Baltic Way 1994 · Problem 1

Solution: Note that

$$(x \circ y) \circ z = x + y + z - x y - y z - x z + x y z = (x-1)(y-1)(z-1) + 1.$$

Hence

$$(x \circ y) \circ z + (y \circ z) \circ x + (z \circ x) \circ y = 3((x-1)(y-1)(z-1) + 1).$$

Now, if the required equality holds we have $(x-1)(y-1)(z-1) = -1$. There are only four possible decompositions of $-1$ into a product of three integers. Thus we have four such triples, namely $(0, 0, 0)$, $(0, 2, 2)$, $(2, 0, 2)$ and $(2, 2, 0)$.