素数公式找到了，试除法寻找1亿位大素数

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\(2^k-1\)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:\(\frac{a}{2^k-1}＝c，\frac{3a}{2^k+1}＝y\)

太阳

\(例1:k＝89，\frac{4^k-1}{3d}＝a，d取最大值，62020897\times18584774046020617\)
\(\frac{4^{89-1}}{3}＝179\times62020897\times18584774046020617\times618970019642690137449562111\)
\(a＝179\times618970019642690137449562111，\frac{a}{2^{89}-1}＝179\)
\(判断:2^{89}-1是素数\)

太阳

试除法，d取最大值，可以寻找梅霖素数，寻找素数公式花费很长的时间，终于找到了

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，\(\frac{4^k-1}{3d}＝a，\frac{3a}{2^k+1}＝c\)，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:\(\frac{a}{2^k-1}＝c，\frac{3a}{2^k+1}＝y\)

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:\(\frac{a}{2^k-1}＝c，\frac{3a}{2^k+1}＝y\)

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\(\left( \frac{2^k+1}{3}\right)\)它们素因子互相转换
结论:\(\frac{a}{2^k-1}\ne c，\frac{3a}{2^k+1}\ne y\)