Question 203, 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 $x<y$, let $p=\frac{x^{3}-y}{1+x y}$, find all integer values that $p$ can take.
Question 203, Solution: If $py>y-x^{3}>0$, then $p=\frac{x^{3}-y}{1+x y}$ is not an integer.
Take $x=2, y=8$, then we have $p=\frac{x^{3}-y}{1+x y}=0$, so $p$ can be 0.
If $p=1$, then $\frac{x^{3}-y}{1+x y}=1$, rearranging gives $y=\frac{x^{3}-1}{x+1}=x^{2}-x+1-\frac{2}{x+1}$. Noting that $y$ is a positive integer, hence $x=1$, and thus $y=0$, which contradicts the fact that $y$ is a positive integer.

When $p \geq 2$, from $p=\frac{x^{3}-y}{1+x y}$, we know $y=\frac{x^{3}-p}{p x+1}$. At this point, we take $x=p^{3}$, then $y=p^{5}-p>p^{3}$, so $p$ can take all positive integers not less than 2.
In summary, the set of all numbers that $p$ can take is $\{p \mid p \in N, p \neq 1\}$.