<s>[INST] Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$. 
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.

We have that $xy^2 + y + 1 \mid x^2y + x + y \implies xy^2 + y + 1 \mid y(x^2y + x + y) - x(xy^2 + y + 1)$
$\iff xy^2 + y + 1 \mid y^2 - x \implies xy^2 + y + 1 \mid xy^2 + y + 1 - x(y^2 - x)$
$\iff xy^2 + y + 1 \mid x^2 + y + 1 \implies xy^2 + y + 1 \le x^2 + y + 1$
$\iff xy^2 \le x^2 \iff y^2 \le x$. And I don't know what to do next.
This problem is adapted from a recent competition.
 [/INST] OP did most of the work by showing that

*

*$xy^2 + y + 1 \mid x - y^2$.

*$ xy^2 + y+ 1 \mid x^2 + y + 1 $

*Since $x^2 + y + 1 > 0$ hence $x^2 + y + 1 \geq xy^2 + y + 1$ so $x \geq y^2$.

To complete the solution using these ideas, suppose $ x - y^2 > 0 $, then $ x-y^2 \geq xy^2 + y + 1$.
We can verify that this is not possible for any positive integer $y$, since $xy^2 + y + 1 > xy^2 \geq x > x - y^2$.
Hence, we must have $ x - y^2 = 0$.
Plugging this back in, we observe that all $ x = y^2$ are solutions to the divisibility condition.
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