Questions in category: 初等数论 (Elementary Number Theory)

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## 1. 2089 和 8209

Posted by haifeng on 2024-05-19 22:29:09 last update 2024-05-19 23:04:42 | Answers (0) | 收藏

8209 is a prime

2089 is a prime

89208029 is a prime

82908029 is a prime

80209289 is a prime

80292809 is a prime

82092089 is a prime

28002899 is a prime

20882909 is a prime

22088909 is a prime

22098809 is a prime

92280809 is a prime

## 2. [Def] 离散对数

Posted by haifeng on 2024-01-23 08:58:44 last update 2024-01-23 09:05:54 | Answers (1) | 收藏

$a^n\equiv b\pmod p ,$

2024年九省联考的一道题目是这样的.

1.  求 $a^{p-1}\pmod p$,  其中 $a=2$, $p=11$.

2.  证明,  对任意 $b,c\in X_p$, 下式成立

$\log_a (bc \mod p)\equiv \log_a b+\log_a c\pmod{p-1}.$

3.  已知 $n=\log_a b$, 对于 $x\in X_p$, $k\in\{1,2,\ldots,p-2\}$. 设 $y_1=a^k$, $y_2=xb^k$, 证明:

$x\equiv y_2 y_1^{n(p-2)}\pmod p .$

## 3. 求正整数 $m$ 的所有可能的约数.

Posted by haifeng on 2024-01-09 09:53:19 last update 2024-01-09 10:08:47 | Answers (0) | 收藏

$d=p_1^{s_1}p_2^{s_2}\cdots p_n^{s_n},$

$\tau(m)=\prod_{i=1}^{n}(t_i+1)=(t_1+1)(t_2+1)\cdots(t_n+1).$

References:

[1] A. K. 苏什凯维奇  著,  叶乃膺  译 《数论初等教程》

## 4. 引理. 有 $d$ 个整数 $u_1,u_2,\ldots,u_d$, 则存在 $u_{i_1}+\cdots+u_{i_t}$ $(1\leqslant i_1 < \cdots < i_t\leqslant d)$, 使得 $d|(u_{i_1}+\cdots+u_{i_t})$.

Posted by haifeng on 2023-05-15 22:42:30 last update 2023-05-15 22:44:36 | Answers (1) | 收藏

Hint. 使用鸽巢原理.

References:

[1] Norbert Hegyvari, On the representation of integers as sums of distinct terms from a fixed set.  ACTA ARITHMETICA, XCII.2 (2000)

## 5. 双平方和问题的解数公式

Posted by haifeng on 2023-03-19 21:59:21 last update 2023-03-19 22:29:23 | Answers (1) | 收藏

$n=2^\gamma\cdot p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_s^{\alpha_s}\cdot q_1^{\beta_1}q_2^{\beta_2}\cdots q_t^{\beta_t}$

$\delta(n)=\biggl[\prod_{i=1}^{s}(\alpha_i+1)\biggr]\biggl[\prod_{j=1}^{t}\frac{(-1)^{\beta_j}+1}{2}\biggr],$

$\varepsilon(n)=\biggl[\frac{(-1)^{\gamma}+1}{2}\biggr]\biggl[\prod_{i=1}^{s}\frac{(-1)^{\alpha_i}+1}{2}\biggr]\biggl[\prod_{j=1}^{t}\frac{(-1)^{\beta_j}+1}{2}\biggr],$

$\xi(n)=(-1)^{\gamma}\cdot\Biggl[\dfrac{(-1)^{\prod_{i=1}^{s}(\alpha_i+1)}-1}{2}\Biggr]\biggl[\prod_{j=1}^{t}\frac{(-1)^{\beta_j}+1}{2}\biggr].$

$\delta(1)=1,\quad\varepsilon(1)=1,\quad\xi(1)=-1.$

$\begin{split} r_2(n)&=\mathrm{Card}\{(x,y)\in\mathbb{Z}^2\ :\ x^2+y^2=n\}\\ &=4\delta(n). \end{split}$

$\begin{split} r_2^{+}(n)&=\mathrm{Card}\{(x,y)\in\mathbb{Z}_{+}^2\ :\ x^2+y^2=n\}\\ &=\delta(n)-\varepsilon(n). \end{split}$

$\begin{split} r_{2*}^{+}(n)&=\mathrm{Card}\{(x,y)\in\mathbb{Z}_{+}^2\ :\ x^2+y^2=n\quad\text{且}\ x\leqslant y\}\\ &=\frac{\delta(n)+\xi(n)}{2}. \end{split}$

Remark:

