Diophantine with non-zero solutions Find all non-zero integers $a,b$ such that:
$(ab)^2 + ab - (a-b)^3 + a^3 + b^3 = 0$ 

I have tried using parity, got nowhere. Tried finding some bounding inequality, got nowhere. I took a small look at divisibility but it looked relatively bashy. Any ideas?
 A: As other answers have said, the problem reduces to solving $a^2b+3a^2+a-3ab+2b^2=0$. It's easy to see that $(a,b)=(-1,-1)$ is a solution. If we write $a=A-1$ and $b=B-1$ and let the algebraic dust settle, we wind up with the nicer-looking quadratic
$$2B^2-A(5-A)B+A^2=0$$
for which the quadratic formula gives
$$B={A(5-A)\pm A\sqrt{(5-A)^2-16}\over4}$$
This will turn out to be an integer (for integer $A$) if and only if the discriminant is a square, i.e., if and only if
$$(A-5-k)(A-5+k)=16$$
for an integer $k$, which we may assume is non-negative.  The factorizations of $16$ that give integer values for $A-5$ and $k$ are $2\cdot8$, $4\cdot4$, $(-4)\cdot(-4)$, and $(-8)\cdot(-2)$, giving $A-5=(2+8)/2=5$, $(4+4)/2=4$, $(-4-4)/2=-4$, and $(-8-2)/2=-5$, for $A=10$, $9$, $1$, and $0$, or $a=9$, $8$, $0$ and $-1$, with corresponding values $B=-20$, $-9$, $1$ and $0$, or $b=-21$, $-10$, $0$ and $-1$.  So there are three non-zero solutions:
$$(a,b)=(-1,-1),\quad (8,-9),\quad\text{and}\quad (9,-21)$$
A: Hint: Your equation can be written as
$$b(a^2b+3a^2-3ab+b^2+a)=0$$
and since $b\ne 0$ we get
$$a^2b+3a^2+a-3ab+2b^2=0$$
A: I have not calculated it all the way to the end, but there is a clear strategy that certainly works. This is probably not the shortest proof, try to find simplifications if you want. 
Expand all the brackets and simplify, you obtain: $2b^2+(a^2-3a)b+(a+3a^2)=0$. 
Solve for $b$: the discriminant $D= (a^2-3a)^2-8(a+3a^2)$ must be a perfect square. Clearly $a+3a^2> 0$, so $\sqrt{D}< a^2-3a$. 
It is not hard to provide lower estimations. Say if $\sqrt{D}\leq a^2-3a-20$, then this yields $41\geq (a-4)^2$. So $\sqrt{D}\leq a^2-3a-20$ is only possible for finitely many values of $a$, namely $-2\leq a\leq 10$. 
If this is not the case, then $\sqrt{D}= a^2-3a-i$ for some $1\leq i \leq 19$. 
So if you substitute all values $-2\leq a\leq 10$, you get concrete univariate equations that you can solve for $b$. 
Then you substitute all values $1\leq i \leq 19$, calculate the solutions of the equations $(a^2-3a-i)^2= (a^2-3a)^2-8(a+3a^2)$ (don't worry, these are 19 quadratic eqations, not quartic, as the equation simplifies to $-2i(a^2-3a)+i^2= -8(a+3a^2)$), which again yields only finitely many possible values for $a$, and then proceed as before. 
