指数
设m>1m>1m>1为整数,aaa是与mmm互素的正整数,则使得ae≡1(mod m)a^e\equiv1\left(\mod m\right)ae≡1(modm)成立的最小正整数eee叫做aaa对模mmm的指数,记作ordm(a)\mathrm{ord}_{m}{\left(a\right)}ordm(a)
原根
若e=φ(m)e=\varphi{\left(m\right)}e=φ(m),则aaa为模mmm的原根
ad≡1(mod m)⟺ordm(a)∣da^d\equiv1\left(\mod m\right)\Longleftrightarrow\mathrm{ord}_{m}{\left(a\right)}\mid dad≡1(modm)⟺ordm(a)∣d
设ppp为奇素数,p−12\frac{p-1}{2}2p−1为素数,若a≢0,1,−1(mod p)a\not\equiv0,1,-1\left(\mod p\right)a≡0,1,−1(modp),则ordp(a)=p−12\mathrm{ord}_{p}{\left(a\right)}=\frac{p-1}{2}ordp(a)=2p−1或p−1p-1p−1
b≡a(mod m)⟹ordm(b)=ordm(a)b\equiv a\left(\mod m\right)\Longrightarrow \mathrm{ord}_{m}{\left(b\right)}=\mathrm{ord}_{m}{\left(a\right)}b≡a(modm)⟹ordm(b)=ordm(a)
a−1a≡1(mod m)⟹ordm(a−1)=ordm(a)a^{-1}a\equiv1\left(\mod m\right)\Longrightarrow \mathrm{ord}_{m}{\left(a^{-1}\right)}=\mathrm{ord}_{m}{\left(a\right)}a−1a≡1(modm)⟹ordm(a−1)=ordm(a)
1=a0,a,⋯ ,aordm(a)−11=a^0,a,\cdots,a^{\mathrm{ord}_{m}{\left(a\right)}-1}1=a0,a,⋯,aordm(a)−1
ad≡ak(mod m)⟺d≡k(mod ordm(a))a^d\equiv a^k\left(\mod m\right)\Longleftrightarrow d\equiv k\left(\mod \mathrm{ord}_{m}{\left(a\right)}\right)ad≡ak(modm)⟺d≡k(modordm(a))
ordm(ad)=ordm(a)(d,ordm(a))\mathrm{ord}_{m}{\left(a^d\right)}=\frac{\mathrm{ord}_{m}{\left(a\right)}}{\left(d,\mathrm{ord}_{m}{\left(a\right)}\right)}ordm(ad)=(d,ordm(a))ordm(a)
设ggg为模mmm的原根,则gdg^dgd为模mmm的原根⟺(d,φ(m))=1\Longleftrightarrow \left(d,\varphi{\left(m\right)}\right)=1⟺(d,φ(m))=1
设k∣ordm(a)k\mid \mathrm{ord}_{m}{\left(a\right)}k∣ordm(a),则使得ordm(ad)=k,1≤d≤ordm(a)\mathrm{ord}_{m}{\left(a^d\right)}=k,1\leq d\leq \mathrm{ord}_{m}{\left(a\right)}ordm(ad)=k,1≤d≤ordm(a)成立的正整数ddd满足ordm(a)k∣d\frac{\mathrm{ord}_{m}{\left(a\right)}}{k}\mid dkordm(a)∣d,且共有φ(k)\varphi{\left(k\right)}φ(k)个这样的ddd
模mmm有原根⟹\Longrightarrow⟹模mmm有φ(φ(m))\varphi{\left(\varphi{\left(m\right)}\right)}φ(φ(m))个不同的原根
(ordm(a),ordm(b))=1⟺ordm(a⋅b)=ordm(a)⋅ordm(b)\left(\mathrm{ord}_{m}{\left(a\right)},\mathrm{ord}_{m}{\left(b\right)}\right)=1\Longleftrightarrow \mathrm{ord}_{m}{\left(a\cdot b\right)}=\mathrm{ord}_{m}{\left(a\right)}\cdot \mathrm{ord}_{m}{\left(b\right)}(ordm(a),ordm(b))=1⟺ordm(a⋅b)=ordm(a)⋅ordm(b)
根据ad≡1(mod m)⟺ordm(a)∣da^d\equiv1\left(\mod m\right)\Longleftrightarrow\mathrm{ord}_{m}{\left(a\right)}\mid dad≡1(modm)⟺ordm(a)∣d,求出mmm的因数,挨个验证
最后更新于3年前