最后更新于3年前
设m>1m>1m>1为整数,aaa是与mmm互素的正整数,若xn≡a(mod m)x^n\equiv a\left(\mod m\right)xn≡a(modm)有解,则aaa为对模mmm的nnn次剩余
STEP1: 验证有解(n,φ(m))∣indga\left(n,\varphi{\left (m\right)}\right)\mid \mathrm{ind}_{g}{a}(n,φ(m))∣indga,ggg为模mmm的原根解数为(n,φ(m))\left(n,\varphi{\left (m\right)}\right)(n,φ(m))STEP2: 等价同余式等价于nindgx≡indga(mod φ(m))n{\mathrm{ind}}_{g}{x}\equiv \mathrm{ind}_{g}{a}\left(\mod \varphi{\left (m\right)}\right)nindgx≡indga(modφ(m))STEP3: 查指标表解出nindgxn{\mathrm{ind}}_{g}{x}nindgx,解出x(mod m)x\left(\mod m\right)x(modm)
STEP1: 验证有解
(n,φ(m))∣indga\left(n,\varphi{\left (m\right)}\right)\mid \mathrm{ind}_{g}{a}(n,φ(m))∣indga,ggg为模mmm的原根
解数为(n,φ(m))\left(n,\varphi{\left (m\right)}\right)(n,φ(m))
STEP2: 等价同余式
等价于nindgx≡indga(mod φ(m))n{\mathrm{ind}}_{g}{x}\equiv \mathrm{ind}_{g}{a}\left(\mod \varphi{\left (m\right)}\right)nindgx≡indga(modφ(m))
STEP3: 查指标表解出nindgxn{\mathrm{ind}}_{g}{x}nindgx,解出x(mod m)x\left(\mod m\right)x(modm)
STEP1: 等价同余式等价于xindgn≡indga(mod φ(m))x{\mathrm{ind}}_{g}{n}\equiv \mathrm{ind}_{g}{a}\left(\mod \varphi{\left (m\right)}\right)xindgn≡indga(modφ(m))STEP2: 查指标表解出x(mod φ(m))x\left(\mod \varphi{\left (m\right)}\right)x(modφ(m))
STEP1: 等价同余式
等价于xindgn≡indga(mod φ(m))x{\mathrm{ind}}_{g}{n}\equiv \mathrm{ind}_{g}{a}\left(\mod \varphi{\left (m\right)}\right)xindgn≡indga(modφ(m))
STEP2: 查指标表解出x(mod φ(m))x\left(\mod \varphi{\left (m\right)}\right)x(modφ(m))