yz+∧·(y+z)=0
xz+∧·(x+z)=0
xy+∧·(x+y)=0
Prin insumare obtinem yz+xz+xy+∧·(2x+2y+2z)=0 <=> yz+xz+xy+2∧·(x+y+z)=0 dar deoarece xy+yz+zx=12 => ∧·(x+y+z)=-6 => ∧·(y+z)=-6-∧z =>daca inlocuim in prima egalitate avem in final z·(y-∧)=6. Prin analogie => y·(x-∧)=6 si respectiv x·(z-∧)=6 => pentru x,y,z∈N\{0} => ∧=-1 => z·(y+1)=6, y·(x+1)=6, z·(y+1)=6 => x=y=z=2 => S={(2;2;2} .
