定义:设m∈Z+a,b∈Zm\in\Z^+\quad a,b\in\Zm∈Z+a,b∈Z,若a,ba,ba,b被mmm除所得到的余数相同,则称aaa与bbb模mmm同余,记作a≡b(modm)a\equiv b\pmod ma≡b(modm).否则称aaa与bbb模mmm不同余,记作a̸≡b(modm)a\not\equiv b\pmod ma̸≡b(modm).
定理:设m∈Z+a,b∈Zm\in\Z^+\quad a,b\in\Zm∈Z+a,b∈Z,则a≡b(modm)a\equiv b\pmod ma≡b(modm)当且仅当∃q∈Za=mq+b\exists q\in\Z \quad a=mq+b∃q∈Za=mq+b当且仅当m∣(a−b)m\mid (a-b)m∣(a−b).
定理:同余是一种等价关系.
即1.反身性:a≡a(modm)a\equiv a\pmod ma≡a(modm).
2.对称性:若a≡b(modm)a\equiv b\pmod ma≡b(modm),则b≡a(modm)b\equiv a\pmod mb≡a(modm).
3.传递性:若a≡b(modm)b≡c(modm)a\equiv b\pmod m \quad b\equiv c\pmod ma≡b(modm)b≡c(modm),则a≡c(modm)a\equiv c\pmod ma≡c(modm).
定理:关于同余的一些性质:设a1≡b1(modm)a2≡b2(modm)a_1\equiv b_1\pmod m\quad a_2\equiv b_2\pmod ma1≡b1(modm)a2≡b2(modm),则
1.a1±a2≡b1±b2(modm)a_1\pm a_2\equiv b_1\pm b_2\pmod ma1±a2≡b1±b2(modm).
2.ca1≡cb1(modm)∀c∈Zca_1\equiv cb_1\pmod m\quad \forall c\in\Zca1≡cb1(modm)∀c∈Z.
3.ka1≡kb1(modkm)∀k∈Z+ka_1\equiv kb_1\pmod{km}\quad \forall k\in\Z^+ka1≡kb1(modkm)∀k∈Z+.
4.a1a2≡b1b2(modm)a_1a_2\equiv b_1b_2\pmod ma1a2≡b1b2(modm).
一般地,若ak≡bk(modm)k=1,2,⋯ ,na_k\equiv b_k\pmod m\quad k=1,2,\cdots,nak≡bk(modm)k=1,2,⋯,n,则
1.∑k=1nak≡∑k=1nbk(modm)\displaystyle \sum_{k=1}^{n}{a_k}\equiv \sum_{k=1}^{n}{b_k}\pmod mk=1∑nak≡k=1∑nbk(modm).
2.∏k=1nak≡∏k=1nbk(modm)\displaystyle \prod_{k=1}^{n}{a_k}\equiv \prod_{k=1}^{n}{b_k}\pmod mk=1∏nak≡k=1∏nbk(modm).
2’.若a≡b(modm)a\equiv b\pmod ma≡b(modm),则an≡bn(modm)∀n∈Z+a^n\equiv b^n\pmod m \quad \forall n\in\Z^+an≡bn(modm)∀n∈Z+.
推论:设ai,bi(0≤i≤n),u,v∈Za_i,b_i(0\le i \le n),u,v\in\Zai,bi(0≤i≤n),u,v∈Z,若ai≡bi(modm)∀0≤i≤n,u≡v(modm)a_i\equiv b_i\pmod m \quad \forall 0\le i \le n,u\equiv v\pmod mai≡bi(modm)∀0≤i≤n,u≡v(modm),则∑i=0maiui≡∑j=0nbjvj(modm)\displaystyle \sum_{i=0}^{m}{a_iu^i}\equiv\sum_{j=0}^{n}{b_jv^j}\pmod mi=0∑maiui≡j=0∑nbjvj(modm)特别地,对于整系数多项式f(x)=anxn+⋯+a1x+a0f(x)=a_nx^n+\cdots+a_1x+a_0f(x)=anxn+⋯+a1x+a0,有f(n)≡f(v)(modm)f(n)\equiv f(v)\pmod mf(n)≡f(v)(modm).
定理:同余对除法的性质:
1.若a≡b(modm),d∣m,d>0a\equiv b\pmod m,d\mid m,d>0a≡b(modm),d∣m,d>0,则a≡b(modd)a\equiv b\pmod da≡b(modd).
2.若a1a2≡b1b2(modm),a2≡b2(modm)a_1a_2\equiv b_1b_2\pmod m,a_2\equiv b_2\pmod ma1a2≡b1b2(modm),a2≡b2(modm)且(a2,m)=1(a_2,m)=1(a2,m)=1,则a1≡b1(modm)a_1\equiv b_1\pmod ma1≡b1(modm).
3.若a≡b(modm),d∣(a,b,m),d>0a\equiv b\pmod m,d\mid (a,b,m),d>0a≡b(modm),d∣(a,b,m),d>0,则ad≡bd(modmd)\dfrac{a}{d}\equiv\dfrac{b}{d}\pmod{\dfrac{m}{d}}da≡db(moddm).
4.若(a,m)=1(a,m)=1(a,m)=1,则∃b∈Z\exists b\in\Z∃b∈Z,使得ab≡1(modm)ab\equiv1\pmod mab≡1(modm).
定义:若(a,m)=1(a,m)=1(a,m)=1,称满足ab≡1(modm)ab\equiv1\pmod mab≡1(modm)的整数bbb为aaa对模mmm的逆,记为a−1a^{-1}a−1.
定理:同余式a≡b(modmi)i=1,2,⋯ ,na\equiv b\pmod{m_i} \quad i=1,2,\cdots,na≡b(modmi)i=1,2,⋯,n同时成立当且仅当a≡b(mod[m1,m2,⋯ ,mn])a\equiv b\pmod {[m_1,m_2,\cdots,m_n]}a≡b(mod[m1,m2,⋯,mn]).