Jacobi符号的定义和基本性质

前言

  本文与博文《Legendre符号的定义和基本性质》有一定的内在关联性。

定义

  设$m\in {\mathbb{Z}}_{>1}$是奇数，$m=p_1{p}_2\cdots p_r$，其中$\forall i$，$p_i$是素数（也是奇素数），且可以重复；$a\in \mathbb{Z}$。定义$a$对$m$的Jacobi符号为$\left(\frac{a}{m}\right)$：
$$\left(\frac{a}{m}\right)=\left(\frac{a}{p_1}\right)\left(\frac{a}{p_2}\right)\cdots \left(\frac{a}{p_r}\right)$$
其中$\left(\frac{a}{p_i}\right)$是$a$对奇素数$p_i$的Legendre符号。

注
  整数$a$对$m$的Jacobi符号$\left(\frac{a}{m}\right)$中，只要求$m$是大于$1$的奇数，而没有和Legendre符号一样要求$m$是素数。这是使用Jacobi符号的一个优势。

基本性质

  设$m\in {\mathbb{Z}}_{>1}$是奇数，$m=p_1{p}_2\cdots p_r$，其中$\forall i$，$p_i$是素数（也是奇素数），$a,b\in \mathbb{Z}$。

1. $\left(\frac{1}{m}\right)=1$

证明
$$\left(\frac{1}{m}\right)=\left(\frac{1}{p_1}\right)\times \left(\frac{1}{p_2}\right)\times \cdots \times \left(\frac{1}{p_r}\right)=1\times 1\times \cdots \times 1=1$$

2. $\left(\frac{-1}{m}\right)={\left(-1\right)}^{\frac{m-1}{2}}=\{\begin{array}{rl}& 1,m\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4\\ & -1,m\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4\end{array}$

证明

1. 第一步，先证明若$p_1,p_2,\cdots ,p_r$是奇数，则$$\frac{p_1-1}{2}+\frac{p_2-1}{2}+\cdots +\frac{p_r-1}{2}\equiv \frac{\left(\prod _{i=1}^r{p}_i\right)-1}{2}\equiv \frac{m-1}{2}\mathrm{m}\mathrm{o}\mathrm{d}2$$
   当$s=1$时，显然有
   $$\frac{p_1-1}{2}\equiv \frac{p_1-1}{2}\mathrm{m}\mathrm{o}\mathrm{d}2$$
   假设$\sum _{i=1}^s\frac{p_i-1}{2}\equiv \frac{\left(\prod _{i=1}^s{p}_i\right)-1}{2}\mathrm{m}\mathrm{o}\mathrm{d}2$，考虑$\frac{\left(\prod _{i=1}^{s+1}{p}_i\right)-1}{2}-\sum _{i=1}^{s+1}\frac{p_i-1}{2}$：
   $$\begin{array}{rl}& \frac{\left(\prod _{i=1}^{s+1}{p}_i\right)-1}{2}-\sum _{i=1}^{s+1}\frac{p_i-1}{2}\\ & =\frac{p_{s+1}\left(\prod _{i=1}^s{p}_i\right)-1}{2}-\left(\sum _{i=1}^s\frac{p_i-1}{2}\right)-\frac{p_{s+1}-1}{2}\\ & \equiv \frac{p_{s+1}\left(\prod _{i=1}^s{p}_i\right)-1}{2}-\frac{\left(\prod _{i=1}^s{p}_i\right)-1}{2}-\frac{p_{s+1}-1}{2}\mathrm{m}\mathrm{o}\mathrm{d}2\\ & =\frac{p_{s+1}\left(\prod _{i=1}^s{p}_i\right)-\left(\prod _{i=1}^s{p}_i\right)-p_{s+1}+1}{2}\\ & =\frac{\left(\left(\prod _{i=1}^s{p}_i\right)-1\right)\left(p_{s+1}-1\right)}{2}\end{array}$$
   由于 ∀ i ∈ { 1 , 2 , ⋯   , s + 1 } \forall i\in \left\{ 1,2,\cdots ,s+1 \right\} ∀i∈{ 1,2,⋯,s+1}，$p_i$是奇数，因此$\prod _{i=1}^s{p}_i,p_{s+1}$也是奇数，$\left(\prod _{i=1}^s{p}_i\right)-1,p_{s+1}-1$均为偶数，$\left(\left(\prod _{i=1}^s{p}_i\right)-1\right)\left(p_{s+1}-1\right)$的素因子分解形式中存在素因子$2$且其指数至少为$2$，因此有
   $$\begin{array}{rl}& \frac{\left(\prod _{i=1}^{s+1}{p}_i\right)-1}{2}-\sum _{i=1}^{s+1}\frac{p_i-1}{2}\equiv \frac{\left(\left(\prod _{i=1}^s{p}_i\right)-1\right)\left(p_{s+1}-1\right)}{2}\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}2\\ & \\ & \Rightarrow \sum _{i=1}^{s+1}\frac{p_i-1}{2}\equiv \frac{\left(\prod _{i=1}^{s+1}{p}_i\right)-1}{2}\mathrm{m}\mathrm{o}\mathrm{d}2\end{array}$$
   由第一数学归纳法，结论得证。

