[ Quadratic equations. Vieta's theorem ]

Solve the system of equations:

$$
\begin{aligned}
& x^{2}+y^{2}-2 z^{2}=2 a^{2} \\
& x+y+2 z=4\left(a^{2}+1\right) \\
& z^{2}-x y=a^{2}
\end{aligned}
$$
$\left(4\left(a^{2}+1\right)-2 z\right)^{2}=(x+y)^{2}=x^{2}+y^{2}+2 x y=2 z^{2}+2 a^{2}+2\left(z^{2}-a^{2}\right)=4 z^{2}$, from which $16\left(a^{2}+1\right)^{2}=16\left(a^{2}+1\right) z$, that is $z=$ $a^{2}+1$. Now the second and third equations can be written as:

$x+y=2\left(a^{2}+1\right)$

$x y=a^{4}+a^{2}+1$.

Solving this system reduces to solving the quadratic equation $t^{2}-2\left(a^{2}+1\right) t+a^{4}+a^{2}+1=0$; solving it, we find $x=a^{2} \pm a+1, y=a^{2} \mp a+1$.

Answer

$\left(a^{2}+a+1, a^{2}-a+1, a^{2}+1\right), \quad\left(a^{2}-a+1, a^{2}+a+1, a^{2}+1\right)$.

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