1. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
x+y & =1, \\
x-y & =a, \\
-4 a x+4 y & =z^{2}+4
\end{aligned}
$$

for the unknowns $x, y, z$ and the real parameter $a$.
1. By adding the first and second equations of the given system, we get $2 x=1+a$, and by subtracting the second equation from the first, we get $2 y=1-a$. Therefore,

$$
x=\frac{1}{2}(1+a), \quad y=\frac{1}{2}(1-a) \text {. }
$$

Substituting $x$ and $y$ into the third equation of the original system, we get the equation

$$
-2 a(1+a)+2(1-a)=z^{2}+4 \text { or } z^{2}+2 a^{2}+4 a+2=0
$$

which we can rewrite as

$$
z^{2}+2(a+1)^{2}=0 \text {. }
$$

Both addends on the left side of the last equation are non-negative numbers. Their sum is 0 precisely when $z=0, a=-1$. Substituting these values into (1) gives $x=0, y=1$.

Conclusion: The given system of equations has a solution only for $a=-1$, and that solution is $x=0, y=1, z=0$. Verification is not necessary in this process.

For a complete solution, award 6 points, of which 2 points for correctly expressing $x$ and $y$ in terms of the parameter $a$ from the first two equations, 3 points for solving equation (2) that arises from substituting these values into the third equation, and 1 point for providing the correct answer.