TITLE: Find $x;y$ such that: $\frac{x^2+y^3}{xy-1} \in \mathbb{Z}$
QUESTION [6 upvotes]: Let $x,y \in \mathbb{Z}$ such that: $$\frac{x^2+y^3}{xy-1} \in \mathbb{Z}$$ Find $x,y$

I don't have any ideas about this problem.

REPLY [3 votes]: Set $(x\rightarrow y, y\rightarrow -x).$ 
Say $\frac{y^2-x^3}{-xy-1}=t.$ Then you have the parametric family
of elliptic curves $y^2+txy=x^3-t.$ For instance if you set $t=80$ you get the 
$(-79, -1),(-120843, -1289),(17723, -1289). $ This gives (using your original notation) the solution
$x=17723,y=-1289.$ So you can find all the $(x,y)$ if you fix the ratio.
Here are some solutions (for various $t$) $(xy\not =0,\ y\not = -1)$ following the previous
 (27, -10),(305, -52),(3, 1),(611, -86),(29, -16),(4, -6),(2, 2),(6, 2), 
(259, -57), (700, -102),(17, 2),(660, -51),(15, -2),(2239, -141),(23, -4),
(143, -19),(138, -20),(7, -3),(46, 17),(107, 17),(1, 2),(17, 2),(14, -5),
(27, -8),(545, -64),(1359, -119), (-18881, -481),(5894, -481),(-20601, -516),
(6669, -516),(-417201030, -551613),(402307479, -551613),(6, 13),(371, 13),
(-2968, -93),(271, -93),(122, 46),(798, 46),(2, 6),(118, 6).

Of course $(x,y)$ are infinitely many ( for instance take $y=-1$) and I can't see any pattern for $(x,y)$ unless you manage to express the solutions of $E_t:y^2+txy=x^3-t,$ with respect to $t.$ Also if you set $t=a^3$ then you get 
the family of points $(x,y)=(a,-a^4).$