模m的阶定义设\(m>1,(a,m)=1\)则使得：\[a^d\equiv1(modm)\]成立的最小正整数\(d_0\)称为\(a\)模\(m\)的阶记为\(\delta_m(a)\)性质设\(m>1,\quadn>1,\quad(a,m)=1\)1.若\(\delta_m(a)=n\),则\(a^k\equiv1(modm)\)且\(n|k\)2.\(a\equivb(modm),\de