To prove the necessity, we need to show that $4 \nmid n$ and $n$ has no prime factors of the form $4k+3$. In this case, we must have $(x, y) = (x, n) = (y, n) = 1$. If $4 \mid n$, then we have
$$x^{2} + y^{2} \equiv 0 \pmod{4}$$

This is impossible when $(x, 2) = (y, 2) = 1$. Therefore, we must have $4 \nmid n$. If there exists a prime $p \equiv 3 \pmod{4}$ such that $p \mid n$, then we have $(p, x) = (p, y) = 1$ and
$$x^{2} + y^{2} \equiv 0 \pmod{p}$$

Following the proof of the necessity in Theorem 2, we can deduce that $p \equiv 1 \pmod{4}$, which is a contradiction.

For sufficiency, $1 = 1^{2} + 0^{2}$ and $2 = 1^{2} + 1^{2}$, so the conclusion is obviously true. We will use induction to prove that when $n = p^{\alpha}$ and $p \equiv 1 \pmod{4}$,
$$x^{2} + y^{2} = p^{\alpha}, \quad (x, y) = 1$$

has a solution. When $\alpha = 1$, this is Theorem 1, so there is a solution. Assume the conclusion holds for $\alpha = k$. We will prove that the conclusion also holds for $\alpha = k + 1$. Suppose
$$\begin{array}{ll} x_{1}^{2} + y_{1}^{2} = p, & (x_{1}, y_{1}) = 1 \\ x_{k}^{2} + y_{k}^{2} = p^{k}, & (x_{k}, y_{k}) = 1 . \end{array}$$

From this and equation (5), we have
$$\begin{array}{l} (x_{1} x_{k} + y_{1} y_{k})^{2} + (x_{1} y_{k} - y_{1} x_{k})^{2} = p^{k+1} \\ (x_{1} x_{k} - y_{1} y_{k})^{2} + (x_{1} y_{k} + y_{1} x_{k})^{2} = p^{k+1} \end{array}$$

At this point,
$$d_{1} = (x_{1} x_{k} + y_{1} y_{k}, x_{1} y_{k} - y_{1} x_{k}) = 1$$

or
$$d_{2} = (x_{1} x_{k} - y_{1} y_{k}, x_{1} y_{k} + y_{1} x_{k}) = 1$$

at least one of them holds. If not, $d_{1} > 1$ and $d_{2} > 1$, then from equations (10) and (11), we know $p \mid d_{1}$ and $p \mid d_{2}$. Therefore, $p \mid 2 x_{1} x_{k}$, so at least one of $p \mid x_{1}$ or $p \mid x_{k}$ must hold, but from equations (8) and (9), this is impossible. Therefore, when $d_{1} = 1$, from equation (10), the conclusion holds for $\alpha = k + 1$. When $d_{2} = 1$, from equation (11), the conclusion holds for $\alpha = k + 1$. This proves that for any $\alpha \geq 1$, the indeterminate equation (7) always has a solution.

Next, we prove that if $(n_{1}, n_{2}) = 1$ and
$$\begin{array}{ll} x_{1}^{2} + y_{1}^{2} = n_{1}, & (x_{1}, y_{1}) = 1 \\ x_{2}^{2} + y_{2}^{2} = n_{2}, & (x_{2}, y_{2}) = 1 \end{array}$$

then when $n = n_{1} n_{2}$, the indeterminate equation (6) must have a solution satisfying $(x, y) = 1$. From equation (5), we have
$$\left(x_{1} x_{2} + y_{1} y_{2}\right)^{2} + \left(x_{1} y_{2} - y_{1} x_{2}\right)^{2} = n_{1} n_{2}$$

If $d = \left(x_{1} x_{2} + y_{1} y_{2}, x_{1} y_{2} - y_{1} x_{2}\right) > 1$, then there must be a prime $q \mid d$. Therefore,
$$\begin{array}{l} x_{1} x_{2} \equiv -y_{1} y_{2} \pmod{q}, \\ x_{1} y_{2} \equiv y_{1} x_{2} \pmod{q} \end{array}$$

At this point, $q \nmid x_{1}$. Otherwise, if $q \mid x_{1}$, from the above two equations, we get $q \mid y_{1} y_{2}$ and $q \mid y_{1} x_{2}$. From $(x_{1}, y_{1}) = 1$, we deduce $q \mid y_{2}$ and $q \mid x_{2}$, which contradicts $(x_{2}, y_{2}) = 1$. Similarly, we can prove that $q$ cannot divide $y_{1}$, $x_{2}$, or $y_{2}$. From equations (15) and (16), we easily get
$$\begin{array}{l} x_{1} \left(x_{2}^{2} + y_{2}^{2}\right) \equiv 0 \pmod{q}, \\ x_{2} \left(x_{1}^{2} + y_{1}^{2}\right) \equiv 0 \pmod{q} \end{array}$$

From this and $q \nmid x_{1} x_{2}$, we get (note: $q$ is a prime)
$$x_{1}^{2} + y_{1}^{2} \equiv x_{2}^{2} + y_{2}^{2} \equiv 0 \pmod{q}$$

From this and equations (12) and (13), we get $n_{1} \equiv n_{2} \equiv 0 \pmod{q}$, which contradicts $(n_{1}, n_{2}) = 1$. Therefore, $d = 1$. From this and equation (14), we have proven the desired conclusion.

Combining all the above discussions, we have proven the sufficiency. Proof complete.