同余基础
同余定义
称$m\in {{\mathbb{Z}}_{>0}}$为模。$\forall a,b\in \mathbb{Z}$，若用$m$去除$a,b$所得余数相同，则称$a,b$对模$m$同余，记作$a\equiv b\left( \bmod m \right)$；若余数不同，则称$a,b$对模$m$不同余，记作$a\not{\equiv }b\left( \bmod m \right)$。
Th3.1.1：$\forall a,b\in \mathbb{Z},\text{ }a,b$对模$m$同余$\Leftrightarrow$$\left. m \right|a-b$，即$\exists t\in \mathbb{Z},\text{ }a=b+mt$。
证明：
设$a=m{{q}_{1}}+{{r}_{1}},\text{ }b=m{{q}_{2}}+{{r}_{2}},\text{ }0\le {{r}_{1}},{{r}_{2}}<m$。
- $a\equiv b\left( \bmod m \right)\text{ }\Rightarrow \text{ }{{r}_{1}}={{r}_{2}}\text{ }\Rightarrow \text{ }a-b=m\left( {{q}_{1}}-{{q}_{2}} \right),\text{ }{{q}_{1}}-{{q}_{2}}\in \mathbb{Z}\text{ }\Rightarrow \text{ }\left. m \right|a-b$
- $\left. m \right|a-b\text{ }\Rightarrow \text{ }\left. m\text{ } \right|\text{ }\left[ m\left( {{q}_{1}}-{{q}_{2}} \right)+\left( {{r}_{1}}-{{r}_{2}}
\right) \right]$，因此$\exists t\in \mathbb{Z},\text{ }s.t.$$$\begin{matrix}
m\left( {{q}_{1}}-{{q}_{2}} \right)+\left( {{r}_{1}}-{{r}_{2}} \right)=mt \\
\Rightarrow \text{ }{{r}_{1}}-{{r}_{2}}=m\left( t-{{q}_{1}}+{{q}_{2}} \right),\text{ }t-{{q}_{1}}+{{q}_{2}}\in \mathbb{Z} \\
\Rightarrow \left. m \right|\left( {{r}_{1}}-{{r}_{2}} \right) \\
\end{matrix}$$另一方面，$0\le {{r}_{1}},{{r}_{2}}<m\text{ }\Rightarrow \text{ }-m<{{r}_{1}}-{{r}_{2}}<m$。因此有${{r}_{1}}-{{r}_{2}}=0\text{ }\Rightarrow \text{ }{{r}_{1}}={{r}_{2}}$，即$a\equiv b\left( \bmod m \right)$。
Th3.1.2：若$\forall {{\alpha }_{1}},{{\alpha }_{2}},...,{{\alpha }_{k}}\in {{\mathbb{Z}}_{\ge 0}},\text{ }{{A}_{{{\alpha }_{1}}{{\alpha }_{2}}...{{\alpha }_{k}}}}\equiv {{B}_{{{\alpha }_{1}}{{\alpha }_{2}}...{{\alpha }_{k}}}}\left( \bmod m \right)$，$\forall i\in \left\{ 1,2,...,k \right\}$，${{x}_{i}}\equiv {{y}_{i}}\left( \bmod m \right)$，则有$$\sum\limits_{{{\alpha }_{1}},\cdots ,{{\alpha }_{k}}}^{{}}{{{A}_{{{\alpha }_{1}}\cdots {{\alpha }_{k}}}}{{x}_{1}}^{{{\alpha }_{1}}}\cdots {{x}_{k}}^{{{\alpha }_{k}}}}\equiv \sum\limits_{{{\alpha }_{1}},\cdots ,{{\alpha }_{k}}}^{{}}{{{B}_{{{\alpha }_{1}}\cdots {{\alpha }_{k}}}}{{y}_{1}}^{{{\alpha }_{1}}}\cdots {{y}_{k}}^{{{\alpha }_{k}}}}\left( \bmod m \right)$$特别地，若$\forall i\in \left\{ 0,1,...,n \right\},{{a}_{i}}\equiv {{b}_{i}}\left( \bmod m \right)$，则$${{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+\cdots +{{a}_{0}}\equiv {{b}_{n}}{{x}^{n}}+{{b}_{n-1}}{{x}^{n-1}}+\cdots {{b}_{0}}\left( \bmod m \right)$$
证明：
$\forall {{\alpha }_{1}},{{\alpha }_{2}},...,{{\alpha }_{k}}\in {{\mathbb{Z}}_{\ge 0}},\text{ }\forall i\in \left\{ 1,2,...,k \right\}$，有
$$\left\{ \begin{aligned}
& {{A}_{{{\alpha }_{1}}\cdots {{\alpha }_{k}}}}\equiv {{B}_{{{\alpha }_{1}}\cdots {{\alpha }_{k}}}}\left( \bmod m \right) \\
& {{x}_{i}}\equiv {{y}_{i}}\left( \bmod m \right) \\
\end{aligned} \right.$$
由性质5)有
$${{A}_{{{\alpha }_{1}}\cdots {{\alpha }_{k}}}}{{x}_{1}}^{{{\alpha }_{1}}}\cdots {{x}_{k}}^{{{\alpha }_{k}}}\equiv {{B}_{{{\alpha }_{1}}\cdots {{\alpha }_{k}}}}{{y}_{1}}^{{{\alpha }_{1}}}\cdots {{y}_{k}}^{{{\alpha }_{k}}}\left( \bmod m \right)$$
再由性质4)即可得
$$\sum\limits_{{{\alpha }_{1}},\cdots ,{{\alpha }_{k}}}^{{}}{{{A}_{{{\alpha }_{1}}\cdots {{\alpha }_{k}}}}{{x}_{1}}^{{{\alpha }_{1}}}\cdots {{x}_{k}}^{{{\alpha }_{k}}}}\equiv \sum\limits_{{{\alpha }_{1}},\cdots ,{{\alpha }_{k}}}^{{}}{{{B}_{{{\alpha }_{1}}\cdots {{\alpha }_{k}}}}{{y}_{1}}^{{{\alpha }_{1}}}\cdots {{y}_{k}}^{{{\alpha }_{k}}}}\left( \bmod m \right)$$
上式中令$k=1$，${{\alpha }_{1}}=0,1,...,n$，${{a}_{{{\alpha }_{1}}}}={{A}_{{{\alpha }_{1}}}},\text{ }{{b}_{{{\alpha }_{1}}}}={{B}_{{{\alpha }_{1}}}},y_1=x_1$，则立即可得
$$\sum\limits_{{{\alpha }_{1}}=0}^{n}{{{a}_{{{\alpha }_{1}}}}{{x}_{1}}^{{{\alpha }_{1}}}}\equiv \sum\limits_{{{\alpha }_{1}}=0}^{n}{{{b}_{{{\alpha }_{1}}}}{{x}_{1}}^{{{\alpha }_{1}}}}\left( \bmod m \right)$$
即
$$\sum\limits_{i=0}^{n}{{{a}_{i}}{{x}^{i}}}\equiv \sum\limits_{i=0}^{n}{{{b}_{i}}{{x}^{i}}}\left( \bmod m \right)$$
同余性质
1. $a\equiv a\left( \bmod m \right)$
证明：由同余定义直接可得。
2. $a\equiv b\left( \bmod m \right)\text{ }\Rightarrow \text{ }b\equiv a\left( \bmod m \right)$
