Find $x;y$ such that: $\frac{x^2+y^3}{xy-1} \in \mathbb{Z}$

Question

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.

Answer 1

[I realized I was only looking for positive solution here. You can take this approach to find negative values, too.]

A start of an approach.

If $D=xy-1$ then $xy\equiv 1\pmod D$ and you need $x^2\equiv -y^3\pmod {D},$ so:

$$x^5=x^3\cdot x^2\equiv -(xy)^3\equiv -1\pmod{D}.$$ Obviously, $x\not\equiv -1\pmod{xy-1}$ except for $y=1$ or $x=1$ an $y=2,3.$ so for $x>3,$ $-1$ must have a no trivial fifth root module $D$, which means $D$ must have a prime divisor $p\equiv 1\pmod 5$.

Basically:

$$\frac{x^5+1}{xy-1} =\frac{x^3(x^2+y^3)-(x^3y^3-1)}{xy-1}=x^3\frac{x^2+y^3}{xy-1}+(x^2y^2+xy+1)$$

This shows that there is a bound on $y$, given $x$ - specifically, $xy-1\leq x^5+1$ or $y\leq x^4+\frac{2}{x}$.

Since $x$ and $xy-1$ are relatively prime, we have that $\frac{x^5+1}{xy-1}$ is an integer if and only if $\frac{x^2+y^3}{xy-1}$ is an integer.

Similarly, $\frac{y^5+1}{xy-1}$ is an integer if and only if $\frac{x^2+y^3}{xy-1}$ is an integer, because:

$$\frac{y^5+1}{xy-1} = \frac{y^2(x^2+y^3)-(x^2y^2-1)}{xy-1}=y^2\frac{x^2+y^3}{xy-1} - (xy+1)$$

This shows the if $(x,y)$ is a solution, then $(y,x)$ is a solution, which is not obvious from the initial statement.

So, given $x$, we need to find a $y$ so that $x^{5}+1$ is divisible by $xy-1$. For example, $x=2$ and $y=1,2,6, 17$.

Given $x=3$, $x^{5}+1=244=2\cdot{122}$. So $(x,y)=(3,1)$ or $(x,y)=(3,41)$.

Checking $x=4$ gives $x^5+1=1025=25\cdot 41$. There is no factor $\equiv -1\pmod 4$.

Checking $x=5$, $x^5+1=2\cdot 3\cdot 521$ has no factor $\equiv -1\pmod{5}$.

$6^5+1=7\cdot 11\cdot 101$, so $xy-1=11,77,101,707$ so $y=2,13,17,118$.

I'm not seeing a pattern, aside from the fact that it seems like if $(x,y)=(a,b)$ is a solution, then $(x,y)=(b,a)$ is a solution.

If $(x,y)$ is a solution, then let $M=\frac{y^5+1}{xy-1}$. Then $M\equiv -1\pmod{y}$, so letting $$(x',y')=\left(y,\frac{M+1}{y}\right)=\left(y,\frac{y^4+x}{xy-1}\right)$$ gives another solution.

This gives us a way to create chains:

$$(1,2),(2,17),(17,2531),\dots$$ or: $$(2,1),(1,3),(3,41),(41,23162),\dots$$ or: $$(2,2),(2,6),(6,118),(118,274226),\dots$$

In particular, it is clear this yields an infinite set of solutions.

If you start with the solutions where $x=2$ and allow the transformations:

$$\begin{align}(x,y)&\to (y,x)\\ (x,y)&\to \left(x,\frac{x^4+y}{xy-1}\right) \end{align}$$

Does this give all solutions? It seems to miss $(x,y)=(6,13)$ and $(x,y)=(6,17)$.

Looking for negative solutions, we see that $x=-1$ allows any $y$.

$x=-2$ then $-2y-1$ must be a factor of $31$, or $y=15,-16$.

From $(0,y)$, a solution, we apply the transformation above to get $(y,-y^4)$, and then $\left(-y^4,-(y+y^6+y^{10})\right)$...

Answer 2

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 $(x,y)=(a,-a^4).$