Có bao nhiêu số nguyên x sao cho tồn tại số thực y thỏa mãn...

Đặt
$\Rightarrow \left\{ \begin{aligned}
& x+y={{3}^{t}} \\
& {{x}^{2}}+2{{y}^{2}}={{4}^{t}} \\
\end{aligned} \right.$$\Rightarrow {{4}^{t}}={{x}^{2}}+\dfrac{{{y}^{2}}}{1/2}\ge \dfrac{{{\left( x+y \right)}^{2}}}{1+\dfrac{1}{2}}=\dfrac{{{3}^{2t}}}{\dfrac{3}{2}}=\dfrac{2}{3}{{.9}^{t}}\Rightarrow {{3.4}^{t}}\ge {{2.9}^{t}}\Rightarrow {{\left( \dfrac{4}{9} \right)}^{t}}\ge \dfrac{2}{3}\Rightarrow t\le \dfrac{1}{2}\left\{ \begin{aligned}
& 0<x+y\le \sqrt{3} \\
& 0<{{x}^{2}}+2{{y}^{2}}\le 2 \\
\end{aligned} \right.{{x}^{2}}+2{{y}^{2}}\le 2\Rightarrow 0\le {{x}^{2}}\le 2\Rightarrow -\sqrt{2}\le x\le \sqrt{2}\Rightarrow x\in \left\{ 0;\pm 1 \right\}xx=0\left\{ \begin{aligned}
& y={{3}^{t}} \\
& 2{{y}^{2}}={{4}^{t}} \\
\end{aligned} \right.\Rightarrow {{2.9}^{t}}={{4}^{t}}\Rightarrow {{\left( \dfrac{4}{9} \right)}^{t}}=2\Leftrightarrow t={{\log }_{\dfrac{4}{9}}}2\Rightarrow y={{3}^{{{\log }_{\dfrac{4}{9}}}2}}x=1\left\{ \begin{aligned}
& y={{3}^{t}}-1 \\
& 2{{y}^{2}}={{4}^{t}}-1 \\
\end{aligned} \right.\Rightarrow 2.{{\left( {{3}^{t}}-1 \right)}^{2}}={{4}^{t}}-1 \left( * \right)t=0\left( * \right) \Rightarrow y=1x=-1\left\{ \begin{aligned}
& y={{3}^{t}}+1 \\
& 2{{y}^{2}}={{4}^{t}}-1 \\
\end{aligned} \right.y={{3}^{t}}+1\Rightarrow y>1\Rightarrow {{y}^{2}}>2\Rightarrow 2{{y}^{2}}>4\Rightarrow {{4}^{t}}-1>4\Rightarrow t>{{\log }_{4}}5>1x\in \left\{ 0;-1 \right\}y{{\log }_{3}}\left( x+y \right)={{\log }_{4}}\left( {{x}^{2}}+2{{y}^{2}} \right)$.