证明：由同余定义直接可得。
3. $a\equiv b\left( \bmod m \right),\text{ }b\equiv c\left( \bmod m \right)\text{ }\Rightarrow \text{ }a\equiv c\left( \bmod m \right)$
证明：
$a\equiv b\left( \bmod m \right)\text{ }\Rightarrow \text{ }\exists {{q}_{a}},{{q}_{b}},{{r}_{1}}\in \mathbb{Z},\text{ }s.t.$
$$\left\{ \begin{aligned}
& a=m{{q}_{a}}+{{r}_{1}} \\
& b=m{{q}_{b}}+{{r}_{1}} \\
\end{aligned} \right.,\text{ }0\le {{r}_{1}}<m$$
$b\equiv c\left( \bmod m \right)\text{ }\Rightarrow \text{ }\exists {{q}_{b}}',{{q}_{c}},{{r}_{2}}\in \mathbb{Z},\text{ }s.t.$
$$\left\{ \begin{aligned}
& b=m{{q}_{b}}'+{{r}_{2}} \\
& c=m{{q}_{c}}+{{r}_{2}} \\
\end{aligned} \right.,\text{ }0\le {{r}_{2}}<m$$
由$b=m{{q}_{b}}+{{r}_{1}}=m{{q}_{b}}'+{{r}_{2}},\text{ }0\le {{r}_{1}},{{r}_{2}}<m$和Th1.1.4得${{q}_{b}}={{q}_{b}}',\text{ }{{r}_{1}}={{r}_{2}}$。因此有$a\equiv c\left( \bmod m \right)$。
4. i)${{a}_{1}}\equiv {{b}_{1}}\left( \bmod m \right),\text{ }{{a}_{2}}\equiv {{b}_{2}}\left( \bmod m \right)\text{ }\Rightarrow \text{ }{{a}_{1}}+{{a}_{2}}\equiv {{b}_{1}}+{{b}_{2}}\left( \bmod m \right)$；<br/>
ii)$a+b\equiv c\left( \bmod m \right)\text{ }\Rightarrow \text{ }a\equiv c-b\left( \bmod m \right)$
证明：
- 由${{a}_{1}}\equiv {{b}_{1}}\left( \bmod m \right),\text{ }{{a}_{2}}\equiv {{b}_{2}}\left( \bmod m \right)$和Th3.1.1，得$\exists {{t}_{1}},{{t}_{2}}\in \mathbb{Z},\text{ }s.t.$
$$\begin{matrix}
{{a}_{1}}={{b}_{1}}+m{{t}_{1}},\text{ }{{a}_{2}}={{b}_{2}}+m{{t}_{2}} \\
\Rightarrow \left( {{a}_{1}}+{{a}_{2}} \right)=\left( {{b}_{1}}+{{b}_{2}} \right)+m\left( {{t}_{1}}+{{t}_{2}} \right) \\
\end{matrix}$$由Th3.1.1，即有${{a}_{1}}+{{a}_{2}}\equiv {{b}_{1}}+{{b}_{2}}\left( \bmod m \right)$。
- $c-b=c+\left( -b \right)$，由性质1)有$c-b\equiv c+\left( -b \right)\left( \bmod m \right)$
又$a+b\equiv c\left( \bmod m \right)$，由性质4)-i)有$c+\left( -b \right)\equiv a+b+\left( -b \right)\left( \bmod m \right)$
$a+b+\left( -b \right)=a$，由性质1)有$a+b+\left( -b \right)\equiv a\left( \bmod m \right)$
最后由性质3)的传递性由$c-b\equiv a\left( \bmod m \right)$。
5. ${{a}_{1}}\equiv {{b}_{1}}\left( \bmod m \right),\text{ }{{a}_{2}}\equiv {{b}_{2}}\left( \bmod m \right)\text{ }\Rightarrow \text{ }{{a}_{1}}{{a}_{2}}\equiv {{b}_{1}}{{b}_{2}}\left( \bmod m \right)$<br/>
特别地，有$a\equiv b\left( \bmod m \right)\text{ }\Rightarrow \text{ }ak\equiv bk\left( \bmod m \right)$
证明：
由${{a}_{1}}\equiv {{b}_{1}}\left( \bmod m \right),\text{ }{{a}_{2}}\equiv {{b}_{2}}\left( \bmod m \right)$和Th3.1.1有，$\exists {{t}_{1}},{{t}_{2}}\in \mathbb{Z}$，使得
$$\begin{matrix}
{{a}_{1}}={{b}_{1}}+m{{t}_{1}},\text{ }{{a}_{2}}={{b}_{2}}+m{{t}_{2}} \\
\Rightarrow {{a}_{1}}{{a}_{2}}={{b}_{1}}{{b}_{2}}+m\left( {{b}_{1}}{{t}_{2}}+{{b}_{2}}{{t}_{1}}+m{{t}_{1}}{{t}_{2}}
\right),\text{ }{{b}_{1}}{{t}_{2}}+{{b}_{2}}{{t}_{1}}+m{{t}_{1}}{{t}_{2}}\in \mathbb{Z} \\
\Rightarrow {{a}_{1}}{{a}_{2}}\equiv {{b}_{1}}{{b}_{2}}\left( \bmod m \right) \\
\end{matrix}$$
在此基础上，令${{a}_{1}}=a,\text{ }{{b}_{1}}=b,$${{a}_{2}}={{b}_{2}}=k\in \mathbb{Z}$，则有
$$ak\equiv bk\left( \bmod m \right)$$
6. $\left\{ \begin{aligned} & a\equiv b\left( \bmod m \right) \\ & a={{a}_{1}}d,\text{ }b={{b}_{1}}d,\text{ }\left( d,m \right)=1 \\ \end{aligned}
\right.\text{ }\Rightarrow \text{ }{{a}_{1}}\equiv {{b}_{1}}\left( \bmod m \right)$
证明：
由$a\equiv b\left( \bmod m \right)$和Th3.1.1有$\left. m \right|a-b$。
又$a-b=d\left( {{a}_{1}}-{{b}_{1}} \right)$，$\left( d,m \right)=1$，由Th1.2.7有$\left. m \right|{{a}_{1}}-{{b}_{1}}$，即${{a}_{1}}\equiv {{b}_{1}}\left( \bmod m \right)$。
7. i)$a\equiv b\left( \bmod m \right),k\in {{\mathbb{Z}}_{\neq 0}}\text{ }\Rightarrow \text{ }ak\equiv bk\left( \bmod \left(m\left|k\right| \right)\right)$；<br/>
ii)$a\equiv b\left( \bmod m \right)\text{ }\Rightarrow \text{ }$$\forall d\in \left\{ \left. d\text{ } \right|\text{ }\left. d \right|a\text{ }\And \text{ }\left. d \right|b\text{ }\And \text{ }\left. d \right|m \right\}$，$\frac{a}{\left| d \right|}\equiv \frac{b}{\left| d \right|}\left( \bmod \frac{m}{\left| d \right|} \right)$
