2 (1) Let $m, n$ be coprime positive integers, $m, n>1$. Let $a$ be an integer coprime to $m n$. Suppose the orders of $a$ modulo $m$ and modulo $n$ are $d_{1}, d_{2}$, respectively, then the order of $a$ modulo $m n$ is $\left[d_{1}, d_{2}\right]$;
(2) Find the order of 3 modulo $10^{4}$.
2. (1) Let the order of $a$ modulo $mn$ be $r$. From $a^{r} \equiv 1(\bmod m n)$, we can deduce $a^{r} \equiv 1(\bmod m)$ and $a^{r} \equiv 1(\bmod n)$. Therefore, $d_{1} \mid r$ and $d_{2} \mid r$, which implies $\left[d_{1}, d_{2}\right] \mid r$. On the other hand, from $a^{d_{1}} \equiv 1(\bmod m)$ and $a^{d_{2}} \equiv 1(\bmod n)$, we can infer $a^{\left[d_{1}, d_{2}\right]} \equiv 1(\bmod m)$ and $a^{\left[d_{1}, d_{2}\right]} \equiv 1(\bmod n)$. Since $(m, n)=1$, it follows that $a^{\left[d_{1}, d_{2}\right]} \equiv 1(\bmod m n)$, thus $r \mid \left[d_{1}, d_{2}\right]$. Combining both results, we conclude $r=\left[d_{1}, d_{2}\right]$.
(2) Direct calculation shows that the order of 3 modulo $2^{4}$ is 4. It is also easy to see that the order of 3 modulo 5 is 4, hence by part (1) of Example 5, the order of 3 modulo $5^{4}$ is $4 \times 5^{3}$. Therefore, by part (1) of this problem, the order of 3 modulo $10^{4}$ is $[4, 4 \times 5^{3}]=500$.