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 |