2. 基于上面的结论，$\exists k\in \mathbb{Z},\text{ s}\text{.t}\text{.}$
   $$\frac{m-1}{2}=\frac{\left(\prod _{i=1}^r{p}_i\right)-1}{2}=2k+\sum _{i=1}^r\frac{p_i-1}{2}$$
   因此有
   $$\begin{array}{rl}& \left(\frac{-1}{m}\right)=\left(\frac{-1}{p_1}\right)\left(\frac{-1}{p_2}\right)\cdots \left(\frac{-1}{p_r}\right)\\ & ={\left(-1\right)}^{\frac{p_1-1}{2}}{\left(-1\right)}^{\frac{p_2-1}{2}}\cdots {\left(-1\right)}^{\frac{p_r-1}{2}}\\ & ={\left(-1\right)}^{\sum _{i=1}^r\frac{p_i-1}{2}}\\ & ={\left(-1\right)}^{\sum _{i=1}^r\frac{p_i-1}{2}+2k}\\ & ={\left(-1\right)}^{\frac{m-1}{2}}\end{array}$$

3. 由于$m$是奇数，所以只存在$m\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4$和$m\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4$两种情况。

   1. 当$m\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4$时，$\exists k\in \mathbb{Z},\text{ s}\text{.t}\text{. }m=4k+1$。此时
      $$\left(\frac{-1}{m}\right)={\left(-1\right)}^{\frac{m-1}{2}}={\left(-1\right)}^{\frac{4k+1-1}{2}}={\left(-1\right)}^{2k}=1$$
   2. 当$m\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4$时，$\exists k\in \mathbb{Z},\text{ s}\text{.t}\text{. }m=4k+3$。此时
      $$\left(\frac{-1}{m}\right)={\left(-1\right)}^{\frac{m-1}{2}}={\left(-1\right)}^{\frac{4k+3-1}{2}}={\left(-1\right)}^{2k+1}=-1$$
      
      

3. 若$a\equiv b\mathrm{m}\mathrm{o}\mathrm{d}m$，则$\left(\frac{a}{m}\right)=\left(\frac{b}{m}\right)$

证明
$$\left(\frac{a}{m}\right)=\left(\frac{a}{p_1}\right)\left(\frac{a}{p_2}\right)\cdots \left(\frac{a}{p_r}\right)=\left(\frac{b}{p_1}\right)\left(\frac{b}{p_2}\right)\cdots \left(\frac{b}{p_r}\right)=\left(\frac{b}{m}\right)$$

4-1. $\left(\frac{ab}{m}\right)=\left(\frac{a}{m}\right)\left(\frac{b}{m}\right)$

证明
$$\begin{array}{rl}& \left(\frac{ab}{m}\right)=\left(\frac{ab}{p_1}\right)\times \left(\frac{ab}{p_2}\right)\times \cdots \times \left(\frac{ab}{p_r}\right)\\ & =\left(\left(\frac{a}{p_1}\right)\left(\frac{b}{p_1}\right)\right)\times \left(\left(\frac{a}{p_2}\right)\left(\frac{b}{p_2}\right)\right)\times \cdots \times \left(\left(\frac{a}{p_r}\right)\left(\frac{b}{p_r}\right)\right)\\ & =\left(\left(\frac{a}{p_1}\right)\left(\frac{a}{p_2}\right)\cdots \left(\frac{a}{p_r}\right)\right)\times \left(\left(\frac{b}{p_1}\right)\left(\frac{b}{p_2}\right)\cdots \left(\frac{b}{p_r}\right)\right)\\ & =\left(\frac{a}{m}\right)\times \left(\frac{b}{m}\right)\end{array}$$

