查看“︁School:李煌數學研究院/超越方程之研究”︁的源代码

== 一 ==
n次方程<math>x^{n}+{{(\frac{\sqrt{p+4}-\sqrt{p}}{2})}^\frac{n-2}{n}}x=\frac{1}{p^{\frac{n}{2n-2}}},n\ne 1,n\ne 0 </math>存在李煌解形式
<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}</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\in\mathbb{R}</math>

* 例如 n=<math>\ln7,p=1</math>時，超越方程<math>x^{\ln7}+{{(\frac{\sqrt{5}-1}{2})}^\frac{\ln7-2}{\ln7}}x=1</math>存在李煌解形式<math>x=\left(\frac{{(\sqrt{5}-1)}^{2}}{4} \right)^{\frac{1}{\ln7}}</math>
* 例如 n=<math>\cos7,p=12</math>時，超越方程<math>x^{\cos7}+{{({2-\sqrt{3}})}^\frac{\cos7-2}{\cos7}}x=\frac{1}{{12}^{\frac{\cos7}{2\cos7-2}}}</math>存在李煌解形式<math>x=\left(\frac{{({4}-2\sqrt{3})}^{2}}{4\times{12^{(\frac{\cos7}{2\cos7-2})}}} \right)^{\frac{1}{\cos7}}</math>
* 例如 <math>n=13i,p=5</math>時，超越方程<math>x^{13i}+{{(\frac{{3}-\sqrt{5}}{2})}^\frac{13i-2}{13i}}x=\frac{1}{5^{\frac{13i}{26i-2}}} </math>存在李煌解形式<math>x=\left(\frac{{({3}-\sqrt{5})}^{2}}{4\times{5^{(\frac{13i}{26i-2})}}} \right)^{\frac{1}{13i}}</math>

==二==

如果<math>c=b+2,b\in Z, 2\nmid b
</math> 成立

则必须满足整除关系

<math>16 \mid  \bigg(-b^{4}c^{4}+124b^{4}c^{2}-2212b^{4}-4b^{3}c^{5}-22b^{3}c^{4}
</math>
<math>
+724b^{3}c^{2}+2b^{2}c^{6}- 62b^{2}c^{5}-149b^{2}c^{4}
</math>
<math>
+940b^{2}c^{3}+2030b^{2}c^{2}-12708b^{2}
</math>
<math>
+12bc^{7}+22bc^{6}-214bc^{5}+1548bc^{3}
</math>
<math>
+2754bc^{2}-9c^{8}+30c^{7}+63c^{6}-314c^{4}+942c^3\bigg)
</math>

==三==

如果<math>a^n+b^n=c^n,2\nmid b,2\nmid c,n \in Z,n>1, a\in Z, b\in Z, c\in Z,
</math> 成立

则必须满足整除关系

<math>16 \mid  \bigg(6{c^{3n}}+2{c^{4n}}-4{c^{5n}}+27{c^{6n}}+6{c^{7n}}-9{c^{8n}}-4{c^{2n}}
   </math>
<math> {+22{b^{2n}}{c^{2n}}}+{28{b^{2n}}{c^{3n}}}+{2{b^{2n}}{c^{6n}}}+{12{b^{2n}}}{-41{b^{2n}}{c^{4n}}}{-38{b^{2n}}{c^{5n}}}
</math>
<math> {-4{b^n}{c^{3n}}}{-12{b^n}{c^{4n}}}{-14{b^n}{c^{5n}}}+{14{b^n}{c^{6n}}}+{12{b^n}{c^{7n}}}{-6{b^n}{c^{2n}}}
</math>
<math> {-52{b^{4n}}}+{44{b^{4n}}{c^{2n}}}+{24{b^{4n}}{c^{3n}}}{-{b^{4n}}{c^{4n}}}+{52{b^{3n}}{c^{2n}}}{-14{b^{3n}}{c^{4n}}}{-4{b^{3n}}{c^{5n}}}\bigg)
</math>

==四==

<math>{\bigg({a+b}\bigg)}^{n}={ab{(a+b)}^{{n-2}}}+{a{(a+b)}^{{n-1}}}+{{b^2{(a+b)}^{{n-2}}}}
</math>

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