Diophantine Equation on squares and cubes Find all integer solutions to: $(a^2+b)(b^2+a)=(a-b)^3$ 
I've found some of the trivial cases, just finding difficulty proving the existence (or not existence) of others. Perhaps taking $mod5$ or something?
 A: (EDITED)
$$(a^2+b)(b^2+a)-(a-b)^3 = b(a^2 b+3 a^2-3 a b+2 b^2+a)$$
The factor $b$ means $a=arbitrary, b=0$ are solutions.
The curve $a^2 b+3 a^2-3 a b+2 b^2+a=0$ has genus $0$ and rational parametrization
$$ a = {\frac {-2{s}^{2}}{ \left( 2\,s+1 \right)  \left( s+1 \right) }}, b = -{
\frac { \left( 4\,s+1 \right) s}{(2s+1)^2}}
$$
If $s = S/T$ with $S, T$ coprime integers, we get
$$ \eqalign{a &=  -1 + \frac{2T}{S+T} - \frac{T}{2S+T}\cr
            b &= -1 + \frac{3T}{2(2S+T)} - \frac{T^2}{2(2S+T)^2} \cr}$$
If $2S+T$ is divisible by any prime, this can't be an integer.  So the only possible cases are: 
 $2S+T=\pm 1$, which leads to the integer solutions $(a,b) = (-1,-1), (0,0), (8,-10), (9,-21)$.
EDIT: Oops, you could also have $2S+T = \pm 2$ if $T$ is divisible by $4$.
This leads to the additional solution $(9,-6)$.
A: If $b\neq 0$ we have to solve a quadratic equation for $b$
$$a+3 a^2-3 a b+a^2 b+2 b^2=0$$
Since the discriminant must be a perfect square (a necessary condition), we get
$(-8+a) a (1+a)^2$ and therefore $a(a-8)$ must be a perfect square, so
$$a(a-8)=m^2$$
for some $m$. Rewriting this gives 
$$(a-4-m)(a-4+m)=16$$
There are finitely many possible values for $a$, namely
$$a=-1,0,8,9$$
and 
$$(a,b)=\{(-1,-1),(0,0),(8,-10),(9,-21),(9,-6)\}$$
