数论小猜想

蔡家雄

设 p, p+2, p+6, p+8 是四生素数，

所谓 完美四生素数立方数，要很大，才可能有解，

p^3=a^3+b^3+c^3

(p+2)^3=a^3+b^3+c^3，a, b, c 可以是 合数，

(p+6)^3=a^3+b^3+c^3，a, b, c 可以是 合数，

(p+8)^3=a^3+b^3+c^3，a, b, c 可以是 合数，

需要四个式子同时有解，才叫 完美四生素数立方数，，，，

Treenewbee

10000以内有6组：
[191, [39, 146, 156]], [193, [25, 68, 190]], [197, [53, 58, 194]], [199, [6, 127, 180]]
[1871, [693, 809, 1785]], [1873, [6, 900, 1801]], [1877, [699, 743, 1803]], [1879, [300, 1087, 1746]]
[3251, [1028, 2355, 2724]], [3253, [525, 2128, 2910]], [3257, [297, 1926, 3014]], [3259, [72, 1315, 3186]]
[3461, [744, 941, 3426]], [3463, [135, 2316, 3076]], [3467, [248, 275, 3466]], [3469, [1723, 2367, 2859]]
[5651, 5653, 5657, 5659]
[9431, 9433, 9437, 9439]

Treenewbee

蔡家雄 发表于 2023-1-31 15:58
求解或反驳，
若 40k+31,  40k+33,  40k+37,  40k+39 是四生素数，

只有大约1/5的k满足条件

40k+31,  40k+33,  40k+37,  40k+39 是四生素数的k:
k={{191,1871,9431,18911,22271,101111,157271,187631,247991,294311,295871,299471,301991,346391,397751,412031,467471,500231,585911,597671,633791,666431,715151,734471,739391,768191,795791,833711,837071,857951,881471,889871,907391,930071,959471,978071,1027751,1043591,1093991,1100831,1121831,1146791,1156031,1157831,1163711,1210871,1228391,1233431,1257071,1285511,1298111,1322591,1381271,1451831,1468631,1524071,1584431,1588751,1678751,1681871,1748471,1798631,1834031,1837391,1857671,1904471,1954151,1959311,2007911,2075831,2079191,2083511,2159231,2176631,2202791,2223671,2266631,2290031,2303591,2362271,2418671,2546231,2552111,2576591,2594951,2822711,2927591,2952791,2989031,2990831}

1/(40k+33) 具有完全循环节长d＝ 40k＋32的k:

k={191,1871,157271,294311,397751,412031,500231,666431,715151,739391,768191,795791,833711,857951,1093991,1121831,1163711,1210871,1233431,1524071,1584431,1837391,1857671,1954151,2079191,2418671,2822711,2927591,2990831}

Treenewbee

40k+31={191,1871,157271,294311,397751,412031,500231,666431,715151,739391,768191,795791,833711,857951,1093991,1121831,1163711,1210871,1233431,1524071,1584431,1837391,1857671,1954151,2079191,2418671,2822711,2927591,2990831}

Treenewbee

四元素数的k：
{4,46,235,472,556,2527,3931,4690,6199,7357,7396,7486,7549,8659,9943,10300,11686,12505,14647,14941,15844,16660,17878,18361,18484,19204,19894,20842,20926,21448,22036,22246,22684,23251,23986,24451,25693,26089,27349,27520,28045,28669,28900,28945,29092,30271,30709,30835,31426,32137,32452,33064,34531,36295,36715,38101,39610,39718,41968,42046,43711,44965,45850,45934,46441,47611,48853,48982,50197,51895,51979,52087,53980,54415,55069,55591,56665,57250,57589,59056,60466,63655,63802,64414,64873,70567,73189,73819,74725,74770,76471,76909,77536,77581,80185,82723,83521,84361,85942,87805,87826,94951,96352,98788,99358}

1/(40k+33) 具有完全循环节长d＝ 40k＋32的k
{4,46,472,556,2527,6199,7357,7486,8659,9943,10300,11686,12505,15844,17878,18361,19204,20842,20926,22036,22246,22684,23251,23986,24451,25693,26089,27349,27520,28045,28669,28945,29092,30271,30835,31426,32137,32452,34531,36295,36715,38101,39718,43711,44965,45850,45934,46441,47611,48853,48982,50197,52087,53980,54415,55069,55591,57250,60466,63655,64414,64873,70567,73189,73819,74725,74770,76471,77581,80185,82723,83521,84361,85942,87805,87826,94951,98788}

蔡家雄

68711097^3=a^3+b^3+c^3

68711097^4=d^4+e^4+f^4  

Treenewbee

蔡家雄 发表于 2023-2-1 22:28
68711097^3=a^3+b^3+c^3

68711097^4=d^4+e^4+f^4

\[68711097^3=23642528^3+39896766^3+62800465^3\]
\[=35849268^3+37840894^3+60744593^3\]
\[=14565486^3+39896766^3+63644841^3\]
\[68711097^4=10409096^4+42878560^4+65932985^4\]

蔡家雄

素数倒数最大循环节长定理

设 k 为非负整数，

若 30k+7 和 120k+29 都是素数，

则 1/(120k+29) 具有最大循环节长d= 120k+28.


蔡氏完全循环节问题

由 10 是 17，257，65537 的原根，

设 m= 16,  256,  65536,

设 k 为正整数，

若 30k+1 和 m*(30k+1)+1 都是素数，

则 1/(m*(30k+1)+1) 具有最大循环节长d= m*(30k+1).

设 k 为非负整数，

若 30k+7 和 m*(30k+7)+1 都是素数，

则 1/(m*(30k+7)+1) 具有最大循环节长d= m*(30k+7).

蔡家雄

Mathematica 计算软件，

能判断 10 是否为 质数P的原根，P 不超过 120 位的数字，

能判断 一个不超过 5000 位数字的大奇数 是否为 质数，

蔡家雄

素数倒数最大循环节长定理

设 k 为非负整数，

若 30k+7 和 120k+29 都是素数，

则 1/(120k+29) 具有最大循环节长d= 120k+28.


蔡氏完全循环节问题

设 n>=3,

设 P 和 2^n*P+1 都是素数，

且 10^(2^n) -1 不能被 2^n*p+1 整除，

若 2^n*P+1 ≡ 17或33（mod  40），

则 10 是 2^n*P+1 的原根，

则 1/(2^n*p+1) 具有最大循环节长d= 2^n*p .


蔡氏完全循环节问题（特例）

由 10 是 17，257，65537 的原根，

设 m= 16,  256,  65536,

设 k 为正整数，

若 30k+1 和 m*(30k+1)+1 都是素数，

则 1/(m*(30k+1)+1) 具有最大循环节长d= m*(30k+1).

设 k 为非负整数，

若 30k+7 和 m*(30k+7)+1 都是素数，

则 1/(m*(30k+7)+1) 具有最大循环节长d= m*(30k+7).