Czech-Polish-Slovak Match

We will use the following algebraic formula:
$$(x^2 + y^2 + z^2)^2 = (x^2 + y^2 - z^2)^2 + (2xz)^2 + (2yz)^2.$$
This means that if a positive integer can be represented as a sum of 3 squares, then so can its square. Consequently, if we put
$$\begin{aligned} (x_0, y_0, z_0) &= (1, 1, 1) \\ (x_{n+1}, y_{n+1}, z_{n+1}) &= (x_n^2 + y_n^2 - z_n^2, 2x_nz_n, 2y_nz_n), \end{aligned}$$
then by a trivial induction we obtain that $x_n^2 + y_n^2 + z_n^2 = 3^{2n}$.
It remains to show that $\gcd(x_n, y_n, z_n) = 1$ for all $n$ and we proceed by induction.
Suppose $\gcd(x_n, y_n, z_n) = 1$, but $x_{n+1}, y_{n+1}, z_{n+1}$ have some common prime divisor $p$. Observe that since $3^{2n}$ is odd, but $2x_nz_n$ and $2y_nz_n$ are even, we have that $x_n^2 + y_n^2 - z_n^2$ is odd, so $p \ne 2$. Hence $p|x_nz_n$ and $p|y_nz_n$. Then either $p|z_n$, or $p|x_n$ and $p|y_n$. In the latter case we could infer from $p|x_n^2 + y_n^2 - z_n^2$ that in fact also $p|z_n$, which contradicts the assumption that $\gcd(x_n, y_n, z_n) = 1$. Hence we are left with the first case: $p|z_n$.
Since $p|x_n^2 + y_n^2 - z_n^2$ and $p|z_n$, we also have that $p|x_n^2 + y_n^2 + z_n^2 = 3^{2n}$. Hence in fact $p = 3$. But the only quadratic residues modulo 3 are 0 and 1, so the two possibilities for a sum of three squares to be divisible by 3 is that either all or none of them is divisible by 3. The former case is excluded by the assumption that $\gcd(x_n, y_n, z_n) = 1$ and the latter by $p|z_n$.
That's all. □