Notice, when $(a, m)=1$, this conclusion is precisely Euler's theorem, hence, this conclusion can be regarded as a generalization of Euler's theorem.

If one of $a, m$ equals 1, the proposition is obvious, so we assume $a, m$ are both greater than 1.

Let $m=m_{1} \cdot m_{2}$, where the prime factors of $m_{1}$ are all prime factors of $a$, and $\left(a, m_{2}\right)=1$. Consequently, $\left(m_{1}, m_{2}\right)=1$. To prove (2) holds, we only need to prove respectively:
$$\begin{array}{l} a^{m} \equiv a^{m-q(m)}\left(\bmod m_{2}\right) \\ a^{m} \equiv a^{m-q(m)}\left(\bmod m_{1}\right) \end{array}$$

Since (3) $\Leftrightarrow a^{\varphi(m)} \equiv 1\left(\bmod m_{2}\right)$ (because $\left(a, m_{2}\right)=1$), and $\varphi(m)=\varphi\left(m_{1}\right.$, $\left.m_{2}\right)=\varphi\left(m_{1}\right) \varphi\left(m_{2}\right)$ (the proof of this conclusion is given in the exercises). Using Euler's theorem, we know $a^{q\left(m_{2}\right)} \equiv 1\left(\bmod m_{2}\right)$, hence $\left(a^{\left.q \phi m_{2}\right)}\right)^{q\left(m_{1}\right)} \equiv 1\left(\bmod m_{2}\right)$, i.e.,
$$a^{\phi(m)} \equiv 1\left(\bmod m_{2}\right),$$

so, (3) holds.
For (4), we only need to prove: $m_{1} \mid a^{m-\alpha(m)}$. For this, we only need to prove: for any prime factor $p$ of $m_{1}$, we have
$$v_{p}\left(m_{1}\right) \leqslant v_{p}(a)(m-\varphi(m))$$

In fact, by the definition of $m_{1}$, we know $v_{p}\left(m_{1}\right)=v_{p}(m) \geqslant 1, v_{p}(a) \geqslant 1$, thus, we have
$$\begin{aligned} v_{p}\left(m_{1}\right) & =v_{p}(m) \leqslant 2^{v_{p}(m)-1} \leqslant p_{\rho}^{v_{\rho}(m)-1} \\ & \leqslant(p-1) p^{v_{\rho}(m)-1} \\ & \leqslant(p-1) p^{v_{\rho}(m)-1} \varphi\left(\frac{m}{p^{v_{\rho}(m)}}\right) \\ & =p^{v_{\rho}(m)} \varphi\left(\frac{m}{p^{v_{\rho}(m)}}\right)-p^{v_{\rho}(m)-1} \varphi\left(\frac{m}{p^{v_{\rho}(m)}}\right) \\ & \leqslant p^{v_{\rho}(m)} \varphi\left(\frac{m}{p_{\rho}^{v_{\rho}(m)}}\right)-\varphi\left(p^{v_{\rho}(m)}\right) \varphi\left(\frac{m}{p^{v_{\rho}(m)}}\right) \\ & =p^{v_{\rho}(m)} \varphi\left(\frac{m}{p_{\rho}^{v_{\rho}(m)}}\right)-\varphi(m) \\ & \leqslant p^{v_{\rho}(m)} \frac{m}{p_{\rho}^{v_{\rho}(m)}}-\varphi(m) \\ & =m^{-} \varphi \varphi(m) \\ & \leqslant v_{p}(a)(m-\varphi(m)) \end{aligned}$$

So, (5) holds, and consequently, (4) holds.
In summary, the proposition holds.