Solving the nonlinear Diophantine equation $x^2-3x=2y^2$ How can I solve (find all the solutions) the nonlinear Diophantine equation $x^2-3x=2y^2$? 
I included here what I had done so far. Thanks for your help.
Note: The equation above can be rewritten into $x^2-3x-2y^2=0$ which is quadratic in $x$. By quadratic formula we have the following solutions for $x$.
\begin{equation}
x=\frac{3\pm\sqrt{9+8y^2}}{2}
\end{equation}
I want $x$ to be a positive integer so I will just consider:
\begin{equation}
x=\frac{3+\sqrt{9+8y^2}}{2}
\end{equation}   
From here I don't know how to proceed but after trying out values for $1\leq y\leq 1000$ I only have the following $y$ that yields a positive integer $x$.
\begin{equation}
y=\{0,3,18,105,612\}.
\end{equation}
Again, thanks for any help.   

I'll solve it in integers instead. $x^2-3x=2y^2$, by the quadratic formula, is equivalent to 
$$x=\frac{3\pm\sqrt{9+8y^2}}{2}$$
So i.e. the problem is equivalent to solving $8y^2+9=m^2$ in integers.
$8y^2\equiv m^2\pmod{3}$, so $(y,m)=\left(3y_1,3m_1\right)$ for some $y_1,m_1\in\Bbb Z$, because $h^2\equiv \{0,1\}\pmod{3}$ for any integer $h$.
The problem is equivalent to $m_1^2-8y_1^2=1$ with $m_1,y_1\in\Bbb Z$. This is a Pell's Equation and has infinitely many solutions given exactly by $\pm\left(3+\sqrt{8}\right)^n=x_n+y_n\sqrt{8}$ for $n\ge 0$ (because $(m_1,y_1)=(3,1)$ is the minimal non-trivial solution), so for example $\pm\left(3+\sqrt{8}\right)^2=\pm17\pm6\sqrt{8}$ and $(m_1,y_1)=(\pm 17,\pm 6)$ is another solution.
In general, Pell's Equation $x^2-dy^2=1$ with $d$ not a square has infinitely many solutions; which are given exactly by $\pm\left(x_m+ y_m\sqrt{d}\right)^n=x_n+y_n\sqrt{d}$, $n\ge 0$, where $(x_m,y_m)$ is the minimal solution, i.e. the solution with $x_m,y_m$ positive integers and minimal value of $x_m+ y_m\sqrt{d}$.
