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?
 A: 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 $b^2+ab+2=0$.
A: This is not a complete solution, but it could give an idea on how to proceed. Add to both sides $ab$ and you get
$$
\frac{a^2b^2(a+b)^2}{4} + ab = (a + b)^2 + 2.
$$
If $ab \geq 2$ and $a+b \neq 0$ then you have
$$\frac{a^2b^2}{4} \leq 1 \iff -2 \leq ab \leq 2.$$
Therefore $ab = 2$ in this case.
If $ab < 2$ and $a + b \neq 0$ you get
$$
\frac{a^2b^2}{4} > 1 \iff ab > 2 \;\; \text{or} \;\; ab < -2.
$$
Hence, you have that $ab < -2$ could be a solution but I don't see how to go on in this case.  It remains to look at $a + b = 0$, in this case you get back $ab = 2$.
