll}
\left(\frac{a}{b}\right), & a \equiv 0,1(\bmod 4), \\
-\left(\frac{a}{b}\right), & a \equiv 2,3(\bmod 4)
\end{array}\right.$$ | 2. Taking $4 \mid a$ as an example. Let $a=2^{a} n, a \geqslant 2,2 \nmid n$. Using the reciprocity law, we get
$$\begin{aligned}
\left(\frac{a}{2 a+b}\right) & =\left(\frac{2^{a}}{2 a+b}\right)\left(\frac{b}{n}\right)(-1)^{(k-1)(b-1) / 4} \\
\left(\frac{a}{b}\right) & =\left(\frac{2^{a}}{b}\right)\left(\frac{b}{n}\rig... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,486 |
4. Prove: The integer $a$ is a quadratic residue modulo every prime if and only if $a=b^{2}$ (Assume the following result: Let $m \geqslant 1,(d, m)=1$, then there must be a prime $p \equiv d(\bmod m))$.
| 4. The sufficiency is obvious. For necessity, we use proof by contradiction. Suppose $a=b^{2} a_{1}, a_{1} \neq 1$ and $a_{1}$ is not a square number, i.e., $a_{1}=$ $\pm 2^{\alpha_{0}} p_{1} \cdots p_{r}\left(\alpha_{0}=0,1 ; p_{i}\right.$ are distinct odd primes). Let $d_{r}$ be a quadratic non-residue modulo $p_{r}$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,488 |
6. Let $\left(\frac{D}{n}\right)$ be the Kronecker symbol from the previous problem. Prove that,
(i) for a given $D$, there always exists an $n$ such that $\left(\frac{D}{n}\right)=-1$;
(ii) $\left(\frac{D}{|D|-1}\right)=\frac{D}{|D|}$ | 6. Let $D=2^{l} k, 2 \nmid k$. (i) Discuss in three cases: $l=0$, $l$ is odd, and $l$ is even, and use the Chinese Remainder Theorem; (ii) Use (a) and (b) from Question 5. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,490 |
Theorem 1 Let $p \nmid a_{n}$. If the $n$-th degree congruence equation (2) has $k$ distinct solutions
$$x \equiv c_{1}, \cdots, c_{k}(\bmod p)$$
then, there must exist a unique pair of integer coefficient polynomials $g_{k}(x)$ and $r_{k}(x)$, such that
$$f(x)=\left(x-c_{1}\right) \cdots\left(x-c_{k}\right) g_{k}(x)+... | Prove uniqueness: If there are also $\bar{g}_{k}(x), \bar{r}_{k}(x)$ such that
$$f(x)=\left(x-c_{1}\right) \cdots\left(x-c_{k}\right) \bar{g}_{k}(x)+p \cdot \bar{r}_{k}(x)$$
then we have
$$\left(x-c_{1}\right) \cdots\left(x-c_{k}\right)\left(g_{k}(x)-\bar{g}_{k}(x)\right)=p\left(\bar{r}_{k}(x)-r_{k}(x)\right) .$$
If ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,493 |
Inference 3 (i) If the number of solutions of the congruence equation (2) is $>n$, then it must be that $p \mid a_{j}, 0 \leqslant j \leqslant n$.
(ii) Let the integer coefficient polynomials $f_{1}, f_{2}$ have degrees less than $p$. If $f_{1}$ and $f_{2}$ are equivalent modulo $p$, then they are certainly congruent m... | Prove by contradiction. If the conclusion does not hold, then there must be a $d, 0 \leqslant d \leqslant n$, such that $p \mid a_{j}$, $d < j \leqslant n$, and $p \nmid a_{d}$. In this case, the number of solutions to the congruence equation (2) is the same as the number of solutions to the congruence equation
$$a_{d}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,495 |
Example 1 Determine whether the congruence equation $2 x^{3}+5 x^{2}+6 x+1 \equiv 0(\bmod 7)$ has three solutions. | Solve: Here the coefficient of the first term is 2. By making an identity transformation, we can know that the original equation has the same solutions as
$$4\left(2 x^{3}+5 x^{2}+6 x+1\right) \equiv x^{3}-x^{2}+3 x-3 \equiv 0(\bmod 7)$$
Performing polynomial division, we get
$$x^{7}-x=\left(x^{3}-x^{2}+3 x-3\right)\l... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,498 |
Example 2 Let $p>2$ be a prime, $p \nmid d$. Find the necessary and sufficient condition for the quadratic congruence equation
$$x^{2}-d \equiv 0(\bmod p)$$
to have 2 solutions. | Since
$$\begin{aligned}
x^{p-1}-1 & =\left(x^{2}\right)^{(p-1) / 2}-d^{(p-1) / 2}+d^{(p-1) / 2}-1 \\
& =\left(x^{2}-d\right) q(x)+d^{(p-1) / 2}-1,
\end{aligned}$$
by Theorem 5, the necessary and sufficient condition for the number of solutions to be 2 is
$$d^{(\rho-1) / 2}-1 \equiv 0(\bmod p)$$
Since $p>2$, the congr... | d^{(p-1)/2} \equiv 1 \pmod{p} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,499 |
Theorem 6 (i) The necessary and sufficient condition for the number of solutions of the congruence equation (2) to be $p$ is
$$f(x)=\left(x^{p}-x\right) g(x)+p \cdot r(x)$$
where the integer-coefficient polynomial $r(x)$ has a degree $<p$, i.e., $f(x)$ is congruent to zero modulo $p$;
(ii) The necessary and sufficient... | First, we prove (i). The sufficiency follows from equation (17) and Fermat's Little Theorem. By polynomial division, we have
$$f(x)=\left(x^{p}-x\right) g(x)+s(x)$$
where $s(x)$ is an integer-coefficient polynomial of degree $<p$. Therefore, $f(x)$ is congruent to $s(x)$ modulo $p$. By the number of solutions of the c... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,500 |
Example 3 Simplify the congruence equation
$$21 x^{18}+2 x^{15}-x^{10}+4 x-3 \equiv 0(\bmod 7)$$ | Solve: First, remove the terms with coefficients that are multiples of 7 to get
$$2 x^{15}-x^{10}+4 x-3 \equiv 0(\bmod 7)$$
Perform polynomial division (19) $(p=7)$ to get
$$\begin{aligned}
2 x^{15}-x^{10}+4 x-3= & \left(x^{7}-x\right)\left(2 x^{8}-x^{3}+2 x^{2}\right) \\
& +\left(-x^{4}+2 x^{3}+4 x-3\right)
\end{alig... | x^{4}-2 x^{3}-4 x+3 \equiv 0(\bmod 7) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,501 |
$$\begin{aligned}
f(x)= & 3 x^{14}+4 x^{13}+2 x^{11}+x^{9}+x^{6}+x^{3} \\
& +12 x^{2}+x \equiv 0(\bmod 5) .
