Problem 2316

(IMO Shortlist 1998 N1/Problem 4) The solutions are $(x, y) = (11t^2, 11t), (7, 1), (121, 1), (10, 2)$ where $t \in \mathbb{Z}^+$.
The condition means $xy^2 + y + 11 \mid x^2y + x + y$. Since
$$y(x^2y + x + y) - x(xy^2 + y + 11) = y^2 - 11x,$$
we have
$$xy^2 + y + 11 \mid y^2 - 11x.$$
* If $y^2 - 11x > 0$, then we have $xy^2 + y + 11 \le y^2 - 11x < y^2$. But then $xy^2 + y + 11 > xy^2 \ge y^2$. This is a contradiction.
* If $y^2 - 11x = 0$, then we can write $y = 11t$ for some $t \in \mathbb{Z}^+$. This yields $x = 11t^2$. Now, we check that
$$\frac{(11t^2)^2(11t) + (11t^2) + (11t)}{(11t^2)(11t)^2 + (11t) + 11} = \frac{11t(121t^4 + t + 1)}{11(121t^4 + t + 1)} = t \in \mathbb{Z},$$
and so all pairs $(x, y) = (11t^2, 11t)$ are solutions.
* If $y^2 - 11x < 0$, then we have $xy^2 + y + 11 \le 11x - y^2 < 11x$. It follows that $11x > xy^2 + y + 11 > xy^2$, and hence $y^2 < 11$. This implies $y = 1, 2, 3$.
- When $y = 1$, we need $x + 12 \mid x^2 + x + 1$. This is equivalent to
$$x + 12 \mid (-12)^2 + (-12) + 1 = 133 = 7 \times 19.$$
The solutions are $(x, y) = (7, 1), (121, 1)$.
- When $y = 2$, we need $4x + 13 \mid 2x^2 + x + 2$. Since $4x + 13$ is odd, this is the same as $4x + 13 \mid 8(2x^2 + x + 2)$. As
$$8(2x^2 + x + 2) = (4x + 13)(4x - 11) + 159,$$
this means $4x + 13 \mid 159 = 3 \times 53$. The only solution is $(x, y) = (10, 2)$.
- When $y = 3$, we need $9x + 14 \mid 3x^2 + x + 3$. Since $3 \nmid 9x + 14$, this is the same as $9x + 14 \mid 27(3x^2 + x + 3)$. As
$$27(3x^2 + x + 3) = (9x + 14)(9x - 11) + 235,$$
this means $9x + 14 \mid 235 = 5 \times 47$. There is no solution.
Therefore, we conclude that the solutions are those listed at the beginning.