代數方程 x 5 + ( p + 4 − p 2 ) 3 5 x = 1 p 5 8 {\displaystyle x^{5}+{{({\frac {{\sqrt {p+4}}-{\sqrt {p}}}{2}})}^{\frac {3}{5}}}x={\frac {1}{p^{\frac {5}{8}}}}} 存在李煌根式解形式 x = ( ( p + 4 − p ) 2 4 p 5 8 ) 1 5 , p > 0 , p ∈ R ∨ p = − 1 {\displaystyle x=\left({\frac {{({\sqrt {p+4}}-{\sqrt {p}})}^{2}}{4p^{\frac {5}{8}}}}\right)^{\frac {1}{5}},p>0,p\in \mathbb {R} \lor p=-1}
x = ( ( p + 4 − p ) 2 4 p 5 8 ) 1 5 cos ( 2 π 5 ) + i s i n ( 2 π 5 ) , p < 0 , p ≠ − 1 , p ∈ R {\displaystyle x={\frac {\left({\frac {{({\sqrt {p+4}}-{\sqrt {p}})}^{2}}{4p^{\frac {5}{8}}}}\right)^{\frac {1}{5}}}{\cos({\frac {2\pi }{5}})+isin({\frac {2\pi }{5}})}},p<0,p\neq -1,p\in \mathbb {R} }
註明： 以上僅爲特殊五次方程之解，並非壹般五次方程之解！
推廣
代數方程 x n + ( p + 4 − p 2 ) n − 2 n x = 1 p n 2 n − 2 {\displaystyle x^{n}+{{({\frac {{\sqrt {p+4}}-{\sqrt {p}}}{2}})}^{\frac {n-2}{n}}}x={\frac {1}{p^{\frac {n}{2n-2}}}}} 存在李煌根式解形式 x = ( ( p + 4 − p ) 2 4 p n 2 n − 2 ) 1 n , p > 0 , p ∈ R ∨ p = − 1 {\displaystyle x=\left({\frac {{({\sqrt {p+4}}-{\sqrt {p}})}^{2}}{4p^{\frac {n}{2n-2}}}}\right)^{\frac {1}{n}},p>0,p\in \mathbb {R} \lor p=-1}
x = ( ( p + 4 − p ) 2 4 p n 2 n − 2 ) 1 n cos ( 2 π n ) + i s i n ( 2 π n ) , p < 0 , p ≠ − 1 , p ∈ R {\displaystyle x={\frac {\left({\frac {{({\sqrt {p+4}}-{\sqrt {p}})}^{2}}{4p^{\frac {n}{2n-2}}}}\right)^{\frac {1}{n}}}{\cos({\frac {2\pi }{n}})+isin({\frac {2\pi }{n}})}},p<0,p\neq -1,p\in \mathbb {R} }
代數方程 x 7 = 1 {\displaystyle x^{7}=1} 存在李煌根式解
x 7 = 1 {\displaystyle x_{7}=1}
x 1 = 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 − 2184 + 540 3 i + 168 ( − 2548 + 588 3 i ) 1 3 − 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 − 2184 + 540 3 i + 168 ( − 2548 + 588 3 i ) 1 3 {\displaystyle x_{1}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}-2184+540{\sqrt {3}}i+168(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}-2184+540{\sqrt {3}}i+168(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}}}}}}
x 2 = 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 − 2184 + 540 3 i + 168 ( − 2548 + 588 3 i ) 1 3 − 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 − 2184 + 540 3 i + 168 ( − 2548 + 588 3 i ) 1 3 {\displaystyle x_{2}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}-2184+540{\sqrt {3}}i+168(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}-2184+540{\sqrt {3}}i+168(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}}}}}}
x 3 = 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 + 336 − 1344 3 i − 84 ( − 2548 + 588 3 i ) 1 3 − 84 3 ( − 2548 + 588 3 i ) 1 3 i − 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 + 336 − 1344 3 i − 84 ( − 2548 + 588 3 i ) 1 3 − 84 3 ( − 2548 + 588 3 i ) 1 3 i {\displaystyle x_{3}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+336-1344{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+336-1344{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}}}
x 4 = 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 + 336 − 1344 3 i − 84 ( − 2548 + 588 3 i ) 1 3 − 84 3 ( − 2548 + 588 3 i ) 1 3 i − 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 + 336 − 1344 3 i − 84 ( − 2548 + 588 3 i ) 1 3 − 84 3 ( − 2548 + 588 3 i ) 1 3 i {\displaystyle x_{4}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+336-1344{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+336-1344{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}}}
x 5 = 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 + 1848 + 840 3 i − 84 ( − 2548 + 588 3 i ) 1 3 + 84 3 ( − 2548 + 588 3 i ) 1 3 i − 3 ( − 2548 + 588 3 i ) 1 3 + − 15 ( − 2548 + 588 3 i ) 2 3 + 1848 + 840 3 i − 84 ( − 2548 + 588 3 i ) 1 3 + 84 3 ( − 2548 + 588 3 i ) 1 3 i {\displaystyle x_{5}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+1848+840{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+1848+840{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}}}
x 6 = 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 + 1848 + 840 3 i − 84 ( − 2548 + 588 3 i ) 1 3 + 84 3 ( − 2548 + 588 3 i ) 1 3 i − 3 ( − 2548 + 588 3 i ) 1 3 − − 15 ( − 2548 + 588 3 i ) 2 3 + 1848 + 840 3 i − 84 ( − 2548 + 588 3 i ) 1 3 + 84 3 ( − 2548 + 588 3 i ) 1 3 i {\displaystyle x_{6}={\frac {3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+1848+840{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}{-3(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}-{\sqrt {-15(-2548+588{\sqrt {3}}i)^{\frac {2}{3}}+1848+840{\sqrt {3}}i-84(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}+84{\sqrt {3}}(-2548+588{\sqrt {3}}i)^{\frac {1}{3}}i}}}}}
[1]
代數方程 x n + ( p + 4 − p 2 ) n − 2 n x = 1 p n 2 n − 2 {\displaystyle x^{n}+{{({\frac {{\sqrt {p+4}}-{\sqrt {p}}}{2}})}^{\frac {n-2}{n}}}x={\frac {1}{p^{\frac {n}{2n-2}}}}} 存在李煌根式解形式 x = ( ( p + 4 − p ) 2 4 p n 2 n − 2 ) 1 n , p > 0 , p ∈ R ∨ p = − 1 {\displaystyle x=\left({\frac {{({\sqrt {p+4}}-{\sqrt {p}})}^{2}}{4p^{\frac {n}{2n-2}}}}\right)^{\frac {1}{n}},p>0,p\in \mathbb {R} \lor p=-1} x = ( ( p + 4 − p ) 2 4 p n 2 n − 2 ) 1 n cos ( 2 π n ) + i s i n ( 2 π n ) , p < 0 , p ≠ − 1 , p ∈ R {\displaystyle x={\frac {\left({\frac {{({\sqrt {p+4}}-{\sqrt {p}})}^{2}}{4p^{\frac {n}{2n-2}}}}\right)^{\frac {1}{n}}}{\cos({\frac {2\pi }{n}})+isin({\frac {2\pi }{n}})}},p<0,p\neq -1,p\in \mathbb {R} }
<<School:李煌數學研究院
存在李煌根式解形式
存在根式解
存在根式解
存在李煌根式解形式
存在李煌根式解
存在李煌根式解形式
