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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? $$(p,q)=(\dfrac{5}{3},3),\rm{or},(\dfrac{5}{3},-\dfrac{1}{3})$$
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    $\begingroup$ Eliminating the radicals for your second equation yields $$q^2 p^4 + \left(-4 q^3 + 4 q\right) p^3 - 2 q^2 p^2 + \left(4 q^3 - 4 q\right) p + \left(4 q^4 - 7 q^2 + 4\right)=0,$$ which is the equation of a singular plane curve. There are packages such as Sage which will desingularize and compute the genus. (If you're interested in this subject, you should learn to use one.) If the genus is $g\ge2$, then there are finitely many solutions. If $g=1$, you can find a Weierstrass model and compute the rank. If $g=0$, there are lots of solutions. $\endgroup$ – Joe Silverman Jun 7 '15 at 16:59
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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.

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  • $\begingroup$ It's Nice answer! $r^8-64r^6+32r^4+256$ is some wrong,because when $p=\dfrac{5}{3}$(i,e,$r=4/3)$ is not such it $\endgroup$ – math110 Jun 8 '15 at 5:07
  • $\begingroup$ But I didn't use that polynomial to determine the rational solutions, I used it to get the genus of $C$, and so deduce finiteness of the solution set. (In fact, the zeros of the polynomial don't give rational solutions at all, since then by $p^2-r^2=1$ the corresponding $r$-values would have to have degree $\leq 2$ over $\mathbb{Q}$, whereas I found that they have degree $4$...) $\endgroup$ – RP_ Jun 8 '15 at 16:25
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Taking the equation in Joe Silverman's comment as the defining equation and asking Magma:

> A := AffineSpace(Rationals(), 2);

> C := Curve(A, q^2*p^4 + (-4*q^3+4*q)*p^3 - 2*q^2*p^2 + (4*q^3-4*q)*p + 4*q^4-7*q^2+4);

Genus(C);

3

> IsHyperelliptic(C);

true Hyperelliptic Curve defined by y^2 = x^8 + 4*x^7 + 6*x^6 + 4*x^5 + x^4 + 4*x^2 + 4*x + 1 over Rational Field ...

> _, C1, toC1 := $1;

> J := Jacobian(C1);

> RankBound(J);

3

it appears that your curve is (birational to) a hyperelliptic curve of genus 3. By Faltings, this implies that there are only finitely many rational points on it. Since the Mordell-Weil group (the group of rational points on its Jacobian variety) is likely to have rank 3, it is probably hard to actually provably determine all of the solutions. Searching for rational points on the hyperelliptic model produces 12 points of small height, leading to the 8 solutions $$(p,q) = (\pm 1, \pm 1), \pm(5/3, 3), \pm(5/3, -1/3)$$ where of course not all make sense with respect to the original equation(s).

It is very likely that these are all the solutions (points on higher genus curves with small coefficients tend to be small), but — as already mentioned — I expect this to be hard to prove.

ADDED later: The Galois group of the polynomial $f(x)$ defining the hyperelliptic curve is quite small (order 48). So "Elliptic Curve Chabauty" methods could apply. Any takers?

Also. I would like to mention that this confirms René's computations that exhibit the curve as a double cover of a conic isomorphic to the projective line ramified in eight points.

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