证明：
- $a\equiv b\left( \bmod m \right)$，由Th3.1.1，$\exists t\in \mathbb{Z},\text{ }s.t.$$a=b+mt$。
式子两边同乘以$k\in {{\mathbb{Z}}_{\ne 0}}$得$ak=bk+\left( mk \right)t=bk+\left( m\left| k \right| \right)\left[ sgn \left( k \right)t \right]$，
其中$ak,bk\in \mathbb{Z},\text{ }m\left| k \right|\in {{\mathbb{Z}}_{>0}},\text{ }sgn \left( k \right)t\in \mathbb{Z}$。因此有$ak\equiv bk\left( \bmod \left( m\left| k \right| \right) \right)$。
- $a\equiv b\left( \bmod m \right)$，由Th3.1.1，$\exists t\in \mathbb{Z},\text{ }s.t.$$a=b+mt$
$\forall d\in \left\{ \left. d\text{ } \right|\text{ }\left. d \right|a\text{ }\And \text{ }\left. d \right|b\text{ }\And \text{ }\left. d \right|m,\text{ }m\in {{\mathbb{Z}}_{>0}} \right\}$，有$d\ne 0$。因此有$\frac{a}{\left| d \right|}=\frac{b}{\left| d \right|}\text{+}\frac{m}{\left| d \right|}t$
且其中$\frac{a}{\left| d \right|},\frac{b}{\left| d \right|}\in \mathbb{Z},\text{ }\frac{m}{\left| d \right|}\in {{\mathbb{Z}}_{>0}}$，因此有$\frac{a}{\left| d \right|}\equiv \frac{b}{\left| d \right|}\left( \bmod \frac{m}{\left| d \right|} \right)$。
8. $\forall i\in \left\{ 1,2,...,k \right\},a\equiv b\left( \bmod {{m}_{i}} \right)\text{ }\Rightarrow \text{ }a\equiv b\left( \bmod \left[ {{m}_{1}},{{m}_{2}},...,{{m}_{k}} \right] \right)$
证明：
- $a=b$时，性质是显然的。
- 当$a\ne b$时，$\forall i\in \left\{ 1,2,...,k \right\}$，$a\equiv b\left( \bmod {{m}_{i}} \right)$，显然有$a\equiv b\left( \bmod \left[ {{m}_{1}} \right] \right)$即$a\equiv b\left( \bmod {{m}_{1}} \right)$。
假设对某个$n\in \left\{ 1,2,...,k-1 \right\}$有$a\equiv b\left( \bmod \left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right] \right)$，由Th3.1.1有$\exists t\in {{\mathbb{Z}}_{\ne 0}},\text{ }s.t.$
$$a=b+\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right]t\text{ }\left( 1 \right)$$
$a\equiv b\left( \bmod {{m}_{n+1}} \right)$，由Th3.1.1有$\text{ }\exists s\in {{\mathbb{Z}}_{\ne 0}},\text{ }s.t.\text{ }$
$$a=b+{{m}_{n+1}}s\text{ }\left( 2 \right)$$
另一方面，由$\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right]\text{ }\!\!\And\!\!\text{ }\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n+1}} \right]$的求解过程，有
$$\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n+1}} \right]=\left[ \left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right],{{m}_{n+1}} \right]=\frac{\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right]{{m}_{n+1}}}{\left( \left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right],{{m}_{n+1}} \right)}\text{ }\left( 3 \right)$$
由Th1.2.4-vi)，$\exists p,q\in \mathbb{Z},\text{ }s.t.$
$$\left( \left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right],{{m}_{n+1}} \right)=\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right]p+{{m}_{n+1}}q\text{ }\left( 4 \right)$$
由$\left( 1 \right)\left( 2 \right)$有
$$\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right]=\frac{a-b}{t}\text{ }\left( 5 \right)$$
$${{m}_{n+1}}=\frac{a-b}{s}\text{ }\left( 6 \right)$$
由$\left( 3 \right)\left( 4 \right)$有
$$\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n+1}} \right]=\frac{\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right]{{m}_{n+1}}}{\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n}} \right]p+{{m}_{n+1}}q}\text{ }\left( 7 \right)$$
将$\left( 5 \right)\left( 6 \right)$代入$\left( 7 \right)$得到
$$\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n+1}} \right]=\frac{\frac{a-b}{t}\times \frac{a-b}{s}}{\frac{a-b}{t}p+\frac{a-b}{s}q}=\frac{a-b}{ps+qt}$$
$$\Rightarrow a=b+\left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n+1}} \right]\left( ps+qt \right),\text{ }ps+qt\in \mathbb{Z}$$
因此由Th3.1.1有$\left. \left[ {{m}_{1}},{{m}_{2}},...,{{m}_{n+1}} \right]\text{ } \right|a-b$。
由数学第一归纳法，即证得本条性质。
9. $a\equiv b\left( \bmod m \right),\text{ }\left. d \right|m\text{ }\Rightarrow \text{ }a\equiv b\left( \bmod \left|d \right|\right)$
证明：
$a\equiv b\left( \bmod m \right)$，由Th3.1.1，$\exists t\in \mathbb{Z},\text{ }s.t.$$a=b+mt$。
