欧拉定理与欧拉函数

欧拉定理:

当a 与 p互质的时候则有 $a^{\varphi(p)} \equiv 1 \ (mod \ p)$
通项公式及证明：
若$n = p ^ k ,p$为质数,则$\varphi (p^k) = p ^k - p ^{k - 1}$
当一个数不包含质因子$p$时就能与$n$互质，
小于等于$n$的数中包含质因子$p$的只有$p^{k-1}$ 个，他们是:
$p, 2*p, 3* p, ...,p ^{k - 1} ∗p$，把他们去除即可.

由唯一分解定理可得: $n = p_1 ^{a_1} p_2 ^{a_2}p_3 ^{a_3}...p_k ^{a_k}$
则 $\varphi (n) = \varphi(p_1^{a_1})\varphi(p_2^{a_2})\varphi(p_3^{a_3})...\varphi(p_k^{a_k})$

根据上述$\varphi (p^k) = p ^k - p ^{k - 1}$可得:
$ \varphi (p) = p^k(1 - $${1}\over {p^k}$)

则 $\varphi (n) = \varphi(p_1^{a_1})\varphi(p_2^{a_2})\varphi(p_3^{a_3})...\varphi(p_k^{a_k})$可化为
$\ \ \ \ \varphi (n) = p_1 ^{a_1}(1 - \frac {1} {p_1}) p_2 ^{a_2}(1 - \frac {1} {p_2})p_3 ^{a_3}(1 - \frac {1} {p_3})...p_k ^{a_k}(1 - \frac {1} {p_k})$
$\ \ \ \ \ \ \ \ \ \ \ = n (1 - \frac {1} {p_1})(1 - \frac {1} {p_2})(1 - \frac {1} {p_3})...(1 - \frac {1} {p_k})$

欧拉函数

$\varphi(n) or \phi(n)$
表示小于n的正整数与n互质的数的个数.
显然有:
当n为质数时 $\varphi(n)$

当n为奇数时 $\varphi(2n) = \varphi(n)$
证明：
$\because$欧拉函数为积性函数.
$\therefore \varphi(2n) = \varphi(2) \ast \varphi(n)$
$\because \varphi(2)=1$
$\therefore \varphi(2n) = \varphi(n)$

证明欧拉函数的积性证明.

条件是m与n互质
可以得到$\phi(mn) = \phi(m) \ast \phi(n)$
证明：
$m = p_1^{a_1}p_2^{a_2}...p_k^{a_k}$
$\phi (m) = m(1- \frac {1}{p_1})(1- \frac {1}{p_2})...(1- \frac {1}{p_k})$
$n = p_1'^{a_1'}p_2'^{a_2'}...p_k'^{a_k'}$
$\phi(n) = n(1- \frac {1}{p_1'})(1- \frac {1}{p_2'})...(1- \frac {1}{p_k'})$
$\because m与n互质$
$\therefore p_1,p_2...p_k与p_1'p_2'...p_k'$两两互不相同
$\therefore \phi(mn) = mn(1- \frac {1}{p_1})(1- \frac {1}{p_2})...(1- \frac {1}{p_k})(1- \frac {1}{p_1'})(1- \frac {1}{p_2'})...(1- \frac {1}{p_k'})$
$\therefore \phi(mn) = \phi(m) \ast \phi(n)$