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

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## 1. 目前已知的梅森素数(Mersenne primes)

Posted by haifeng on 2019-09-21 09:47:10 last update 2019-09-21 15:05:06 | Answers (0) | 收藏

No. $p$ $M_p$ isPrime
1 2 $2^2-1=3$ Y
2 3 $2^3-1=7$ Y
3 5 $2^5-1=31$ Y
4 7 $2^7-1=127$ Y
11 $2^{11}-1=2047$ N
5 13 $2^{13}-1=8191$ Y
6 17 $2^{17}-1=131071$ Y
7 19 $2^{19}-1=524287$ Y
23 $2^{23}-1=8388607$ N
29 $2^{29}-1=536870911$ N
8 31 $2^{31}-1=2147483647$ Y

No. $p$ $M_p$ Value
9 61 $2^{61}-1$ 2305843009213693951
10 89 $2^{89}-1$ 618970019642690137449562111
11 107 $2^{107}-1$ 162259276829213363391578010288127
12 127 $2^{127}-1$ 170141183460469231731687303715884105727
13 521 $2^{521}-1$
14 607 $2^{607}-1$
15 1279 $2^{1279}-1$
16 2203 $2^{2203}-1$
17 2281 $2^{2281}-1$
18 3217 $2^{3217}-1$
19 4253 $2^{4253}-1$
20 4423 $2^{4423}-1$

>> isprime(2^3217-1)
in> isprime(2^3217-1)
in> 2^3217-1

out> 259117086013202627776246767922441530941818887553125427303974923161874019266586362086201209516800483406550695241733194177441689509238807017410377709597512042313066624082916353517952311186154862265604547691127595848775610568757931191017711408826252153849035830401185072116424747461823031471398340229288074545677907941037288235820705892351068433882986888616658650280927692080339605869308790500409503709875902119018371991620994002568935113136548829739112656797303241986517250116412703509705427773477972349821676443446668383119322540099648994051790241624056519054483690809616061625743042361721863339415852426431208737266591962061753535748892894599629195183082621860853400937932839420261866586142503251450773096274235376822938649407127700846077124211823080804139298087057504713825264571448379371125032081826126566649084251699453951887789613650248405739378594599444335231188280123660406262468609212150349937584782292237144339628858485938215738821232393687046160677362909315071
2^3217-1 is a Mersenne number, we use Lucas-Lehmer test.
2^3217-1 is a Mersenne prime.

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

References:

https://www.mersenne.org/primes/

## 2. 证明 $8\overbrace{33\cdots3}^{2n}=1\overbrace{66\cdots 67}^{n}\times 4\overbrace{99\cdots 9}^{n}$.

Posted by haifeng on 2019-09-11 08:32:22 last update 2019-09-11 09:42:23 | Answers (0) | 收藏

$\dfrac{\frac{5}{6}}{\frac{3}{6}}=\frac{5}{3}\Rightarrow\ 0.833333\cdots=1.666666\cdots\times 0.5$

$0.833\cdots 33=1.66\cdots 67\times 0.499\cdots 9$

$8\overbrace{33\cdots3}^{2n}=1\overbrace{66\cdots 67}^{n}\times 4\overbrace{99\cdots 9}^{n}$

[Hint] 使用归纳法即可证明.

## 3. 关于阶乘

Posted by haifeng on 2019-07-28 23:12:54 last update 2019-07-28 23:34:22 | Answers (0) | 收藏

$4!+1==25==5^2$

$5!+1==121==11^2$

$7!+1==5041==71^2$

$11!+1==39916801$ is a prime

References:

Richard K. Guy, Unsolved Problems in Number Theory, D25 Equations involving factorial $n$.

## 4. 求方程 $x^{341-1}\equiv 1\pmod{341}$ 的解, 这里 $x$ 为正整数.

Posted by haifeng on 2019-07-10 11:32:12 last update 2019-07-10 11:32:47 | Answers (0) | 收藏

>> solve(x^340mod341==1,x,1,340)
in> solve(x^340@341~1,x,1,340)
ans>> x=1
ans>> x=2
ans>> x=4
ans>> x=8
ans>> x=15
ans>> x=16
ans>> x=23
ans>> x=27
ans>> x=29
ans>> x=30
ans>> x=32
ans>> x=35
ans>> x=39
ans>> x=46
ans>> x=47
ans>> x=54
ans>> x=58
ans>> x=60
ans>> x=61
ans>> x=63
ans>> x=64
ans>> x=70
ans>> x=78
ans>> x=85
ans>> x=89
ans>> x=91
ans>> x=92
ans>> x=94
ans>> x=95
ans>> x=97
ans>> x=101
ans>> x=108
ans>> x=109
ans>> x=116
ans>> x=120
ans>> x=122
ans>> x=123
ans>> x=125
ans>> x=126
ans>> x=128
ans>> x=139
ans>> x=140
ans>> x=147
ans>> x=151
ans>> x=153
ans>> x=156
ans>> x=157
ans>> x=159
ans>> x=163
ans>> x=170
ans>> x=171
ans>> x=178
ans>> x=182
ans>> x=184
ans>> x=185
ans>> x=188
ans>> x=190
ans>> x=194
ans>> x=201
ans>> x=202
ans>> x=213
ans>> x=215
ans>> x=216
ans>> x=218
ans>> x=219
ans>> x=221
ans>> x=225
ans>> x=232
ans>> x=233
ans>> x=240
ans>> x=244
ans>> x=246
ans>> x=247
ans>> x=249
ans>> x=250
ans>> x=252
ans>> x=256
ans>> x=263
ans>> x=271
ans>> x=277
ans>> x=278
ans>> x=280
ans>> x=281
ans>> x=283
ans>> x=287
ans>> x=294
ans>> x=295
ans>> x=302
ans>> x=306
ans>> x=309
ans>> x=311
ans>> x=312
ans>> x=314
ans>> x=318
ans>> x=325
ans>> x=326
ans>> x=333
ans>> x=337
ans>> x=339

