19.2.25 ** Positive integers $x, y$, satisfying $x<y$. Let
$$
P=\frac{x^{3}-y}{1+x y},
$$

find all integer values that $P$ can take.
When $y=x^{3}$ (for example, $x=2, y=8$), $P=0$.
When $y>x^{3}$, $|P|=\frac{y-x^{3}}{1+xy}$ is a positive integer. In other words, for any integer $k$ greater than 1, let $x=k^{3}, y=k\left(k^{4}-1\right)$, then $P=k$.
Finally, if $P=1$, then from (2),
$$
y=\frac{x^{3}-1}{x+1}=x^{2}-x+1-\frac{2}{x+1} .
$$

Thus, $x+1 \mid 2, x=1, y=0$, which contradicts that $y$ is a positive integer.
Therefore, $P$ can be 0 and all integers greater than 1.