SUPERMAN2 wrote: Find all positive integers $ x,y,z,t$ such that $ x,y,z$ are pairwise coprime and $ (x + y)(y + z)(z + x) = xyzt$. $ xyz|(x+y)(y+z)(z+x) \Leftrightarrow \begin{cases} x|(x+y)(y+z)(z+x) \\ y|(x+y)(y+z)(z+x) \\ z|(x+y)(y+z)(z+x) \end{cases}$ $ \Leftrightarrow \begin{cases} x|y+z \\ y|z+x \\ z|x+y \end{cases} \Leftrightarrow xyz|x+y+z$ WLOG, $ x \le y \le z$. $ xyz|x+y+z \le 3z \Rightarrow x=1$ $ yz|1+y+z \ge yz \Rightarrow (y-1)(z-1) \le 2 \Rightarrow y=1$ or $ 2$ Case 1: $ y=1 \Rightarrow z|2+z \Rightarrow z=1,2 \Rightarrow (x,y,z,t)=(1,1,1,8), (1,1,2,9)$ as solutions. Case 2: $ y=2 \Rightarrow 2z|3+z \Rightarrow z=3 \Rightarrow (x,y,z,t)=(1,2,3,10)$ as solution. So $ (x,y,z,t)=(1,1,1,8),(1,1,2,9),(1,2,1,9),(2,1,1,9),(1,2,3,10),(1,3,2,10),(2,1,3,10),(2,3,1,10),(3,1,2,10),(3,2,1,10)$ are the only solutions.