Find the solutions of the diophantine equation $(x^2-y^2)(z^2-w^2)=2xyzw$

Question

Let $x,y,z,w$ be postive integers. Find all solutions of: $$(x^2-y^2)(z^2-w^2)=2xyzw$$

This gives: $$\left(\dfrac{x}{y}-\dfrac{y}{x}\right)\left(\dfrac{z}{w}-\dfrac{w}{z}\right)=2$$

$$\left(p-\dfrac{1}{p}\right)\left(q-\dfrac{1}{q}\right)=2$$

Answer 1

Given,

$$\Big(p-\frac{1}{p}\Big)\,r=2\tag1$$ $$q-\frac{1}{q} = r\tag2$$

Solve for $p$ in $(1)$ and $q$ in $(2)$,

$$p=\frac{1\pm\sqrt{r^2+1}}{r}$$

$$q=\frac{r\pm\sqrt{r^2+4}}{2}$$

The discriminant should be a square. We have to find rational number $r=u/v$ such that,

$$u^2+v^2 = w_1^2\\ u^2+4v^2 = w_2^2$$

But $d=4$ is not a concordant form/number.

Thus, there is no solution $u,v$, and your original equation has no integer solutions.

Answer 2

There are NO solutions in rationals or integers.

The equation \begin{equation*} \left( p-\frac{1}{p} \right) \left( q-\frac{1}{q} \right) =2 \end{equation*} is equivalent to the quadratic \begin{equation*} p^2 + \frac{2q}{1-q^2} p - 1=0 \end{equation*}

For this to have rational solutions the discriminant must be a rational square, so there must exist $D \in \mathbb{Q}$ such that \begin{equation*} D^2=q^4-q^2+1 \end{equation*}

The quartic has a rational point when $q=0$, so is birationally equivalent to an elliptic curve. The curve is \begin{equation*} V^2=U^3-7U^2+12U=U(U-3)(U-4) \end{equation*} with \begin{equation*} q=\frac{V}{2U-6} \end{equation*}

Pari-gp gives $7$ finite torsion points, $(0,0)$, $(3,0)$, $(4,0)$, $(2, \pm 2)$ and $(6,\pm 6)$. Denis Simon's ellrank package give rank $0$, so the torsion points are the only rational points.

The torsion points give $q=0$, $q=\pm 1$ or $q$ undefined, all of which are unacceptable values for a positive integer solution.