<s>[INST] 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.
 [/INST] 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.
</s>