\end{aligned}$$
Example 4 Simplify the congruence equation
$$\begin{aligned}
f(x)= & 3 x^{14}+4 x^{13}+2 x^{11}+x^{9}+x^{6}+x^{3} \\
& +12 x^{2}+x \equiv 0(\bmod 5) .
\end{aligned}$$ | From the identical congruence (3) $(p=5)$, we can obtain
$$\begin{array}{ll}
x^{14} \equiv x^{6} \equiv x^{2}(\bmod 5), & x^{13} \equiv x^{5} \equiv x(\bmod 5) \\
x^{11} \equiv x^{3}(\bmod 5), & x^{9} \equiv x^{5} \equiv x(\bmod 5) \\
x^{6} \equiv x^{2}(\bmod 5) &
\end{array}$$
Therefore, the original congruence equat... | x \equiv 0,1,2(\bmod 5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,502 |
10. Let $a, b$ be positive integers, and there exist integers $x, y$ such that $a x + b y = 1$. Prove:
(i) $[a, b] = a b$;
(ii) $(a c, b) = (c, b)$. | 10. (i) From Example 3 of §2, we know that if $a|l, b| l$, then $a b \mid l$; (ii) If $d|a c, d| b$, then $d \mid a c x+b y c=c$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,503 |
Theorem 7 If $n \mid p-1$, then the necessary and sufficient condition for the congruence equation (22) to have a solution is
$$a^{(p-1) / n} \equiv 1(\bmod p)$$
and when there is a solution, the number of solutions is $n$. | Necessity: If $x_{0}$ is a solution of (22), then from $p \nmid a$ we know $p \nmid x_{0}$. This, together with Fermat's Little Theorem, leads to
$$a^{(p-1) / n} \equiv\left(x_{0}^{n}\right)^{(p-1) / n} \equiv x_{0}^{p-1} \equiv 1(\bmod p)$$
Sufficiency: If equation (23) holds, then we have
$$\begin{aligned}
x^{p-1}-1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,504 |
Theorem 8 If $n \nmid p-1$, then the necessary and sufficient condition for the congruence equation (22) to have a solution is that the congruence equation
$$x^{k} \equiv a(\bmod p), \quad p \nmid a$$
has a solution, where $k=(n, p-1)$, and the number of solutions of both is the same. That is, the necessary and suffic... | Proof: By Theorem 5 of Chapter 1, §3, there exist positive integers $r, s$ such that
$$k=r \cdot n-s \cdot(p-1)$$
If (22) has a solution $x=c$, then $x=c^{n / k}$ is a solution to (25); conversely, if (25) has a solution $x=e$, then by equation (27),
$$\left(e^{r}\right)^{n}=e^{k} \cdot e^{s(p-1)} \equiv a(\bmod p)$$
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,505 |
2. Find all solutions to the following congruence equations $f(x) \equiv 0(\bmod p)$, and express $f(x)$ in the form given by Theorem 1 of §8 and the previous problem:
(i) $f(x)=14 x^{5}-6 x^{4}+8 x^{3}+6 x^{2}-13 x+5, p=7$;
(ii) $f(x)=8 x^{4}+3 x^{3}+x+9, p=7$;
(iii) $f(x)=x^{7}+10 x^{6}+x^{5}+20 x^{4}+8 x^{3}-18 x^{2... | 2. $f(x)$ can be expressed in the following forms:
(i) $(x-1)(x+2)\left(x^{2}+2\right)+7\left(2 x^{5}-x^{4}+x^{3}+x^{2}-2 x+1\right)$;
(ii)
$$\begin{array}{l}
(x-1)\left(x^{3}+4 x^{2}+4 x+5\right)+7\left(x^{4}+2\right) \\
\quad=(x-1)^{2}\left(x^{2}-x+2\right)+7\left(x^{3}+x+1\right)
\end{array}$$
(iii)
$$\begin{array}{... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,508 |
3. Let prime $p \nmid a_{n}$, and the congruence equation $a_{n} x^{n}+\cdots+a_{0} \equiv 0(\bmod p)$ has exactly $n$ solutions: $x \equiv c_{1}, \cdots, c_{n}(\bmod p)$. Also set
$$\sigma_{1}=\sum_{i=1}^{n} c_{i}, \quad \sigma_{2}=\sum_{1 \leqslant i \neq j \leqslant n} c_{i} c_{j}, \quad \cdots, \quad \sigma_{n}=c_{... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Algebra | proof | Yes | Yes | number_theory | false | 740,509 |
4. Use Theorem 5 of § 8 to prove:
(i) $2 x^{3}-x^{2}+3 x+11 \equiv 0(\bmod 5)$ has 3 solutions;
(ii) $x^{6}-4 x^{5}+6 x^{4}+6 x^{3}+3 x^{2}-2 x+3 \equiv 0(\bmod 13)$ has 6 solutions. | 4. (i) The solution to the original congruence equation is the same as that of the congruence equation $x^{3}+2 x^{2}-x+3 \equiv 0(\bmod 5)$.
