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求证题

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发表于 2023-5-5 07:05 | 显示全部楼层 |阅读模式
本帖最后由 太阳 于 2023-5-5 07:07 编辑

已知:奇数\(a>0\),\(c>0\),\(a=3c\),\(k>0\),\(\frac{4^a+2^a+1}{73}\ne k\),\(t>a^2m^2\)
\(\left( 4^a+2^a+1\right)\)的最小质因数是\(m\),\(\frac{4^a+2^a+1}{m}\)的最小质因数是\(t\)
\(\frac{4^a+2^a+1}{mt}\)的最大质因数是\(y\)
求证:\(y>\sqrt{\frac{4^a+2^a+1}{mt}}\)
已知:整数\(a>0\),\(c>0\),\(m>0\),\(t>0\),\(a=c^2\),\(a\ne m\)
方程\(\frac{\left( 2^a+1\right)^2+3}{\left( 2^m+1\right)^2+3}=t\),有唯一的正整数解,\(\sqrt{a}=m\)
\(\frac{\left( 2^a+1\right)^2+3}{\left( 2^m+1\right)^2+3}\)的最大质因数是\(y\)
求证:\(y>\sqrt{\frac{\left( 2^a+1\right)^2+3}{\left( 2^m+1\right)^2+3}}\)
已知:整数\(a>0\),\(c>0\),\(m>0\),\(t>0\)
\(a=\left( 3^c+1\right)^2\),方程\(\frac{\left( 2^a+1\right)^2+3}{\left( 2^m+1\right)+3}=t\)
求证:方程\(\frac{\left( 2^a+1\right)^2+3}{\left( 2^m+1\right)+3}=t\),最小正整数解,\(m=3^c+1\)
例1:\(a=81\),\(\frac{4^{81}+2^{81}+1}{73}\ne k\)
\(\left( 4^{81}+2^{81}+1\right)\)的最小质因数是487
\(\frac{4^{81}+2^{81}+1}{487}\)的最小质因数是16753783618801
\(\frac{4^{81}+2^{81}+1}{487\times16753783618801}\)的最大质因数是3712990163251158343>\(\sqrt{\frac{4^{81}+2^{81}+1}{487\times3712990163251158343}}\)
例2:\(a=100\),\(m\ne100\),方程\(\frac{\left( 2^a+1\right)^2+3}{\left( 2^m+1\right)^2+3}=t\),有唯一的正整数解,\(\sqrt{a}=m=10\)
\(\frac{\left( 2^{100}+1\right)^2+3}{\left( 2^{10}+1\right)^2+3}\)的最大质因数是170886618823141738081830950807292771648313599433>\(\sqrt{\frac{\left( 2^{100}+1\right)^2+3}{\left( 2^{10}+1\right)^2+3}}\)
\(a=16,100,6724,532900\),具备条件\(a=\left( 3^c+1\right)^2\)
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