Now, today, my friend tell me this problem was posted by American Mathematical Monthly (Vol. 111, No. 2 Feb., 2004), p. 165 by Wu wei Chao ,and It is said that this problem is unsolved, until now. Find the diophantine-equation rational points


$$2\sqrt{\dfrac{q^2-1}{p^2-1}}=\sqrt{q}(\sqrt{p+1}-\sqrt{p-1})$$
or
$$\dfrac{1}{\sqrt{p-1}}-\dfrac{1}{\sqrt{p+1}}=\sqrt{\dfrac{q}{q^2-1}}$$



> 
> Question (1):
>  before I conjectured this diophantine-equations have only a finite number of solutions? 
> 
> 
> Question (2): Can we find all solution?
> 
> 
> 


Now about question 1, René has proved it. this equation have only a finite number of solutions.


But the Question 2, doesn't seem to be simple to solve.


* Can anyone complete the question $(2)$? Other words, How to prove this equation only has the following two solutions?
$$(p,q)=(\dfrac{5}{3},3),\rm{or},(\dfrac{5}{3},-\dfrac{1}{3})$$



Find all rational solutions of this diophantine-equation?
The number of rational solutions to your equation is finite. In short: your equation defines a genus $3$ curve, as follows from a straightforward computation and an application of Riemann--Hurwitz; finally, by Faltings' theorem, the number of rational points on a curve of genus $>1$ is finite.


One shows this as follows. Your equation defines a smooth projective curve $C$ whose function field $K$ is generated by $p$ and $q$. Take the second equation
$$
\frac{1}{\sqrt{p-1}} - \frac{1}{\sqrt{p+1}} = \sqrt{\frac{q}{q^2-1}}.
$$
Squaring it, we get
$$
\frac{q}{q^2-1} = \frac{2p}{p^2-1}-\frac{2}{\sqrt{p^2-1}}
$$
showing that $r:=\sqrt{p^2-1}$ is in $K$. Put differently, $C$ maps to the rational curve $C\_0$ given by
$$
p^2-r^2=1.
$$ 
In fact, since as we've seen $q/(q^2-1) = f$ with $f=2p/r^2-2/r$, the curve $C$ is a double cover of $C\_0$ (and therefore hyperelliptic). We determine the number of branch points. 


Rewriting the last equation, we find
$$
( qf ) ^2- (qf) - f^2 = 0,
$$
which is a quadratic (in variable $qf$) with discriminant $1+4f^2$, so $C \to C\_0$ is ramified over the points on $C\_0$ where $f = \pm i/2$. This last equality expands to $4(p-r)=\pm ir^2$, which gives $p+r=1/(p-r)=\mp 4i/r^2$, so $r=[(p+r)-(p-r)]/2=\mp (2i/r^2 + ir^2/8),$ giving four solutions for $r$ for each choice of sign, each corresponding to a unique value of $p$, so eight branch points $(p,r)$ in total. (Indeed, the eight values of $r$ are given by the zeros of $r^8 - 64r^6 + 32r^4 + 256
$, which sage factors as $(r^4 - 8r^3 + 16)(r^4 + 8r^3 + 16)$, so we indeed have $8$ distinct solutions.) Therefore $C$ has genus $\left\lfloor(8-1)/2\right\rfloor=3$ by Riemann--Hurwitz, which proves that your equation has finitely many solutions by Faltings' theorem.


