I am trying to solve the following quadratic diophantine equation in $\mathbb Z[T]$:
$$((T+1)X+TY-1-Z)((T+1)X+TY-1+Z)=24XY$$
One has the following trivial solutions:
$(X,Y,Z)=(0,Y,\pm(1-TY))$, $(X,0,\pm(1-(T+1)X))$. Can one describe all the solutions of this equation (at least an algorithm to obtain all of them)?


Thanks in advance for any answer.



Quadratic Diophantine equation in $\mathbb Z[T]$
If the number $T$ is set by the problem statement. Then in the equation.


$$((T+1)x+Ty-1-z)((T+1)x+Ty-1+z)=24xy$$


The solutions can be written as.


$$x=\pm{s}(p((6-T-T^2)s\pm{T})+1)$$


$$y=p((T+1)s\mp1)$$


$$z=(pT\mp1)(s(T+1)\mp1)-p(s(T+1)\mp2)(T\mp{s}(T^2+T-6))$$


$$\*\*\*$$


Symmetric solution to the previous one.


$$x=p(Ts\mp1)$$


$$y=\pm{s}(p((6-T-T^2)s\pm(T+1))+1)$$


$$z=p(s(T^2+T-6)(\pm{Ts}-2)\pm(T+1))\mp{Ts}+1$$


$p , s - $ any polynomials.


