Diophantine Equation on squares and cubes

Question

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?

Answer 1

(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 $(9,-6)$.

Answer 2

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

$$(a,b)=\{(-1,-1),(0,0),(8,-10),(9,-21),(9,-6)\}$$