4-2. $\left(\frac{\prod _{i=1}^n{a}_i}{m}\right)=\prod _{i=1}^n\left(\frac{a_i}{m}\right)$

证明
  思想同4-1。

5. 若$\gcd \left(b,m\right)=1$，则$\left(\frac{a{b}^2}{m}\right)=\left(\frac{a}{m}\right)$

证明
$$\begin{array}{rl}& \begin{array}{rl}& \gcd \left(b,m\right)=1\\ & m=p_1{p}_2\cdots p_r\end{array}\}\Rightarrow \forall i,\gcd \left(b,p_i\right)=1\\ & \Rightarrow \forall i,p_i\cancel{|}b\\ & \Rightarrow \left(\frac{b}{p_i}\right)=\pm 1,{\left(\frac{b}{p_i}\right)}^2=1\end{array}$$
基于此，有
$$\left(\frac{a{b}^2}{m}\right)=\left(\frac{a}{m}\right){\left(\frac{b}{m}\right)}^2=\left(\frac{a}{m}\right)\times 1=\left(\frac{a}{m}\right)$$

6. $\left(\frac{2}{m}\right)={\left(-1\right)}^{\frac{m^2-1}{8}}=\{\begin{array}{rl}& 1,m\equiv \pm 1\mathrm{m}\mathrm{o}\mathrm{d}8\\ & -1,m\equiv \pm 3\mathrm{m}\mathrm{o}\mathrm{d}8\end{array}$

证明

1. 第一步，先证明若$p_1,p_2,\cdots ,p_r$是奇数，则
   $$\frac{{p_1}^2-1}{8}+\frac{{p_2}^2-1}{8}+\cdots +\frac{{p_r}^2-1}{8}\equiv \frac{{\left(\prod _{i=1}^r{p}_i\right)}^2-1}{8}\mathrm{m}\mathrm{o}\mathrm{d}2$$
   当$s=1$时，显然有
   $$\frac{{p_1}^2-1}{8}\equiv \frac{{p_1}^2-1}{8}\mathrm{m}\mathrm{o}\mathrm{d}2$$
   假设$\sum _{i=1}^s\frac{{p_i}^2-1}{8}\equiv \frac{{\left(\prod _{i=1}^s{p}_i\right)}^2-1}{8}\mathrm{m}\mathrm{o}\mathrm{d}2$，考虑$\frac{{\left(\prod _{i=1}^{s+1}{p}_i\right)}^2-1}{8}-\sum _{i=1}^{s+1}\frac{{p_i}^2-1}{8}$：
   $$\begin{array}{rl}& \frac{{\left(\prod _{i=1}^{s+1}{p}_i\right)}^2-1}{8}-\sum _{i=1}^{s+1}\frac{{p_i}^2-1}{8}\\ & =\frac{{p_{s+1}}^2{\left(\prod _{i=1}^s{p}_i\right)}^2-1}{8}-\sum _{i=1}^s\frac{{p_i}^2-1}{8}-\frac{{p_{s+1}}^2-1}{8}\\ & \equiv \frac{{p_{s+1}}^2{\left(\prod _{i=1}^s{p}_i\right)}^2-1}{8}-\frac{{\left(\prod _{i=1}^s{p}_i\right)}^2-1}{8}-\frac{{p_{s+1}}^2-1}{8}\mathrm{m}\mathrm{o}\mathrm{d}2\\ & =\frac{\left({p_{s+1}}^2-1\right)\left({\left(\prod _{i=1}^s{p}_i\right)}^2-1\right)}{8}\end{array}$$
   由于 ∀ i ∈ { 1 , 2 , ⋯   , s + 1 } \forall i\in \left\{ 1,2,\cdots ,s+1 \right\} ∀i∈{ 1,2,⋯,s+1}，$p_i$是奇数，因此$\prod _{i=1}^s{p}_i,p_{s+1}$也是奇数。再由于任意一个奇数$2k+1$的平方满足
   $${\left(2k+1\right)}^2=4k^2+4k+1\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4$$
   因此有
   $$\begin{array}{rl}& {p_{s+1}}^2-1\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}4\\ & {\left(\prod _{i=1}^s{p}_i\right)}^2-1\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}4\end{array}$$
   即$\left({p_{s+1}}^2-1\right)\left({\left(\prod _{i=1}^s{p}_i\right)}^2-1\right)$的素因子分解形式中存在素因子2且其指数至少为4，而$8=2^3$，因此有
   $$\begin{array}{rl}& \frac{{\left(\prod _{i=1}^{s+1}{p}_i\right)}^2-1}{8}-\sum _{i=1}^{s+1}\frac{{p_i}^2-1}{8}\equiv \frac{\left({p_{s+1}}^2-1\right)\left({\left(\prod _{i=1}^s{p}_i\right)}^2-1\right)}{8}\equiv 0\mathrm{m}\mathrm{o}\mathrm{d}2\\ & \\ & \Rightarrow \sum _{i=1}^{s+1}\frac{{p_i}^2-1}{8}\equiv \frac{{\left(\prod _{i=1}^{s+1}{p}_i\right)}^2-1}{8}\mathrm{m}\mathrm{o}\mathrm{d}2\end{array}$$
   由第一数学归纳法，结论得证。

