How can I handle the quartic diophantine equation in three variables $x$, $y$ and $z$ $$x^4-x^2=y^2-z^2$$ in general, i.e, does exists a (three-variable) parametric solution?
What I've tried is modifying Brahmagupta's identity, which yields a parametric solution to $x^2+y^2=u^2+v^2$
$$\left(ac-bd\right)^2 + \left(ad+bc\right)^2 = \left(ac+bd\right)^2 + \left(ad-bc\right)^2$$
Reorganizing, there are two options:
$$\left(ad+bc\right)^2 - \left(ad-bc\right)^2 = \left(ac+bd\right)^2 - \left(ac-bd\right)^2 \tag{1}$$
$$ \left(ad+bc\right)^2 - \left(ac+bd\right)^2 = \left(ac+bd\right)^2 - \left(ad+bc\right)^2 \tag{2}$$
For $(1)$, we can simplify as $\left(\alpha+\beta\right)^2 - \left(\alpha-\beta\right)^2 = \left(\gamma+\delta\right)^2 - \left(\gamma-\delta\right)^2$. Looking at the left side, we try to get
$$ \alpha+\beta = x^2 \quad \quad \alpha-\beta=x$$ Then, $$\alpha = \frac{1}{2}x(x + 1) \quad \quad \beta=\frac{1}{2} (x-1) x$$
By parity ($x$ or $(x + 1)$ is even, similar to $(x-1)$ and $x$ ), we can obtain the straightforward (but trivial) one-parameter solution(s)
$$a=\frac{n}{2} \quad d=n+1 \quad b=\frac{n}{2} \quad c=n-1$$
$$\left(n^2\right)^2 - \left(n\right)^2 = (n^2)^2-(n)^2$$
The results are similar if you move around the $\frac{1}{2}$ and the polynomial terms on the variables $a,b,c,d$ (example $a=\frac{n+1}{2}, d=n$).
For $(2)$, we have
$$\left(ad+bc\right)=x^2 \quad \quad \left(ac+bd\right)=x$$
but a continuation isn't clear. I've also tried finding solution tuples using a Python script, and the number of solutions is vast.
Note that $x^4-x^2\equiv0\pmod{4}$ for all $x$. Factoring the right hand side as $$x^4-x^2=(y+z)(y-z),\tag{1}$$ we have $y+z\equiv y-z\pmod{2}$, and so both are even. So this yields a factorization of $\frac{x^4-x^2}{4}$.
Conversely, for any factorization $\frac{x^4-x^2}{4}=uv$ we have a solution $$(y,z)=\left(u+v,u-v\right),$$ to equation $(1)$. So you are looking for a parametrization of all factorizations of numbers of the form $\frac{x^4-x^2}{4}$.
One ad-hoc way to do so is to take any integer $x$ and any divisor $u$ of $\frac{x^4-x^2}{4}$, and set $v:=\frac{x^4-x^2}{4u}$, and set $y:=u+v$ and $z=u-v$. That is to say $$\left\{\left(x,u+\frac{x^4-x^2}{4v},u-\frac{x^4-x^2}{4v}\right):\ u,x\in\Bbb{Z},\ u\text{ divides }\frac{x^4-x^2}{4}\right\}.$$