Pirkuliyev Rovsen wrote: Solve in integer the equation $\frac{1}{2}(x+y)(y+z)(x+z)+(x+y+z)^{3}=1-xyz$ $\iff$ $(2x+y+z)(x+2y+z)(x+y+2z)=2$ and so : $2x+y+z=2$ and $x+2y+z=x+y+2z=1$ $\implies$ $(x,y,z)=(1,0,0)$ $2x+y+z=2$ and $x+2y+z=x+y+2z=-1$ $\implies$ $(x,y,z)=(2,-1,-1)$ $2x+y+z=-2$ and $x+2y+z=-1$ and $x+y+2z=1$ $\implies$ no integer solution Hence the solutions $(1,0,0)$ and $(2,-1,-1)$ and permutations