Finding shortcuts in solving Diophantine equation Well, I'm trying to solve the following Diophantine equation in the two unknowns $x,y$, for known values of $a,b,c,d,g \in \mathbb{Z}$ satisfying $a,b,g > 0$ and $c,d < 0$:
$$\frac{x(x+1)(xb+c)}{a}=y(yd+g)$$

Question: are there shortcuts to take for possible solutions? Or must I take a brute force approach in order to solve this?

 A: There are, in fact, $6$ obvious rational solutions
\begin{equation*}
(0,0) \hspace{0.5cm} (0,-g/d) \hspace{0.5cm} (-1,0) \hspace{0.5cm} (-1,-g/d) \hspace{0.5cm} (-c/b,0) \hspace{0.5cm} (-c/b,-g/d)
\end{equation*}
The presence of an $x^3$ term and a $y^2$ term immediately suggests an elliptic curve. Straightforward algebra shows the the equation is equivalent to
\begin{equation*}
-v^2+u^3+4ad(b+c)u^2+16a^2bcd^2u+16a^4b^2d^2g^2
\end{equation*}
with
\begin{equation*}
x=\frac{u}{4abd} \hspace{2cm} y=\frac{v-4a^2bdg}{8a^2bd^2}
\end{equation*}
The obvious rational points come from $u=0$, $u=-4abd$, $u=-4acd$. For varying values of the parameters, the behaviour of the curve is quite varied. If the torsion subgroup is small the number of rational points suggests the rank is greater than $0$.
For example $a=1$, $b=2$, $c=-3$ $d=-4$ and $g=5$ gives the curve $v^2=u^3+16u^2-1536u+25600$ which has no finite torsion points but rank $2$. One of the generators is $(32,160)$ which, when doubled, gives $(-16,224)$ leading to the solution $(1/2,3/2)$ of the original problem.
A: For $(a,b,c,d,g)=(1,2,-3,-4,5)$, with suitable conditions on $(x,y)$, there are more solutions given below:
$$(x,y)=(-18,-54)$$
$$(x,y)=\left(-\frac{13}8,\frac{65}{32}\right)$$
