Given that $a$ and $b$ are positive integers and $a > b$, from equation (28) we have $x = 2ab > 0$, $y = a^2 - b^2 > 0$, $z = a^2 + b^2 > 0$, which means $x, y, z$ satisfy the conditions $x > 0$, $y > 0$, $z > 0$, and $2 \mid x$ in equation (27). From equation (28), we have
$$\begin{aligned} x^2 + y^2 & = 4a^2b^2 + (a^2 - b^2)^2 = a^4 + 2a^2b^2 + b^4 \\ & = (a^2 + b^2)^2 = z^2 \end{aligned}$$

Therefore, equation (28) is a positive integer solution that satisfies equation (21). Now, let $(x, y) = d$, then $d \mid x$, $d \mid y$, $d^2 \mid x^2$, $d^2 \mid y^2$, and $d^2 \mid (x^2 + y^2)$. Thus, from equation (21), we get $d^2 \mid z^2$, which implies $d \mid z$. Let $y = dl$ and $z = dm$, then from equation (28) we have
$$\begin{array}{l} 2a^2 = a^2 - b^2 + a^2 + b^2 = y + z = d(l + m) \\ 2b^2 = a^2 + b^2 - (a^2 - b^2) = z - y = d(m - l) \end{array}$$

Therefore, $d \mid 2a^2$ and $d \mid 2b^2$, which implies $d \mid 2(a^2, b^2)$. Since $(a, b) = 1$ in equation (28), we have $(a^2, b^2) = 1$. From $d \mid 2(a^2, b^2)$ and $(a^2, b^2) = 1$, we get $d \mid 2$. Since $a - b = a + b - 2b$ and $2 \nmid (a + b)$ in equation (28), we have
$$2 \nmid (a + b), \quad 2 \nmid (a - b),$$

which means $2 \nmid (a^2 - b^2)$. From equation (28), we have $2 \nmid y$. Since $2 \nmid y$, $d \mid 2$, and $(x, y) = d$, we get $d = 1$, i.e., $(x, y) = 1$. Therefore, equation (28) satisfies all the conditions in equation (27).

Furthermore, when $x, y, z$ are any positive integer solutions that satisfy all the conditions in equation (27) for equation (21), $x, y, z$ can be expressed using the formulas in equation (28) (proof see Exercise 9). Hence, the theorem is proved.