Prove that if $x, y$ are not coprime, then there exists a prime $p$ such that $p \mid x, p \mid y$. By (1), we know $p \mid z^{2}$. From this and Theorem 1 of Chapter 1, §5, we conclude $p \mid z$. However, this contradicts $(x, y, z)=1$. Similarly, we can prove $(y, z)=1$ and $(z, x)=1$. By $(x, y)=1$, we know that $x, y$ cannot both be even. $x, y$ cannot both be odd either. Because if they were both odd, it would imply $4 \nmid x^{2}+y^{2}$ and $z$ would be even. But by (1), we know
$$4 \mid z^{2}=x^{2}+y^{2}$$

which is a contradiction. Therefore, $x, y$ must be one odd and one even, i.e., equation (5) holds.