nayel wrote: Determine the ordered systems $(x,y,z)$ of positive rational numbers for which $x+\frac{1}{y},y+\frac{1}{z}$ and $z+\frac{1}{x}$ are integers. If we write $x=\frac{p}{q}$, $(p,q)$ coprime, $y=\frac{r}{s}$, $(r,s)$ coprime, $z=\frac{t}{u}$, $(t,u)$ coprime, we have : $\frac{p}{q}+\frac{s}{r}=\frac{pr+qs}{qr}$ integer $\implies$ $q|pr$ and so $q|r$ and also $r|qs$ and so $r|q$ and so $q=r$ and $q|(p+s)$ So we have $q=r$, $s=t$ and $u=p$ and : $q|(p+s)$, and so $q|(p+q+s)$ $s|(q+p)$, and so $s|(p+q+s)$ $p|(s+q)$, and so $p|(p+q+s)$ And so $pqs|(p+q+s)$ and very few solutions (since normally $pqs>p+q+s$) : $(1,1,1)$, $(1,1,2)$ and $(1,2,3)$ And so the solutions for $(x,y,z)$ : $(1,1,1)$, $(1,\frac{1}{2},2)$, $(\frac{1}{2},2,1)$, $(2,1,\frac{1}{2})$, $(\frac{1}{2},\frac{2}{3},3)$, $(3,\frac{1}{2},\frac{2}{3})$, $(\frac{2}{3},3,\frac{1}{2})$, $(\frac{1}{3},\frac{3}{2},2)$, $(2,\frac{1}{3},\frac{3}{2})$, $(\frac{3}{2},2,\frac{1}{3})$

nice solution. also the same as mine. can this be solved in any other way?