$$x^{5}-x=\left(x^{2}-2 x\right)\left(x^{3}+2 x^{2}-x+3\right)+5\left(x^{3}-5 x^{2}+5 x\right) .$$
(ii) $x \equiv 0(\bmod 13)$ is not a root. $x^{12}-1=\left(x^{6}-4 x^{5}+6 x^{4... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,510 |
5. 求下列同余方程的等价同余方程:
(i) $3 x^{11}+3 x^{8}+5 \equiv 0(\bmod 7)$;
(ii) $4 x^{20}+3 x^{13}+2 x^{7}+3 x-2 \equiv 0(\bmod 5)$;
(iii) $2 x^{15}-3 x^{10}+8 x^{6}+7 x^{5}+6 x^{3}+2 x-8 \equiv 0(\bmod 7)$;
(iv) $2 x^{17}+5 x^{16}+3 x^{14}+5 x^{12}+6 x^{10}+2 x^{9}+5 x^{8}+9 x^{7}+22 x^{6}$ $+3 x^{4}+6 x^{3}-5 x^{2}+12 x+3 \equiv... | 5. (i) $x^{5}+x^{2}-3 \equiv 0(\bmod 7)$;
(ii) $x^{4}-2 x^{3}-x+2 \equiv 0(\bmod 5)$;
(iii) $x^{6}-3 x^{4}+x^{3}+2 x-1 \equiv 0(\bmod 7)$;
(iv) $x^{10}+4 x^{9}-x^{8}-x^{6}+x^{4}+x^{3}+2 x-5 \equiv 0(\bmod 11)$. | null | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,511 |
Theorem 1 Let $p$ be a prime number. Then, for any $n$-variable integer coefficient polynomial $f\left(x_{1}, \cdots, x_{n}\right)$ not equivalent to zero modulo $p$, there must exist a unique $n$-variable integer coefficient polynomial $f^{*}\left(x_{1}, \cdots, x_{n}\right)$ such that (i) $f$ and $f^{*}$ are equivale... | Proof: When $n=1$, by Theorem 6 of §8, the conclusion holds (why). Assume the theorem holds for $n=k$. When $n=k+1$, repeatedly use Fermat's Little Theorem $x^{p} \equiv x(\bmod p)$ for each variable $x_{j}$ of $f$ until all exponents are less than $p$, and replace each coefficient $a$ of a monomial in $f$ with $a^{*} ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,517 |
Theorem 3 (Chevalley's Theorem) Let $n$ be a positive integer, $p$ a prime, and $f\left(x_{1}, \cdots, x_{n}\right)$ an $n$-variable integer polynomial with degree $d$ less than $n$. Then, if the congruence equation
$$f\left(x_{1}, \cdots, x_{n}\right) \equiv 0(\bmod p)$$
is solvable, it has at least two distinct solu... | Prove by contradiction. If the congruence equation (6) has only one solution \( x_{j} \equiv a_{j}(\bmod p), 1 \leqslant j \leqslant n \), then consider
\[ F\left(x_{1}, \cdots, x_{n}\right)=1-f\left(x_{1}, \cdots, x_{n}\right)^{p-1}. \]
By Fermat's Little Theorem, we have
\[ F\left(x_{1}, \cdots, x_{n}\right)=\left\{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,519 |
3. (i) 给出两个二元二次型 $f(x, y)$, 使得 $f(x, y) \equiv 0(\bmod 5)$ 仅有显然解 $x \equiv y \equiv 0(\bmod 5)$ ；
(ii) 给出两个三元三次型 $f(x, y, z)$, 使得 $f(x, y, z) \equiv 0(\bmod 2)$仅有显然解 $x \equiv y \equiv z \equiv 0(\bmod 2)$. | $\begin{array}{l}\text { 3. (i) } x^{2}+3 y^{2}, x^{2}-x y+y^{2} \text {. (ii) } x^{3}+y^{3}+z^{3}+x^{2} y+y^{2} z+z^{2} x+x y z \text {; } \\ x y z+x y(1+z)+y z+z x(1+y)+x(1+y)(1+z)+y(1+z)(1+x) \\ +z(1+x)(1+y) .\end{array}$ | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,523 |
4. Let $a \equiv d \equiv 4(\bmod 9), b \equiv 0(\bmod 3)$ and $c \equiv \pm 1(\bmod 3)$. Prove:
The indeterminate equation
$$a x^{3}+3 b x^{2} y+3 c x y^{2}+d y^{3}=z^{3}$$
has only the trivial solution $x=y=z=0$. | 4. Let's assume $(x, y, z)=1$. If there are non-trivial solutions, consider the congruence equation modulo 9, and prove that $(3, x y)=1$ and $z \equiv a x+b y(\bmod 3)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,524 |
12. Let $g, l$ be given positive integers. Prove:
(i) There exist integers $x, y$ such that $(x, y)=g,[x, y]=l$ if and only if $g \mid l$;
(ii) There exist positive integers $x, y$ such that $(x, y)=g, x y=l$ if and only if $g^{2} \mid l$. | 12. (i) Sufficiency: Let $l=c g$, take $x=g, y=c g$; (ii) Sufficiency: Let $l=d g^{2}$, take
$$x=g, \quad y=d g .$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,526 |
5. Prove: The indeterminate equation
$$(7 a+1) x^{3}+(7 b+2) y^{3}+(7 c+4) z^{3}+(7 d+1) x y z=0$$
has only the trivial solution $x=y=z=0$. | 5. Let's assume $(x, y, z)=1$, and note that $u^{3} \equiv 0, \pm 1(\bmod 7)$. If there are non-trivial solutions, consider the congruence equation modulo 7, and discuss the cases $(z, 7)=1,(z, 7)=7$ separately. | proof | Algebra | proof | Yes | Yes | number_theory | false | 740,527 |
13. Find all positive integer pairs $(a, b)$ that satisfy $(a, b)=10,[a, b]=100$. | 13. $(a / 10, b / 10)=1,[a / 10, b / 10]=10 . a / 10, b / 10$ can only take the values $1,2,5,10$, and satisfy the above conditions. Therefore, we get the following table:
