Example 5 Solve the congruence equation
$$f(x)=2 x^{7}-x^{5}-3 x^{3}+6 x+1 \equiv 0(\bmod 5) .$$
Solve using the constant congruence identity (3). By polynomial division, we get
$$\begin{aligned}
f(x) & =\left(2 x^{2}-1\right)\left(x^{5}-x\right)-x^{3}+5 x+1 \\
& \equiv\left(2 x^{2}-1\right)\left(x^{5}-x\right)-x^{3}+1(\bmod 5) .
\end{aligned}$$

Therefore, the original congruence equation is equivalent to
$$x^{3} \equiv 1(\bmod 5)$$

and has the same number of solutions. Direct computation shows that the solution is $x \equiv 1(\bmod 5)$.