Parametrization of the rational $x$ where both $\sqrt{1+x^2}$ and $\sqrt{9+x^2}$ are rational

Question

I could parametrize the rational $x$ where $\sqrt{1+x^2}$ is rational as: $y^2-x^2=1,\ y=\frac{a^2+b^2}{a^2-b^2},\ x=\frac{2ab}{a^2-b^2}$. $\sqrt{9+x^2}$ is similar. However, I cannot find $x$ where "both" are rational. Is there any method except brute force? Thanks.

Answer 1

Here are my thoughts. This not a complete answer, in the end one has to find rational points on a cubic curve (elliptic curve), which I don't know much about.

Setting $t = x-y$, you obtain a rational parametrization of $y^2 - x^2 = 1$ as $$x(t) = \frac{t^2-1}{2t}, \quad y(t) = -\frac{t^2+1}{2t}, \quad t \neq 0.$$ Similarly for $s = x-y$ you obtain a parametrization of $y^2 - x^2 = 9$ as $$x(s) = \frac{s^2-9}{2s}, \quad y(s) = - \frac{s^2+9}{2s}, \quad s \neq 0.$$ If you now set $x(t) = x(s)$, you obtain the cubic equation $$\tag{*}\label{eq}st^2 - s = s^2 t - 9t.$$ So any rational point $(s,t) \in \mathbb Q^2$ which satisfies \eqref{eq} gives you a solution. There is an obvious solution $(s,t) = (0,0)$, but we cannot plug this into $x(s)$ or $x(t)$.