\begin{tabular}{|c|cccc|}
\hline$a / 10$ & 1 & 10 & 2 & 5 \\
\hline$b / 10$ & 10 & 1 & 5 & 2 \\
\hline
\end{tabular} | (a, b) = (10, 100), (100, 10), (20, 50), (50, 20) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,537 |
Property U Let $a^{-1}$ be the inverse of $a$ modulo $m$, i.e., $a^{-1} a \equiv 1(\bmod m)$. We have
$$\delta_{m}\left(a^{-1}\right)=\delta_{m}(a)$$ | Prove that the sufficient and necessary condition for $a^{d} \equiv 1(\bmod m)$ to hold is $\left(a^{-1}\right)^{d} \equiv 1(\bmod m)$ immediately follows. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,541 |
Property VII Let $k$ be a non-negative integer, then
$$\delta_{m}\left(a^{k}\right)=\frac{\delta_{m}(a)}{\left(\delta_{m}(a), k\right)}$$
Moreover, in a reduced residue system modulo $m$, there are at least $\varphi\left(\delta_{m}(a)\right)$ numbers whose index modulo $m$ is equal to $\delta_{m}(a)$. | Let $\delta=\delta_{m}(a), \delta^{\prime}=\delta /(\delta, k), \delta^{*}=\delta_{m}\left(a^{k}\right)$. Equation (3) is to prove $\delta^{*}=\delta^{\prime}$. By definition, we know
$$a^{k \delta^{*}} \equiv 1(\bmod m), \quad a^{k \delta^{\prime}} \equiv 1(\bmod m)$$
Thus, by property II, we get
$$\delta\left|k \del... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,542 |
Property VIII $\delta_{m}(a b)=\delta_{m}(a) \delta_{m}(b)$ holds if and only if
$$\left(\delta_{m}(a), \delta_{m}(b)\right)=1$$ | Let $\delta^{\prime}=\delta_{m}(a), \delta^{\prime \prime}=\delta_{m}(b), \delta=\delta_{m}(a b), \eta=\left[\delta_{m}(a), \delta_{m}(b)\right]$.
Sufficiency We have
So $\delta^{\prime} \mid \delta \delta^{\prime \prime}$. From this and $\left(\delta^{\prime}, \delta^{\prime \prime}\right)=1$, it follows that $\delta... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,543 |
Property IX (i) If $n \mid m$, then $\delta_{n}(a) \mid \delta_{m}(a)$;
(ii) If $\left(m_{1}, m_{2}\right)=1$, then
$$\delta_{m_{1} m_{2}}(a)=\left[\delta_{m_{1}}(a), \delta_{m_{2}}(a)\right] .$$ | Proof (i) can be directly derived from Property II. From (i), we have $\delta^{*} \mid \delta_{m_{1} m_{2}}(a)$, where $\delta^{*}=\left[\delta_{m_{1}}(a), \delta_{m_{2}}(a)\right]$. On the other hand, it is clear that $a^{\theta^{*}} \equiv 1\left(\bmod m_{j}\right), j=1,2$. From this and $\left(m_{1}, m_{2}\right)=1$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,544 |
Property X Let $\left(m_{1}, m_{2}\right)=1$, then for any $a_{1}, a_{2}$, there must be an $a$ such that
$$\delta_{m_{1} m_{2}}(a)=\left[\delta_{m_{1}}\left(a_{1}\right), \delta_{m_{2}}\left(a_{2}\right)\right] .$$ | Consider the system of congruences
$$x \equiv a_{1}\left(\bmod m_{1}\right), \quad x \equiv a_{2}\left(\bmod m_{2}\right) .$$
By the Chinese Remainder Theorem (Theorem 1, §3, Chapter 4), this system of congruences has a unique solution:
$$x \equiv a\left(\bmod m_{1} m_{2}\right)$$
It is clear that $\delta_{m_{1}}(a)=... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,545 |
Property XII The necessary condition for the existence of a primitive root modulo $m$ is
$$m=1,2,4, p^{a}, 2 p^{a},$$
where $p$ is an odd prime. | Prove that when $m$ does not belong to the cases listed in formula (9), there must be
$$m=2^{\alpha}(\alpha \geqslant 3), \quad 2^{\alpha} p_{1}^{\alpha_{1} \cdots} p_{r}^{\alpha}(\alpha \geqslant 2, r \geqslant 1)$$
or
$$2^{\alpha} p_{1}^{\alpha_{1}} \cdots p_{r}^{\alpha}(\alpha \geqslant 0, r \geqslant 2),$$
where ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,549 |
1. Let $m=5,11,12,13,14,15,17,19,20,21,23,36,40,63$.
(i) List the exponent table modulo $m$;
(ii) If $m$ has a primitive root, find all primitive roots in the smallest positive residue system modulo $m$ and the smallest positive primitive root modulo $m$;
(iii) If $m$ does not have a primitive root, find all integers i... | 1. $m=5,11,12,13,14,15,17,19,20,21,23,36,40,63$ index table.
\begin{tabular}{|c|rrrc|}
\hline$a$ & -2 & -1 & 1 & 2 \\
\hline$\delta_{5}(a)$ & 4 & 2 & 1 & 4 \\
\hline
\end{tabular}
$\varphi(5)=\lambda(5)=4$. Primitive roots: 2,3.
