2. For integers $x, y, z$ it holds that $x^{2}+y-z=10, x^{2}-y+z=22$. Find the smallest possible value of the expression $x^{2}+y^{2}+z^{2}$.
2. By adding the given equations, we get

$$
\left(x^{2}+y-z\right)+\left(x^{2}-y+z\right)=2 x^{2}=10+22=32,
$$

thus $x^{2}=16$ or $x= \pm 4$. If we substitute this result back into the first or second of the given equations, in both cases we get $z=y+6$.

Now let's substitute both obtained equations into the expression we need to minimize:

$x^{2}+y^{2}+z^{2}=16+y^{2}+(y+6)^{2}=2 y^{2}+12 y+52=2\left(y^{2}+6 y+26\right)=2\left((y+3)^{2}+17\right)$.

Since $(y+3)^{2} \geq 0$, the smallest possible value of the given expression is $2 \cdot 17=34$. This value is achieved by the modified expression for $y=-3$, from which we get $z=y+6=3$, so the found numbers $x, y$, and $z$ are indeed integers, as required by the problem.

For a complete solution, award 6 points. For deriving both relationships $x^{2}=16$ and $z=y+6$, award 1 point each, 1 point for rewriting the expression in the problem in terms of one unknown, 2 points for transforming it into an expression with a square, and 1 point for finding the minimum 34. However, if it is not stated that the value 34 is indeed achieved by the expression, deduct 1 point.