素数公式找到了，完美的证明

太阳

已知:整数\(a＞0，c＞0，d＞0，\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)
\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)，素数\(k＞0，m＞0\)
求证:\(2^k-1＝m\)
证明:假设\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(4^k-1＝\left( 2^k-1\right)\times\left( 2^k+1\right)＝f_1\times g_1\times h_1\times\cdots\times t_n\)
因为\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
所以\(\left( 2^k-1\right)\)和\(\left( \frac{2^k+1}{3}\right)\)它们的素因子互相转换
必定有\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
如果\(\frac{a}{2^k-1}＝c\)，只有一种情况，\(\left( 2^k-1\right)\)是素数，\(\left( \frac{2^k+1}{3}\right)\)是合数
所以\(\left( 2^k-1\right)\)是素数，命题得证
(1):\(\left( 2^k-1\right)\)是素数，\(\left( \frac{2^k+1}{3}\right)\)是素数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
(2):\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
(3):\(\left( 2^k-1\right)\)是素数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
结论:\(\frac{a}{2^k-1}＝c，\frac{3a}{2^k+1}＝y\)
已知:整数\(a＞0，c＞0，d＞0，\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)
\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)，素数\(k＞0，m＞0\)
求证:\(\frac{2^k+1}{3}＝m\)
证明:假设\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(4^k-1＝\left( 2^k-1\right)\times\left( 2^k+1\right)＝f_1\times g_1\times h_1\times\cdots\times t_n\)
因为\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)
所以\(\left( 2^k-1\right)\)和\(\left( \frac{2^k+1}{3}\right)\)它们的素因子互相转换
必定有\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
如果\(\frac{3a}{2^k+1}＝c\)，只有一种情况，\(\frac{2^k+1}{3}\)是素数，\(\left( 2^k-1\right)\)是合数
所以\(\left( \frac{2^k+1}{3}\right)\)是素数，命题得证
(1):\(\left( 2^k-1\right)\)是素数，\(\left( \frac{2^k+1}{3}\right)\)是素数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
(2):\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
(3):\(\left( 2^k-1\right)\)是素数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)
结论:\(\frac{a}{2^k-1}＝c，\frac{3a}{2^k+1}＝y\)
命题重点:\(\left( 2^k-1\right)\)和\(\left( \frac{2^k+1}{3}\right)\)
\(2^k-1＝fg，\frac{2^k+1}{3}＝ht\)，素数\(f＞0\)，\(t＞0\)
(1):设\(f＞h\)，结论:\(t＞g\)，(2):设\(h＞f\)，结论:\(g＞t\)
\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(4^k-1＝\left( 2^k-1\right)\times\left( 2^k+1\right)＝f_1\times g_1\times h_1\times\cdots\times t_n\)
因为\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
所以\(\left( 2^k-1\right)\)和\(\left( \frac{2^k+1}{3}\right)\)它们的素因子互相转换
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)

太阳

已知:整数\(a＞0，c＞0，d＞0，\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)
\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)，素数\(k＞0，m＞0\)
求证:\(2^k-1＝m\)
证明:假设\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(4^k-1＝\left( 2^k-1\right)\times\left( 2^k+1\right)＝f_1\times g_1\times h_1\times\cdots\times t_n\)
因为\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
所以\(\left( 2^k-1\right)\)和\(\left( \frac{2^k+1}{3}\right)\)它们的素因子互相转换
必定有\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
如果\(\frac{a}{2^k-1}＝c\)，只有一种情况，\(\left( 2^k-1\right)\)是素数，\(\left( \frac{2^k+1}{3}\right)\)是合数
所以\(\left( 2^k-1\right)\)是素数，命题得证
(1):\(\left( 2^k-1\right)\)是素数，\(\left( \frac{2^k+1}{3}\right)\)是素数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
(2):\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
(3):\(\left( 2^k-1\right)\)是素数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
结论:\(\frac{a}{2^k-1}＝c，\frac{3a}{2^k+1}＝y\)
已知:整数\(a＞0，c＞0，d＞0，\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)
\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)，素数\(k＞0，m＞0\)
求证:\(\frac{2^k+1}{3}＝m\)
证明:假设\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(4^k-1＝\left( 2^k-1\right)\times\left( 2^k+1\right)＝f_1\times g_1\times h_1\times\cdots\times t_n\)
因为\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)
所以\(\left( 2^k-1\right)\)和\(\left( \frac{2^k+1}{3}\right)\)它们的素因子互相转换
必定有\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
如果\(\frac{3a}{2^k+1}＝c\)，只有一种情况，\(\frac{2^k+1}{3}\)是素数，\(\left( 2^k-1\right)\)是合数
所以\(\left( \frac{2^k+1}{3}\right)\)是素数，命题得证
(1):\(\left( 2^k-1\right)\)是素数，\(\left( \frac{2^k+1}{3}\right)\)是素数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
(2):\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
(3):\(\left( 2^k-1\right)\)是素数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)
结论:\(\frac{a}{2^k-1}＝c，\frac{3a}{2^k+1}＝y\)
命题重点:\(\left( 2^k-1\right)\)和\(\left( \frac{2^k+1}{3}\right)\)
\(2^k-1＝fg，\frac{2^k+1}{3}＝ht\)，素数\(f＞0\)，\(t＞0\)
(1):设\(f＞h\)，结论:\(t＞g\)，(2):设\(h＞f\)，结论:\(g＞t\)
\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(4^k-1＝\left( 2^k-1\right)\times\left( 2^k+1\right)＝f_1\times g_1\times h_1\times\cdots\times t_n\)
因为\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
所以\(\left( 2^k-1\right)\)和\(\left( \frac{2^k+1}{3}\right)\)它们的素因子互相转换
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
\(例1:k＝89，\frac{4^k-1}{3d}＝a，d取最大值，62020897\times18584774046020617\)
\(\frac{4^{89-1}}{3}＝179\times62020897\times18584774046020617\times618970019642690137449562111\)
\(a＝179\times618970019642690137449562111，\frac{a}{2^{89}-1}＝179\)
\(判断:2^{89}-1是素数\)

