<s>[INST] Find all pairs such that $x^2 y + x + y$ is divisible by $xy^2 + y + 7$ 
Find all pairs $(x, y)$ of positive integers such that $x^2 y + x + y$ is divisible by $xy^2 + y + 7$. If there are too many to write, write a generic form.

I was thinking of rewriting the divisibility in a simpler way such as $xy^2 + y + 7\mid x^2 y + x + y \implies xy^2 + y + 7\mid x^2y-xy^2+x-y$ but I am not sure if that helps. Is there an easier way?
 [/INST] We have $x=7n^2,y=7n$ as a class of solutions because then $n(xy^2+7+y)=x^2y+y+x$.
There is one other solution $x=11,y=1$, giving $xy^2+7+y=19,x^2y+y+x=7\cdot19$.
We have $y(x^2y+x+y)-x(xy^2+y+7)=y^2-7x$, so $xy^2+y+7$ must divide $y^2-7x$.
One possibility is $y^2=7x$. That implies 7 divides $y$ and hence also $x$. So we get the solutions in the first line.
Otherwise, since $y^2-7x<y^2<xy^2+y+7$ we must have $y^2<7x$. But then if $y\ge3$ we have $xy^2+y+7>9x>7x-y^2>0$ which is impossible if $xy^2+y+7$ is a factor of $7x-y^2$. So $y=1$ or 2. It is easy to check that $y=1$ gives the solution $x=11,y=1$, whilst $y=2$ does not give a solution.
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