原根与指标

$$e_{p}(a) = (a^{e}\equiv 1\pmod{p}的最小指数 e, e \geqslant 1)$$

- $e_{p}(a)$总是整除 $p-1$

$$\begin{align*}\because\  &a^{e_{p}(a)}\equiv 1\pmod{p}\\ &假设 a^{n}\equiv 1\pmod{p}, G=\gcd(e_{p}(a), n)\\\therefore\ &e_{p}(a)u - nv = G\\&a^{e_{p}(a)u} \equiv 1 \pmod{p}\\&a^{e_{p}(a)u}=a^{nv+G} \equiv a^{G}\pmod{p}\\&a^{G}\equiv 1 \pmod{p}\\\therefore\ &G=e_p(a)\\\because\ &a^{p-1}\equiv 1\pmod{p}\\\therefore\ &e_{p}(a) 整除 p-1 \end{align*}$$

$$对于那些e_{p}(g)=p-1的数g称为模p的原根$$

- 每个素数$p$都有原根，并且有$\phi(p-1)$个原根

 

 

- 对于素数$p$的原根$a$，$a，a^{2}，a^{3}，\cdots，a^{p-2}，a^{p-1}\pmod{p}$都是不同的，否则如果$a^{i}\equiv a^{j}\pmod{p}$，$a^{i-j}\equiv 1\pmod{p}$

$$对于原根g所对应的以1~p-1为指数的每个数a，a所对应的指标为其指数，记为I(a)$$

- $I(ab) \equiv I(a) + I(b)\pmod{p-1}$

- $I(a^{k}) \equiv kI(a)\pmod{p-1}$

$$\begin{align*}\because\ & g^{I(ab)}\equiv ab\equiv g^{I(a)}g^{I(b)}\equiv g^{I(a)+I(b)}\pmod{p}\\\therefore\ &g^{I(ab)-I(a)-I(b)}\equiv 1 \pmod{p}\\\because\ &g为原根\\\therefore\ &I(ab)-I(a)-I(b)整除p-1\\&I(ab) \equiv I(a) + I(b)\pmod{p-1}\end{align*}$$