查看“︁School:李煌數學研究院/伽羅華群論之n次方程研究”︁的源代码

== 一 ==
代數方程<math>x^{5}+{{(\frac{\sqrt{p+4}-\sqrt{p}}{2})}^\frac{3}{5}}x=\frac{1}{p^{\frac{5}{8}}} </math>存在李煌根式解形式
<math>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</math>

<math>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\ne -1,p\in\mathbb{R}</math>

* 例如 p=-12時，代數方程<math>x^{5}+{{(\frac{\sqrt{-8}-\sqrt{-12}}{2})}^\frac{3}{5}}x=\frac{1}{{(-12)}^{\frac{5}{8}}} </math>存在根式解<math>x=\frac{\left(\frac{{(\sqrt{-8}-\sqrt{-12})}^{2}}{4{(-12)}^{\frac{5}{8}}}\right)^{\frac{1}{5}}}{\cos(\frac{2\pi}{5})+isin(\frac{2\pi}{5})}</math>
* 例如 p=5時，代數方程<math>x^{5}+{{(\frac{{3}-\sqrt{5}}{2})}^\frac{3}{5}}x=\frac{1}{5^{\frac{5}{8}}} </math>存在根式解<math>x=\left(\frac{{(3-\sqrt{5})}^{2}}{4\times 5^{\frac{5}{8}}}\right)^{\frac{1}{5}}</math>

'''注明：'''
以上仅爲特殊五次方程之解，並非壹般五次方程之解！

'''推广'''

代數方程<math>x^{n}+{{(\frac{\sqrt{p+4}-\sqrt{p}}{2})}^\frac{n-2}{n}}x=\frac{1}{p^{\frac{n}{2n-2}}} </math>存在李煌根式解形式
<math>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</math>

<math>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\ne -1,p\in\mathbb{R}</math>

== 二 ==
代數方程<math>x^{7}=1</math>存在李煌根式解

<math>x_7=1</math>

<math>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}}  }             }</math>

<math>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}}  }             }</math>

<math>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  } }</math>


<math>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  } }</math>

<math>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  } }</math>

<math>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  } }</math>

=== 研究难度对比 ===
[http://zh.wikipedia.org/wiki/%E4%B8%83%E9%82%8A%E5%BD%A2]

== 三 ==

代數方程<math>x^{n}+{{(\frac{\sqrt{p+4}-\sqrt{p}}{2})}^\frac{n-2}{n}}x=\frac{1}{p^{\frac{n}{2n-2}}} </math>存在李煌根式解形式
<math>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</math>
<math>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\ne -1,p\in\mathbb{R}</math>

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