Prove that expanding both sides immediately yields the proof.
(10) tells us that if $m_{1}, m_{2}$ are two positive integers, both of which can be expressed as the sum of two squares of integers, then $m_{1}, m_{2}$ must also be expressible as the sum of two squares of integers. Let $n$ be any given positive integer with the factorization
$$n=m^{2} n_{1}, \mu\left(n_{1}\right) \neq 0$$

Here $\mu$ is the Möbius function, defined as
$$\mu(r)=\left\{\begin{array}{l} 1, \text { when } r=1, \\ (-1)^{\prime}, \text { when } r \text { is the product of } l \text { distinct primes, } \\ 0, \text { when } r \text { is divisible by the square of a prime. } \end{array}\right.$$

Thus (11) means that $m^{2}$ is the largest square factor of $\boldsymbol{n}$, and $n_{1}$ no longer contains any square factors. Since $m^{2}=m^{2}+o^{2}$ is clearly the sum of the squares of two integers $m$ and $o$, by Lemma 3, as long as $n_{1}$ can be expressed as the sum of two squares of integers, then $\boldsymbol{n}$ must also be expressible as the sum of two squares of integers. Since $n_{1}$ has no square factors, there must be a factorization
$$n_{1}=p_{1} \cdots p_{s}, s \geqslant 1, p_{1}<\cdots<p_{s}$$

If each $p_{i}(1 \leqslant i \leqslant s)$ is a prime of the form $4 k+1$, or the smallest $p_{1}=2=1^{2}+1^{2}$, then by Theorem 2 of the previous section, each $p_{i}(1 \leqslant i \leqslant s)$ can be expressed as the sum of two positive squares of integers. Repeatedly applying Lemma 3 to these sums of squares, it is easy to see that $n$ must be the sum of two squares of integers.

Conversely, we naturally ask: if $n$ can be expressed as the sum of two squares of integers, must each prime factor in the factorization (13) of $\boldsymbol{n}_{1}$ be a prime of the form $4 k+1$?

Assume some coefficient $p_{11}=4 k+3, p_{0} \mid n_{1}$, is an odd number. Let $(x, y)=d$, then $x=d x_{1}, y=d y_{1},\left(x_{1}, y_{1}\right)=1, \mathrm{~J}$, so
$$n=d^{2}\left(x_{1}^{2}+y_{1}^{2}\right)$$
If $p_{0} \mid y_{1}$, this contradicts $\left(x_{1} \cdot y_{1}\right)=1$, i.e.,
$$x_{1}^{2}+y_{1}^{2}=0\left(\bmod p_{0}\right), p_{0}+x_{1} y_{1}.$$

From $p_{0}+x_{1}$, there must be some $x_{2}, p_{10}+x_{2}$ such that
$$x_{1} x_{2} \equiv 1\left(\bmod p_{0}\right)$$

Using (15) and (16), we get
$$\begin{aligned} 0 & \equiv x_{2}^{2}\left(x_{1}^{2}+y_{1}^{2}\right)=\left(x_{1} x_{2}\right)^{2}+\left(x_{2} y_{1}\right)^{2} \\ & \equiv 1+\left(x_{2} y_{1}\right)^{2}\left(\bmod p_{0}\right) \end{aligned}$$

This indicates that -1 must be a quadratic residue of $p_{0}$, which contradicts $p_{0}=3(\bmod 4)$. In summary, we have proven the following conclusion.