勾股数新公式

蔡家雄

蔡氏完全循环节问题

若 3^(2n)+2^(2n+1) 是素数，

则 10 是素数 3^(2n)+2^(2n+1) 的原根。

简记为  \(g(3^{2n}+2^{2n+1})=10\) .

Treenewbee

若 3^n+2^(n+1) 是素数，

则 3 是素数 3^n+2^(n+1) 的最小原根。
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{{1,3},{2,3},{3,3},{4,3},{5,5},{6,3},{9,5},{11,5},{12,3},{15,5},{17,5},{22,3},{32,3},{33,5},{35,5},{36,3},{46,3},{47,5},{59,5},{63,5},{80,5}}

Treenewbee

蔡家雄 发表于 2023-3-19 22:06
蔡氏完全循环节问题

若 3^(2n)+2^(2n+1) 是素数，

若 3^(2n)+2^(2n+1) 是素数，

则 10 是素数 3^(2n)+2^(2n+1) 的原根。

------------------------------------
{{{2,10},{3,12},{4,10},{5,13},{6,10},{9,11},{11,14},{12,10},{15,19},{17,11},{22,10},{32,10},{33,13},{35,14},{36,10},{46,10},{47,11},{59,11},{63,20},{80,10}}

蔡家雄

若 3^n+2^(n+1) 是素数，

则 5 是素数 3^n+2^(n+1) 的原根。

简记为  \(g(3^n+2^{n+1})=5\) .

蔡家雄

蔡氏完全循环节问题

若 3^(2n)+2^(2n+1) 是素数，

则 10 是素数 3^(2n)+2^(2n+1) 的原根。

简记为  \(g(3^{2n}+2^{2n+1})=10\) .

已知 2n=2, 4, 6, 12, 22, 32, 36, 46, 80, ...

Treenewbee

蔡家雄 发表于 2023-3-20 10:03
若 3^n+2^(n+1) 是素数，

则 5 是素数 3^n+2^(n+1) 的原根。

n<200时，结论正确：

{{1,5},{2,5},{3,5},{4,5},{5,5},{6,5},{9,5},{11,5},{12,5},{15,5},{17,5},{22,5},{32,5},{33,5},{35,5},{36,5},{46,5},{47,5},{59,5},{63,5},{80,5},{101,5},{154,5},{159,5},{173,5}}

Treenewbee

蔡家雄 发表于 2023-3-20 13:04
蔡氏完全循环节问题

若 3^(2n)+2^(2n+1) 是素数，

n<100时结论正确：

{{1, 10}, {2, 10}, {3, 10}, {6, 10}, {11, 10}, {16, 10}, {18, 10}, {23, 10}, {40, 10}, {77, 10}}

蔡家雄

蔡氏完全循环节问题

若 3^(2n)+2^(2n+1) 是素数，

则 10 是素数 3^(2n)+2^(2n+1) 的原根。

简记为  \(g(3^{2n}+2^{2n+1})=10\) .

有 2n=2, 4, 6, 12, 22, 32, 36, 46, 80, 154, 236, 250, 992, ......

Treenewbee

蔡家雄 发表于 2023-3-20 21:05
若 (2n+1)^2+4 是素数，

且 2n+1 不能被 3 整除，

{{2,2},{3,2},{6,2},{8,2},{17,2},{18,2},{23,2},{32,2},{33,2},{36,2},{42,2},{47,2},{48,3},{51,2},{57,2},{62,2},{68,2},{77,2},{81,2},{83,2},{96,2},{101,2},{107,2},{108,2},{116,2},{117,2},{122,2},{126,2},{132,2},{137,2},{138,2},{143,2},{146,2},{153,2},{156,2},{158,2},{173,2},{186,2},{192,2},{201,2},{203,2},{212,2},{213,2},{227,2},{231,2},{237,2},{243,2},{251,2},{266,2},{273,2},{288,2},{296,2},{302,2},{303,2},{306,2},{311,2},{332,2},{333,2},{338,2},{342,2},{351,2},{353,2},{356,2},{371,2},{372,2},{377,2},{381,2},{383,2},{392,2},{393,2},{402,2},{411,2},{413,2},{422,3},{423,2},{426,2},{441,2},{452,3},{467,2},{476,2},{483,2},{486,2},{491,2},{498,2},{513,3},{527,3},{528,2},{536,3},{537,2},{543,2},{552,2},{558,2},{561,2},{572,2},{578,2},{587,2},{588,2},{591,2},{597,2},{617,2},{618,2},{623,2},{627,3},{642,2},{656,2},{657,2},{662,2},{668,2},{672,2},{681,2},{683,3},{696,2},{702,2},{707,2},{711,3},{723,2},{728,2},{737,2},{753,2},{758,2},{761,2},{762,2},{771,2},{792,2},{797,2},{806,2},{813,2},{816,3},{818,2},{821,2},{822,2},{836,2},{852,2},{861,2},{876,2},{878,2},{881,2},{902,2},{926,2},{932,3},{941,2},{942,3},{948,2},{962,2},{977,2},{978,3},{993,2}}

果然是：非2，即3，没大于3的 .结论错误，那是因为n不够大：
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Treenewbee

若 (2n+1)^2+4 是素数，

且 2n+1 不能被 3 整除，

则 3 是素数 (2n+1)^2+4 的原根。
------------------------------------------------
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