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.   
 A: 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 $x_m+ y_m\sqrt{d}$.
A: It is evident that equation.
$$x^2-3x=2y^2$$
This is a Pell equation. And we have to use solutions which are determined by the sequence. The next value from the previous one.
$$p_2=3p_1+4s_1$$
$$s_2=2p_1+3s_1$$
If we use the sequence where the first number. $(p_1 ;s_1) - (1 ; 1)$
Decisions will be.
$$x=6s^2$$
$$y=3ps$$
If we use the sequence where the first number.  $(p_1 ;s_1) - (3 ; 2)$
Decisions will be.
$$x=3p^2$$
$$y=3ps$$
A: Multiply through by $4$ and complete the square, $(2x-3)^2 - 8 y^2 = 9,$ so that $(2x-3)^2 + y^2 \equiv 0 \pmod 3.$ It follows that both $(2x-3)$ and $y$ are divisible by $3,$ therefore $x$ as well. Write $x=3t, y=3v.$ So far $(2t-1)^2 - 8 v^2 = 1.$ Let $u=2t-1,$ so $u^2 - 8 v^2 = 1.$ 
...if we write $$ x = \frac{3(1+u)}{2} $$ and $$ y = 3v, $$
we get $$ u^2 - 8 v^2 = 1.  $$ This gives a sequence of pairs, $u=1,3,17,99,...$ and $v=0,1,6,35...$
see https://oeis.org/A001541  and  https://oeis.org/A001109
These obey
$$ u_{n+1} = 3 u_n + 8 v_n,  $$
$$ v_{n+1} =  u_n + 3 v_n.  $$ By Cayley-Hamiton
$$ u_{n+2} = 6 u_{n+1} - u_n,  $$
$$ v_{n+2} = 6 v_{n+1} - v_n.  $$ Then
$$ x_{n+2} = 6 x_{n+1} - x_n - 6,  $$
$$ y_{n+2} = 6 y_{n+1} - y_n.  $$  The (non-negative) $(x,y)$ pairs begin
$$ (3,0) $$ 
$$ (6,3) $$ 
$$ (27,18) $$ 
$$ (150,105) $$ 
$$ (867,612) $$ 
$$ (5046,3567) $$ 
