院系:李煌数学研究院/伽罗华群论之n次方程研究

一

代数方程$x^5+{(\frac{\sqrt{p+4}-\sqrt{p}}{2})}^{\frac{3}{5}}x=\frac{1}{p^{\frac{5}{8}}}$存在李煌根式解形式 $x={\left(\frac{{(\sqrt{p+4}-\sqrt{p})}^2}{4p^{\frac{5}{8}}}\right)}^{\frac{1}{5}},p>0,p\in ℝ\vee p=-1$

$x=\frac{{\left(\frac{{(\sqrt{p+4}-\sqrt{p})}^2}{4p^{\frac{5}{8}}}\right)}^{\frac{1}{5}}}{\cos (\frac{2\pi }{5})+isin(\frac{2\pi }{5})},p<0,p\ne -1,p\in ℝ$

• 例如 p=-12时，代数方程$x^5+{(\frac{\sqrt{-8}-\sqrt{-12}}{2})}^{\frac{3}{5}}x=\frac{1}{{(-12)}^{\frac{5}{8}}}$存在根式解$x=\frac{{\left(\frac{{(\sqrt{-8}-\sqrt{-12})}^2}{4{(-12)}^{\frac{5}{8}}}\right)}^{\frac{1}{5}}}{\cos (\frac{2\pi }{5})+isin(\frac{2\pi }{5})}$
• 例如 p=5时，代数方程$x^5+{(\frac{3-\sqrt{5}}{2})}^{\frac{3}{5}}x=\frac{1}{5^{\frac{5}{8}}}$存在根式解$x={\left(\frac{{(3-\sqrt{5})}^2}{4\times 5^{\frac{5}{8}}}\right)}^{\frac{1}{5}}$

注明： 以上仅为特殊五次方程之解，并非壹般五次方程之解！

推广

代数方程$x^n+{(\frac{\sqrt{p+4}-\sqrt{p}}{2})}^{\frac{n-2}{n}}x=\frac{1}{p^{\frac{n}{2n-2}}}$存在李煌根式解形式 $x={\left(\frac{{(\sqrt{p+4}-\sqrt{p})}^2}{4p^{\frac{n}{2n-2}}}\right)}^{\frac{1}{n}},p>0,p\in ℝ\vee p=-1$

$x=\frac{{\left(\frac{{(\sqrt{p+4}-\sqrt{p})}^2}{4p^{\frac{n}{2n-2}}}\right)}^{\frac{1}{n}}}{\cos (\frac{2\pi }{n})+isin(\frac{2\pi }{n})},p<0,p\ne -1,p\in ℝ$

二

代数方程$x^7=1$存在李煌根式解

$x_7=1$

$x_1=\frac{3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}+\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}-2184+540\sqrt{3}i+168(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}}}{-3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}+\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}-2184+540\sqrt{3}i+168(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}}}$

$x_2=\frac{3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}-\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}-2184+540\sqrt{3}i+168(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}}}{-3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}-\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}-2184+540\sqrt{3}i+168(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}}}$

$x_3=\frac{3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}+\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}+336-1344\sqrt{3}i-84(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}-84\sqrt{3}(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}i}}{-3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}+\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}+336-1344\sqrt{3}i-84(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}-84\sqrt{3}(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}i}}$

$x_4=\frac{3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}-\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}+336-1344\sqrt{3}i-84(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}-84\sqrt{3}(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}i}}{-3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}-\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}+336-1344\sqrt{3}i-84(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}-84\sqrt{3}(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}i}}$

$x_5=\frac{3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}+\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}+1848+840\sqrt{3}i-84(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}+84\sqrt{3}(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}i}}{-3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}+\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}+1848+840\sqrt{3}i-84(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}+84\sqrt{3}(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}i}}$

$x_6=\frac{3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}-\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}+1848+840\sqrt{3}i-84(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}+84\sqrt{3}(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}i}}{-3(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}-\sqrt{-15(-2548+588\sqrt{3}i{)}^{\frac{2}{3}}+1848+840\sqrt{3}i-84(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}+84\sqrt{3}(-2548+588\sqrt{3}i{)}^{\frac{1}{3}}i}}$

研究难度对比

[1]

三

代数方程$x^n+{(\frac{\sqrt{p+4}-\sqrt{p}}{2})}^{\frac{n-2}{n}}x=\frac{1}{p^{\frac{n}{2n-2}}}$存在李煌根式解形式 $x={\left(\frac{{(\sqrt{p+4}-\sqrt{p})}^2}{4p^{\frac{n}{2n-2}}}\right)}^{\frac{1}{n}},p>0,p\in ℝ\vee p=-1$ $x=\frac{{\left(\frac{{(\sqrt{p+4}-\sqrt{p})}^2}{4p^{\frac{n}{2n-2}}}\right)}^{\frac{1}{n}}}{\cos (\frac{2\pi }{n})+isin(\frac{2\pi }{n})},p<0,p\ne -1,p\in ℝ$

<<School:李煌数学研究院