Determine all pairs of $(x,y)$ such that $\frac{x^2y+x+y}{xy^2+y+11}$ is an integer Here is a problem that I got:


  Determine all pairs of positive integer $(x,y)$ such that $\dfrac{x^2y+x+y}{xy^2+y+11}$ is an integer.


My attempt:

First notice that $(11, 11)$ is a solution. And if $\dfrac{x^2y+x+y}{xy^2+y+11}$ is an integer, then $xy^2+y+11|x^2y+x+y$. Hence $x^2y-xy^2+x-11|xy^2+y+11$.

And then I'm stuck from here. Can anyone help me? Thanks.
Hint: Notice that $$xy^2+y+11 \mid x(xy^2+y+11)-y(x^2y+x+y)=11x-y^2$$

Then you have three cases to check: $11x-y^2=0$, $11x-y^2&lt;0$, and $11x-y^2&gt;0$.
From $\dfrac{x^2 y+x+y}{x y^2+y+11}=k$ we get

$x^2 y+x+y=k x y^2+k y+11 k\quad(*)$

from which we deduce that

$x^2 y=k x y^2$ and then $x=ky$

plug this in $(*)$

$k^2 y^3+k y+y=k^2 y^3+k y+11 k$

that is $y=11k$ and then $x=11k^2$

Thus $(11k^2;\;11k)$ for any $k\in\mathbb{N};\;k&gt;0$ satisfy the given equation

There are three more solutions that do not enter in the previous formula

$(7;\;1)$ which gives $3$

$(10;\;2)$ which gives $4$

$(121;\;1)$ which gives $111$

BTW

plugging $(11k^2;\;11k)$  in the given equation gives $k$ as result

hope this helps