$\left. d \right|m\text{ }\Rightarrow \text{ }\exists s\in \mathbb{Z},\text{ }s.t.\text{ }m=ds=\left| d \right|\times sgn \left( d \right)s$。
因此有$a=b+\left| d \right|\times \left[ sgn \left( d \right)st \right],\text{ }sgn \left( d \right)st\in \mathbb{Z}$，$a\equiv b\left( \bmod \left| d \right| \right)$。
10. $a\equiv b\left( \bmod m \right)\text{ }\Rightarrow \text{ }$${{G}_{1}}:=\left\{ \left. d\text{ } \right|\text{ }\left. d \right|a\text{ }\And \text{ }\left. d \right|m \right\}=\left\{ \left. d\text{ } \right|\text{ }\left. d \right|b\text{ }\And \text{ }\left. d \right|m \right\}=:{{G}_{2}}$。特别地，有$\left( a,m \right)=\left( b,m \right)$
证明：
$a\equiv b\left( \bmod m \right)$，由Th3.1.1，$\exists t\in \mathbb{Z}$，使得$a=b+mt$。
- $\forall d\in {{G}_{1}},\text{ }\left. d \right|a\text{ }\!\!\And\!\!\text{ }\left. d \right|m$，由Th1.1.3，有$\left. d \right|a-mt=b$$\Rightarrow \left. d \right|b\text{ }\And \text{ }\left. d \right|m$，
因此有$d\in {{G}_{2}},\text{ }{{G}_{1}}\subseteq {{G}_{2}}$。
- $\forall d\in {{G}_{2}}$，$\left. d \right|b\text{ }\!\!\And\!\!\text{ }\left. d \right|m$，由Th1.1.3，有$\left. d \right|b+mt=a\text{ }\Rightarrow \text{ }\left. d \right|a\text{ }\And \text{ }\left. d \right|m$，
因此有$d\in {{G}_{1}},\text{ }{{G}_{2}}\subseteq {{G}_{1}}$。
- 综上，有${{G}_{1}}={{G}_{2}}$。特别地有$\max {{G}_{1}}=\max {{G}_{2}}$即$\left( a,m \right)=\left( b,m \right)$。
同余应用
检查某种形式下的一个整数是否含有某个因数（充要条件）
- $\forall a\in \mathbb{Z}$，$a={{a}_{n}}{{10}^{n}}+{{a}_{n-1}}{{10}^{n-1}}+...+{{a}_{0}},\text{ }0\le \left| {{a}_{i}} \right|<10,\text{ }i=0,1,...,n$
- $\left. 3 \right|a\text{ }\Leftrightarrow \text{ }\left. 3 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}}$
证明：
$10\equiv 1\left( \bmod 3 \right)$，由Th3.1.2有
$${{a}_{n}}{{10}^{n}}+{{a}_{n-1}}{{10}^{n-1}}+...+{{a}_{0}}\equiv {{a}_{n}}{{1}^{n}}+{{a}_{n-1}}{{1}^{n-1}}+...+{{a}_{0}}\left( \bmod 3 \right)$$
又$a=\sum\limits_{i=0}^{n}{{{a}_{i}}{{10}^{i}}},\text{ }\sum\limits_{i=0}^{n}{{{a}_{i}}}=\sum\limits_{i=0}^{n}{{{a}_{i}}{{1}^{i}}}$，由性质1), 3)有
$$a\equiv \sum\limits_{i=0}^{n}{{{a}_{i}}}\left( \bmod 3 \right)$$
$\left. 3 \right|3$，在此基础上，由性质10)有$\left. 3 \right|a\text{ }\Rightarrow \text{ }\left. 3 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}},\text{ }\left. 3 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}}\text{ }\Rightarrow \text{ }\left. 3 \right|a$，即有
$$\left. 3 \right|a\text{ }\Leftrightarrow \text{ }\left. 3 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}}$$
- $\left. 9 \right|a\text{ }\Leftrightarrow \text{ }\left. 9 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}}$
证明：参照结论1同理可证。
- $\left. 11 \right|a\text{ }\Leftrightarrow \text{ }\left. 11 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}$
证明：
$$10=11\times 0+10,\text{ }-1=11\times \left( -1 \right)+10\text{ }\Rightarrow \text{ 10}\equiv -\text{1}\left( \bmod 11 \right)$$
由Th3.1.2有，
$$\sum\limits_{i=0}^{n}{{{a}_{i}}{{10}^{i}}}\equiv \sum\limits_{i=0}^{n}{{{a}_{i}}{{\left( -1 \right)}^{i}}}\left( \bmod 11 \right)$$
又$a=\sum\limits_{i=0}^{n}{{{a}_{i}}{{10}^{i}}}$，由性质1),3)有
$$a\equiv \sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}\left( \bmod 11 \right)$$
$\left. 11 \right|11$，在此基础上，由性质10)，有$\left. 11 \right|a\text{ }\Rightarrow \text{ }\left. 11 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}},\text{ }\left. 11 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}\Rightarrow \left. 11 \right|a$，即
$$\left. 11 \right|a\text{ }\Leftrightarrow \text{ }\left. 11 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}$$
- 例子：
①$a=5874192$，$\sum\limits_{i=0}^{n}{{{a}_{i}}}=36,\text{ }\left. 3 \right|36,\text{ }\left. 9 \right|36\text{ }\Rightarrow \text{ }\left. 3 \right|a,\text{ }\left. 9 \right|a$；
②$a=435693,\text{ }\sum\limits_{i=0}^{n}{{{a}_{i}}}=30,\text{ }\left. 3 \right|30$，$9$不整除$30$，因此$\left. 3 \right|a$且$9$不整除$a$。
- $\forall a\in \mathbb{Z}$，$a={{a}_{n}}{{100}^{n}}+{{a}_{n-1}}{{100}^{n-1}}+...+{{a}_{0}},\text{ }0\le \left| {{a}_{i}} \right|<100,\text{ }i=0,1,...,n$。
