Solve the system of equations for $x, y \in \mathbb R$. $$\large \left\{ \begin{align} x^3 - y^3 = (y + 1)x^2 - (x - 1)y^2\\ x^2 + 4\sqrt{y + 4} = 2x + y + 7 \end{align}\right.$$
(Ahem.) This problem is provided to you by a recent competition. Could you please check if my solution is correct? And if there are any other solutions that are more feasible, please post them.
$y \ge -4$ is the given condition.
We have that $$x^3 - y^3 = (y + 1)x^2 - (x - 1)y^2 \iff (x - y)(x^2 + xy + y^2) = (x - y)xy + (x^2 + y^2)$$
$$\iff (x - y)(x^2 + y^2) = x^2 + y^2 \iff \left[ \begin{align} x^2 + y^2 = 0\\ x - y = 1\end{align}\right.$$
If $x^2 + y^2 = 0$ then $x = y = 0$. Plugging $x = y = 0$ in $x^2 + 4\sqrt{y + 4} = 2x + y + 7$, we have that $0^2 + 4\sqrt{0 + 4} = 2 \cdot 0 + 0 + 7 \implies 8 = 7$, which is false.
$\implies x - y = 1 \iff y = x - 1$.
In addition, $$x^2 + 4\sqrt{y + 4} = 2x + y + 7 \iff x^2 - 2x + 1 = (y + 4) - 4\sqrt{y + 4} + 4$$
$$\iff (x - 1)^2 = (\sqrt{y + 4} - 2)^2 \iff \pm(x - 1) = \sqrt{y + 4} - 2 \iff \pm y = \sqrt{y + 4} - 2$$
$$\iff \pm y + 2 = \sqrt{y + 4} \iff (y \pm 2)^2 = y + 4 \iff \left[ \begin{align} y^2 + 3y = 0\\ y^2 - 5y = 0\end{align} \iff y \in \{0, -3, 5\}\right.$$
$$\implies (x, y) \in \{(1,0), (-2, -3), (6, 5)\}$$
Plugging each possible answer in $x^2 + 4\sqrt{y + 4} = 2x + y + 7$, we have that $x = 1$ and $y = 0$.