Solve the system of equations below, listing all un-ordered triplets of integers:
x^2 −(y+z+yz)x+ (y+z)yz = 0
y^2 −(z+x+zx)y+ (z+x)zx = 0
z^2 −(x+y+xy)z+ (x+y)xy = 0
Let p=y+z and q=y+z
Then the 1st equation becomes x^2-(p+q)*x+p*q=0
equivalent to (x-p)*(x-q)=0
which implies x1=p x2=q i.e. x=y+z or x=y*z
By symmetry two other equations force y=x+z or y=x*z
and z=x+y or z=x*y.
Clearly, out of 8 possible combinations of "binary triplets" only
(0,0,0) and (1,1,1) qualify.