\(\Leftrightarrow n^3-5n=2\left(2^{m-1}-5\right)\)
\(\Leftrightarrow n\left(n^2-5\right)=2\left(2^{m-1}-5\right)\)
\(\Rightarrow\left[{}\begin{matrix}n⋮2\\n^2-5⋮2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}n=2k\left(k\in n,k>0\right)\\n^2-5=2k\left(k\in N\right)\end{matrix}\right.\)
-TH1: \(2k\left(4k^2-5\right)=2\left(2^{m-1}-5\right)\)
\(\Leftrightarrow4k^3-5k=2^{m-1}-5\)
Có: \(2^{m-1}-5⋮3\)\(\Rightarrow\left[{}\begin{matrix}k⋮3\\4k^2\equiv1\left(mod3\right)\end{matrix}\right.\)
Vậy \(2^{m-1}=4k^3-5k+5\)
...
-TH2:\(\Rightarrow2^{m-1}-5⋮2\Rightarrow m=1\)
=> Ko tìm đc m.
Nguyễn Việt Lâm Giải giúp TH1.
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