Finding the integer solutions Find all integer solutions of
$$(a + b^2)(a^2 + b) = (a − b)^3.$$
Obviously $b = 0$ is one. But how to get other solutions?

On simplification, $$2b^3+(a^2-3a)b^2+b(a+3a^2)=0$$
If $$b\ne 0, 2b^2+(a^2-3a)b+(a+3a^2)=0 $$ which is a quadratic equation in $b$
As $b$ is integer, the discriminant must be perfect square.
So, $$(a^2-3a)^2-4\cdot2\cdot(a+3a^2)=a^4-6a^3-15a^2-8a=f(a)$$(say)
Clearly, $f(0)=0=f(-1)$ so, $a^4-6a^3-15a^2-8a=a(a+1)(a^2-7a-8)=a(a+1)^2(a-8)$
So, we need $a(a-8)$ to be perfect square $\implies a(a-8)\ge 0\implies a\le0$ or $a\ge 8$.
Now, $$a^2-8a=(a-4)^2-16\iff 4^2=(a-4)^2-(a^2-8a)$$
Observe that 
$n^2-(n-1)^2=2n-1$ is odd $\ne 16$
$n^2-(n-2)^2\le 16 \iff n \le5$
$n^2-(n-3)^2=6n-9$ is odd
$n^2-(n-4)^2=8n-16\le 16\iff n\le \frac{25}8$ 
$4^2=5^2-3^2,4^2-0^2$ 
So, we need 
$(i)a^2-8a=3^2$ or $(a-4)^2=5^2 $ both imply $a=-1,9$
or $(ii)a^2-8a=0\implies a=0,8$
There is no more solutions in integers. 
