\(...\Leftrightarrow2^x=3^y+1\)
Với y chẵn, đặt y=2k ( \(k\in Z^+\) )
\(3^y=3^{2k}=\left(3k\right)^2\equiv1\left(mod4\right)\)
\(\Rightarrow3^y+1\equiv2\left(mod4\right)\)
\(\Rightarrow2^x\equiv2\left(mod4\right)\) ; giả sử \(x\ge2\Rightarrow2^x\equiv0\left(mod4\right)\)
\(\Rightarrow x< 2\Rightarrow x=1\Rightarrow3^y=1\Rightarrow y=0\) ( không thỏa mãn )
Xét y lẻ
\(3^y=3^{26}+1\left(k\in Z^+\right)\)
Ta có
\(3^{2k+1}=3\cdot3^{2k}=3\cdot9k\)
\(\left(8a+1\right)\left(8a+1\right)=64a^2+16a+1\equiv1\left(mod8\right)\)
=> Tích của các số khi chia cho 8 dư 1 là một số chia cho 8 dư 1
\(\Rightarrow3\cdot9k\equiv3\left(mod8\right)\)
\(\Rightarrow3^y+1\equiv4\left(mod8\right)\)
\(\Rightarrow2^x\equiv4\left(mod8\right)\)
\(x\ge3\Rightarrow2^x\equiv0\left(mod8\right)\)
Nếu
\(\left[{}\begin{matrix}x=1\Rightarrow2^x=2\equiv2\left(mod8\right)\left(ktm\right)\\x=2\Rightarrow2^x=4\equiv4\left(mod8\right)\left(tm\right)\Rightarrow y=1\left(tm\right)\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
