8.39 Find all integer solutions \(x\), \(y\) that satisfy the equation \(x^{2}+x=y^{4}+y^{3}+y^{2}+y\).
(1st All-Soviet Union Mathematical Olympiad, 1967)
【Solution】Given the equation can be transformed into
$$
\begin{array}{l}
4 x^{2}+4 x+1=4 y^{4}+4 y^{3}+4 y^{2}+4 y+1, \\
(2 x+1)^{2}=\left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1, \\
(2 x+1)^{2}=\left(2 y^{2}+y\right)^{2}+2\left(2 y^{2}+y\right)+1-y^{2}+2 y, \\
(2 x+1)^{2}=\left(2 y^{2}+y+1\right)^{2}-\left(y^{2}-2 y\right) .
\end{array}
$$

We first find the integer solutions $y$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
3 y^{2}+4 y+1>0 \\
y^{2}-2 y>0
\end{array}\right.
$$

It is easy to find that $y$ is an integer not equal to $-1, 0, 1, 2$.
(1) When $y \neq -1, 0, 1, 2$ and $y \in \mathbb{Z}$,
Since at this time $3 y^{2}+4 y+1>0, y^{2}-2 y>0$.
From (1) and (2), we get
$$
\left(2 y^{2}+y\right)^{2}<(2 x+1)^{2}<\left(2 y^{2}+y+1\right)^{2} \text {. }
$$

Since there are no perfect squares between the adjacent squares $\left(2 y^{2}+y\right)^{2}$ and $\left(2 y^{2}+y+1\right)^{2}$, $(2 x+1)^{2}$ cannot be a perfect square, i.e., the original equation has no integer solutions at this time.
(2) When $y=-1, 0, 1, 2$,
For $y=0$, from the original equation, we get $x=0$ or $x=-1$;
For $y=-1$, from the original equation, we get $x=0$ or $x=-1$;
For $y=1$, from the original equation, we get $x=\frac{-1 \pm \sqrt{17}}{2} \notin \mathbb{Z}$;
For $y=2$, from the original equation, we get $x=5$ or $x=-6$.
Therefore, the given equation has six sets of solutions.
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x = 0 , } \\
{ y = 0 , }
\end{array} \quad \left\{\begin{array} { l } 
{ x = - 1 , } \\
{ y = 0 , }
\end{array} \quad \left\{\begin{array}{l}
x=0, \\
y=-1,
\end{array}\right.\right.\right. \\
\left\{\begin{array} { l } 
{ x = - 1 , } \\
{ y = - 1 , }
\end{array} \quad \left\{\begin{array} { l } 
{ x = 5 , } \\
{ y = 2 , }
\end{array} \quad \left\{\begin{array}{l}
x=-6, \\
y=2 .
\end{array}\right.\right.\right. \\
\end{array}
$$