Solve $32x^2 -y^2 = 448$ I am trying to find all integer solutions to the following equation:
$$32x^2 - y^2 = 448$$
This is what I have tried so far:
The equation describes a hyperbola, and so I try the usual trick of intersecting the curve with a line of rational slope to find rational solutions first.
Knowing the point (4,8) satisfies the equation, I solve the following system:
$$\left\{ 
\begin{array}{c}
32x^2 - y^2 = 448 \\ 
y = m(x - 4) + 8 \\
\end{array}
\right.$$
After a bunch of algebra, I get:
$$x = \frac{-4m^2+16m-128}{32-m^2}$$
$$y = \frac{8m^2-256m+256}{32-m^2}$$
Finally, substituting $m = \frac{u}{v}$, I get:
$$x = \frac{-4u^2+16uv-128v^2}{32v^2-u^2}$$
$$y = \frac{8u^2-256uv+256v^2}{32v^2-u^2}$$
Cool, with any choice of $u$ and $v$, I get a rational solution.
But since cancelling the denominators does not work, I do not know how to continue to get integer solutions only.
Is this perhaps not the right way to go?
Any help would be much appreciated.

Note that you can simplify and substitute in the following way:
$$\begin{array}{lll}
y^2 = 2^5(x^2-14)&&\text{substitute } y = 2^2z \\
z^2 = 2(x^2-14)&&\text{substitute } z = 2a\\
2a^2 = x^2-14&& \\
2(a^2+7) = x^2&&\text{substitute } x = 2b\\
a^2+7 = 2b^2&&\text{substitute } a = 2c+1\\
2(c^2+c + 2) = b^2&&\text{substitute } b = 2d\\
2d^2 = c^2 + c + 2&&\\
2(d-1)(d+1) = c(c+1)&&\\
\end{array}$$
so that $y = 16c+8$ and $x=4d$. This might work better for your approach.
