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?
On
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.
If $x=y^2-y$ so $$y^4-y^2=(y^4-y^2)^2-2,$$ which gives $$y^4-y^2=-1,$$ which is impossible, or $$y^4-y^2=2.$$ Can you end it now?
I got that the second case: $$\sqrt{x+y}+y+2=0$$ is impossible.
Indeed, $y\leq-2$ and $$x+y=(y+2)^2$$ or $$x=y^2+3y+4,$$ which gives $$(y^2+3y+4)(y^2+y)=(y^4-y^2)^2-2$$ or $$y^8-2y^6-4y^3-7y^2-4y-2=0$$ or $$y^8-4y^6+2y^6-8y^4+8y^4-32y^2-4y^3+25y^2-100-4y+98=0$$ or $$(y^2-4)(y^6+2y^4+8y^2+25)-4y^3-4y+98=0,$$ which is impossible for $y\leq-2$.