89*. Find all pairs of integers \(x\) and \(y\) that satisfy the equation

\[
x^{2} + x = y^{4} + y^{3} + y^{2} + y
\]
89. Answer: $(0 ;-1),(-1 ;-1),(0 ; 0),(-1 ; 0),(5 ; 2)$, we get

Multiplying both sides of the equation by 4 and adding 1 to each,

$(2 x+1)^{2}=\left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1=\left(2 y^{2}+y+1\right)^{2}-\left(y^{2}-2 y\right)$. If $y$ is an integer and different from $-1,0,1$ and 2, then $3 y^{2}+4 y+1>0$ and $y^{2}-2 y>0$ in this case

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

These inequalities mean that $(2 x+1)^{2}$ lies between two consecutive squares, which is impossible for integers $x$. Substituting $y=-1, y=0, y=1$ and $y=2$ into the equation, we find the answer.