院系:李煌数学研究院/高次代数方程x^n=1的根式解的解析表达式

李煌公式

• 方程xⁿ=1满足n =3k+2, n>3,n,k是整数,之李煌根式解为：

  $x_{1}={\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}+{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}$

  $x_{2}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}+{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{2}$

  $x_{3}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}+{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{3}$

  $...$

  $x_{n}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}+{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{n}=1$

• 方程xⁿ=1满足n =3k+1, n>3,n,k是整数,之李煌根式解为：

  $x_{1}={\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}-{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}$

  $x_{2}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}-{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{2}$

  $x_{3}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}-{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{3}$

  $...$

  $x_{n}={{\Bigg (}{\frac {{-{\frac {1}{2}}}-{\frac {i{\sqrt {3}}}{2}}}{(-{\frac {1}{2}}-{{\frac {i{\sqrt {3}}}{2}})}^{\frac {1}{n}}}}{\Bigg )}}^{n}=1$

来源

• 《南昌理工学院学报》.李煌

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