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

Đặt . Khi đó, ta có
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}}}{\dfrac{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\{ -1;0;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=0X=-1\left\{ \begin{aligned}
& y={{3}^{t}}+1 \\
& 2{{y}^{2}}={{4}^{t}}-1 \\
\end{aligned} \right.y={{3}^{t}}+1\Rightarrow y>1{{X}^{2}}+2{{y}^{2}}\le 2\Rightarrow 0\le 2{{y}^{2}}\le 2\Rightarrow 2{{y}^{2}}\le 2\Rightarrow {{y}^{2}}\le 1\Rightarrow \left| y \right|\le 1y>1X=-1X\in \left\{ 0;1 \right\}x\in \left\{ -1;0 \right\}y2{{\log }_{3}}\left( x+y+1 \right)={{\log }_{2}}\left( {{x}^{2}}+2x+2{{y}^{2}}+1 \right)$.