- $\left. 101 \right|a\text{ }\Leftrightarrow \text{ }\left. 101 \right|\sum\limits_{i=0}^{n}{\left(-1\right)^i{{a}_{i}}}$
证明：
$100=101\times 0+100,\text{ }-1=101\times \left( -1 \right)+100\text{ }\Rightarrow \text{ }100\equiv -1\left( \,\bmod \,101 \right)$，由Th3.1.2有
$$\sum\limits_{i=0}^{n}{{{a}_{i}}{{100}^{i}}}\equiv \sum\limits_{i=0}^{n}{{{a}_{i}}{{\left( -1 \right)}^{i}}}\left( \,\bmod \,101 \right)$$
又$a=\sum\limits_{i=0}^{n}{{{a}_{i}}{{100}^{i}}}$，由性质1), 3)有
$$a\equiv \sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}\left( \,\bmod \,101 \right)$$
$\left. 101 \right|101$，在此基础上，由性质10)有$\left. 101 \right|a\text{ }\Rightarrow \text{ }\left. 101 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}},\text{ }\left. 101 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}\text{ }\Rightarrow \text{ }\left. 101 \right|a$，即
$$\left. 101 \right|a\text{ }\Leftrightarrow \text{ }\left. 101 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}$$
- $\forall a\in \mathbb{Z}$，$a={{a}_{n}}{{1000}^{n}}+{{a}_{n-1}}{{1000}^{n-1}}+...+{{a}_{0}},\text{ }0\le \left| {{a}_{i}} \right|<1000,\text{ }i=0,1,...,n$。
- $\left. 7 \right|a\text{ }\Leftrightarrow \text{ }\left. 7 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}$
证明：
$1000=7\times 142+6,\text{ }-1=7\times \left( -1 \right)+6\text{ }\Rightarrow \text{ }1000\equiv -1\left( \bmod 7 \right)$，由Th3.1.2有
$$\sum\limits_{i=0}^{n}{{{a}_{i}}{{1000}^{i}}}\equiv \sum\limits_{i=0}^{n}{{{a}_{i}}{{\left( -1 \right)}^{i}}}\left( \bmod 7 \right)$$
又$a=\sum\limits_{i=0}^{n}{{{a}_{i}}{{1000}^{i}}}$，由性质1),3)有
$$a\equiv \sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}\left( \bmod 7 \right)$$
$\left. 7 \right|7$，在此基础上，由性质10)得$\left. 7 \right|a\text{ }\Rightarrow \text{ }\left. 7 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}},\text{ }\left. 7 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}\Rightarrow \left. 7 \right|a$即
$$\left. 7 \right|a\text{ }\Leftrightarrow \text{ }\left. 7 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}$$
- $\left. 11 \right|a\text{ }\Leftrightarrow \text{ }\left. 11 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}$
证明：参照情形1同理可证。
- $\left. 13 \right|a\text{ }\Leftrightarrow \text{ }\left. 13 \right|\sum\limits_{i=0}^{n}{{{\left( -1 \right)}^{i}}{{a}_{i}}}$
证明：参照情形1同理可证。
- $\left. 37 \right|a\text{ }\Leftrightarrow \text{ }\left. 37 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}}$
证明：
$1000=37\times 27+1,\text{ }1=37\times 0+1\text{ }\Rightarrow \text{ 1000}\equiv \text{1}\left( \bmod 37 \right)$，由Th3.1.2有
$$\sum\limits_{i=0}^{n}{{{a}_{i}}{{1000}^{i}}}\equiv \sum\limits_{i=0}^{n}{{{a}_{i}}{{1}^{i}}}\left( \bmod 37 \right)$$
又$a=\sum\limits_{i=0}^{n}{{{a}_{i}}{{1000}^{i}}},\text{ }\sum\limits_{i=0}^{n}{{{a}_{i}}{{1}^{i}}}=\sum\limits_{i=0}^{n}{{{a}_{i}}}$，由性质1），3）有
$$a\equiv \sum\limits_{i=0}^{n}{{{a}_{i}}}\left( \bmod 37 \right)$$
$\left. 37 \right|37$，在此基础上，由性质10）有$\left. 37 \right|a\text{ }\Rightarrow \text{ }\left. 37 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}},\text{ }\left. 37 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}}\text{ }\Rightarrow \text{ }\left. 37 \right|a$，即有
$$\left. 37 \right|a\text{ }\Leftrightarrow \text{ }\left. 37 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}}$$
- 例子：
①$a=637693=637\times {{1000}^{1}}+693\times {{1000}^{0}},\text{ }{{\left( -1 \right)}^{1}}\times 637+{{\left( -1 \right)}^{0}}693=56$，而$\left. 7 \right|56$，$11\And 13$不整除$56$，因此$7$是$a$的因数，$11\And 13$不是$a$的因数；
②$$\begin{matrix}
a=75\text{ }312\text{ }289=75\times {{1000}^{2}}+312\times {{1000}^{1}}+289\times {{1000}^{0}} \\
{{\left( -1 \right)}^{2}}\times 75+{{\left( -1 \right)}^{1}}\times 312+{{\left( -1 \right)}^{0}}\times 289=52 \\
\end{matrix}$$$\left. 13 \right|52$，而$7\And 11$不整除$52$，因此$13$是$a$的因数，$7\And 11$不是$a$的因数。
- $\forall a\in \mathbb{Z}$，$a={{a}_{n}}{{10000}^{n}}+{{a}_{n-1}}{{10000}^{n-1}}+...+{{a}_{0}},\text{ }0\le \left| {{a}_{i}} \right|<10000,\text{ }i=0,1,...,n$。
