Making $x+\frac1y$, $y+\frac1z$, and $z+\frac1x$ all integer.

Solution 1:

Frankly, it's just a matter of working through the algebra...

As @Michael suggests, we may WLOG assume that $x$ is a positive integer, since the equations are invariant under cyclic permutations. Put $x=a \in \mathbb{Z}^+$. Then since $a+\frac{1}{y}$ is an integer, $\frac{1}{y}$ is an integer so we may write $y=\frac{1}{b}$, where $b \in \mathbb{Z}, b \not =0$. Now let $y+\frac{1}{z}=\frac{1}{b}+\frac{1}{z}=c, z+\frac{1}{x}=z+\frac{1}{a}=d$, where $c, d \in \mathbb{Z}$.

We have $z=\frac{1}{c-\frac{1}{b}}=\frac{b}{bc-1}$, so $\frac{b}{bc-1}+\frac{1}{a}=d$, so $ab+(bc-1)=ad(bc-1)$, so $ab=(ad-1)(bc-1)$.

Note that $ab \not =0$ so $ad-1 \not =0, bc-1 \not =0$. We have $a \mid (ad-1)(bc-1)$ and $\gcd(a, ad-1)=1$ so $a \mid bc-1$. We have $bc-1 \mid ab$ and $\gcd(bc-1, b)=1$ so $bc-1 \mid a$. Thus $bc-1= \pm a$, and so $ad-1= \pm b$.

Case 1: We take the positive sign, so that $bc-1=a, ad-1=b$. Then we get $c=\frac{a+1}{b}=\frac{a+1}{ad-1}$. Since $a>0$ we have $|a+1|=a+1>0$ so since $c \in \mathbb{Z}$, $a+1=|a+1| \geq |ad-1| \geq |ad|-1=a|d|-1$, so $2 \geq a(|d|-1)$.

If $|d| \geq 2$ then $2 \geq a(|d|-1)\geq a$, so $a=1, 2$.

If $a=1$, then $bc=2, d=b+1$. We get $(b, c)=(1, 2), (2, 1), (-1, -2), (-2, -1)$. Note $b=-1, -2$ give $d=0, -1$, contradicting $|d| \geq 2$. Thus $(b, c)=(1, 2), (2, 1)$. Note $x=a=1, y=\frac{1}{b}, z=d-\frac{1}{a}=b+1-1=b$. Thus $(x, y, z)=(1, 1, 1), (1, \frac{1}{2}, 2)$. (Easily checked to work)

If $a=2$ then $bc=3, b=2d-1$. We get $(b, c)=(1, 3), (3, 1), (-1, -3), (-3, -1)$. Note $b=1, -1, -3$ gives $d=1, 0, -1$, contradicting $|d| \geq 2$. Thus $b=3, c=1, x=a=2, y=\frac{1}{b}=\frac{1}{3}, z=d-\frac{1}{a}=\frac{b+1}{2}-\frac{1}{2}=\frac{b}{2}=\frac{3}{2}$. Thus $(x, y, z)=(2, \frac{1}{3}, \frac{3}{2})$. (Easily checked to work)

Otherwise $d=0, \pm 1$.

$d=0$ gives $b=-1, c=-(a+1), x=a, y=\frac{1}{b}=-1, z=d-\frac{1}{a}=-\frac{1}{a}$. We thus get an infinite family of solutions $(x, y, z)=(a, -1, -\frac{1}{a}), a \in \mathbb{Z}^+$. (Easily checked to work)

$d=-1$ gives $b=-(a+1), c=\frac{a+1}{ad-1}=\frac{a+1}{-a-1}=-1, x=a, y=\frac{1}{b}=-\frac{1}{a+1}, z=d-\frac{1}{a}=-1-\frac{1}{a}=-\frac{a+1}{a}$. We thus get an infinite family of solutions $(x, y, z)=(a, -\frac{1}{a+1}, -\frac{a+1}{a}), a \in \mathbb{Z}^+$. (Easily checked to work)

$d=1$ gives $b=a-1, c=\frac{a+1}{a-1}=1+\frac{2}{a-1}$ so $(a-1) \mid 2$ so $a-1=1, 2$ since $a-1 \geq 0$. This gives $a=2, 3$, so $(a, b, c)=(2, 1, 3), (3, 2, 2)$. This gives $(x, y, z)=(2, 1, \frac{1}{2}), (3, \frac{1}{2}, \frac{2}{3})$. (Easily checked to work)

Case 2: We take the negative sign, so that $bc-1=-a, ad-1=-b$, so $b=1-ad, c=\frac{1-a}{b}=\frac{1-a}{1-ad}=\frac{a-1}{ad-1}$.

If $a=1$, then $bc=0, b=1-d$. Since $b \not =0$ , $c=0$, so $d=1-b, x=a=1, y=\frac{1}{b}, z=d-\frac{1}{a}=1-b-1=-b$. This gives $(x, y, z)=(1, \frac{1}{b}, -b), b \in \mathbb{Z}, b \not =0$.

Otherwise $a>1$, so that $a-1>0$, so since $c \in \mathbb{Z}$, $a-1=|a-1| \geq |ad-1 \geq |ad|-1=a|d|-1$ so $0 \geq a(|d|-1)$ so $|d| \leq 1$, so $d=0, \pm 1$.

If $d=0$, then $b=1, c=1-a$. We get $(x, y, z)=(a, 1, -\frac{1}{a}), a \in \mathbb{Z}^+$, which is easily checked to work.

If $d=-1$, then $b=1+a, c=\frac{a-1}{-1-a}=-\frac{a-1}{a+1}=-1+\frac{2}{a+1}$. However $a+1>2$, so $c$ is not an integer. Thus no solutions for $d=-1$.

If $d=1$, then $b=1-a, c=1$. We get $(x, y, z)=(a, -\frac{1}{a-1}, \frac{a-1}{a}), a \in \mathbb{Z}^+$, which is easily checked to work.

In summary: The solutions above are $(x, y, z)=(1, 1, 1), (1, \frac{1}{2}, 2), (2, \frac{1}{3}, \frac{3}{2}), (a, -1, -\frac{1}{a}), (a, -\frac{1}{a+1}, -\frac{a+1}{a}), (2, 1, \frac{1}{2}), (3, \frac{1}{2}, \frac{2}{3}),(1, \frac{1}{b}, -b), (a, 1, -\frac{1}{a}), (a, -\frac{1}{a-1}, \frac{a-1}{a})$, where $a \in \mathbb{Z}^+$ and $b \in \mathbb{Z}, b \not =0$. To get all solutions, we take these and their cyclic permutations. Note that $(2, 1, \frac{1}{2})$ is a cyclic permutation of $(1, \frac{1}{2}, 2)$, and $(a, 1, -\frac{1}{a})$ is included in $(-b, 1, \frac{1}{b})$, which is a cyclic permutation of $(1, \frac{1}{b}, -b)$. Thus we can simplify our list of solutions:

All solutions are given by $(x, y, z)=(1, 1, 1), (1, \frac{1}{2}, 2), (2, \frac{1}{3}, \frac{3}{2}), (a, -1, -\frac{1}{a}), (a, -\frac{1}{a+1}, -\frac{a+1}{a}), (3, \frac{1}{2}, \frac{2}{3}),(1, \frac{1}{b}, -b), (a, -\frac{1}{a-1}, \frac{a-1}{a})$ and their cyclic permutations, where $a \in \mathbb{Z}^+$ and $b \in \mathbb{Z}, b \not =0$ (and for the last family of solutions, $a \not =1$).