递降法 证明$x^3+y^3+z^3=0,(x,y,z\inZ)$则$xyz=0$

hbghlyj

第75页Proof of Theorem 13.0.1
递降的前一步：

Since $(x+y)+\omega(x+\omega y)+\omega^2\left(x+\omega^2 y\right)=0$, we have (after a little rearrangement)
\[\tag{13.6}
\left(x^{\prime}\right)^3+\mu\left(y^{\prime}\right)^3=u^{\prime} \lambda^{3(n-1)}\left(z^{\prime}\right)^3,
\]
where $x^{\prime}=z_2, y^{\prime}=z_3, \mu=\omega u_3 / u_2, z^{\prime}=z_1$ and $u^{\prime}=-u_1 / \omega u_2$.

(13.6)是怎么得出的？

hbghlyj

\begin{gather*}
x+y=\lambda^{3 n-2} u_1 z_1^3 \\
x+\omega y=\lambda u_2 z_2^3 \\
x+\omega^2 y=\lambda u_3 z_3^3\\
\omega(x+\omega y)+\omega^2\left(x+\omega^2 y\right)=-(x+y)
\end{gather*}
将前3式代入第4式，
\[\omega\lambda u_2 z_2^3+\omega^2\lambda u_3 z_3^3=-\lambda^{3 n-2} u_1 z_1^3
\]
除以$\omega\lambda u_2$，
\[
\left(z_2\right)^3+\frac{\omega u_3}{u_2}\left(z_3\right)^3=-\frac{u_1}{\omega u_2}\lambda^{3(n-1)}\left(z_1\right)^3,
\]
由$\mu=\frac{\omega u_3}{u_2}$
\[
\left(z_2\right)^3+\mu\left(z_3\right)^3=-\frac{u_1}{\omega u_2}\lambda^{3(n-1)}\left(z_1\right)^3,
\]
终于得到了

hbghlyj

如果是5次的，还能用这个方法吗？
$x^5+y^5+z^5=0,(x,y,z\inZ)$