2. 基于上面的结论，$\exists k\in \mathbb{Z},\text{ s}\text{.t}\text{.}$
   $$\frac{m^2-1}{8}=\frac{\left(\prod _{i=1}^r{p_i}^2\right)-1}{8}=2k+\sum _{i=1}^r\frac{{p_i}^2-1}{8}$$
   因此有
   $$\begin{array}{rl}& \left(\frac{2}{m}\right)=\left(\frac{2}{p_1}\right)\left(\frac{2}{p_2}\right)\cdots \left(\frac{2}{p_r}\right)\\ & ={\left(-1\right)}^{\frac{{p_1}^2-1}{8}}{\left(-1\right)}^{\frac{{p_2}^2-1}{8}}\cdots {\left(-1\right)}^{\frac{{p_r}^2-1}{8}}\\ & ={\left(-1\right)}^{\sum _{i=1}^r\frac{{p_i}^2-1}{8}}\\ & ={\left(-1\right)}^{\sum _{i=1}^r\frac{{p_i}^2-1}{8}+2k}\\ & ={\left(-1\right)}^{\frac{m^2-1}{8}}\end{array}$$

3. 由于$m$是奇数，因此只有$m\equiv \pm 1\mathrm{m}\mathrm{o}\mathrm{d}8$和$m\equiv \pm 3\mathrm{m}\mathrm{o}\mathrm{d}8$的情况。

   1. 当$m\equiv \pm 1\mathrm{m}\mathrm{o}\mathrm{d}8$时，$\exists k\in \mathbb{Z},\text{ s}\text{.t}\text{. }m=8k\pm 1$，此时有
      $$\left(\frac{2}{m}\right)={\left(-1\right)}^{\frac{m^2-1}{8}}={\left(-1\right)}^{\frac{{\left(8k\pm 1\right)}^2-1}{8}}={\left(-1\right)}^{\frac{64k^2\pm 16k}{8}}={\left(-1\right)}^{8k^2\pm 2k}=1$$
   2. 当$m\equiv \pm 3\mathrm{m}\mathrm{o}\mathrm{d}8$时，$\exists k\in \mathbb{Z},\text{ s}\text{.t}\text{. }m=8k\pm 3$，此时有
      $$\left(\frac{2}{m}\right)={\left(-1\right)}^{\frac{m^2-1}{8}}={\left(-1\right)}^{\frac{{\left(8k\pm 3\right)}^2-1}{8}}={\left(-1\right)}^{\frac{64k^2\pm 48k+8}{8}}={\left(-1\right)}^{8k^2\pm 6k+1}=-1$$
      
      

7. 设$m,n\in {\mathbb{Z}}_{>1}$是奇数，则$\left(\frac{n}{m}\right)={\left(-1\right)}^{\frac{m-1}{2}\cdot \frac{n-1}{2}}\left(\frac{m}{n}\right)=\{\begin{array}{rl}& \left(\frac{m}{n}\right),m\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4或n\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4\\ & -\left(\frac{m}{n}\right),m\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4且n\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4\end{array}$

