School:李煌數學研究院/超越方程之研究

一

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

• 例如 n=$\ln 7,p=1$時，超越方程$x^{\ln 7}+{(\frac{\sqrt{5}-1}{2})}^{\frac{\ln 7-2}{\ln 7}}x=1$存在李煌解形式$x={\left(\frac{{(\sqrt{5}-1)}^2}{4}\right)}^{\frac{1}{\ln 7}}$
• 例如 n=$\cos 7,p=12$時，超越方程$x^{\cos 7}+{(2-\sqrt{3})}^{\frac{\cos 7-2}{\cos 7}}x=\frac{1}{{12}^{\frac{\cos 7}{2\cos 7-2}}}$存在李煌解形式$x={\left(\frac{{(4-2\sqrt{3})}^2}{4\times 12^{(\frac{\cos 7}{2\cos 7-2})}}\right)}^{\frac{1}{\cos 7}}$
• 例如 $n=13i,p=5$時，超越方程$x^{13i}+{(\frac{3-\sqrt{5}}{2})}^{\frac{13i-2}{13i}}x=\frac{1}{5^{\frac{13i}{26i-2}}}$存在李煌解形式$x={\left(\frac{{(3-\sqrt{5})}^2}{4\times 5^{(\frac{13i}{26i-2})}}\right)}^{\frac{1}{13i}}$

二

如果$c=b+2,b\in Z,2\nmid b$ 成立

则必须满足整除关系

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

三

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

则必须满足整除关系

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

四

${\left(a+b\right)}^n=ab{(a+b)}^{n-2}+a{(a+b)}^{n-1}+b^2{(a+b)}^{n-2}$

<<School:李煌數學研究院