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[数论] $n\mid2^n-8$

本帖最后由 hbghlyj 于 2020-1-19 20:32 编辑

红色的是尚未证明的
(1)有无穷多正整数满足$n\mid2^n-8$
证明:取n=3p,p是奇素数,由费马小定理$p\mid(8^{p-1}-1)$,而$8^{p-1}-1\equiv(-1)^{p-1}-1=0\pmod3$,所以$3p\mid8(8^{p-1}-1)$
(2)有无穷多正整数满足$n\mid2^n+2$
In[10]:= Select[Range[1000],Divisible[2^#-8,#]&]
Out[10]={1,2,3,4,8,9,15,21,33,39,51,57,63,69,87,93,111,123,129,141,159,177,183,195,201,213,219,237,248,249,267,291,303,309,315,321,327,339,381,393,399,411,417,447,453,471,489,501,519,537,543,573,579,591,597,633,669,681,687,693,699,717,723,731,753,771,789,807,813,819,831,843,849,879,921,933,939,951,993}
In[11]:= Differences[%]
Out[11]= {1,1,1,4,1,6,6,12,6,12,6,6,6,18,6,18,12,6,12,18,18,6,12,6,12,6,18,11,1,18,24,12,6,6,6,6,12,42,12,6,12,6,30,6,18,18,12,18,18,6,30,6,12,6,36,36,12,6,6,6,18,6,8,22,18,18,18,6,6,12,12,6,30,42,12,6,12,42}
去掉开始的5个,大多数n都是3的倍数
In[2]:= Do[ If[ PowerMod[ 2, n, n ] + 2 == n, Print[n]], {n, 2, 1500000000, 4} ]
2
6
66
946
8646
180246
199606
265826
383846
1234806
3757426
9880278
14304466
23612226
27052806
43091686
63265474
66154726
69410706
81517766
106047766
129773526
130520566
149497986
大多数个位都是6
A006517
有无穷多正整数满足$n\mid2^n+a$,求所有的整数a

本帖最后由 hbghlyj 于 2020-1-16 15:14 编辑

$n>1,n\mid2^n+2$,求证$n\equiv2\pmod4$
证明:假设n=4k,$4k\mid2^{4k}+2,2k\mid2^{4k-1}+1$,矛盾。
假设n为奇数(n>1),设$n=m2^k+1$,其中$k\geq1$,m为奇数,则n$\mid2^{m2^k}+1$,所以n的每一个素因子p$\mid2^{m2^k}+1$,$2^{k+1}\mid\delta_p(2)$,从而$2^{k+1}\mid p-1$。因此,$n=m 2^k+1$是形如$t 2^{k+1}+1$的素数之积,$2^{k+1}\mid n-1$,矛盾。

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本帖最后由 hbghlyj 于 2020-1-16 15:16 编辑

求所有$n\in\mathbb N+,n\mid2^{n-1}+1$
根据2#,只有n=1

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本帖最后由 hbghlyj 于 2020-1-19 20:33 编辑

(2)存在无穷多正整数n使得${\rm{n}}|{2^{\rm{n}}} + 2$
首先加强命题,存在无穷多n使得${\rm{n}}|{2^{\rm{n}}} + 2$且${\rm{n}} - 1|{2^{\rm{n}}} + 1$.n=2显然满足.下面证明将n换成$2^n+2$,证明它满足以上两式. ${\rm{n}} - 1\left| {{2^{\rm{n}}} + 1,\therefore {2^{{\rm{n}} - 1}} + 1} \right|{2^{{2^{\rm{n}}} + 1}} + 1,{2^{\rm{n}}} + 2|{2^{{2^{\rm{n}}} + 2}} + 2$
显然$2||n,\therefore {4^{\frac{{\rm{n}}}{2}}} + 1|{4^{{2^{{\rm{n}} - 1}} + 1}} + 1$所以存在无穷多n使得${\rm{n}}|{2^{\rm{n}}} + 2$

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本帖最后由 hbghlyj 于 2020-1-19 21:01 编辑

(3)数列$\{a_n\},a_n=2^n-3$含有无穷多个相互互素的整数
对k用数学归纳法.k为相互互素的整数的个数.$a_2=1,a_3=5,a_4=13$故k=3成立.假设k时成立,$2^{n_1}-3,2^{n_2}-3,\cdots,2^{n_k}-3$中所有素因子记为$p_1,p_2,\cdots,p_r$,$2^{p_i-1}\equiv1\pmod{p_i}$,令$n_{k+1}=(p_1-1)(p_2-1)\cdots(p_k-1),2^{n_{k+1}}-1\equiv0\pmod{p_1p_2\cdots p_k},2^{n_{k+1}}-3\equiv -2\pmod{p_1p_2\cdots p_k},(2^{n_{k+1}}-3,p_i)=1\Rightarrow (2^{n_{k+1}}-3,2^{n_i}-3)=1$
加强:数列$\{a_n\},a_n=2^n-3$含有无穷多项与先前的各项都互素,但不存在一项与之后的无穷项都互素

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本帖最后由 hbghlyj 于 2022-3-4 15:46 编辑

https://dms.umontreal.ca/~andrew/PDF/Notices1.pdf
第2页:
We define composite number $n$ to be a base 2 pseudoprime if $n$ divides $2^n-2$.

脚注3:
Malo proved, in 1903, that there are infinitely many odd composite base 2 pseudoprimes, by showing that if $n=ab$(with $a,b>1$) is such a number, then so is $n'=2^n-1$ and then we get the sequence $n$, $2^n-1,2^{2^n-1}-1,2^{2^{2^n-1}-1}-1,\dots$ of base 2 pseudoprimes by iterating this observation.
This is proved by observing that, since $a$ divides $n$ which divides $n'-1$, thus $x^a-1$ divides $x^n-1$, which divides $x(x^{n'-1}-1)=x^{n'}-x$, and so, in particular with $x=2$, we get that $2^a-1$ divides $2^n-1=n'$ which divides $2^{n'}-2$.

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本帖最后由 hbghlyj 于 2022-3-4 16:10 编辑

回复 6# hbghlyj
这个网站好生奇怪...每个网页上只放着一个PDF...网站就是互相链接的一些PDF
Andrew Granville (one of the three mathematicians who, in 1992, proved that there are infinitely many Carmichael numbers)

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