Example 2 Let the odd function $f(x)$ have the domain $\mathbf{R}$. It satisfies $f(x)+f(x+2)=a, f(1)=0$, where $a$ is a constant. Try to determine how many roots the equation $f(x)=0$ has at least in the interval $(-3,7)$.

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Analysis: Deriving new equations from abstract equations to investigate the periodicity of $f(x)$.
Solution: In $f(x)+f(x+2)=a$, replace $x$ with $x+2$, yielding $f(x+2)+f(x+4)=a$.
Therefore, $f(x)=f(x+4)$, so 4 is a period of $f(x)$. Thus, $f(5)=f(1)=0, f(3)=$ $f(-1)=-f(1)=0, f(4)=f(0)=0$. Therefore, $-1,0,1,3,4,5$ are 6 roots of the equation $f(x)=0$ within $(-3,7)$.

On the other hand, $f(-2)=f(-2+4)=f(2)=-f(-2)$. Therefore, $f(-2)=0, f(2)=0$, so $f(6)=f(2+4)=f(2)=0$. Therefore, $-2,2,6$ are also 3 roots of the equation $f(x)=0$ within $(-3,7)$.
Thus, the equation $f(x)=0$ has at least 9 roots within $(-3,7)$.
! Evaluation: This problem utilizes the periodicity and parity of the function by assigning values flexibly, thereby solving the problem quickly.