\begin{tabular}{|c|rrrrrrrrrr|}
\hline$a$ & -5 & -4 & -3 & -2 & -1 & 1 & 2 & 3 & 4 & 5 \\
\... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,550 |
2. Find $\delta_{41}(10), \delta_{43}(7), \delta_{55}(2), \delta_{65}(8), \delta_{91}(11), \delta_{69}(4), \delta_{231}(5)$. | 2. In sequence: $5,6,20,4,12,11,30$.
| 5,6,20,4,12,11,30 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,551 |
3. Let $\lambda(m)$ be given by (7) of $\S 1$.
(i) Find all positive integers $m$ such that $\lambda(m)=1,2,3,4,5,6,7,8,12$;
(ii) When $(m, n)=1$, $\lambda(m n)=[\lambda(m), \lambda(n)]$;
(iii) Let $d$ be a given positive integer, and $m$ be the largest positive integer $n$ such that $\lambda(n)=d$. Prove that if $\lam... | 3. (i)
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$\lambda(m)$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 12 \\
\hline \multirow{3}{*}{$m$} & \multirow{3}{*}{1,2} & $4,8,3$ & \multirow{3}{*}{ none } & $16,5,10$ & \multirow{3}{*}{ none } & 7,14 & \multirow{3}{*}{ none } & 32,96 & $2^{a_{0}} 3^{a_{1}} 7^{a_{2}} 13$, \\
\hline &... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,552 |
6. Let $m \geqslant 1, n \geqslant 1$ and $(n, \varphi(m))=1$. Prove: When $x$ runs through the reduced residue system modulo $m$, $x^{n}$ also runs through the reduced residue system modulo $m$. | 6. If $\left(x_{1} x_{2}, m\right)=1, x_{1}^{n} \equiv x_{2}^{n}(\bmod m),\left(x_{1} x_{2}^{-1}\right)^{n} \equiv 1(\bmod m)$. Let the order of $x_{1} x_{2}^{-1}$ modulo $m$ be $\delta$, then $\delta \mid(n, \varphi(m))$, so $\delta=1$, i.e., $x_{1} \equiv x_{2}(\bmod m)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,555 |
7. Let prime $p>2$. Prove: $\delta_{p}(a)=2$ if and only if $a \equiv$ $-1(\bmod p)$. Does this conclusion hold for composite moduli? | 7. The sufficiency is obvious. If $a \neq 1(\bmod p), a^{2} \equiv 1(\bmod p)$, then it must be that $a \equiv -1(\bmod p)$, i.e., the necessity holds. The condition is not necessary for composite moduli. For example, $\delta_{12}( \pm 5)=2$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,556 |
8. Let $p$ be a prime, $\delta_{p}(a)=3$. Prove: $\delta_{p}(1+a)=6$.
| 8. From $\delta_{p}(a)=3$, we get $a \neq \pm 1(\bmod p), a^{2}+a+1 \equiv 0(\bmod p)$. Therefore, $1+a \neq 1(\bmod p),(1+a)^{2} \equiv 1+2 a+a^{2} \equiv a \not \equiv 1(\bmod p),(1+a)^{3} \equiv-1(\bmod p)$. Hence, $\delta_{p}(1+a)=6$ | 6 | Number Theory | proof | Yes | Yes | number_theory | false | 740,557 |
9. If $\delta_{m}(a)=m-1$, then $m$ is a prime.
untranslated text remains the same as requested. | 9. From $m-1=\delta_{m}(a) \mid \varphi(m)$, we deduce $\varphi(m)=m-1$, which completes the proof. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,558 |
15. Find all positive integer solutions \(a, b, c\) that satisfy \((a, b, c)=10,[a, b, c]=100\). | 15. $(a / 10, b / 10, c / 10)=1,[a / 10, b / 10, c / 10]=10 . a / 10, b / 10, c / 10$ can only take the values $1,2,5,10$ and satisfy the above two conditions. There are three possible scenarios: (i) $a / 10=b / 10=c / 10$, which is impossible; (ii) Two of the three are equal, and the other is uniquely determined by th... | 36 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,559 |
10. Let $p$ be a prime, $\delta_{p}(a)=h$. Prove:
(i) If $2 \mid h$, then $a^{h / 2} \equiv -1 \pmod{p}$;
(ii) If $4 \mid h$, then $\delta_{p}(-a)=h$;
(iii) If $2 \mid h, 4 \nmid h$, then $\delta_{p}(-a)=h / 2$. | 10. (i) From $p \nmid a^{h / 2}-1, p \mid a^{h}-1=\left(a^{h / 2}-1\right)\left(a^{h / 2}+1\right)$, we get the result.
(ii) From (i), we have $(-a)^{h / 2} \equiv a^{h / 2} \equiv-1(\bmod p)$. Let $h^{\prime}=\delta_{p}(-a)$. We have $h / 2=$ $\delta_{p}\left(a^{2}\right)=\delta_{p}\left((-a)^{2}\right)=h^{\prime} /\l... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,560 |
12. If the prime $p \equiv 3(\bmod 4)$, then $g$ is a primitive root modulo $p$ if and only if
$$\delta_{p}(-g)=(p-1) / 2$$ | 12. Necessity follows from part (iii) of question 10. If $\delta_{p}(-g)=(p-1) / 2$, then $g^{(p-1) / 2} \equiv-1$ $(\bmod p)$. From this and $\delta_{p}(g) /\left(\delta_{p}(g), 2\right)=\delta_{p}\left(g^{2}\right)=\delta_{p}\left((-g)^{2}\right)=\delta_{p}(-g) /\left(\delta_{p}(-g), 2\right)=$ $(p-1) / 2$, it follow... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,562 |
15. Let the $n$-th Fermat number be $F_{n}=2^{2^{n}}+1$. Prove:
(i) $\delta_{F_{n}}(2)=2^{n+1}$;
(ii) If a prime $p \mid F_{n}$, then $\delta_{p}(2)=2^{n+1}$;
(iii) A prime factor $p$ of $F_{n}$ satisfies $p \equiv 1\left(\bmod 2^{n+1}\right)$;
(iv) If $F_{n}$ is a prime and $n>1$, then 2 is not a primitive root modulo... | 15. (i) $F_{n} \mid 2^{2^{n+1}}-1$, so $\delta_{F_{n}}(2) \mid 2^{n+1}$. From this and $F_{n} \nmid 2^{2^{l}}-1, l \leqslant n$, it follows that $\delta_{F_{n}}(2)=2^{n+1}$.
