Solving the Diophantine Equation $Axy^2 + Bxz^2 + Cy + Dz = 0$

Question

While working on a problem, I obtained the system

$$ \begin{align} ub_3+a_4 t=1\\ ut(2p^2+q^2 )-a_4 b_3=r \end{align} \tag{1} $$

where $p,q,r$ are fixed constants and $u, b_3, a_4, t$ are unknown integers.

I tried eliminating a term by multiplying the equations by $t(2p^2 + q^2)$ and $b_3$ respectively

$$ \begin{align} utb_3 (2p^2+q^2 )+a_4 t^2 (2p^2+q^2 )=t(2p^2+q^2 ) \\ utb_3 (2p^2+q^2 )-a_4 b_3^2=rb_3 \end{align} \tag{2} $$

and subtracted to obtain

$$ a_4 t^2 (2p^2+q^2 )+a_4 b_3^2=t(2p^2+q^2 )-rb_3 \tag{3} $$

This is a diophantine equation of the form

$$ Axy^2 + Bxz^2 + Cy + Dz = 0 \tag{4} $$

Are there any methods for solving this type of equation? Please provide references to solution methods.

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In general, I am looking for methods to solve the system

$$ \begin{align} Awx + Byz = C\\ Dwz + Exy = F \end{align} \tag{5} $$

which was reduced to the form in Eqn. $(4)$.

Answer 1

$\begin{cases} ub_3+a_4 t=1\\ ut(2p^2+q^2)-a_4 b_3=r \end{cases} \\ \overset{\displaystyle\not{t}}{\implies} \bigg(2a_4 b_3+r\bigg)^2+(2p^2+q^2)\bigg(2b_3u-1\bigg)^2 = (2p^2+q^2)+r^2$

it is "quadratic Thue".

And like

$Axy^2 + Bxz^2 + Cy + Dz = 0 \implies A (2 B x z + D)^2 + B (2 A x y + C)^2 = A D^2 + B C^2$

Answer 2

$Axy^2+Bxz^2+Cy+Dz=0 \iff x(Ay^2+Bz^2)=-Cy-Dz $

In your situation, $A \geq 0$ and $B=1$, so you only have $Ay^2+Bz^2=0$ in a few specific cases you should be able to sort out. Outside of those cases: $x=\frac {-Cy-Dz}{Ay^2+Bz^2}$ and you can check that this yields all the solutions.

From that you have to find $y$ and $z$ so that $Ay^2+Bz^2$ divides $Cy+Dz$, but as far as I know there is no streamlined way to solve that.

Answer 3

►Your initial system has the form (we can change $-$ by $+$)$$xy+zw=1\\axw+zy=b$$ Four unknowns and only two equations so there are many solutions (maybe not integers). Taking $(y,w)$ fixed coprimes, for example $(4,5)$ the first equation has infinitely many integer solutions and you have to test if some of them are compatible with the second equation. This suggest a way to find out solutions perhaps.

►► Your general system $(5)$ leads, by elimination of one unknown, to an equation of the form $(4)$. You do have, by elimination of $w$ $$\begin{align} Awx + Byz = C\\ Dwz + Exy = F \end{align}\Rightarrow AEyx^2-DByz^2=AFx-DCz$$ and the same happens if you eliminate any other unknown: you get an equation like $(4)$.

►►►Let the equation $$Axy^2 + Bxz^2 + Cy + Dz = 0 \tag{4}$$ It is a cubic and find its integer solutions could be very hard or impossible, even when there exists integers solutions.