（扩展）欧拉定理

欧拉定理

结论

设 \(a, m \in N^{+}\) 且 \(\gcd(a, m) = 1\)，有 \(a^{\varphi(m)} \equiv 1(\bmod m)\)

证明

设 \(r_1, r_2, ..., r_{\varphi(m)}\) 是模 \(m\) 意义下的简化剩余系，则 \(ar_1, ar_2, ..., ar_{\varphi(m)}\) 也是模 \(m\) 意义下的一个简化剩余系。所以 \(\prod\limits_{i = 1}^{\varphi(m)} r_i \equiv \prod\limits_{i = 1}^{\varphi(m)} ar_i \equiv a^{\varphi(m)} \prod\limits_{i = 1}^{\varphi(m)} r_i(\bmod m)\)。约去 \(\prod\limits_{i = 1}^{\varphi(m)} r_i\) 可得 \(a^{\varphi(m)} \equiv 1(\bmod m)\)

当 \(m\) 为素数时，因为 \(\varphi(m) = m - 1\)，代入欧拉定理可得费马小定理

扩展欧拉定理

结论

\[a^b \equiv \begin{cases} \begin{aligned} &a^{b\ \bmod\ \varphi(m)}, \qquad & \gcd(a, m) = 1 \\ &a^{b}, & \gcd(a, m) \neq 1, b < \varphi(m) \\ &a^{b\ \bmod\ \varphi(m) + \varphi(m)}, & \gcd(a, m) \neq 1, b \geq \varphi(m) \end{aligned} \end{cases} \]

证明

\(\gcd(a, m) = 1\)

由普通欧拉定理得 \(a^{\varphi(m)} \equiv 1 \pmod m\)

设 \(b = k \varphi(m) + r, 0 \leq r < \varphi(m)\)

因为 \(a^b \equiv a^{k \varphi(m)} \times a^r \pmod m\)

又因为 \(a^{\varphi(m)} \equiv 1 \pmod m\)

所以 \(a^b \equiv a^r \equiv a^{b\ \bmod\ \varphi(m)} \pmod m\)

\(\gcd(a, m) \neq 1, b < \varphi(m)\)

证明略。

\(\gcd(a, m) \neq 1, b \geq \varphi(m)\)

对于任意质数 \(p\)，设 \(m = p^r \times s\)，其中 \(r\) 为 \(m\) 含质因子 \(p\) 的个数，即 \(\gcd(p, s) = \gcd(p^r, s) = 1\)

根据普通欧拉定理知 \(p^{\varphi(s)} \equiv 1 \pmod s\)

又根据欧拉函数的性质知 \(\varphi(m) = \varphi(p^r) \times \varphi(s)\)

所以 \(p^{\varphi(m)} \equiv p^{\varphi(p^r) \times \varphi(s)} = (p^{\varphi(s)})^{\varphi(p^r)} \equiv 1 \pmod s\)

---

设 \(p^{\varphi(m)} = k \times s + 1\)，

\[\begin{aligned} p^{\varphi(m) + r} &= (k \times s + 1) \times p^r \\ &= k \times s \times p^r + p^r \\ &= k \times m + p^r \\ \end{aligned} \]

\(\therefore p^{\varphi(m) + r} \equiv p^r \pmod m\)

---

\[\begin{aligned} \because \forall x \geq r, p^x &\equiv p^{x - r} \times p^r \qquad &\pmod m\\ &\equiv p^{x - r} \times p^{\varphi(m) + r} &\pmod m\\ &\equiv p^{x + \varphi(m)} &\pmod m \\ \end{aligned} \]

【结论 \(1\)】\(\therefore \forall x \geq r, p^x \equiv p^{x + k \times \varphi(m)} \pmod m\)，其中 \(k \in \mathbb{N}\)

---

\[\begin{aligned} &\because p^r \mid m \\ &\therefore \varphi(m) \geq \varphi(p^r) \\ &\because \varphi(p^r) \geq r, b \geq \varphi(m) \\ &\therefore b \geq \varphi(m) \geq \varphi(p^r) \geq r &\end{aligned} \]

设 \(b = k \times \varphi(m) + y\)，其中 \(0 \leq y < \varphi(m)\)

\[b = \varphi(m) + (k - 1) \times \varphi(m) + y \\ \therefore \varphi(m) + y \geq r, k - 1 \in \mathbb{N} \]

由结论 \(1\) 知 \(p^b \equiv p^{\varphi(m) + y} \equiv p^{b\ \bmod\ \varphi(m) + \varphi(m) } \pmod m\)

---

设 \(a = \prod\limits_{i = 1}^m p_i\)，其中 \(p_i\) 为质数，有

\[\begin{aligned} a^b &= \prod\limits_{i = 1}^m p_i^b \qquad &\pmod m\\ &\equiv \prod\limits_{i = 1}^m p_i^{b\ \bmod\ \varphi(m) + \varphi(m)} &\pmod m\\ &\equiv (\prod\limits_{i = 1}^m p_i)^{b\ \bmod\ \varphi(m) + \varphi(m)} &\pmod m\\ &\equiv a^{b\ \bmod\ \varphi(m) + \varphi(m)} &\pmod m \end{aligned} \]