8. Find the integer pairs $(x, y)$ that satisfy the condition $x^{4}+x^{3}+x^{2}+x=y^{2}+y$.

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8. Multiplying both sides of the conditional equation by 4, we have
$$
4 x^{4}+4 x^{3}+4 x^{2}+4 x=4 y^{2}+4 y \text {. }
$$

By completing the square, we get
$$
4 x^{4}+4 x^{3}+4 x^{2}+4 x+1=(2 y+1)^{2} \text {. }
$$

Notice that
$$
\left(2 x^{2}+x\right)^{2}=4 x^{4}+4 x^{3}+x^{2}
$$

and $\left(2 x^{2}+x+1\right)^{2}=4 x^{4}+4 x^{3}+5 x^{2}+2 x+1$
are two consecutive square numbers, and
$$
2 x^{2}+x=2\left(x+\frac{1}{4}\right)^{2}-\frac{1}{8}
$$

as a real-valued function, its minimum integer value is 0.
Thus, under the condition that $x$ takes integer values,
$$
\left|2 x^{2}+x\right| \leqslant\left|2 x^{2}+x+1\right| \text {. }
$$

Hence, there are only two cases.
$$
\begin{array}{l}
\text { (1) }(2 y+1)^{2}=4 x^{4}+4 x^{3}+4 x^{2}+4 x+1 \\
\leqslant 4 x^{4}+4 x^{3}+x^{2} \text {. }
\end{array}
$$
$$
\begin{array}{l}
\text { (2) }(2 y+1)^{2}=4 x^{4}+4 x^{3}+4 x^{2}+4 x+1 \\
\geqslant 4 x^{4}+4 x^{3}+5 x^{2}+2 x+1
\end{array}
$$

If (1) holds, then
$$
3 x^{2}+4 x+1 \leqslant 0,-1 \leqslant x \leqslant-\frac{1}{3}, x=-1 \text {. }
$$

Thus, $(2 y+1)^{2}=1$, which gives $y=0$ or $y=-1$, leading to the solutions $(-1,0),(-1,-1)$.
If (2) holds, then
$$
x^{2}-2 x \leqslant 0,0 \leqslant x \leqslant 2 \text {. }
$$

Hence, $x=0,1,2$.
When $x=0$, $y=-1$ or 0, thus the solutions are $(0,0),(0,-1)$.
When $x=1$, $y^{2}+y-4=0$, $y$ has no integer solutions.
When $x=2$, $y^{2}+y-30=0$, $y=-6$ or 5. The solutions are $(2,-6),(2,5)$.
In summary, the solutions that satisfy the conditions are
$$
(-1,0),(-1,-1),(0,0),(0,-1) \text {, }
$$