4. 166 Find all integer solutions \(x, y\) that satisfy \(x^{2}+x=y^{4}+y^{3}+y^{2}+y\).

 untranslated text:
4. 166 求满足 $x^{2}+x=y^{4}+y^{3}+y^{2}+y$ 的所有整数 $x, y$.

 translated text:
4. 166 Find all integer solutions \(x, y\) that satisfy \(x^{2}+x=y^{4}+y^{3}+y^{2}+y\).
【Solution】Multiply both sides of the equation by 4, then add 1 to get

That is
$$
\begin{array}{c}
(2 x+1)^{2}=\left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1, \\
(2 x+1)^{2}=\left(2 y^{2}+y+1\right)^{2}-\left(y^{2}-2 y\right) .
\end{array}
$$

If $y$ is an integer different from $-1, 0, 1$, and 2, then
$$
3 y^{2}+4 y+1>0 \text { and } y^{2}-2 y>0 \text {. }
$$

In this case, we have
$$
\left(2 y^{2}+y\right)^{2}<(2 x+1)^{2}<\left(2 y^{2}+y+1\right)^{2},
$$

Therefore, $2 x+1$ lies between the integers $2 y^{2}+y$ and $2 y^{2}+y+1$, so $2 x+1$ cannot be an integer, and thus $x$ cannot be an integer. This means that when $y$ takes an integer value different from $-1, 0, 1, 2$, the original equation has no integer solutions.

Substituting $y=-1, y=0, y=1, y=2$ into the original equation, we get the solutions of the original equation:
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x = 0 , } \\
{ y = - 1 ; } \\
{ x = - 6 , }
\end{array} \left\{\begin{array} { l } 
{ x = - 1 , } \\
{ y = - 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x = 0 , } \\
{ y = 0 ; }
\end{array} \left\{\begin{array} { l } 
{ x = - 1 , } \\
{ y = 0 ; }
\end{array} \left\{\begin{array}{l}
x=5, \\
y=2 ;
\end{array}\right.\right.\right.\right.\right. \\
\left\{\begin{array}{l}
x=2 .
\end{array}\right. \\
\hline
\end{array}
$$