Example 2 Let $a, b_{1}, b_{2}, \cdots, b_{n}, c_{1}, c_{2}, \cdots, c_{n} \in \mathbf{R}$, such that
$$
x^{2 n}+a \sum_{i=1}^{2 n-1} x^{i}+1=\prod_{i=1}^{n}\left(x^{2}+b_{i} x+c_{i}\right)
$$

holds for any real number $x$. Find the values of $c_{1}, c_{2}, \cdots, c_{n}$.
Proof of a lemma first.
Lemma Let $p(z)=z^{2 n}+a \sum_{i=1}^{2 n-1} z^{i}+1$. Then the polynomial $p(z)$ has at least $n-1$ complex roots on the upper unit circle.
Proof In fact, if $p(z)=0$, then
$$
-a=\frac{\left(z^{2 n}+1\right)(z-1)}{z^{2 n}-z} \text {. }
$$

Let $z=\omega^{2}$.
Thus, $-a=\frac{\left(\omega^{4 n}+1\right)\left(\omega^{2}-1\right)}{\omega^{4 n}-\omega^{2}}$
$$
=\frac{\left(\omega^{2 n}+\omega^{-2 n}\right)\left(\omega-\omega^{-1}\right)}{\omega^{2 n-1}-\omega^{-(2 n-1)}} \text {. }
$$

Take $\omega=\mathrm{e}^{\mathrm{i} \theta}$, we get
$$
-a=\frac{2 \cos 2 n \theta \cdot \sin \theta}{\sin (2 n-1) \theta}=\frac{\sin (2 n+1) \theta}{\sin (2 n-1) \theta}-1 .
$$

It suffices to prove: for each real number $a$, the equation about $\theta$
$$
f(\theta)=\sin (2 n+1) \theta+(a-1) \sin (2 n-1) \theta=0
$$

has at least $n-1$ solutions in the interval $\left(0, \frac{\pi}{2}\right)$.
If $a=1$, then when $\theta_{k}=\frac{k \pi}{2 n+1}(k=1,2, \cdots, n)$, we have $f\left(\theta_{k}\right)=0$.
If $a \neq 1$, for $\theta_{k}=\frac{k \pi}{2 n+1}(k=1,2, \cdots, n)$, note that,
$$
\begin{array}{l}
(k-1) \pi<\theta_{k}<k \pi, \\
f\left(\theta_{k}\right) f\left(\theta_{k+1}\right)<0,
\end{array}
$$

i.e., $f(\theta)=0$ has at least one solution in the interval $\left(\theta_{k}, \theta_{k+1}\right)$.

In summary, $p(z)=0$ has at least $n-1$ roots in the interval $(0, \pi)$.
Back to the original problem.
Let the $n-1$ roots of $p(z)=0$ on the upper unit circle be $z_{1}, z_{2}, \cdots, z_{n-1}$. Then their conjugate complex numbers $\overline{z_{1}}, \overline{z_{2}}, \cdots$, $\overline{z_{n-1}}$ are also its roots.
Thus, $x^{2}+b_{k} x+c_{k}=\left(x-z_{k}\right)\left(x-\overline{z_{k}}\right)$.
Therefore, $c_{k}=z_{k} \overline{z_{k}}=1(k=1,2, \cdots, n-1)$.
Since $c_{1} c_{2} \cdots c_{n}=1$, we have $c_{n}=1$, i.e., $c_{1}=c_{2}=\cdots=c_{n}=1$.