9 Find the integer solutions of the indeterminate equation $\left(x^{2}-y^{2}\right)^{2}=1+16 y$.

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9. From the equation, we know $y \geqslant 0$, and $x \neq y$. Without loss of generality, let $x \geqslant 0$.

If $x>y$, then

we get
$$\begin{aligned}
1+16 y & =\left(x^{2}-y^{2}\right)^{2} \\
& \geqslant\left[(y+1)^{2}-y^{2}\right]^{2} \\
& =(2 y+1)^{2} \\
& =4 y^{2}+4 y+1 \\
0 & \leqslant y \leqslant 3 .
\end{aligned}$$

By
$$y=0,1,2,3,$$

we obtain the integer solutions of the original equation as
$$(x, y)=(1,0),(4,3)$$

If $x<y$, then

we get
$$\begin{aligned}
1+16 y & =\left(y^{2}-x^{2}\right)^{2} \\
& \geqslant\left[y^{2}-(y-1)^{2}\right]^{2} \\
& =(2 y-1)^{2} \\
& =4 y^{2}-4 y+1, \\
0 & \leqslant y \leqslant 5
\end{aligned}$$

By
$$y=0,1,2,3,4,5,$$

we obtain the integer solutions of the original equation as
$$(x, y)=(4,5)$$

When $x \leqslant 0$, we get
$$(x, y)=(-1,0),(-4,3),(-4,5)$$

In summary, the solutions of the equation are $(x, y)=( \pm 1,0),( \pm 4,3),( \pm 4,5)$.