<s>[INST] Find the solutions of the diophantine equation $(x^2-y^2)(z^2-w^2)=2xyzw$ Let $x,y,z,w$ be postive integers. Find all solutions of:
$$(x^2-y^2)(z^2-w^2)=2xyzw$$
This gives: 
$$\left(\dfrac{x}{y}-\dfrac{y}{x}\right)\left(\dfrac{z}{w}-\dfrac{w}{z}\right)=2$$
$$\left(p-\dfrac{1}{p}\right)\left(q-\dfrac{1}{q}\right)=2$$
 [/INST] 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.
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