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素数公式,找两个整数解,破解1亿位大素数,难度不大

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发表于 2024-5-24 17:37 | 显示全部楼层 |阅读模式
已知:整数\(a>0\),\(b>0\),\(m>0\),\(t>0\),\(a=\frac{c-1}{2}\)
\(b=a^2-2c\),奇数\(c>9\),\(n>0\),素数\(p>0\)
方程\(\frac{a\times\left( a+1\right)\times\left( b-c+1\right)^n+m^2}{t}-c=0\),有整数解
方程\(\frac{a\times\left( a+1\right)\times\left( b+c-1\right)^n+m^2}{t}-c=0\),有整数解
求证:\(c=p\)
已知:整数\(a>0\),\(b>0\),\(m>0\),\(t>0\),\(a=\frac{c-1}{2}\)
\(b=a^2-2c\),奇数\(c>9\),\(n>0\),素数\(p>0\)
方程\(\frac{a\times\left( a+1\right)\times\left( b-c-1\right)^n+m^2}{t}-c=0\),有整数解
方程\(\frac{a\times\left( a+1\right)\times\left( b+c+1\right)^n+m^2}{t}-c=0\),有整数解
求证:\(c=p\)
已知:整数\(a>0\),\(b>0\),\(m>0\),\(t>0\),\(a=\frac{c-1}{2}\)
\(b=a^2-2c\),奇数\(c>9\),\(n>0\),素数\(p>0\)
方程\(\frac{a\times\left( a+1\right)\times\left( b-1\right)^n+m^2}{t}-c=0\),有整数解
方程\(\frac{a\times\left( a+1\right)\times\left( b+1\right)^n+m^2}{t}-c=0\),有整数解
求证:\(c=p\)
已知:整数\(a>0\),\(b>0\),\(m>0\),\(t>0\),\(a=\frac{c-1}{2}\)
\(b=a^2-2c\),奇数\(c>9\),\(n>0\),素数\(p>0\)
方程\(\frac{a\times\left( a+1\right)\times\left( 2b-2\right)^n+m^2}{t}-c=0\),有整数解
方程\(\frac{a\times\left( a+1\right)\times\left( 2b+2\right)^n+m^2}{t}-c=0\),有整数解
求证:\(c=p\)
已知:整数\(a>0\),\(b>0\),\(m>0\),\(t>0\),\(u>0\),\(y>0\)
\(\frac{c}{y}\ne u\),\(a=\frac{c-1}{2}\),\(b=a^2-2c\),奇数\(c>9\),\(n>0\),素数\(p>0\)
方程\(\frac{a\times\left( a+1\right)\times\left( by-y\right)^n+m^2}{t}-c=0\),有整数解
方程\(\frac{a\times\left( a+1\right)\times\left( by+y\right)^n+m^2}{t}-c=0\),有整数解
求证:\(c=p\)
已知:整数\(a>0\),\(b>0\),\(h>0\),\(k>0\),\(m>0\),\(t>0\)
\(u>0\),\(y>0\),\(c>m\),\(\frac{c}{3}\ne h\),\(\frac{c}{5}\ne k\),\(\frac{c}{y}\ne u\)
\(a=\frac{c-1}{2}\),\(b=a^2-2c\),奇数\(c>9\),\(n>0\),素数\(p>0\)
方程\(\frac{a\times\left( a+1\right)\times\left( by-y\right)^n+m^2}{t}-c=0\),有整数解
方程\(\frac{a\times\left( a+1\right)\times\left( by+y\right)^n+m^2}{t}-c=0\),有整数解
求证:\(c=p\)
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