Solve the following equation if $x$ and $y$ are integers:

$$
x^{2}+x=y^{4}+y^{3}+y^{2}+y
$$
$$
x^{2}+x=y^{4}+y^{3}+y^{2}+y
$$

Multiply both sides of the equation by 4 and add 1 to both sides. This way, the left side becomes a perfect square. Try to make the right side a perfect square as well. Thus, we arrive at the equation

$$
(2 x+1)^{2}=\left(2 y^{2}+y+1\right)^{2}-\left(y^{2}-2 y\right)=\left(2 y^{2}+y\right)^{2}+\left(3 y^{2}+4 y+1\right)
$$

If $y$ is an integer and not equal to -1, 0, 1, or 2, then from the two forms of the right side of the equation,

$$
\left(2 y^{2}+y\right)^{2}<(2 x+1)^{2}<\left(2 y^{2}+y+1\right)^{2}
$$

thus $(2 x+1)^{2}$ falls between two consecutive square numbers, so it cannot be a perfect square.

If $y=-1$ or 0, then from (1) we get $x^{2}+x=0$, so $x=0$ or $x=-1$.

If $y=1$, then from (1) we get $x^{2}+x=4$, and no such integer $x$ exists.

If $y=2$, then from (1) we get $x^{2}+x=30$, so $x=5$ or $x=-6$.

Therefore, the equation (1) has six solutions in the integers, which are: $(-1, -1)$, $(-1, 0)$, $(0, -1)$, $(0, 0)$, $(5, 2)$, and $(-6, 2)$.

Remark. We can also arrive at the solution by considering the discriminant of the quadratic equation (1) in $x$.