Let $f$ and $S$ as follows.
$f(x)=x-\frac{1}{x}$, $S=f(\mathbb{Q}-\{0\})$.
For given $a,b\in\mathbb{Q}$, I want to find(or parametrize, if possible) $x\in\mathbb{Q}$ such that
$(f(x)+a\in S)\wedge (f(x)+b\in S)$
In other words, I want to find rational $x$s such that $x-\frac{1}{x}+a=r_a-\frac{1}{r_a}$ and $x-\frac{1}{x}+b=r_b-\frac{1}{r_b}$ for some rational $r_a$ and $r_b$.
My approach is as follows. Since $r-\frac{1}{r}$ is a parametrization of rational $y$ in $z^2=y^2+4$, this is equivalent to finding rational $x$s such that $(x^2+ax-1)^2+4x^2$ and $(x^2+bx-1)^2+4x^2$ are both rational squares. Using Mordell's_birational_transformation, we can convert quartic curves into elliptic curves. However, it is extremely hard to find common rational $x$ points of two elliptic curves.
Can anyone give different approaches? Thank you.
Edit:
Another approach: $a=y-\frac{1}{y}-(x-\frac{1}{x})$. We may generate rational $(x,y,z)$ on the surface $z-\frac{1}{z}=(1-\frac{b}{a})(x-\frac{1}{x})+\frac{b}{a}(y-\frac{1}{y})$. However, I don't know how to achieve this.
Edit: Any ideas for at least $a=-b$?
