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 $$\begin{array}{rl}(x_0,y_0,z_0)& =(1,1,1)\\ (x_{n+1},y_{n+1},z_{n+1})& =(x_n^2+y_n^2-z_n^2,2x_n{z}_n,2y_n{z}_n),\end{array}$$ 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 $2x_n{z}_n$ and $2y_n{z}_n$ are even, we have that $x_n^2+y_n^2-z_n^2$ is odd, so $p\ne 2$. Hence $p|x_n{z}_n$ and $p|y_n{z}_n$. 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. □