- $\left. 101 \right|a\text{ }\Leftrightarrow \text{ }\left. 101 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}}$
证明：
$10000=101\times 99+1,\text{ }1=101\times 0+1\text{ }\Rightarrow \text{ }10000\equiv 1\left( \bmod 101 \right)$，由Th3.1.2有
$$\sum\limits_{i=0}^{n}{{{a}_{i}}{{10000}^{i}}}\equiv \sum\limits_{i=0}^{n}{{{a}_{i}}{{1}^{i}}}\left( \bmod 101 \right)$$
又$a=\sum\limits_{i=0}^{n}{{{a}_{i}}{{10000}^{i}}},\text{ }\sum\limits_{i=0}^{n}{{{a}_{i}}{{1}^{i}}}=\sum\limits_{i=0}^{n}{{{a}_{i}}}$，由性质1), 3)有
$$a\equiv \sum\limits_{i=0}^{n}{{{a}_{i}}}\left( \bmod 101 \right)$$
$\left. 101 \right|101$，在此基础上，由性质10)有$\left. 101 \right|a\text{ }\Rightarrow \text{ }\left. 101 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}},\text{ }\left. 101 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}}\text{ }\Rightarrow \text{ }\left. 101 \right|a$，即
$$\left. 101 \right|a\text{ }\Leftrightarrow \text{ }\left. 101 \right|\sum\limits_{i=0}^{n}{{{a}_{i}}}$$
验算整数计算结果的正确性（结果正确的必要条件之一：弃九法）
- 验算若干个整数乘积的结果
考虑$m$个整数
$$\begin{matrix}
{{a}^{\left( 1 \right)}}={{a}_{{{n}_{1}}}}^{\left( 1 \right)}{{10}^{{{n}_{1}}}}+{{a}_{{{n}_{1}}-1}}^{\left( 1 \right)}{{10}^{{{n}_{1}}-1}}+...+{{a}_{0}}^{\left( 1 \right)},\text{ }0\le \left| {{a}_{i}}^{\left( 1 \right)} \right|<10 \\
{{a}^{\left( 2 \right)}}={{a}_{{{n}_{2}}}}^{\left( 2 \right)}{{10}^{{{n}_{2}}}}+{{a}_{{{n}_{2}}-1}}^{\left( 2 \right)}{{10}^{{{n}_{2}}-1}}+...+{{a}_{0}}^{\left( 2 \right)},\text{ }0\le \left| {{a}_{i}}^{\left( 2 \right)} \right|<10 \\
...... \\
{{a}^{\left( m \right)}}={{a}_{{{n}_{m}}}}^{\left( m \right)}{{10}^{{{n}_{m}}}}+{{a}_{{{n}_{m}}-1}}^{\left( m \right)}{{10}^{{{n}_{m}}-1}}+...+{{a}_{0}}^{\left( m \right)},\text{ }0\le \left| {{a}_{i}}^{\left( m \right)} \right|<10 \\
\end{matrix}$$并设我们计算出来的这$m$个整数的乘积为
$$P={{p}_{t}}{{10}^{t}}+{{p}_{t-1}}{{10}^{t-1}}+...+{{p}_{0}},\text{ }0\le \left| {{p}_{i}} \right|<10$$
则有
$$\prod\limits_{j=0}^{m}{\left( \sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}} \right)}\not{\equiv }\sum\limits_{k=0}^{t}{{{p}_{k}}}\left( \bmod 9 \right)\text{ }\Rightarrow \text{ }P\ne \prod\limits_{j=0}^{m}{{{a}^{\left( j \right)}}}$$
证明：
$10\equiv 1\left( \bmod 9 \right)$，由Th3.1.2有$\forall j\in \left\{ 1,2,...,m \right\}$，
$$\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}{{10}^{i}}}\equiv \sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}{{1}^{i}}}$$
又${{a}^{\left( j \right)}}=\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}{{10}^{i}}},\text{ }\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}}=\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}{{1}^{i}}}$，由性质1), 3)有
$${{a}^{\left( j \right)}}\equiv \sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}}\left( \bmod 9 \right)$$
类似可得
$$P\equiv \sum\limits_{k=0}^{t}{{{p}_{k}}}\left( \bmod 9 \right)$$
不断应用性质5)，即可得
$$\prod\limits_{j=0}^{m}{{{a}^{\left( j \right)}}}\equiv \prod\limits_{j=0}^{m}{\left( \sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}} \right)}\left( \bmod 9 \right)$$
若$P=\prod\limits_{j=0}^{m}{{{a}^{\left( j \right)}}}$，则由性质1), 3)可得
$$\sum\limits_{k=0}^{t}{{{p}_{k}}}\equiv \prod\limits_{j=0}^{m}{\left( \sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}} \right)}\left( \bmod 9 \right)$$
其逆否命题亦正确，即
$$\prod\limits_{j=0}^{m}{\left( \sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}} \right)}\not{\equiv }\sum\limits_{k=0}^{t}{{{p}_{k}}}\left( \bmod 9 \right)\text{ }\Rightarrow \text{ }P\ne \prod\limits_{j=0}^{m}{{{a}^{\left( j \right)}}}$$
例子：
$a=28997,\text{ }b=39495$，${{P}_{1}}=1\text{ }145\text{ }236\text{ }415$，${{P}_{2}}=1\text{ }145\text{ }235\text{ }615$
$$\sum\limits_{i=0}^{4}{{{a}_{i}}}\equiv 17\left( \bmod 9 \right),\text{ }\sum\limits_{i=0}^{4}{{{b}_{i}}}\equiv 3\left( \bmod 9 \right),\text{ }\sum\limits_{i=0}^{9}{{{p}_{i}}^{\left( 1 \right)}}\equiv 5\left( \bmod 9 \right),\text{ }\sum\limits_{i=0}^{9}{{{p}_{i}}^{\left( 2 \right)}}\equiv 15\left( \bmod 5 \right)$$
