18.3.47 Find the integer solutions of the equation $x^{2}+x=y^{4}+y^{3}+y^{2}+y$.

 untranslated text:
18.3.47 求方程 $x^{2}+x=y^{4}+y^{3}+y^{2}+y$ 的整数解.
Multiplying both sides of the original equation by 4, and then adding: 1, we get
$$
(2 x+1)^{2}=4 y^{4}+4 y^{3}+4 y^{2}+4 y+1 .
$$

Notice that the left side of (1) is a perfect square. The right side of (1) can be transformed into
$$
4 y^{4}+4 y^{3}+4 y^{2}+4 y+1-\left(2 y^{2}+y\right)^{2}+\left(3 y^{2}+4 y+1\right),
$$

Since $3 y^{2}+4 y+1>0$ has the solution $y>-\frac{1}{3}$ or $y\left(2 y^{2}+y\right)^{2}$. Similarly,
$$
4 y^{4}+4 y^{3}+4 y^{2}+4 y+1=\left(2 y^{2}+y+1\right)^{2}+\left(2 y-y^{2}\right),
$$

Since $2 y-y^{2}2$, or $y(2 x+1)^{2}>\left(2 y^{2}+y\right)^{2} .
$$

Such $x$ obviously does not exist (since $\left(2 y^{2}+y\right)^{2}$ and $\left(2 y^{2}+y+1\right)^{2}$ are consecutive perfect squares). Therefore, we only need to consider the solutions of the original equation when $y=-1,0,1,2$. Substituting into the original equation, we get the solutions $(0,-1),(-1,-1),(0,0),(-1,0),(5,2),(-6,2)$.