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[数论] 存在性证明4

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hbghlyj posted 2019-7-25 16:35 |Read mode
任给相邻两奇数的平方,介于它们之间是否存在四个不同的正整数成比例.
举例:1<2<3<4<6<9,2:3=4:6
9<10<12<20<24<25,10:12=20:24
解:设正整数a,b,2a:3b=4a:6b,令2a<3b,则最小数为2a,最大数为6b
$2{\text{a}} > {\left( {2{\text{n}} - 1} \right)^2},6{\text{b}} < {\left( {2{\text{n}} + 1} \right)^2} \Leftrightarrow {\text{a}} \geqslant \left\lceil {\frac{{{{\left( {2{\text{n}} - 1} \right)}^2}}}{2}} \right\rceil ,{\text{b}} \leqslant \left\lfloor {\frac{{{{\left( {2{\text{n}} + 1} \right)}^2}}}{6}} \right\rfloor $
问题转化为对于任何正整数n,下面的不等式恒成立$2\left\lceil {\frac{{{{\left( {2{\text{n}} - 1} \right)}^2}}}{2}} \right\rceil  \leqslant 3\left\lfloor {\frac{{{{\left( {2{\text{n}} + 1} \right)}^2}}}{6}} \right\rfloor $
然而上面的不等式只对n=1,2成立。第一次尝试失败。

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