Problem 1531

If $a$ or $b$ is $0$, both must be $0$.

Consider the case $a, b \neq 0$. Let $g > 0$ be the G.C.D. of $a$ and $b$, and $a = g a'$, $b = g b'$. Substituting them into the given equation we get
$$ga'^2 (b'^2 - 4g a'^3) = b'^3.$$
Since $a'$ divides $b'^3$ and $a'$ and $b'$ are relatively prime, $a' = \pm 1$. Therefore $a$ divides $b$. Let $b = a c$ ($c \in \mathbb{Z}$). Substituting them into the last equation we get
$$a c^2 = 4a^2 + c^3.$$
Let $d > 0$ be the G.C.D of $a$ and $c$, and $a = dA$, $c = dC$. Substituting them into the last equation we get
$$d C^2 (A - C) = 4A^2.$$
Since $A$ and $C$ are relatively prime, $C^2$ must divide $4$. Therefore $C = -2, -1, 1, 2$.

Since $d = \frac{4A^2}{C^2(A-C)} = \frac{4}{C^2}(A+C) + \frac{4}{A-C}$ is an integer, $A-C$ must divide $4$. By $d > 0$, $A-C$ must be positive. Checking all the possible cases one by one, we get $(A, C) = (-1, -2), (3, -1), (1, -1), (5, 1), (3, 1), (2, 1), (3, 2)$.

From $b = a c$, we get all the possible cases, including $(a, b) = (0, 0)$, as $(a, b) = (0, 0), (-1, 2), (2, -4), (27, -243), (27, 486), (32, 512), (54, 972), (125, 3125)$. So the answer is $8$.