1/ Với \(x=0\Rightarrow y=\pm2\)
\(x=1\Rightarrow y^2=5\Rightarrow\) ko có y nguyên thỏa mãn
Với \(x>1\Rightarrow VT\) lẻ \(\Rightarrow VP\) lẻ \(\Rightarrow y=2k+1\)
\(2^x+2=\left(2k+1\right)^2-1=4k\left(k+1\right)\)
\(\Leftrightarrow2^{x-1}+1=2k\left(k+1\right)\)
Do \(x>1\Rightarrow2^{x-1}\) chẵn \(\Rightarrow VT\) lẻ, mà VP chẵn \(\Rightarrow\) pt vô nghiệm
Vậy pt có nghiệm \(\left\{{}\begin{matrix}x=0\\y=\pm2\end{matrix}\right.\)
2/
- Nếu \(x\) lẻ \(\Rightarrow x=2k+1\Rightarrow VT=2.2^{2k}+57=2.4^k+57\)
Do \(4^k\equiv1\left(mod3\right)\Rightarrow2.4^k\equiv2\left(mod3\right)\Rightarrow\left(2.4^k+57\right)\equiv2\left(mod3\right)\)
Mà \(VP=y^2\), luôn có \(\left[{}\begin{matrix}y^2\equiv0\left(mod3\right)\\y^2\equiv1\left(mod3\right)\end{matrix}\right.\) \(\forall y\in Z\Rightarrow\) pt vô nghiệm
- Nếu x chẵn \(\Rightarrow x=2k\) pt trở thành:
\(57=y^2-\left(2^k\right)^2=\left(y-2^k\right)\left(y+2^k\right)\)
Do x; y nguyên dương \(\Rightarrow y+2^k\ge3\Rightarrow y+2^k=\left\{57;19;3\right\}\)
TH1: \(\left\{{}\begin{matrix}y+2^k=57\\y-2^k=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=29\\2^k=28\end{matrix}\right.\) (loại)
TH2: \(\left\{{}\begin{matrix}y+2^k=19\\y-2^k=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=11\\2^k=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=11\\x=6\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}y+2^k=3\\y-2^k=19\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=11\\2^k=-8\end{matrix}\right.\) \(\Rightarrow x=-6< 0\left(l\right)\)
