12.70 Find the integer solutions of the equation $x^{5}-x^{3}-x^{2}+1=y^{2}$.
(Recommended problem by the Soviet Ministry of Education, 1990)
Given the equation can be transformed as
$$
\begin{array}{c}
x^{2}\left(x^{3}-1\right)-\left(x^{3}-1\right)=y^{2}, \\
\left(x^{3}-1\right)\left(x^{2}-1\right)=y^{2}, \\
(x-1)^{2}(x+1)\left(x^{2}+x+1\right)=y^{2} .
\end{array}
$$

When $x=0$,
$$
y^{2}=1, \quad y= \pm 1 \text {. }
$$

When $x= \pm 1$,
$$
y=0 \text {. }
$$

Assume $x, y$ are integer solutions of the equation, and $x \neq-1,0,1$.
From (1), $y^{2}$ is divisible by $(x-1)^{2}$, so $y$ is divisible by $x-1$, which implies
$$
(x+1)\left(x^{2}+x+1\right)=\left(\frac{y}{x-1}\right)^{2} \text {. }
$$

Since $\frac{y}{x-1}$ is an integer, $(x+1)\left(x^{2}+x+1\right)$ must be a perfect square. Given that
$$
x^{2}+x+1=x(x+1)+1 \text {, }
$$

we know that $x+1$ and $x^{2}+x+1$ have no common divisor greater than 1, i.e., $x+1$ and $x^{2}+x+1$ are coprime.
Therefore, both $x+1$ and $x^{2}+x+1$ must be perfect squares.
Thus, $x+1 \geqslant 0$, and since $x \neq 1,-1,0$, we have $x \geqslant 2$.
Therefore, $\quad x^{2}<x^{2}+x+1<(x+1)^{2}$,
which means $x^{2}+x+1$ lies between $x^{2}$ and $(x+1)^{2}$, i.e., between two consecutive perfect squares, and thus cannot be a perfect square.
This contradiction shows that the equation has no integer solutions for $x \neq-1,0,1$.

Therefore, the equation has only four integer solutions:
$$
(x, y)=(0,1),(0,-1),(1,0),(-1,0) .
$$