(ii) $\delta_{p}(2) \mid \delta_{F_{n}}(2)=2^{n+1}$. Suppose $\delta_{p}(2)=2^{d}, d \leqslant n+1$, i.e., $p \mid 2^{2^{d}}-1, p \nmid 2^{2^{d-1}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,565 |
16. Let $p, q$ be prime numbers. Prove:
(i) If $q \equiv 1(\bmod 4), p=2 q+1$, then 2 is a primitive root modulo $p$;
(ii) If $q \equiv-1(\bmod 4), p=2 q+1$, then -2 is a primitive root modulo $p$;
(iii) If $q \equiv 1(\bmod 2), q>3, p=2 q+1$, then $-3,-4$ are both primitive roots modulo $p$; $\square$
(iv) If $q \equi... | 16. (i) $2^{2 q} \equiv 1(\bmod p)$. It suffices to prove that $2^{2} \not \equiv 1(\bmod p)$ and $2^{q} \not \equiv 1(\bmod p)$. The first equation holds since $p > 3$. If $2^{q} \equiv 1(\bmod p)$, then $2^{q+1} \equiv 2(\bmod p)$, so 2 is a quadratic residue modulo $p$, and $p \equiv \pm 1(\bmod 8)$. However, $p \eq... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,566 |
18. If $3<q \equiv 3(\bmod 4), p=2 q+1$ are all primes, then there are at least three consecutive integers that are primitive roots modulo $p$. Provide two specific examples to illustrate the conclusion. | 18. $\varphi(p)=2 q$. A quadratic residue modulo $p$ cannot be a primitive root. $p \equiv-1(\bmod 8)$, so -1 is a quadratic non-residue. Let $a \neq 1(\bmod p)$ be a quadratic non-residue. $a^{(p-1) / 2} \equiv a^{q} \equiv-1(\bmod p)$, from this and $\delta_{p}(a) \mid 2 q$ we deduce $\delta_{p}(a)=2 q$, meaning $a$ ... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,568 |
16. Find the least common multiple of the following arrays: (i) 198,252 ; (ii) 482,1689 . | 16. (i) $[198,252]=2[99,126]=18[11,14]=2^{2} \cdot 3^{2} \cdot 7 \cdot 11$, the last step uses part (i) of question 10; (ii) $[482,1687]=241[2,7]=2 \cdot 7 \cdot 241$. | 2 \cdot 7 \cdot 241 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,570 |
20. Let $n>1, a>b \geqslant 1$. Prove:
(i) The prime factors of $a^{n}-b^{n}$ must either be of the form $k n+1$, or be factors of $a^{n_{1}}-b^{n_{1}}$, where $n_{1} \mid n, n_{1}<n$;
(ii) The prime factors of $a^{n}+b^{n}$ must either be of the form $2 k n+1$, or be factors of $a^{n_{1}}+b^{n_{1}}$, where $n_{1}<n, n... | 20. (i) Follows from the previous problem.
(ii) Let $p \mid a^{n}+b^{n}$. If $p \mid a$ or $p=2$ then the conclusion is obviously true. If $2<p \nmid a$, then $p \mid c^{n}+1$, where $c \equiv a b^{-1}(\bmod p)$. Let $\lambda$ be the smallest positive integer $s$ such that $c^{s} \equiv-1(\bmod p)$, then $2 \lambda$ is... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,571 |
21. Let $m>1, c \geqslant 1, (a, m)=1$; and let $\delta_{m}\left(a^{c}\right)=d$. Try to determine the condition that $\delta_{m}(a)$ should satisfy and the number of possible values it can take. | 21. $\delta_{m}(a)=\left(\delta_{m}(a), c\right) \cdot d$.
Translates to:
21. $\delta_{m}(a)=\left(\delta_{m}(a), c\right) \cdot d$. | \delta_{m}(a)=\left(\delta_{m}(a), c\right) \cdot d | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,572 |
22. Let $m>1,(a b, m)=1$ and $\lambda=\left(\delta_{m}(a), \delta_{m}(b)\right)$. Prove:
(i) $\lambda^{2} \delta_{m}\left((a b)^{\lambda}\right)=\delta_{m}(a) \delta_{m}(b)$;
(ii) $\lambda^{2} \delta_{m}(a b)=\left(\delta_{m}(a b), \lambda\right) \delta_{m}(a) \delta_{m}(b)$. | 22. (i) $\delta_{m}\left(a^{\lambda}\right)=\delta_{m}(a) / \lambda, \delta_{m}\left(b^{\lambda}\right)=\delta_{m}(b) / \lambda$. Therefore, $\left(\delta_{m}\left(a^{\lambda}\right), \delta_{m}\left(b^{\lambda}\right)\right)=1$.
