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.
Best Answer
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 $$ (3,0) $$ $$ (6,3) $$ $$ (27,18) $$ $$ (150,105) $$ $$ (867,612) $$ $$ (5046,3567) $$