3. Determine the real numbers $a, b$ for which the system

$$
\begin{aligned}
x^{2}+y^{2}+2 z^{2} & =16 \\
x y z^{2}+x y+z^{2} & =a \\
x+y+2 z & =b
\end{aligned}
$$

has exactly one solution in the set of real numbers.

A natural attempt is to completely solve the given system with respect to the parameters $a$, $b$ and from this solution determine for which $a, b$ the system has exactly one solution. However, we believe that this is a rather cumbersome approach in this case. Two unknowns can indeed be eliminated (for example, $x$ and $y$), but the resulting fourth-degree equation with two parameters will not be solvable by students.

Therefore, we will try to use the fact from the beginning that the system is supposed to have exactly one solution.
SOLUTION. Assume that the system has exactly one solution $x=s, y=t, z=u$. Since the unknowns $x$ and $y$ appear in the same form in all equations, it can be guessed and verified that $x=t, y=s$ and $z=u$ is also a solution to the given system. Since the system has a unique solution, it must be that $t=s$, and thus $x=y$. Substituting, we get the system

$$
\begin{gathered}
x^{2}+z^{2}=8, \\
x^{2} z^{2}+x^{2}+z^{2}=a, \\
x+z=\frac{1}{2} b .
\end{gathered}
$$

If $(x, z)$ is some solution to this system, then the triplet $(x, x, z)$ is a solution to the original system. Therefore, if the original system has a unique solution, the new system must also have a unique solution (*). However, this new system is also symmetric with respect to the unknowns $x$ and $z$. Therefore, it will have a unique solution only if $x=z$.

Substituting, we get the system

$$
\begin{aligned}
x^{2} & =4, \\
x^{4}+2 x^{2} & =a, \\
x & =\frac{1}{4} b,
\end{aligned}
$$

which has a unique solution. According to the first equation, this is either $x=2$ (then $b=8, a=24$), or $x=-2$ (then $b=-8, a=24$).

By these considerations, we have arrived at the following conclusion:

If the given system has exactly one solution, then it is only for $a=24, b=8$, and that $x=2$, $y=2, z=2$, or for $a=24, b=-8$, and that $x=-2, y=-2, z=-2$.

We still need to verify whether in these two cases the given system has any other solution (other than the symmetric one we calculated not by equivalent transformations, but by simplification).

Let $a=24, b=8$. Substituting, we get the system

$$
\begin{aligned}
x^{2}+y^{2}+2 z^{2} & =16, \\
x y z^{2}+x y+z^{2} & =24, \\
x+y+2 z & =8 .
\end{aligned}
$$

This system can be solved in several ways. We will present two here.
a) Eliminate the unknowns $x, y$, for example, by first rearranging the equations:

$$
\begin{aligned}
x^{2}+y^{2} & =16-2 z^{2}, \\
x y & =\frac{24-z^{2}}{1+z^{2}}, \\
x+y & =8-2 z .
\end{aligned}
$$

We then get

$$
(8-2 z)^{2}=(x+y)^{2}=x^{2}+y^{2}+2 x y=16-2 z^{2}+2 \cdot \frac{24-z^{2}}{1+z^{2}} .
$$

After rearrangement, we get

$$
3 z^{4}-16 z^{3}+28 z^{2}-16 z=0 .
$$

Given that we know $z=2$ is a root of this equation, we can rearrange it step by step to the form

$$
z(z-2)^{2}(3 z-4)=0 .
$$

From this, it follows that $z=0, z=\frac{4}{3}$, or $z=2$.

If $z=0$, we get

$$
\begin{aligned}
x^{2}+y^{2} & =16, \\
x y & =24, \\
x+y & =8
\end{aligned}
$$

and it is easy to verify that this system has no solution (the numbers $x, y$ would have to be the roots of the quadratic equation $t^{2}-8 t+24=0$, which has a negative discriminant).

If $z=\frac{4}{3}$, we get

$$
\begin{aligned}
x^{2}+y^{2} & =\frac{112}{9}, \\
x y & =8, \\
x+y & =\frac{16}{3}
\end{aligned}
$$

and again it is easy to verify that this system has no solution.

If $z=2$, we get

$$
\begin{aligned}
x^{2}+y^{2} & =8, \\
x y & =4, \\
x+y & =4 .
\end{aligned}
$$

It is easy to see that this system has a unique solution $x=y=2$.

b) A more clever approach uses only the first and third equations and the inequality between the quadratic and arithmetic means:

$$
4=2^{2}=\left(\frac{1}{4}(x+y+z+z)\right)^{2} \leqq \frac{1}{4}\left(x^{2}+y^{2}+z^{2}+z^{2}\right)=4
$$

Equality between the arithmetic and quadratic means occurs precisely when all terms are equal. Hence, $x=y=z=2$.

The case $a=24, b=-8$ is evaluated similarly, and in that case, the solution is also unique.

Answer. The given system has a unique solution for $a=24, b=8$ or $a=24, b=-8$.