# 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. – Joe Silverman Aug 20 '15 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? – Igor Rivin Aug 20 '15 at 15:43
• @IgorRivin don't understand the question. The equation is easy to solve it. – individ Aug 20 '15 at 15:50
• If it is easy to solve, it should be easy to explain how you got the solution... – Igor Rivin Aug 20 '15 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 – joaopa Aug 21 '15 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. – Yemon Choi Sep 16 '15 at 21:26