# Quadratic Diophantine equation in $\mathbb Z[T]$

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)?

• First, it might be easier to first ask for solutions in $\mathbb C[T]$, then if those can be characterized, figure out which ones are in $\mathbb Z[T]$. Second, some motivation for your problem would be nice. (It's easy to make up lots of equations. For example, why 24?) Third, you previously asked a related, albeit easier, question (mathoverflow.net/questions/211874); it's helpful if you indicate this. Aug 20, 2015 at 22:24

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.

• Would you care to elaborate on how you get these? Aug 20, 2015 at 15:43
• @IgorRivin don't understand the question. The equation is easy to solve it. Aug 20, 2015 at 15:50
• If it is easy to solve, it should be easy to explain how you got the solution... Aug 20, 2015 at 15:55
• I checked your two solutions solution Both are correct. But you did not give any proof that there does not exist other solutions. Mathematics is not a matter of trust but proofs Aug 21, 2015 at 6:45
• Downvoting because the question asks for all solutions, and individ does not give any justification for his or her claim that this method produces all the solutions. Sep 16, 2015 at 21:26