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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.

Mike Pierce
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Lê Tấn Khang
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2 Answers2

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[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 $\left(-y^4,-(y+y^6+y^{10})\right)$...

Thomas Andrews
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  • It's an optimistic conjecture - I don't think it is true. – Thomas Andrews May 27 '15 at 19:01
  • The cases where $x=2$ I already completely covered, above: $y=1,2,6,17$. @user2566092 – Thomas Andrews May 27 '15 at 19:04
  • Whoops I missed that about $x = 2$. In any event, maybe there are only finitely many solutions that aren't captured by your conjecture. (I've noticed sometimes that happens with this kind of stuff, e.g. you characterize all "sufficiently large" solutions and so then you only have finitely many "small" cases to check) – user2566092 May 27 '15 at 19:08
  • It seems unlikely to me, @user2566092 . It seems that it shouldn't be too rare that $x^5+1$ is divisible by two numbers $\equiv -1\pmod x$, but the above growth is pretty quick. – Thomas Andrews May 27 '15 at 19:12
  • There are $66$ solutions with $x\leq 2000$ and $x<y$, via brute force computing. – Thomas Andrews May 27 '15 at 19:37
  • This is probably similar to your relation, so I won't add a new answer for it, but I will mention it here: $$ \begin{align} \frac{x^2+y^3}{xy-1} &=\frac{y^5+1}{xy-1},x^2-(xy+1),y^3\tag{1}\ &=\frac{x^5+1}{xy-1},y^3-(x^2y^2+xy+1),x^2\tag{2} \end{align} $$ – robjohn May 29 '15 at 22:00
  • @robjohn yeah, it's really just easier to see it as a question of modular arithmetic, module $xy-1$, and you are just using that $x$ and $y$ are inverses. There will be lots of ways of proving it. – Thomas Andrews May 29 '15 at 22:09
  • @ThomasAndrews: Indeed. It just seemed more direct to me to use this to show that if $\frac{y^5+1}{xy-1}\in\mathbb{Z}$ or $\frac{x^5+1}{xy-1}\in\mathbb{Z}$ then $\frac{x^2+y^3}{xy-1}\in\mathbb{Z}$, and that if $(a,b)$ is a solution, then $(b,a)$ is, too. But in the end, they are all equivalent. – robjohn May 29 '15 at 22:34
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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 $(x,y)=(a,-a^4).$

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