2次の不定方程式(つづき2)

\begin{align*}
&2x^2+3xy-2y^2-3x+4y-5=0 \land x,y\in\mathbb{Z}\\
\Longleftrightarrow~&x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x,y\in\mathbb{Z}\\
\Longleftrightarrow~&x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x,y\in\mathbb{Z} \\
&\land \exists k \in \{0,1,2,\cdots\}[25y^2-50y+49=k^2]\\
\Longleftrightarrow~&x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x,y\in\mathbb{Z} \\
&\land \exists k \in \{0,1,2,\cdots\}[(5y-k-5)(5y+k-5)=-24]\\
\Longleftrightarrow~&x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x,y\in\mathbb{Z} \\
&\land \exists k \in \{0,1,2,\cdots\}\left[\begin{cases}5y-k-5=-2\\5y+k-5=12 \end{cases} \lor \begin{cases}5y-k-5=-4\\5y+k-5=6\end{cases}\right.\\
&\lor \left. \begin{cases}5y-k-5=-6\\5y+k-5=4\end{cases} \lor \begin{cases}5y-k-5=-12\\5y+k-5=2\end{cases} \right]\\
\Longleftrightarrow~&x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x,y\in\mathbb{Z} \\
&\land \exists k \left[\begin{cases}5y-k-5=-2\\5y+k-5=12\\ k\in \{0,1,2,\cdots\}\end{cases} \lor \begin{cases}5y-k-5=-4\\5y+k-5=6 \\ k\in \{0,1,2,\cdots\}\end{cases}\right.\\
&\lor \left. \begin{cases}5y-k-5=-6\\5y+k-5=4 \\ k\in \{0,1,2,\cdots\}\end{cases} \lor \begin{cases}5y-k-5=-12\\5y+k-5=2 \\ k\in \{0,1,2,\cdots\}\end{cases} \right]\\
\Longleftrightarrow~&x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x,y\in\mathbb{Z} \\
&\land \exists k \left[\begin{cases}y=2\\k=7\\ k\in \{0,1,2,\cdots\}\end{cases} \lor \begin{cases}y=\frac{6}{5}\\k=5 \\ k\in \{0,1,2,\cdots\}\end{cases}\right.\\
&\lor \left. \begin{cases}y=\frac{4}{5}\\k=5 \\ k\in \{0,1,2,\cdots\}\end{cases} \lor \begin{cases}y=0\\k=7 \\ k\in \{0,1,2,\cdots\}\end{cases} \right]\\
\Longleftrightarrow~&x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x\in\mathbb{Z} \\
&\land \exists k \left[\begin{cases}y=2\\k=7\\ k\in \{0,1,2,\cdots\}\\y\in\mathbb{Z}\end{cases} \lor \begin{cases}y=\frac{6}{5}\\k=5 \\ k\in \{0,1,2,\cdots\}\\y\in\mathbb{Z}\end{cases}\right.\\
&\lor \left. \begin{cases}y=\frac{4}{5}\\k=5 \\ k\in \{0,1,2,\cdots\}\\y\in\mathbb{Z}\end{cases} \lor \begin{cases}y=0\\k=7 \\ k\in \{0,1,2,\cdots\}\\y\in\mathbb{Z}\end{cases} \right]\\
\Longleftrightarrow~&x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x\in\mathbb{Z} \\
&\land \exists k \left[\begin{cases}y=2\\k=7\\ k\in \{0,1,2,\cdots\}\\y\in\mathbb{Z}\end{cases} \lor \begin{cases}y=0\\k=7 \\ k\in \{0,1,2,\cdots\}\\y\in\mathbb{Z}\end{cases} \right]\\
\Longleftrightarrow~&x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x\in\mathbb{Z} \\
&\land \left( \exists k \begin{cases}y=2\\k=7\\ k\in \{0,1,2,\cdots\}\\y\in\mathbb{Z}\end{cases} \lor \exists k\begin{cases}y=0\\k=7 \\ k\in \{0,1,2,\cdots\}\\y\in\mathbb{Z}\end{cases} \right)\\
\Longleftrightarrow~&\left( x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x\in\mathbb{Z} \land \exists k \begin{cases}y=2\\k=7\\ k\in \{0,1,2,\cdots\}\\y\in\mathbb{Z}\end{cases}\right)\\
&\lor \left( x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4} \land x\in\mathbb{Z} \land \exists k\begin{cases}y=0\\k=7 \\ k\in \{0,1,2,\cdots\}\\y\in\mathbb{Z}\end{cases} \right)\\
\Longleftrightarrow~&\exists k \begin{cases}y=2\\k=7\\ k\in \{0,1,2,\cdots\}\\x,y\in\mathbb{Z}\\x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4}\end{cases} \lor \exists k\begin{cases}y=0\\k=7 \\ k\in \{0,1,2,\cdots\}\\x,y\in\mathbb{Z}\\x=\frac{-3(y-1)\pm \sqrt{25y^2-50y+49}}{4}\end{cases}\\
\Longleftrightarrow~&\exists k \begin{cases}y=2\\k=7\\ k\in \{0,1,2,\cdots\}\\x,y\in\mathbb{Z}\\x=\frac{-3\pm 7}{4}\end{cases} \lor \exists k\begin{cases}y=0\\k=7 \\ k\in \{0,1,2,\cdots\}\\x,y\in\mathbb{Z}\\x=\frac{3 \pm 7}{4}\end{cases}\\
\Longleftrightarrow~&\exists k \begin{cases}y=2\\k=7\\ k\in \{0,1,2,\cdots\}\\x,y\in\mathbb{Z}\\x=1\end{cases} \lor \exists k\begin{cases}y=0\\k=7 \\ k\in \{0,1,2,\cdots\}\\x,y\in\mathbb{Z}\\x=-1\end{cases}\\
\Longleftrightarrow~&\left(\begin{cases}(x,y)=(1,2)\\x,y\in \mathbb{Z}\end{cases} \land \exists k \begin{cases}k=7\\ k\in \{0,1,2,\cdots\}\end{cases}\right) \\
&\lor \left(\begin{cases} (x,y)=(-1,0)\\\\x,y\in \mathbb{Z}\end{cases} \land \exists k\begin{cases}k=7 \\ k\in \{0,1,2,\cdots\}\end{cases}\right)\\
\Longleftrightarrow~&\begin{cases}(x,y)=(1,2)\\x,y\in \mathbb{Z}\end{cases} \lor \begin{cases} (x,y)=(-1,0)\\x,y\in \mathbb{Z}\end{cases}\\
\Longleftrightarrow~&(x,y)=(1,2) \lor (x,y)=(-1,0)
\end{align*}

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