Canadian Mathematical Olympiad

Solution:
Subtracting the second equation from the first gives $x y - x z = 2 z - 2 y$. Factoring $y - z$ from each side and rearranging gives
$$(x + 2)(y - z) = 0$$
so either $x = -2$ or $z = y$.

If $x = -2$, the first equation becomes $-2 y = z + 2 - y$, or $y + z = -2$. Substituting $x = -2$, $y + z = -2$ into the third equation gives $y z = -2 - (-2) = 0$. Hence either $y$ or $z$ is $0$, so if $x = -2$, the only solutions are $(-2, 0, -2)$ and $(-2, -2, 0)$.

If $z = y$ the first equation becomes $x y = -x$, or $x(y + 1) = 0$. If $x = 0$ and $z = y$, the third equation becomes $y^2 = -2 y$ which gives $y = 0$ or $y = -2$. If $y = -1$ and $z = y = -1$, the third equation gives $x = -1$. So if $y = z$, the only solutions are $(0, 0, 0)$, $(0, -2, -2)$ and $(-1, -1, -1)$.

In summary, there are 5 solutions: $(-2, 0, -2)$, $(-2, -2, 0)$, $(0, 0, 0)$, $(0, -2, -2)$ and $(-1, -1, -1)$.

Solution:
Adding $x$ to both sides of the first equation gives
$$x(y + 1) = z - y = (z + 1) - (y + 1) \Rightarrow (x + 1)(y + 1) = z + 1.$$
Similarly manipulating the other two equations and letting $a = x + 1$, $b = y + 1$, $c = z + 1$, we can write the system in the following way.
$$\left\{\begin{array}{l} a b = c \\ a c = b \\ b c = a \end{array}\right.$$
If any one of $a, b, c$ is $0$, then it's clear that all three are $0$. So $(a, b, c) = (0, 0, 0)$ is one solution and now suppose that $a, b, c$ are all nonzero. Substituting $c = a b$ into the second and third equations gives $a^2 b = b$ and $b^2 a = a$, respectively. Hence $a^2 = 1$, $b^2 = 1$ (since $a, b$ nonzero). This gives 4 more solutions: $(a, b, c) = (1, 1, 1)$, $(1, -1, -1)$, $(-1, 1, -1)$ or $(-1, -1, 1)$. Reexpressing in terms of $x, y, z$, we obtain the 5 ordered triples listed in Solution 1.