## 6. 等幂和

Posted by haifeng on 2023-01-30 10:51:37 last update 2023-01-30 21:14:55 | Answers (0) | 收藏

\begin{aligned}
3^3 + ... + 5^3 &= 6^3,\\
3^3 + ... + 22^3 &= 40^3,\\
6^3 + ... + 30^3 &= 60^3,\\
6^3 + ... + 69^3 &= 180^3,\\
11^3 + ... + 14^3 &= 20^3,\\
15^3 + ... + 34^3 &= 70^3,\\
\end{aligned}

$\sum_{k=1}^{n^3}(\frac{n^4-3n^3-2n-2}{6}+k)^3=(\frac{n^5+n^3-2n}{6})^3$

>> solve_n3plus_until_M3_eq_p3(1,1000)
in> solve_n3plus_until_M3_eq_p3(1,1000)
out>
3^3 + ... + 5^3 = 6^3
3^3 + ... + 22^3 = 40^3
6^3 + ... + 30^3 = 60^3
6^3 + ... + 69^3 = 180^3
11^3 + ... + 14^3 = 20^3
11^3 + ... + 109^3 = 330^3
15^3 + ... + 34^3 = 70^3
34^3 + ... + 158^3 = 540^3
213^3 + ... + 365^3 = 1581^3
213^3 + ... + 555^3 = 2856^3
273^3 + ... + 560^3 = 2856^3
291^3 + ... + 339^3 = 1155^3
406^3 + ... + 917^3 = 5544^3
556^3 + ... + 654^3 = 2805^3
646^3 + ... + 798^3 = 3876^3

---------
Total: 15 solutions.

------------------------

>> solve_n3plus_until_M3_eq_p3(10000,11000)
in> solve_n3plus_until_M3_eq_p3(10000,11000)
out>

---------
Total: 0 solutions.

------------------------

References:

## 7. 解方程 $n^3+(n+1)^3+\cdots+(n+m)^3=p^3$

Posted by haifeng on 2022-09-04 13:40:46 last update 2022-09-04 15:52:39 | Answers (0) | 收藏

$n^3+(n+1)^3+\cdots+(n+m)^3=p^3$

>> solve_n3plus_until_M3_eq_p3(1,1000)
in> solve_n3plus_until_M3_eq_p3(1,1000)
out> 3^3 + ... + 5^3 = 6^3
3^3 + ... + 22^3 = 40^3
6^3 + ... + 30^3 = 60^3
6^3 + ... + 69^3 = 180^3
11^3 + ... + 14^3 = 20^3
11^3 + ... + 109^3 = 330^3
15^3 + ... + 34^3 = 70^3
34^3 + ... + 158^3 = 540^3
213^3 + ... + 365^3 = 1581^3
213^3 + ... + 555^3 = 2856^3
273^3 + ... + 560^3 = 2856^3
291^3 + ... + 339^3 = 1155^3
406^3 + ... + 917^3 = 5544^3
556^3 + ... + 654^3 = 2805^3
646^3 + ... + 798^3 = 3876^3

---------
Total: 15 solutions.

>> solve_n3plus_until_M3_eq_p3(1000,2000)
in> solve_n3plus_until_M3_eq_p3(1000,2000)
out>
---------
Total: 0 solutions.

Remark: 问题来源于“数学人交流群”.

## 8. 求方程 $x^3+y^3=z^n$ 的正整数解, 这里 $x,y\in[8,100]$.

Posted by haifeng on 2022-08-24 12:48:48 last update 2022-08-24 12:48:48 | Answers (1) | 收藏

## 9. 求解不定方程 $13x+19y=1$.

Posted by haifeng on 2022-08-10 21:24:33 last update 2022-08-24 13:09:33 | Answers (0) | 收藏

>> IndefiniteEquation(13,19)
in> IndefiniteEquation(13,19)
Solve the equation: 13*x+19*y = 1
19==1*13+6
13==2*6+1
1 2
test: 9==9*2-3*3

test: 9*3-3*2, 结果是 21, 个位数是1, 符合要求.

>>  IndefiniteEquation(19,21)
in> IndefiniteEquation(19,21)
Solve the equation: 19*x+21*y = 1
21==1*19+2
19==9*2+1
1 9
test: 9==1*9-9*0

in> IndefiniteEquation(13,19)
out>
Solve the equation: 13*x+19*y = 1
19==1*13+6
13==2*6+1
1 2
test: -1==9*2-3*3
13*3-19*2 == 1
x = 3+19t
y = -2-13t

-----------end---------

## 10. $2^{7^9}\equiv 2^7\pmod{279}$

Posted by haifeng on 2022-06-13 20:03:38 last update 2022-06-13 20:16:19 | Answers (0) | 收藏

$2^{7^9}\equiv 2^7\pmod{279}$

By haifeng

Question:

$x^{y^z}\equiv x^y\pmod{\overline{xyz}},$

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