Find all possible relationships between $a$ and $b$ 
Find all possible relationships between $a\in \mathbb{R}$ and $b \in \mathbb{R}$ such that
  $$\frac{a^2b^2(a+b)^2}{4}=a^2+ab+b^2+2$$

I can see that if $ab=2$, then we have equality. However, this was through observation only. Are there any other ways of determining such relationships?

To simplify things, lets first substitute $x=ab$ and $y=a+b$, to get
$$\frac{x^2 y^2}{4} = y^2 - x + 2\\
x^2 y^2 = 4y^2 - 4x + 8 \\
(x^2-4) y^2 + 4(x - 2) = 0\\
(x-2)((x+2) y^2 + 4) = 0$$
So either $x=2$, or $(x+2)y^2+4=0$.
So $ab=2$ or $(ab+2)(a+b)^2+4=0$.
The latter case can be rearranged and factored further, using the temporary substitution $z=ab+2$:
$$(ab+2)(a+b)^2+4=0\\
z(a+b)^2+4=0\\
za^2 + zb^2 + 2abz + 4=0\\
za^2 + zb^2 + 2(z-2)z + 4=0\\
za^2 + zb^2 + z^2 + (z-2)^2=0\\
za^2 + zb^2 + z^2 + a^2b^2=0\\
(a^2+z)(b^2+z)=0\\
(a^2+ab+2)(b^2+ab+2)=0$$
This means that the original equation was actually factorisable as 
$$(ab-2)(a^2+ab+2)(b^2+ab+2)=0$$
And we actually have exactly three possible relationships: $ab=2$; $a^2+ab+2=0$; and $b^2+ab+2=0$.