$17\times 3=9\times 5+6,\text{ }5=9\times 0+5\text{ }\Rightarrow \text{ }17\times 3\not{\equiv }5\left( \bmod 9 \right)$
因此有$ab\ne {{P}_{1}}$，${{P}_{1}}$的计算有误。
$17\times 3=9\times 5+6,\text{ }15=9\times 1+6\text{ }\Rightarrow \text{ }17\times 3\equiv 15\left( \bmod 9 \right)$
但是还是有$ab\ne {{P}_{2}}$。
事实上，正确的计算结果是$P=1\text{ }145\text{ }236\text{ }515$。
- 验算若干个整数相加的结果
考虑$m$个整数
$$\begin{matrix}
{{a}^{\left( 1 \right)}}={{a}_{{{n}_{1}}}}^{\left( 1 \right)}{{10}^{{{n}_{1}}}}+{{a}_{{{n}_{1}}-1}}^{\left( 1 \right)}{{10}^{{{n}_{1}}-1}}+...+{{a}_{0}}^{\left( 1 \right)},\text{ }0\le \left| {{a}_{i}}^{\left( 1 \right)} \right|<10 \\
{{a}^{\left( 2 \right)}}={{a}_{{{n}_{2}}}}^{\left( 2 \right)}{{10}^{{{n}_{2}}}}+{{a}_{{{n}_{2}}-1}}^{\left( 2 \right)}{{10}^{{{n}_{2}}-1}}+...+{{a}_{0}}^{\left( 2 \right)},\text{ }0\le \left| {{a}_{i}}^{\left( 2 \right)} \right|<10 \\
...... \\
{{a}^{\left( m \right)}}={{a}_{{{n}_{m}}}}^{\left( m \right)}{{10}^{{{n}_{m}}}}+{{a}_{{{n}_{m}}-1}}^{\left( m \right)}{{10}^{{{n}_{m}}-1}}+...+{{a}_{0}}^{\left( m \right)},\text{ }0\le \left| {{a}_{i}}^{\left( m \right)} \right|<10 \\
\end{matrix}$$并设我们计算出来的这$m$个整数的和为
$$P={{p}_{t}}{{10}^{t}}+{{p}_{t-1}}{{10}^{t-1}}+...+{{p}_{0}},\text{ }0\le \left| {{p}_{i}} \right|<10$$
则有
$$\sum\limits_{j=0}^{m}{\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}}}\not{\equiv }\sum\limits_{k=0}^{t}{{{p}_{k}}}\left( \bmod 9 \right)\text{ }\Rightarrow \text{ }P\ne \sum\limits_{j=0}^{m}{{{a}^{\left( j \right)}}}$$
证明：
$10\equiv 1\left( \bmod 9 \right)$，由Th3.1.2有$\forall j\in \left\{ 1,2,...,m \right\}$，
$$\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}{{10}^{i}}}\equiv \sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}{{1}^{i}}}$$
又${{a}^{\left( j \right)}}=\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}{{10}^{i}}},\text{ }\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}}=\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}{{1}^{i}}}$，由性质1), 3)有
$${{a}^{\left( j \right)}}\equiv \sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}}\left( \bmod 9 \right)$$
类似可得
$$P\equiv \sum\limits_{k=0}^{t}{{{p}_{k}}}\left( \bmod 9 \right)$$
不断应用性质4)，即可得
$$\sum\limits_{j=0}^{m}{{{a}^{\left( j \right)}}}\equiv \sum\limits_{j=0}^{m}{\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}}}\left( \,\bmod \,9 \right)$$
若$P=\sum\limits_{j=0}^{m}{{{a}^{\left( j \right)}}}$，则由性质1), 3)可得
$$\sum\limits_{k=0}^{t}{{{p}_{k}}}\equiv \sum\limits_{j=0}^{m}{\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}}}\left( \,\bmod \,9 \right)$$
其逆否命题亦正确，即
$$\sum\limits_{k=0}^{t}{{{p}_{k}}}\not{\equiv }\sum\limits_{j=0}^{m}{\sum\limits_{i=0}^{{{n}_{j}}}{{{a}_{i}}^{\left( j \right)}}}\left( \,\bmod \,9 \right)\text{ }\Rightarrow \text{ }P\ne \sum\limits_{j=0}^{m}{{{a}^{\left( j \right)}}}$$
注：由于$\left. 3 \right|9$，因此用弃九法验证就已经包含了用弃三法验证。
练习3.1
- 应用检查因数的方法求出下列各数的标准分解式：
1)1535625；2)1158066。
解：
- $$\begin{aligned}
& {{a}^{\left( 0 \right)}}=1535625,\text{ }\sum\limits_{i=0}^{6}{{{a}_{i}}^{\left( 0 \right)}}=27,\text{ }\left. {{3}^{2}} \right|27\text{ }\Rightarrow \text{ }\left. {{3}^{2}} \right|{{a}^{\left( 0 \right)}} \\
& {{a}^{\left( 1 \right)}}=\frac{{{a}^{\left( 0 \right)}}}{{{3}^{2}}}=170625,\text{ }\sum\limits_{i=0}^{5}{{{a}_{i}}^{\left( 1 \right)}}=21,\text{ }\left. 3 \right|21\text{ }\Rightarrow \text{ }\left. 3 \right|{{a}^{\left( 1 \right)}} \\
& {{a}^{\left( 2 \right)}}=\frac{{{a}^{\left( 1 \right)}}}{3}=56875=56\times {{1000}^{1}}+875\times {{1000}^{0}},\text{ }-56+875=819,\text{ }\left. 13 \right|819\text{ }\Rightarrow \text{ }\left. 13 \right|{{a}^{\left( 2 \right)}} \\
& {{a}^{\left( 3 \right)}}=\frac{{{a}^{\left( 2 \right)}}}{13}=4375=4\times {{1000}^{1}}+375\times {{1000}^{0}},\text{ }-4+375=371,\text{ }\left. 7 \right|371\text{ }\Rightarrow \text{ }\left. 7 \right|{{a}^{\left( 3 \right)}} \\
& {{a}^{\left( 4 \right)}}=\frac{{{a}^{\left( 3 \right)}}}{7}=53 \\
\end{aligned}$$$$\Rightarrow 1535625={{a}^{\left( 0 \right)}}={{3}^{3}}\times 7\times 13\times 53$$