(ii) Using (i) and $\delta_{m}\left((a b)^{\lambda}\right)=\delta_{m}(a b) /\left(\delta_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,573 |
23. Let $q, p$ be odd primes, $p=4 q+1$.
(i) Prove: The congruence equation $x^{2} \equiv -1(\bmod p)$ has exactly two solutions, and both are quadratic non-residues modulo $p$; $\square$
(ii) Prove: Among all quadratic non-residues modulo $p$, except for the two solutions of the congruence equation in (i), all are pri... | 23. (i) $p \equiv 1(\bmod 4)$, so there are exactly two solutions $\pm x_{0} .\left(\frac{ \pm x_{0}}{p}\right)=\left(\frac{x_{0}}{p}\right) \cdot\left(\frac{x_{0}}{p}\right)=1$ if and only if $1 \equiv x_{0}^{(p-1) / 2} \equiv x_{0}^{2 q} \equiv x_{0}^{2}(\bmod p)$, so $\left(\frac{x_{0}}{p}\right)=-1$.
(ii) Let $a$ b... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,574 |
25. Let prime $p>2, \delta_{p}(a)=4$. Find the least positive residue of $(a+1)^{4}$ modulo $p$.
| 25. There must be $a^{4} \equiv 1(\bmod p), a^{2} \equiv-1(\bmod p), a^{3} \equiv-a(\bmod p)$. Therefore, $(1+a)^{4}=$ $a^{4}+4 a^{3}+6 a^{2}+4 a+1 \equiv-4(\bmod p)$. The smallest positive residue is $p-4$. | p-4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,576 |
27. Let prime $p>2, g$ be a primitive root of $p$. Prove: There exists a positive integer $k$, such that
$$g^{k+1} \equiv g^{k}+1(\bmod p)$$ | 27. There must be a $k$ such that $g^{h} \equiv g-1(\bmod p), 0 \leqslant h \leqslant p-2$, taking $k=p-1-h$ will suffice. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,578 |
28. Let $p>2$ be a prime, and $g$ be a primitive root of $p$. Prove: for any positive integer $k$, it is impossible to have
$$g^{k+2} \equiv g^{k+1}+1 \equiv g^{k}+2(\bmod p)$$
to hold. | 28. If there is such a $k$, then $g^{k+1}(g-1) \equiv 1(\bmod p), g^{k}(g-1) \equiv 1(\bmod p)$, and thus $g^{k}(g-1)^{2} \equiv 0(\bmod p)$. This contradicts the fact that $g$ is a primitive root. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,579 |
31. (i) Let the prime $p=2^{n}+1(n>1)$. Prove: 3 is a primitive root of $p$;
(ii) Let $m=2^{n}+1(n>1)$. Prove: $m$ is a prime if and only if
$$3^{(m-1) / 2} \equiv-1(\bmod m)$$ | 31. (i) As part of question 15 (vi). (ii) If $m$ is a prime, then $m=2^{2^{h}}+1, h \geqslant 1$. In the proof of question 15 (vi), it has been shown that 3 is a quadratic non-residue modulo $m$, so the necessity holds. If $3^{(m-1) / 2} \equiv -1(\bmod m)$, then $3^{2^{n^{n-1}}} \equiv-1(\bmod m), 3^{2^{n}} \equiv 1(\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,583 |
33. Let $m \geqslant 3$. Prove the following: the arithmetic sequence $1+\operatorname{lm}(l=0,1, \cdots)$ must contain infinitely many primes. (i) The original proposition is equivalent to the statement that the arithmetic sequence must contain at least one prime. (ii) Let $q$ be a prime, $\mathrm{q} \mid m^{m}-1$ and... | 33. (ii) Prove the necessity by contradiction. If $q^{r+1} \mid m^{m}-1$, then it must be that $q \mid m$, which is a contradiction; (iv) The prime factors of $A_{1}, A_{2}$ must be the prime factors of $m^{m}-1$. Let $q$ be a prime factor of $m^{m}-1$, $\delta_{q}(m)=h$. Then, $q \mid m^{s}-1\left(s \in S_{1}\right.$ ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,585 |
Theorem 3 Let $p$ be an odd prime. Then, for any $\alpha \geqslant 1$, there must be a primitive root modulo $p^{\alpha}$. In fact, there exists $\widetilde{g}$ such that for all $\alpha \geqslant 1, \tilde{g}$ is a common primitive root modulo $p^{\alpha}$ and modulo $2 p^{\alpha}$. | Prove the following in several steps:
(i) If $g$ is a primitive root modulo $p^{\alpha+1}(\alpha \geqslant 1)$, then $g$ is certainly a primitive root modulo $p^{\alpha}$. Let $\delta=\delta_{p^{e}}(g)$. By property III of § 1, we know $\delta \mid \varphi\left(p^{a}\right)$. From
$$g^{\delta} \equiv 1\left(\bmod p^{*}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,588 |
Example 1 Find the primitive root of $p=23$.
保留源文本的换行和格式，直接输出翻译结果。
(Note: The last sentence is a note for the translator and should not be included in the final translation. The correct output should be as follows:)
Example 1 Find the primitive root of $p=23$.
| Since the index of $a$ modulo $p$ must be a divisor of $p-1$, to find the index, we just need to compute the residue of $a^{d}$ modulo $p$, where $d \mid p-1$.
Here $p-1=22=2 \cdot 11$, its divisors $d=1,2,11,22$. First, find the index of $a=2$ modulo 23:
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 23) \\
2^{11} \equiv\le... | 5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,589 |
Example 2 Find a primitive root modulo 41. | Solve $41-1=40=2^{3} \cdot 5$, divisors $d=1,2,4,8,5,10,20,40$. Now we will find the orders of $a=2,3, \cdots$.
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 41), 2^{4} \equiv 16(\bmod 41), 2^{5} \equiv-9(\bmod 41) \\
2^{10} \equiv-1(\bmod 41), \quad 2^{20} \equiv 1(\bmod 41)
\end{array}$$
So $\delta_{41}(2) \mid 20$. Since... | 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,590 |
Example 3 Find a primitive root modulo 43. | Solve $43-1=42=2 \cdot 3 \cdot 7$. The divisors $d=1,2,3,7,6,14,21,42$. First, find the index of 2.
