<s>[INST] System of 2 equations Solve the system of equations:

$$
\left\{\begin{matrix}2\sqrt{x+y} = y^2+y-x & \quad(1) \\ x(y^2+y)=(y^4-y^2)^2-2  & \quad(2) \end{matrix}\right.
$$

My attempt:
From $(1)$ I get:
$$(1) \implies (\sqrt{x+y}+y+2)(\sqrt{x+y}-y) = 0$$
So either $\sqrt{x+y}+y+2 = 0 \implies x = y^2+3y+4$ or $\sqrt{x+y}-y = 0 \implies x = y^2-y$. If I replace either of them into $(2)$, It would be very messy since this will become a $8$th-degree polynomial. 
How can I solve this problem?
 [/INST] If we substitute $x=y^2-y$ we can factor $(2)$ as
$$(y^2-2)(y^2+1)(y^4-y^2+1)=0$$
and we get $y=\pm\sqrt2,\pm i,\pm w,\pm w^*$ where $w=e^{i\pi/6}$. Only $y=\sqrt2,i,w,w^*$ lead to equality in $(1)$ under the principal square root. If $x=y^2+3y+4$ we get an irreducible octic in $y$ with two real roots and three complex conjugate root pairs, but all of them do not lead to equality in $(1)$ if the principal square root is taken.
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