Find all $x,y$ so that $\dfrac{x+y+2}{xy-1}$ is an integer.

Question

I am trying to find the integers $x,y$ so that $\dfrac{x+y+2}{xy-1}$ is an integer. What I have done: I suppose there exists $t$ such that $$t=\dfrac{x+y+2}{xy-1}$$ where $xy\neq 1$ then consider the following scenarios: $$x=y$$ $$x>y>0$$ $$x>0>y$$ ,etc. This approach helps me find the solutions but it is very long. Any simpler method?

Answer 1

If $x=y=-1$, then $x+y+2=xy-1=0$. Therefore, some of the solutions below need to be prefaced with "except for $x=y=-1$".

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If $x=y$, then $\frac{x+y+2}{xy-1}=\frac{2}{x-1}$, which says that $$ x=y\in\{0,2,3\}\quad\text{are the only solutions with $x=y$}\tag{1} $$ Furthermore, $$ x+y=-2\quad\text{is a solution}\tag{2} $$ We also have that $$ xy=0\quad\text{is a solution}\tag{3} $$ If $x=-1$ or $y=-1$, $\frac{x+y+2}{xy-1}=-1$. Therefore, $$ (x+1)(y+1)=0\quad\text{is a solution}\tag{4} $$ If $y=1$, then the solutions to $\frac{x+y+2}{xy-1}=\frac{x+3}{x-1}\in\mathbb{Z}$ must have $-3\le x\le5$. Checking gives $$ \{(-3,1),(-1,1),(0,1),(2,1),(3,1),(5,1)\}\tag{5} $$ If $x=1$, then the solutions to $\frac{x+y+2}{xy-1}=\frac{y+3}{y-1}\in\mathbb{Z}$ must have $-3\le y\le5$. Checking gives$$ \{(1,-3),(1,-1),(1,0),(1,2),(1,3),(1,5)\}\tag{6} $$ Since $$ |xy-1|-|x+y+2|\ge(|x|-1)(|y|-1)-4\tag{7} $$ if $x+y\ne-2$ and $(|x|-1)(|y|-1)\ge5$, then $(x,y)$ cannot be a solution. Since all the solutions with $|x|\le1$ or $|y|\le1$ are covered above, we only need to check $2\le|x|,|y|\le5$. Checking these $64$ cases, the only new solutions not counted above are $$ \{(2,5),(5,2)\}\tag{8} $$ Therefore, all solutions are given by $(1)$-$(6)$ and $(8)$.

Answer 2

If $x>3$ and $y>3$, then the denominator is bigger than the nominator because:

$xy-1-x-y-2 = (x-1)(y-1) - 4 > 2 \cdot 2 - 4 = 0 $

So the given number cannot be an integer.

You can look at the other cases manually:
$0 \le x \le 3$ and $0 \le y \le 3$.

If you allow negative x,y then you can apply similar observations, only the denominator gets changed slightly. Think about it.

Answer 3

Wlog $|x|\le |y|$.

$x=0$ leads to $\frac{y+2}{-1}$, which is always an integer.

$x=1$ leads to $\frac{y+3}{y-1}=1+\frac4{y-1}$ which is an integer iff $y-1$ is a (positive or negative) divisor of $4$, so $y\in\{-3,-1,2,3,5\}$ ($y=0$ is excluded by $|y|\ge |x|$)

$x=-1$ leads to $\frac{y+1}{-y-1}=-1$, always an integer

$|x|\ge2$ leads to $|xy-1|\ge 2|y|-1\ge 3$ and $|x+y+2|\le 2|y|+2$, so that $\left|\frac{x+y+2}{xy-1}\right|\le\frac{2|y|+2}{2|y|-1}=1+\frac{3}{2|y|-1}\le 2$ with equality only for $|x|=|y|=2$. So we need only check the special case $|x|=|y|=2$ and otherwise if the quotient can be $-1$ or $0$ or $1$.

• $(2,2)$ leads to quotient $2$; $(2,-2)$ and $(-2,2)$ lead to $-\frac 25$; $(-2,-2)$ leads to $-\frac 23$
• The quotient is $0$ iff $y=-x-2$.
• The quotient is $1$ iff $x+y+2=xy-1$, i.e., $(x-1)(y-1)=4$; the finitely many factorizations give us finitely many solutions $x=y=3$ or $x=2,y=5$ (all else is exclude by $|y|\ge |x|\ge 2$)
• The quotient is $-1$ iff $x+y+2=1-xy$, i.e., $(x+1)(y+1)=0$ - contradicting $|y|\ge|x|\ge 2$.

So in summary, as solutions $(x,y;\frac{x+y+2}{xy-1})$ we have the families $$ (0,t;-t-2), (t,0;-t-2), (-1,t;-1),(t,-1;-1),(t,-2-t;0)$$ and the special solutions $$(1,2;5),(2,1;5),(1,3;3),(3,1;3),(1,5;2),(5,1;2),(2,2;2),(3,3;1),(2,5;1),(5,2;1).$$

Answer 4

For this to happen we need $|x+y+2|\geq |xy-1|$ or $x+y+2=0$.

If $x+y+2=0$ we get solutions $(a,-a-2),(-a-2,a)$.

otherwise we have $|x+y|+2\geq |xy|-1\rightarrow |x|+|y|+3\geq |x||y|$.

Notice that if $x$ or $y$ is $0$ it holds trivially.

We now find all possible values $n,m> 0$ so that $n+m+3\geq nm\iff \frac{n+3}{n-1}\geq m$

If $n=1$ any $m$ works.

If $n=2$ we get $m\leq 5$

If $n=3$ we get $m \leq 3$

If $n=4$ we get $m\leq 2$

If $n=5$ we get $m\leq 2$.

If $n\geq 6$ we get $m=1$.

So we first analyse the case when $|x|=1$.

If $x=1$ we get $y-1$ divides $y+3$, so $y-1$ divides $(y+3)-(y-1)=4$, hence $y-1$ can be $-4,-2,-1,1,2$ or $4$. It is easy to see all of them work. So we get $(1,-3),(1,-1),(1,0),(1,2),(1,3),(1,5)$.

If $x=-1$ we get $-(y-1)$ divides $y-1$. So $y+1$ divides $y-1$ and hence $y+1$ divides $y+1-(y-1)=2$. So $y+1$ can be $-2,-1,1,2$. It is easy to see all of them work. So we get $(-1,-3),(-1,-2),(-1,0),(-1,1)$.

Since $x$ and $y$ play symmetric roles in the equation, this gives all solutions with $|x|=1$ or $|y|=1$.

Now we analyse when $|x|=2$.

If $x=2$ we get $2y-1$ divides $y+4$, we need only try with $y=-5,-4,-3,-2,0,2,3,4,5$. We see only $(2,-4),(2,0),(2,2),(2,5)$ work.

If $x=-2$ we get $-2y-1$ divides $y$, so $2y+1$ divides $y$, and so $2y+1$ divides $2y+1-2(y)=1$. So we only try with $2y+1=1$ and $2y+1=-1$. So $(-1,-1)$ and $(-1,1)$ are solutions.

So we only need to check when $|x|,|y|>2$.

If $|x|=3$ we need only check if $|y|=3$. There are four cases, and out of these only $(3,3)$ works.

This finishes all the cases.

(I'll try to make this a bit cleaner when I get home).