Solve the following system of equations:
$x^{2}+y z=0$,
$v^{2}+y z=0$,
$(x+v) y=2$,
$(x+v) z=-2$.
Subtract equation (2) from (1); we get

$$
x^{2}-v^{2}=(x+v)(x-v)=0
$$

A product is zero if and only if one of its factors is zero. The assumption $x+v=0$ leads to a contradiction, for example, in equation (3) it would result in $0=2$. Thus, we are left with $x-v=0$, which means $x=v$. Therefore, none of $x, y$, and $z$ can be zero.

From equations (3) and (4), substituting $x=v$ and rearranging, we get

$$
x y=1, \quad \text { that is } \quad y=\frac{1}{x}
$$

and

$$
x z=-1, \quad \text { that is } \quad z=-\frac{1}{x}
$$

Substituting these into (1),

$$
x^{2}-\frac{1}{x^{2}}=0, \quad \text { from which } \quad x= \pm 1
$$

If $x=1$, then $y=1, z=-1$ and $v=1$; if $x=-1$, then $y=-1, z=1$ and $v=-1$.

By substitution, we can verify that the obtained values indeed satisfy the system of equations.

Krisztián Gueth (Szombathely, Kanizsai D. Gymnasium, 12th grade)