Example 6 Find the smallest positive integer $n$, such that there exist rational-coefficient polynomials $f_{1}, f_{2}, \cdots, f_{n}$, satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+\cdots+f_{n}^{2}(x) .
$$
(51st IMO Shortlist)
【Analysis】For the case $n=5$,
$$
x^{2}+7=x^{2}+2^{2}+1+1+1 \text {, }
$$

it meets the requirement.
Now we prove that $n \leqslant 4$ does not meet the requirement.
Assume there exist four rational coefficient polynomials $f_{1}(x)$, $f_{2}(x)$, $f_{3}(x)$, $f_{4}(x)$ (which may include the zero polynomial), such that $x^{2}+7=\sum_{i=1}^{4}\left(f_{i}(x)\right)^{2}$, and the degree of the polynomial $f_{i}(x)$ is at most 1.
Let $f_{i}(x)=a_{i} x+b_{i}$. Then
$$
\sum_{i=1}^{4} a_{i}^{2}=1, \sum_{i=1}^{4} a_{i} b_{i}=0, \sum_{i=1}^{4} b_{i}^{2}=7 .
$$

Let $p_{i}=a_{i}+b_{i}$, $q_{i}=a_{i}-b_{i}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{4} p_{i}^{2}=\sum_{i=1}^{4}\left(a_{i}^{2}+2 a_{i} b_{i}+b_{i}^{2}\right)=8, \\
\sum_{i=1}^{4} q_{i}^{2}=\sum_{i=1}^{4}\left(a_{i}^{2}-2 a_{i} b_{i}+b_{i}^{2}\right)=8, \\
\sum_{i=1}^{4} p_{i} q_{i}=\sum_{i=1}^{4}\left(a_{i}^{2}-b_{i}^{2}\right)=-6 .
\end{array}
$$

Let the least common multiple of the denominators of the rational numbers $p_{i}$, $q_{i}$ ($i=1,2,3,4$) be $m$. Then there exist integers $x_{i}$, $y_{i}$ (after converting $p_{i}$, $q_{i}$ to a common denominator $m$, the numerators of $p_{i}$, $q_{i}$), such that
$$
\left\{\begin{array}{l}
\sum_{i=1}^{4} x_{i}^{2}=8 m^{2}, \\
\sum_{i=1}^{4} y_{i}^{2}=8 m^{2}, \\
\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2} .
\end{array}\right.
$$

Assume $m$ is the smallest positive integer satisfying the above system of indeterminate equations.
$$
\text { From } \sum_{i=1}^{4} x_{i}^{2}=8 m^{2} \equiv 0(\bmod 8) \text {, we know } x_{1} \text {, } x_{2} \text {, } x_{3} \text {, }
$$
$x_{4}$ are all even.
$$
\text { From } \sum_{i=1}^{4} y_{i}^{2}=8 m^{2} \equiv 0(\bmod 8) \text {, we know } y_{1} \text {, } y_{2} \text {, } y_{3} \text {, }
$$
$y_{4}$ are all even.
From $\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2} \equiv 0(\bmod 4)$, we know $m$ is also even.
Thus, $\frac{x_{i}}{2}$, $\frac{y_{i}}{2}$, $\frac{m}{2}$ also satisfy the system of indeterminate equations.
This contradicts the minimality of $m$.