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

>> 339^340mod 341
in> 339^340@341

out> 1

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

>> expmo(339,340,341)
in> expmo(339,340,341)
calculate: 339^340(mod 341)
out> 1

## 5. 一些特殊的素数

Posted by haifeng on 2019-06-26 12:44:35 last update 2019-07-06 17:52:05 | Answers (0) | 收藏

229

19
109
1009
10009
1000000009
1000000000000000009
10000000000000000000009

>> factorise(1000000000001)
in> factorise(1000000000001)
73*137*99990001

>> factorise(1000000000000000001)
in> factorise(1000000000000000001)
101*9901*999999000001

>> factorise(10000000000000000001)
in> factorise(10000000000000000001)
11*909090909090909091

>> factorise(100000000000000000001)
in> factorise(100000000000000000001)
73*137*1676321*5964848081

>> factorise(1000000000000000000001)
in> factorise(1000000000000000000001)
7*7*11*13*127*2689*459691*909091

10000000000000000000001 == 89*101*1056689261*1052788969

1000000000000000000000001 == 17*5882353*9999999900000001

>> factorise(10000000000000000000000001)
in> factorise(10000000000000000000000001)
11*251*5051*9091*78875943472201

1000000000000000000000000001 == 7*11*13*19*52579*70541929*14175966169

>> factorise(10000000000000000000000000000001)
in> factorise(10000000000000000000000000000001)
11*909090909090909090909090909091

>> factorise(1000000000000000000000009)
in> factorise(1000000000000000000000009)
53*193*5189*82633*189169*1205257

100000000000000000000000000000000001 == 11*9091*4147571*909091*265212793249617641

1000000000000000000000000000000000001 == 73*137*3169*98641*99990001*3199044596370769

10000000000000000000000000000000000001 == 11*7253*422650073734453*296557347313446299

13
1367
13367
13333333367
13333333333333367
133333333333333333367
13333333333333333333333367
133333333333333333333333333333333333367

in> factorise(1333333333333333367)
7*190476190476190481

## 6. [Def] Carmichael数 (卡米歇尔数)

Posted by haifeng on 2019-06-12 23:29:05 last update 2019-07-10 11:34:36 | Answers (0) | 收藏

$a^{n-1}\equiv 1\pmod n$

Thm. (Carmichael数的判别法) 设 $n$ 为合数, 则 $n$ 是 Carmichael 数当且仅当 $n$ 是奇数且对整除 $n$ 的每个素数 $p$, 满足

• $p^2\not\mid n$
• $p-1\mid n-1$

## 7. 伪素数

Posted by haifeng on 2019-06-12 23:16:46 last update 2019-06-19 14:01:19 | Answers (4) | 收藏

Fermat 小定理: 设 $p$ 是一个素数, 且 $0 < a < p$, 则有 $a^{p-1}\equiv 1\pmod p$.

10000 以内的pseudo prime 有:

341
561
645
1105
1387
1729
1905
2047
2465
2701
2821
3277
4033
4371
4681
5461
6601
7957
8321
8481
8911

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

Total: 21

[10000, 20000] 以内的伪素数

10261, 10585, 11305,12801,13741,13747,13981,14491,15709,15841,16705,18705,18721,19951

## 8. [欧拉-费马定理]

Posted by haifeng on 2019-04-06 20:02:47 last update 2019-04-06 20:02:47 | Answers (0) | 收藏

Thm. 若 $a$ 与 $m$ 互素, 则

$a^{\varphi(m)}\equiv 1\pmod m$

$a^{p-1}\equiv 1\pmod p$

Remark:

[俄] A. K. 苏什凯维奇 著, 叶乃膺 译《数论初等教程》 定理 58.

## 9. 求同余方程 $x^{205}\equiv 3\pmod {1024}$.

Posted by haifeng on 2019-04-06 19:42:31 last update 2019-04-06 20:19:01 | Answers (1) | 收藏

[讨论]

Note: $2^k$ 当 $k > 2$ 时没有元根(primitive root).

## 10. 设 $a,b,c$ 是三个素数, 求方程 $a(a+b)=12768+c$ 的所有解.

Posted by haifeng on 2019-04-04 09:29:04 last update 2019-04-04 09:35:34 | Answers (1) | 收藏

Remark:

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