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For $x,y\in \mathbb{Z}^+,$ when is the following expression an integer?


The associated Diophantine equation is symmetric in $x, y, z$, but I couldn't do anything more with that. I tried several factoring tricks without luck. The best I could do was find three solutions such that $0<x\le y\le z$. They are: $(2,5,8)$, $(2,4,13)$ and $(3,3,7)$.

The expression seems to converge pretty quickly to some non-integer between 1 and 2.

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TonyK, they are correct. – sasha Nov 14 '11 at 19:29
Sorry, my mistake. Comment deleted. – TonyK Nov 14 '11 at 19:53
The observation of symmetry is interesting. The equation is $xyz - (xy+yz+yz) - (x+y+z) + 1 = 0$. The quadratic term can be removed by writing everything in terms of $(x-1,y-1,z-1)$. – zyx Nov 14 '11 at 20:57
"The observation of symmetry is interesting," says zyx. :) – sasha Nov 14 '11 at 21:15
up vote 4 down vote accepted

Since $$ \frac{(1-x)-(1+x)y}{(1+x)+(1-x)y} = \frac{ xy+x+y-1}{xy-x-y-1} = 1 + \frac{2(x+y) }{xy-x-y-1} $$

and $ 2x+2y < xy - x -y - 1 $ if $ 3(x+y) < xy - 1 .$ Suppose $ x\leq y$, then $ 3(x+y) \leq 6y \leq xy-1 $ if $ x\geq 7. $ So all solutions must have $0\leq x< 7 $ so it is reduced to solving $7$ simpler Diophantine equations.

If $x=0 $ then $ \displaystyle z= 1 - \frac{2y}{y+1}$ so the only solutions are $ (0,0,1)$ and $ (0,1,0).$

If $x=1$ then $ \displaystyle z= -y$ so $(1,m,-m)$ is a solution for $ m\geq 1.$

If $x=2$ then $ \displaystyle z = 1 + \frac{4+2y}{y-3}$ which is an integer for $y=1,2,4,5,8,13.$

I will leave you to find the others. Each of the cases are now simple Diophantine equations.

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Note, $2x+2y<xy - x - y -1$ if and only if $(x-3)(y-3)>10$ – Thomas Andrews Nov 14 '11 at 19:48
Ah, the ol' add-and-subtract-the-same-quantity-in-the-numerator trick. Forgot about that! Thanks. And my suspicion about convergence has been confirmed. – sasha Nov 14 '11 at 19:49
You've only proven $x\leq 7$, if $(x,y)=(5,8)$, then you have a solution. – Thomas Andrews Nov 14 '11 at 19:54
And $(x,y)=(4,13)$ – Thomas Andrews Nov 14 '11 at 19:57
@Thomas I realized that, hopefully my edit has fixed that issue. – Ragib Zaman Nov 14 '11 at 19:59

We write out a solution, to check whether something like it can be "borrowed" for a problem set.

The numerator is $1-x-y-xy$, which can be written as $-[(x+1)(y+1)-2]$. The denominator is $-[(x-1)(y-1)-2]$. So our fraction is $$\frac{(x+1)(y+1)-2}{(x-1)(y-1)-2}.$$ We can take $x=1$ or $y=1$. The ratio is then an integer, albeit negative. This gives infinitely many trivial solutions. We will list the non-trivial solutions.

By symmetry we can assume that $x \le y$. Let $x=2$. Then we want $(-1-3y)/(3-y)$ to be an integer. But $$\frac{3y+1}{y-3}=3+\frac{10}{y-3}.$$ So $y-3$ must divide $10$, giving (since $y \ge x$), the solutions $y=2$, $4$, $5$, $8$, and $13$.

Next we deal with $x=3$. A calculation similar to the previous one (but shorter) gives that the only $y\ge 3$ are given by $y=3$ and $y=7$.

Next we deal with $x=4$. We need $\dfrac{5y+3}{3y-5}$ to be an integer. Any common divisor of these two numbers must divide $3(5y+3)-5(3y-5)$, which is $34$. The only divisor of $34$ which is of the form $3y-5$, where $y \ge 4$, is given by $y=13$.

Similarly, if $x=5$, we get the solution $y=8$.

For the rest, we use essentially the analysis of @Ragib Zaman. One needs to verify that $2x+2y< xy-x-y-1$, or equivalently that $(x-3)(y-3)>10$. This is true if $x\ge 6$ and $y\ge 7$. (It fails at $x=y=6$, but that is not a solution.)

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I appreciate the analysis. – sasha Nov 15 '11 at 16:47

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