Bulgaria competition problems, 2022

Answer: $(0, 0)$, $(1, 0)$, $(-1, 0)$, $(9, 4)$, $(-9, -4)$, $(16, 6)$ and $(-16, -6)$. The right hand side can be written in the form $(a - 2b)(a - 3b)^2$. By setting $n = a - 3b$ we obtain $3a = n^2(3a - 6b) = n^2(2n + a)$, and thus $(3 - n^2)a = 2n^3$. When $n = 0$ it follows that $a = 0$ and $b = 0$; when $n = \pm 1$ we obtain $a = \pm 1$ and $b = 0$; when $n = \pm 2$ we obtain $a = \mp 16$ and $b = \mp 6$; when $n = \pm 3$ we obtain $a = \mp 9$ and $b = \mp 4$; when $n = \pm 4$ the left hand side is divisible by $13$ but the right hand side is not; when $n = \pm 5, \pm 6$ the left hand side is divisible by $11$ but the right hand side is not. On other hand for $|n| \ge 7$ since $3 - n^2 \ne 0$,
$$a = \frac{2n^3}{3 - n^2} = \frac{6n}{3 - n^2} - 2n$$
and $n^2 - 3 > 6|n|$ (it is equivalent to $|n|(|n| - 6) > 3$), the number $a$ is not an integer.