证明
  设$m=p_1\cdots p_r,n=q_1\cdots q_s$，则有
$$\begin{array}{rl}& \left(\frac{n}{m}\right)=\prod _{i=1}^r\left(\frac{n}{p_i}\right)=\prod _{i=1}^r\prod _{j=1}^s\left(\frac{q_j}{p_i}\right)\\ & \left(\frac{m}{n}\right)=\prod _{j=1}^s\left(\frac{m}{q_j}\right)=\prod _{j=1}^s\prod _{i=1}^r\left(\frac{p_i}{q_j}\right)\end{array}$$

1. 若$\gcd \left(m,n\right)\ne 1$，则至少存在一对$p_{i_0}=q_{j_0}$，对应地有
   $$\begin{array}{rl}& p_{i_0}|q_{j_0}\Rightarrow \left(\frac{q_{j_0}}{p_{i_0}}\right)=0\Rightarrow \left(\frac{n}{m}\right)=0\\ & q_{j_0}|p_{i_0}\Rightarrow \left(\frac{p_{i_0}}{q_{j_0}}\right)=0\Rightarrow \left(\frac{m}{n}\right)=0\end{array}$$
   故自然成立
   $$\left(\frac{n}{m}\right)={\left(-1\right)}^{\frac{m-1}{2}\cdot \frac{n-1}{2}}\left(\frac{m}{n}\right)$$

2. 若$\gcd \left(m,n\right)=1$，则$\forall i,j$，奇素数$p_i,q_j$满足$p_i\ne q_j$，因此可对Legendre符号$\left(\frac{q_j}{p_i}\right)$使用二次反转定律：
   $$\begin{array}{rl}& \left(\frac{n}{m}\right)=\prod _{i=1}^r\prod _{j=1}^s\left(\frac{q_j}{p_i}\right)=\prod _{i=1}^r\prod _{j=1}^s\left({\left(-1\right)}^{\frac{p_i-1}{2}\cdot \frac{q_j-1}{2}}\left(\frac{p_i}{q_j}\right)\right)\\ & =\left(\prod _{i=1}^r\prod _{j=1}^s{\left(-1\right)}^{\frac{p_i-1}{2}\cdot \frac{q_j-1}{2}}\right)\left(\prod _{j=1}^s\prod _{i=1}^r\left(\frac{p_i}{q_j}\right)\right)\\ & ={\left(-1\right)}^{\left(\sum _{i=1}^r\frac{p_i-1}{2}\right)\left(\sum _{j=1}^s\frac{q_j-1}{2}\right)}\left(\frac{m}{n}\right)\end{array}$$
   由性质2证明过程第一步的结论有
   $$\begin{array}{rl}& \{\begin{array}{rl}& \sum _{i=1}^r\frac{p_i-1}{2}\equiv \frac{\left(\prod _{i=1}^r{p}_i\right)-1}{2}=\frac{m-1}{2}\mathrm{m}\mathrm{o}\mathrm{d}2\\ & \sum _{j=1}^s\frac{q_j-1}{2}\equiv \frac{\left(\prod _{j=1}^s{q}_j\right)-1}{2}=\frac{n-1}{2}\mathrm{m}\mathrm{o}\mathrm{d}2\end{array}\\ & \\ & \Rightarrow \left(\sum _{i=1}^r\frac{p_i-1}{2}\right)\left(\sum _{j=1}^s\frac{q_j-1}{2}\right)\equiv \frac{m-1}{2}\cdot \frac{n-1}{2}\mathrm{m}\mathrm{o}\mathrm{d}2\end{array}$$
   因此$\exists k\in \mathbb{Z},\text{ s}\text{.t}\text{.}$
   $$\left(\sum _{i=1}^r\frac{p_i-1}{2}\right)\left(\sum _{j=1}^s\frac{q_j-1}{2}\right)=2k+\frac{m-1}{2}\cdot \frac{n-1}{2}$$
   基于此有
   $$\begin{array}{rl}& \left(\frac{n}{m}\right)={\left(-1\right)}^{\left(\sum _{i=1}^r\frac{p_i-1}{2}\right)\left(\sum _{j=1}^s\frac{q_j-1}{2}\right)}\left(\frac{m}{n}\right)\\ & ={\left(-1\right)}^{2k+\frac{m-1}{2}\cdot \frac{n-1}{2}}\left(\frac{m}{n}\right)\\ & ={\left(-1\right)}^{\frac{m-1}{2}\cdot \frac{n-1}{2}}\left(\frac{m}{n}\right)\end{array}$$
   