yangchuanju

导弹一颗：
2^83-1=9671406556917033397649407<25>=167*57912614113275649087721<23>
2^83+1=9671406556917033397649409<25>=3*499*1163*2657*155377*13455809771<11>
(2^83+1)/3=499*1163*2657*155377*13455809771<11>
4^83-1=3*499*1163*2657*155377*13455809771<11>*167*57912614113275649087721<23>
√[(2^k+1)/3]=√[(2^83+1)/3]=1795494969538.76
(2^k+1)/3＞d＞√[(2^k+1)/3]，d取最大值
d=13455809771
(4^83-1)/3d=167*57912614113275649087721<23>*499*1163*2657*155377=a
a/(2^k-1)=(167*57912614113275649087721<23>*499*1163*2657*155377)/(167*57912614113275649087721<23>)=499*1163*2657*155377=c
2^83-1=167*57912614113275649087721<23>不是素数！

太阳的素数公式被我的导弹命中了！

朱容仟

yangchuanju 发表于 2023-4-21 17:26
导弹一颗：
2^83-1=9671406556917033397649407=167*57912614113275649087721
2^83+1=9671406556917033397 ...

你就是专门踢馆太阳的，哈哈哈哈 你帮我看看我主页我的帖子，有些式子有没有反例

太阳

yangchuanju:网友应该相信素数公式存在吧

朱容仟

太阳 发表于 2023-4-21 21:33
yangchuanju:网友应该相信素数公式存在吧

太阳你看看我主题  发的帖子哥猜的经验公式

yangchuanju

太阳 发表于 2023-4-21 21:33
yangchuanju:网友应该相信素数公式存在吧

3楼的导弹是专门轰炸你的，d=134...火药未装足，小于平方根，还要乘以1163*2657*155377；相应的a和c都要除掉3因子。
根据你的命题条件，(2^k+1)/3＞d＞√[(2^k+1)/3]，d只与2^k+1有关联呀！
为什么硬拉进2^k-1的因子呀！

(2^k+1)/3＞d＞√[(2^k+1)/3]，d取最大值
d=13455809771*1163*2657*155377
(4^83-1)/3d=167*57912614113275649087721<23>*499=a
a/(2^k-1)=(167*57912614113275649087721<23>*499)/(167*57912614113275649087721<23>)=499=c
2^83-1=167*57912614113275649087721<23>不是素数！

太阳

命题重点:\(\left( 2^k-1\right)\)和\(\left( \frac{2^k+1}{3}\right)\)
\(2^k-1＝fg，\frac{2^k+1}{3}＝ht\)，素数\(f＞0\)，\(t＞0\)
(1):设\(f＞h\)，结论:\(t＞g\)，(2):设\(h＞f\)，结论:\(g＞t\)
\(\left( 2^k-1\right)\)是合数，\(\left( \frac{2^k+1}{3}\right)\)是合数
\(4^k-1＝\left( 2^k-1\right)\times\left( 2^k+1\right)＝f_1\times g_1\times h_1\times\cdots\times t_n\)
因为\(\frac{2^k+1}{3}＞d＞\sqrt{\frac{2^k+1}{3}}\)，\(d\)取最大值，\(\frac{4^k-1}{3d}＝a，\frac{a}{2^k-1}＝c\)
所以\(\left( 2^k-1\right)\)和\(\left( \frac{2^k+1}{3}\right)\)它们的素因子互相转换
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)
yangchuanju:请看:命题重点

太阳

yangchuanju 发表于 2023-4-21 22:54
3楼的导弹是专门轰炸你的，d=134...火药未装足，小于平方根，还要乘以1163*2657*155377；相应的a和c都 ...

素数公式的证明，写的很详细了

太阳

(2^k+1)/3＞d＞√[(2^k+1)/3]，d只与2^k+1是有关系的，d取最大值