5 positive integers $x, y$ satisfy $x<y$, let $P=\frac{x^{3}-y}{1+x y}$, find all integer values that $P$ can take.

The positive integers $x, y$ satisfy $x<y$, and we define $P=\frac{x^{3}-y}{1+x y}$. Determine all possible integer values of $P$.
Solution one: The set of all values that $P$ can take is $\{k \mid k=0$ or $k \geqslant 2, k \in \mathbf{Z}\}$.
On one hand, $P$ can take some of the above values:
Let $x=2, y=8$, clearly $x0$, and $P=\frac{k^{9}-\left(k^{5}-k\right)}{1+k^{3}\left(k^{5}-k\right)}=\frac{k^{9}-k^{5}+k}{k^{8}-k^{4}+1}=k$.
On the other hand, we will prove that $P$ cannot take any other integer values.
If $Px^{3}$, thus $0P^{3}$ holds, which means $P$ can take any integer $\geqslant 2$.
When $P<0$, by $\left|x^{3}-y\right|=y-x^{3}<x y+1 \Rightarrow$ contradiction.
Therefore, the values that $P$ can take are $\{k \mid k \in \mathbf{N}, k=0$ or $k \geqslant 2\}$.
Comment: This problem originates from the discussion of the positive integer solutions of the equation $x^{2}+y^{2}=z(1+x y)$. In the 1991 IMO, there was a problem: Given $(x y+1) \mid\left(x^{2}+y^{2}\right), x, y \in \mathbf{N}^{*}$, prove that $\frac{x^{2}+y^{2}}{1+x y}$ is a perfect square.

Exploring this problem, we find that if $\frac{x^{2}+y^{2}}{1+x y}$ is an integer, then it must be that $\frac{x^{2}+y^{2}}{1+x y}=(x, y)^{2}$, where $(x, y)$ represents the greatest common divisor of $x$ and $y$.