1. Find the integer solutions of the equation $x^{5}-x^{3}-x^{2}+1=y^{2}$.
1. (1) When $x=0$, $y= \pm 1$; when $x=1$, $y=0$; when $x=-1$, $y=0$.
(2) When $x \neq 0$ and $x \neq \pm 1$, $y^{2}=x^{5}-x^{3}-x^{2}+1=x^{3}\left(x^{2}-1\right)-\left(x^{2}-1\right)=\left(x^{2}-1\right)\left(x^{3}-1\right)=(x-1)^{2}(x+1)\left(x^{2}+x+1\right)$.
$$
\begin{array}{l}
\because \quad x-1 \neq 0, \\
\therefore \quad x-1 \mid y .
\end{array}
$$

Let $y=(x-1) t$. Then (1) becomes $t^{2}=(x+1)\left(x^{2}+x+1\right)$.
$$
\because x^{2}+x+1 \equiv(-1)^{2} \equiv 1(\operatorname{mod}(x+1)) \text { and } x+1 \neq 0 \text {, }
$$
$\therefore \quad\left(x+1, x^{2}+x+1\right)=1$, so there exists an integer $u$ such that $u \mid t$ and $u^{2} \equiv x^{2}+x+1$.
However, when $x>0$, $x^{2}<x^{2}+x+1<(x+1)^{2} \Rightarrow x^{2}+x+1$ is not a perfect square. When $x \leqslant-2$, $(x-1)^{2}<x^{2}+x+1<x^{2} \Rightarrow x^{2}+x+1$ is not a perfect square. Therefore, when $x \neq 0, x \neq \pm 1$, $x^{2}+x+1$ is not a perfect square, which contradicts (2).
Combining (1) and (2), we conclude that the integer solutions to the equation $x^{5}-x^{3}-x^{2}+1=y^{2}$ are $(\pm 1,0)(0, \pm 1)$.