12.67 Find all integer solutions to the equation $2 x^{4}+1=y^{2}$.
(China National Training Team Selection Test, 1993)
［Solution］(1) When $y=0$, the equation has no integer solutions.
(2) When $x=0$, $y= \pm 1$, so the equation has solutions
$$
x=0, \quad y=1 \text { and } x=0, \quad y=-1 \text {. }
$$
(3) If $\left(x_{0}, y_{0}\right)$ is a solution to the equation, then $\left(x_{0}, \pm y_{0}\right)$ and $\left(-x_{0}, \pm y_{0}\right)$ are also solutions to the equation, so we only need to consider the natural number solutions of the equation.
We prove that the equation
$$
2 x^{4}+1=y^{2}
$$

has no natural number solutions.
If (1) has a natural number solution $(x, y)$, then $y$ is odd, let $y=2 z+1$, then (1) becomes
$$
x^{4}=2 z(z+1) \text {. }
$$

Therefore, $x$ is even, let $x=2 u$. Thus we have
$$
8 u^{4}=z(z+1) \text {. }
$$
$$
\begin{array}{l} 
\text { Since }(z, z+1)=1, \text { we have } \\
\text { I } .\left\{\begin{array}{l}
z=8 v^{4}, \\
z+1=w^{4}, \\
v w=u .
\end{array}\right. \\
\text { II. }\left\{\begin{array}{l}
z=v^{4}, \\
z+1=8 w^{4}, \\
v w=u .
\end{array} \quad(v, w)=1 .\right.
\end{array}
$$

For case II, we have the equation
$$
8 w^{4}=v^{4}+1 \text {. }
$$

Since
$$
\begin{array}{l}
v^{4} \equiv 0,1 \quad(\bmod 8), \\
8 w^{4} \equiv 0 \quad(\bmod 8) .
\end{array}
$$

Thus case II has no solutions.
For case I, we have
$$
w^{4}=8 v^{4}+1 \text {. }
$$

Thus $w$ is odd, let $w=2 q+1$, then
$$
\begin{aligned}
& (2 q+1)^{4}=8 v^{4}+1, \\
v^{4}= & 2 q^{4}+4 q^{3}+3 q^{2}+q \\
= & q(q+1)\left(2 q^{2}+2 q+1\right) .
\end{aligned}
$$

Since
$$
\left(q, q+1,2 q^{2}+2 q+1\right)=1 \text {, }
$$

we should have
$$
\left\{\begin{array}{l}
q=\alpha^{4} \\
q+1=\beta^{4}
\end{array}\right.
$$
$$
\beta^{4}-\alpha^{4}=1 \text {. }
$$

This equation has no natural number solutions, so (1) has no natural number solutions.
Combining (1), (2), and (3), the known integer solutions to the equation are $(x, y)=(0,1)$, $(0,-1)$.