   

3. 在已经证明了
   $$\left(\frac{n}{m}\right)={\left(-1\right)}^{\frac{m-1}{2}\cdot \frac{n-1}{2}}\left(\frac{m}{n}\right)$$
   的基础上，研究$\left(\frac{n}{m}\right)$与$\left(\frac{m}{n}\right)$的符号关系。
   由于$m,n$均为奇数，因此只有以下情况：$m\equiv 1,3\mathrm{m}\mathrm{o}\mathrm{d}4,n\equiv 1,3\mathrm{m}\mathrm{o}\mathrm{d}4$。

   1. 当$m\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4$时，$\exists k\in \mathbb{Z},\text{ s}\text{.t}\text{.}$$m=4k+1$，此时有
      $$\left(\frac{n}{m}\right)={\left(-1\right)}^{\frac{4k+1-1}{2}\cdot \frac{n-1}{2}}\left(\frac{m}{n}\right)={\left({\left(-1\right)}^{2k}\right)}^{\frac{n-1}{2}}\left(\frac{m}{n}\right)=\left(\frac{m}{n}\right)$$
      同理，$n\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4$时也有$\left(\frac{n}{m}\right)=\left(\frac{m}{n}\right)$
   2. 当$\sim \left(m\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4\vee n\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4\right)$即$m\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4\wedge n\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4$时，$\exists k_1,k_2\in \mathbb{Z},\text{ s}\text{.t}\text{.}$ $m=4k_1+3,n=4k_2+3$，此时有
      $$\begin{array}{rl}& \left(\frac{n}{m}\right)={\left(-1\right)}^{\frac{4k_1+3-1}{2}\cdot \frac{4k_2+3-1}{2}}\left(\frac{m}{n}\right)\\ & ={\left(-1\right)}^{\left(2k_1+1\right)\left(2k_2+1\right)}\left(\frac{m}{n}\right)\\ & ={\left({\left(-1\right)}^{2k_1+1}\right)}^{2k_2+1}\left(\frac{m}{n}\right)\\ & ={\left(-1\right)}^{2k_2+1}\left(\frac{m}{n}\right)\\ & =-\left(\frac{m}{n}\right)\end{array}$$
      因此有
      $$\left(\frac{n}{m}\right)=\{\begin{array}{rl}& \left(\frac{m}{n}\right),m\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4或n\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}4\\ & -\left(\frac{m}{n}\right),m\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4且n\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4\end{array}$$
      
      

8. 设大于1的奇数$m$的标准分解式为$m=p_1\cdots p_r$。若$\gcd \left(a,m\right)=1$，$2\cancel{|}a$，则$\left(\frac{a}{m}\right)={\left(-1\right)}^{\sum _{i=1}^r\sum _{k=1}^{\frac{p_i-1}{2}}\left[\frac{ak}{p_i}\right]}$

证明
$$\left(\frac{a}{m}\right)=\left(\frac{a}{p_1}\right)\cdots \left(\frac{a}{p_r}\right)$$
$m=p_1\cdots p_r$是大于1的奇数，因此$p_1,\cdots ,p_r$都是奇素数。由于$\gcd \left(a,m\right)=1$，因此 ∀ i ∈ { 1 , ⋯   , r } \forall i\in \left\{ 1,\cdots ,r \right\} ∀i∈{ 1,⋯,r}，$\gcd \left(a,p_i\right)=1$。又$2\cancel{|}a$，因此由博文《Legendre符号的定义和基本性质》中的性质8有
$$\left(\frac{a}{p_i}\right)={\left(-1\right)}^{\sum _{k=1}^{\frac{p_i-1}{2}}\left[\frac{ak}{p_i}\right]}$$
于是有
$$\left(\frac{a}{m}\right)=\prod _{i=1}^r{\left(-1\right)}^{\sum _{k=1}^{\frac{p_i-1}{2}}\left[\frac{ak}{p_i}\right]}={\left(-1\right)}^{\sum _{i=1}^r\sum _{k=1}^{\frac{p_i-1}{2}}\left[\frac{ak}{p_i}\right]}$$