Nếu \(x=2k\Rightarrow\left(3^h+1\right)\left(3^h+1\right)=2y\)
\(\Leftrightarrow\left\{{}\begin{matrix}3^k+1=2^a\\3^k-1=2^b\end{matrix}\right.\) \((a,b.là.a,b.thuộc.N;a>b)\)
Ta có
\(2^a-2^b=2\\ \Leftrightarrow2^b\left(2^{a-b}-1\right)=2\\ \Leftrightarrow\left\{{}\begin{matrix}2^{a-b}-1=1\\2^b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2^{a-b}=2\\b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3^h+1=4\\3^h-1=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3^h=3\\2^y=3^2-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}h=1\rightarrow x=2\\y=3\end{matrix}\right.\)
\(\left(+\right)x=2^k+1\\ \left(+\right)3^x-1=3\left(3^{2h}-1\right)+2=3\left(9^h-1^h\right)\)
Vì \(\left(9^h-1^h\right)⋮\left(9-1\right)\Rightarrow2^y:8\left(dư.2\right)\)
\(\Rightarrow2^y=2\Rightarrow y=1\\ 3^x-1=2\Rightarrow3^x=3\Rightarrow x=1\)
Vậy các cặp số nguyên dương \(\left(x,y\right)\) là \(\left(2;3\right)\)
