25. Find the integer solutions of the indeterminate equation $8 x^{4}+1=y^{2}$.

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25. Notice that $(x, y)=(0, \pm 1),( \pm 1, \pm 3)$ are solutions to the equation. If $(x, y)$ is an integer solution to the equation and $|x|>1$, we may assume $y>1$.

From $8 x^{4}=(y-1)(y+1)$, and since $y$ is odd, we can set
$$\left\{\begin{array} { l } 
{ y - 1 = 2 u ^ { 4 } , } \\
{ y + 1 = 4 v ^ { 4 } , }
\end{array} \text { or } \left\{\begin{array}{l}
y-1=4 v^{4}, \\
y+1=2 u^{4},
\end{array}\right.\right.$$

where $u, v \in \mathbf{N}^{*}$. Thus, we always have $u^{4}-2 v^{4}= \pm 1$, and therefore
$$\left(u^{4}-v^{4}\right)^{2}=v^{8} \pm 2 v^{4}+1=v^{8} \pm u^{4}$$

Using the result from Example 5 in Section 4.2 and the conclusion from the previous problem, we know that $u v^{2}\left(u^{4}-v^{4}\right)=0$, hence $u=v$, leading to $u^{4}-2 v^{4}=-1$, i.e., $u=v=1$. This implies $y=3$, leading to $|x|=1$, a contradiction. Therefore, the only solutions to the equation are $(x, y)=(0, \pm 1),( \pm 1, \pm 3)$.