<s>[INST] How to find integer solutions of the equation $x(x+9)=y(y+6)$ where x,y are integers? I found that the equation becomes $$(2x+9)^2-(2y+6)^2=45$$. And $45 = 9^2-6^2 = 7^2-{2^2}$. From here I found these solutions $(x,y)=(0,0), \ (0,-6), \ (-9,0),\ (-9,-6),\ (-1,-2),\ (-1, 4), \ (-8,-2), \ (-8,-4)$.
Does this equation has finite numbers of integer solutions and what would be better approach to solve such problems?
 [/INST] The equation in the title (which is different from the one in your text) can be written as
$$(x^2-y^2)+6(x-y)=0 \implies (x-y)(x+y+6)=0.$$
Now either $x=y$ or $x+y+6=0$.
So it has solutions as $\{(a,a) \, | \, a \in \Bbb{Z}\} \cup \{(a,-6-a)\, |\, a \in \Bbb{Z}\}$.
After you modified your original problem
The equation can be written as
$$(2x+2y+15)(2x-2y+3)=45.$$
Now factor $45=ab$ and solve the system
\begin{align*}
2x+2y+15&=a\\
2x-2y+3&=b
\end{align*}
To get
$$x=\frac{a+b-18}{4} \quad \text{ and } \quad y=\frac{a-b-12}{4}.$$
So test the factors $a,b$ of $45$ such that these quantities are integers.
\begin{align*}
ab&=45\\
a+b & \equiv 2 \pmod{4}\\
a-b & \equiv 0 \pmod{4}\\
\end{align*}
Hopefully you can take it from here and see for example $a=9,b=5$ is a solution.
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