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 43), 2^{3} \equiv 8(\bmod 43), 2^{6} \equiv 8^{2} \equiv 21(\bmod 43) \\
2^{7} \equiv-1(\bmod 43), \quad 2^{14} \equiv 1(\bmod 43)
\end{array}$$
So $\delta_{43}(2)=14$. Next, find... | 3^{42} \equiv 1 + 2 \cdot 43 \pmod{43^2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,591 |
18. Let $n \geqslant 1$. Denote by $\varphi(n)$ the number of integers in the set $\{1,2, \cdots, n\}$ that are coprime to $n$. Prove:
(i) $\varphi(1)=\varphi(2)=1$;
(ii) When $n \geqslant 3$, $2 \mid \varphi(n)$;
(iii) When $n=p$ is a prime, $\varphi(p)=p-1$. | 18. (ii) $(d, n)=1 \Longleftrightarrow(n-d, n)=1$. Let $d_{1}, \cdots, d_{r}$ be all positive integers less than $n / 2$ and coprime to $n$, then $d_{1}, \cdots, d_{r}, n-d_{r}, \cdots, n-d_{1}$ are all positive integers not exceeding $n$ and coprime to $n$, where $n \geqslant 3$. (iii) Derive from the definition of pr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,592 |
4. Try to find a $g$, for all $\alpha \geqslant 1$, it is a primitive root of $p^{\alpha}, 2 p^{\alpha}$:
$$p=11,13,17,19,31,37,53,71$$ | 4. In sequence: $7,7,3,3,3,-5,3,7$.
| 7,7,3,3,3,-5,3,7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,598 |
5. Let $p$ be a prime, $k \geqslant 1$. Prove
$$1^{k}+2^{k}+\cdots+(p-1)^{k} \equiv\left\{\begin{array}{ll}
0(\bmod p), & p-1 \nmid k \\
-1(\bmod p), & p-1 \mid k
\end{array}\right.$$ | 5. Let $g$ be a primitive root modulo $p$. $1^{k}+\cdots+(p-1)^{k} \equiv \sum_{j=1}^{p-1} g^{j k}(\bmod p)$, and from this, along with $g^{k} \not \equiv 1(\bmod p)$ when $p-1 \nmid k$; $g^{k} \equiv 1(\bmod p)$ when $p-1 \mid k$, we can derive the desired conclusion. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,599 |
6. Let $p>2$ be a prime, and the standard prime factorization of $p-1$ is $q_{1}^{\beta_{1}} \cdots q_{r}^{\beta}$. Prove:
(i) For any $j(1 \leqslant j \leqslant r)$, there exists $a_{j}$ whose order modulo $p$ is $q_{j}^{\beta_{j}}$ (do not use the existence of a primitive root modulo $p$); $\square$
(ii) $a_{1} \cdot... | 6. (i) Let the different exponents that $1,2, \cdots, p-1$ can take be $\delta_{1}, \cdots, \delta_{s}$. We have $\tau=$ $\left[\delta_{1}, \cdots \delta_{s}\right] \mid p-1$, and let the prime factorization of $\tau$ be $p_{1}^{q_{1}} \cdots p_{t^{t}}^{q_{1}}$. There must be $a_{j}$ such that the exponent of $a_{j}$ m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,600 |
7. Let $1975 \leqslant n \leqslant 1985$, ask which of these $n$ have primitive roots. | 7. Only $n=1979$ is a prime number. | 1979 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,601 |
Theorem 1 (Division with Remainder) Let $a, b$ be two given integers, $a \neq 0$, then there exists a unique pair of integers $q$ and $r$, satisfying
$$b=q a+r, \quad 0 \leqslant r<|a| .$$
Moreover, $a \mid b$ if and only if $r=0$. | To prove uniqueness, if there are also integers $q^{\prime}$ and $r^{\prime}$ satisfying
$$b=q^{\prime} a+r^{\prime}, \quad 0 \leqslant r^{\prime}<|a|,$$ then from the above equation and Theorem 1 (vi) of § 2, we deduce $|a| \leqslant r^{\prime}-r$. This contradicts $r^{\prime}-r<|a|$. We will prove that there must be ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,603 |
9. Find all primitive roots $g, 1<g<p$ modulo $p: p=19,31,37,53,71$. | 9. Using the result from question 1, take $g$ as the smallest positive primitive root modulo $p$, and then find the smallest positive residue of $g^{k}$ modulo $p$, where $1 \leqslant k \leqslant p-1$, and $(k, p-1)=1$. The results are as follows:
- Modulo 19: $2,3,10,13,14,15$;
- Modulo 31: 3, $11,12,13,17,21,22,24$... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,604 |
11. Let $\lambda(m)$ be given by formula (7) of $\S 1$. Prove: there exists $a$ such that $\delta_{m}(a)=$ $\lambda(m)$, and there are at least $\varphi(\lambda(m))$ pairwise incongruent modulo $m$ values of $a$ with this property. | 11. For the modulus $2^{a_{0}}$, there exists a number $a_{0}$ with an index of $2^{\sigma_{0}}$. For the modulus $p_{j}^{\sigma_{j}}$, there exists a number $a_{j}$ with an index of $\varphi\left(p_{j}^{\circ}\right)$ (i.e., a primitive root). From this and property X of $\S 1$, the desired conclusion follows. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,606 |
Property II Let $g$ be a primitive root modulo $m$, $(a b, m)=1$, then we have
$$\gamma_{m, g}(a b) \equiv \gamma_{m, g}(a)+\gamma_{m, g}(b)(\bmod \varphi(m))$$ | Let $\gamma(c)=\gamma_{m, g}(c)$. We have
$$a b \equiv g^{\gamma(a)} \cdot g^{\gamma(b)} \equiv g^{\left.\gamma_{(a)}\right)+\gamma_{(b)}}(\bmod m)$$
From this and property I, we obtain the desired conclusion. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,608 |
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