- $$\begin{aligned}
& {{a}^{\left( 0 \right)}}=1158066,\text{ }\sum\limits_{i=0}^{6}{{{a}_{i}}^{\left( 0 \right)}}=27,\text{ }\left. {{3}^{2}} \right|27\text{ }\Rightarrow \text{ }\left. {{3}^{2}} \right|{{a}^{\left( 0 \right)}} \\
& {{a}^{\left( 1 \right)}}=\frac{{{a}^{\left( 0 \right)}}}{{{3}^{2}}}=128674=128\times {{1000}^{1}}+674\times {{1000}^{0}},\text{ }-128+674=546,\text{ }\left. 13 \right|546\text{ }\Rightarrow \text{ }\left. 13 \right|{{a}^{\left( 1 \right)}} \\
& {{a}^{\left( 2 \right)}}=\frac{{{a}^{\left( 1 \right)}}}{13}=42=2\times 3\times 7 \\
\end{aligned}$$$$\Rightarrow 1158066={{a}^{\left( 0 \right)}}=2\times {{3}^{3}}\times 7\times 13$$
- 证明$\left. 641 \right|{{2}^{32}}+1$。
证明:
$$\begin{aligned}
& {{2}^{16}}=65536\text{ }\Rightarrow \text{ }{{2}^{16}}\equiv 65536\left( \bmod 641 \right) \\
& \left\{ \begin{aligned}
& 65536=641\times 102+154 \\
& 154<641\text{ }\Rightarrow \text{ }154=641\times 0+154 \\
\end{aligned} \right.\text{ }\Rightarrow \text{ 65536}\equiv \text{154}\left( \bmod 641 \right) \\
\end{aligned}$$因此有${{2}^{16}}\equiv 154\left( \bmod 641 \right)$。
$$\begin{aligned}
& {{2}^{32}}={{\left( {{2}^{16}} \right)}^{2}}\equiv {{154}^{2}}=23716\left( \bmod 641 \right) \\
& \left\{ \begin{aligned}
& 23716=641\times 36+640 \\
& 640<641\text{ }\Rightarrow \text{ }640=641\times 0+640 \\
& -1=641\times \left( -1 \right)+640 \\
\end{aligned} \right.\text{ }\Rightarrow \text{ }23716\equiv 640\equiv -1\left( \bmod 641 \right) \\
\end{aligned}$$因此有${{2}^{32}}\equiv -1\left( \bmod 641 \right)\text{ }\Rightarrow \text{ }{{2}^{32}}+1\equiv 0\left( \bmod 641 \right)\text{ }\Rightarrow \text{ }\left. 641 \right|{{2}^{32}}+1$。
注：
① 该题的第一个思想是将整除转化为同余，体现在$$\left. 641 \right|{{2}^{32}}\text{+1 }\Leftrightarrow \text{ }{{\text{2}}^{32}}+1\equiv 0\left( \bmod 641 \right)\text{ }\Leftrightarrow \text{ }{{\text{2}}^{32}}\equiv -1\left( \bmod 641 \right)$$
② 在①的基础上，该题的第二个思想是不断将大数同绝对值小于除数（641）的整数建立起同余关系，使得计算在笔算层面一直保持在可控量级上；
③ 由于${{10}^{5}}=641\times 156+{{2}^{2}}$，因此有${{10}^{5}}\equiv {{2}^{2}}\left( \bmod 641 \right)\text{ }\Rightarrow \text{ }{{\left( {{10}^{5}} \right)}^{16}}+1\equiv {{\left( {{2}^{2}} \right)}^{16}}+1\left( \bmod 641 \right)$即
$${{10}^{80}}+1\equiv {{2}^{32}}+1\left( \bmod 641 \right)$$
所以也有$\left. 641 \right|{{10}^{80}}+1$。
- 设$a$是任一奇数，则有${{a}^{{{2}^{n}}}}\equiv 1\left( \bmod {{2}^{n+2}} \right)\left( n\ge 1 \right)$。
证明：
令$G=\left\{ \left. 2m+1 \right|m\in \mathbb{Z} \right\}$为全体奇数集合。
- 先证明$n=1$时，$\forall a\in G$，均有${{a}^{{{2}^{1}}}}\equiv 1\left( \bmod {{2}^{1+2}} \right)$即${{a}^{2}}\equiv 1\left( \bmod 8 \right)$。
$a=2m+1\text{ }\Rightarrow \text{ }{{a}^{2}}=4{{m}^{2}}+4m+1=4\left( {{m}^{2}}+m \right)+1$。
$m$是奇数时，${{m}^{2}},m$均为奇数，${{m}^{2}}+m$为偶数，因此$\exists {{q}_{1}}\in \mathbb{Z}$，使得${{m}^{2}}+m=2{{q}_{1}}$，也就有
${{a}^{2}}=8{{q}_{1}}+1\equiv 1\left( \bmod 8 \right)$
$m$是偶数时，${{m}^{2}},m$均为偶数，${{m}^{2}}+m$为偶数，因此$\exists {{q}_{2}}\in {{\mathbb{Z}}}$，使得${{m}^{2}}+m=2{{q}_{2}}$，也就有
${{a}^{2}}=8{{q}_{2}}+1\equiv 1\left( \bmod 8 \right)$
- $\forall a\in G$，假设$n=k$时有${{a}^{{{2}^{k}}}}\equiv 1\left( \bmod {{2}^{k+2}} \right)$，则有
$${{a}^{{{2}^{k}}}}-1\equiv 0\left( \bmod {{2}^{k+2}} \right)\text{ }\Rightarrow \text{ }\left. {{2}^{k+2}} \right|{{a}^{{{2}^{k}}}}-1$$
因此$\exists t\in \mathbb{Z},\text{ }s.t.\text{ }{{a}^{{{2}^{k}}}}-1={{2}^{k+2}}t$，即${{a}^{{{2}^{k}}}}=1+{{2}^{k+2}}t$，于是有
$$\begin{matrix}
{{a}^{{{2}^{k+1}}}}={{\left( {{a}^{{{2}^{k}}}} \right)}^{2}}={{\left( 1+{{2}^{k+2}}t \right)}^{2}}={{2}^{k+3}}\left( {{2}^{k+1}}{{t}^{2}}+t \right)+1,\text{ }{{2}^{k+1}}{{t}^{2}}+t\in \mathbb{Z} \\
\Rightarrow {{a}^{{{2}^{k+1}}}}\equiv 1\left( \bmod {{2}^{k+3}} \right) \\
\end{matrix}$$由数学第一归纳法最终得到，$\forall a\in G$，$\forall n\in {{\mathbb{Z}}_{\ge 1}}$，有
$${{a}^{{{2}^{n}}}}\equiv 1\left( \bmod {{2}^{n+2}} \right)$$
注：当尝试用理论和逻辑推理暂且行不通或者过于复杂时，涉及未知量$n$的证明题可考虑递推思想，考虑用数学归纳法等做法。
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