Japan 2007

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

Consider the case $a,b\ne 0$. Let $g>0$ be the G.C.D. of $a$ and $b$, and $a=g{a}^{\prime }$, $b=g{b}^{\prime }$. Substituting them into the given equation we get $$g{a}^{\prime 2}(b^{\prime 2}-4g{a}^{\prime 3})=b^{\prime 3}.$$ Since $a^{\prime }$ divides $b^{\prime 3}$ and $a^{\prime }$ and $b^{\prime }$ are relatively prime, $a^{\prime }=\pm 1$. Therefore $a$ divides $b$. Let $b=ac$ ($c\in \mathbb{Z}$). 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 $b=ac$, 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 $8$.