\(\Leftrightarrow3^x=y^3+1=\left(y+1\right)\left(y^2-y+1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}y+1=3^m\\y^2-y+1=3^n\end{matrix}\right.\)
Gọi \(d=ƯC\left(y+1;y^2-y+1\right)\)
\(\Rightarrow y\left(y+1\right)-\left(y^2-y+1\right)⋮d\)
\(\Rightarrow2y-1⋮d\Rightarrow2\left(y+1\right)-\left(2y-1\right)⋮d\)
\(\Rightarrow3⋮d\Rightarrow\left[{}\begin{matrix}d=1\\d=3\end{matrix}\right.\)
- Với \(d=1\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y+1=1\\y^2-y+1=3^n\end{matrix}\right.\\\left\{{}\begin{matrix}y+1=3^m\\y^2-y+1=1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=0\Rightarrow x=0\\y=1\left(ktm\right)\end{matrix}\right.\)
- Với \(d=3\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y+1=3\\y^2-y+1=3^n\end{matrix}\right.\\\left\{{}\begin{matrix}y+1=3^m\\y^2-y+1=3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=2\Rightarrow x=2\\y=-1\left(ktm\right)\end{matrix}\right.\)
