齐次三次丢番图方程的特解变换

青青子衿

\begin{align*}
&\qquad\qquad\qquad\quad\,a^3+b^3+c^3=d^3\\
&\qquad\qquad\qquad\quad\alpha^3+\beta^3+\gamma^3=\delta^3\\
\\
&\left\{\begin{split}
\alpha&=a(a+c)u^2-(b-d) (b+d)uv+c(b-d)v^2\\
\beta&=b(a+c)u^2+(a-c) (a+c)uv-d(b-d)v^2\\
\gamma&=c(a+c)u^2+(b-d) (b+d)uv+a(b-d)v^2\\
\delta&=d(a+c)u^2+(a-c) (a+c)uv-b(b-d)v^2
\end{split}\right.\\
\\
&\left\{\begin{split}
a&=-\frac{\alpha(\alpha +\gamma )u^2+(\beta -\delta ) (\beta +\delta )u v+\gamma(\beta -\delta )v^2}{\sqrt{\left[(\alpha +\gamma )u^2+(\beta -\delta )v^2\right]^3}}\\
b&=-\frac{\beta(\alpha +\gamma )u^2-(\alpha -\gamma ) (\alpha +\gamma )u v-\delta(\beta-\delta)v^2}{\sqrt{\left[(\alpha +\gamma )u^2+(\beta-\delta)v^2\right]^3}}\\
c&=-\frac{\gamma(\alpha +\gamma )u^2-(\beta -\delta ) (\beta +\delta )u v+\alpha(\beta -\delta )v^2}{\sqrt{\left[(\alpha +\gamma )u^2+(\beta -\delta )v^2\right]^3}}\\
d&=-\frac{\delta(\alpha +\gamma )u^2-(\alpha -\gamma ) (\alpha +\gamma )u v-\beta   (\beta-\delta)v^2}{\sqrt{\left[(\alpha +\gamma )u^2+(\beta -\delta )v^2\right]^3}}
\end{split}\right.
\end{align*}

青青子衿

\begin{gather*}
a^3+b^3+c^3=d^3\\
\alpha^3+\beta^3+\gamma^3=\delta^3\\
A=a\alpha^2+b\beta^2+c\gamma^2-d\delta^2\\
B=\alpha\,\!a^2+\beta\,\!b^2+\gamma\,\!c^2-\delta\,\!d^2\\
\\
(aA-\alpha\,\!B)^3 + (bA-\beta\,\!B)^3 + (cA-\gamma\,\!B)^3= (dA-\delta\,\!B)^3\\
\end{gather*}

1. {(a*A - \[Alpha]*B), (b*A - \[Beta]*B), (c*A - \[Gamma]*B), (d*
2.        A - \[Delta]*
3.        B), (a*A - \[Alpha]*B)^3 + (b*A - \[Beta]*B)^3 + (c*
4.          A - \[Gamma]*B)^3 - (d*A - \[Delta]*B)^3} /. {A ->
5.      a*\[Alpha]^2 + b*\[Beta]^2 + c*\[Gamma]^2 - d*\[Delta]^2,
6.     B -> \[Alpha]*a^2 + \[Beta]*b^2 + \[Gamma]*c^2 - \[Delta]*
7.        d^2} /. {a -> 3, b -> 4, c -> 5,
8.    d -> 6, \[Alpha] -> 1, \[Beta] -> 8, \[Gamma] -> 6, \[Delta] ->
9.     9} // Factor

复制代码

\begin{align*}
&\qquad\qquad\>\>\left\{
\begin{split}
v_{k+1}&=\operatorname{Prj}_{W}\!\Big[v_k-\alpha_k\nabla_w\Phi(u_k,w)_{w=u_k}\Big],\\
u_k&=\operatorname{Prj}_{W}\!\Big[v_k-\alpha_k\nabla_w\Phi(v_k,w)_{w=v_k}\Big],\\
v_0&\in\,W,\qquad\,k=0,1,2,\cdots.
\end{split}
\right.\\
\\
&\alpha_{k+1}=\min\left\{\alpha_k,\sqrt{\dfrac{2}{3}}\dfrac{\Vert\,u_k-v_k\Vert}{\Vert\nabla_w\Phi(u_k,w)_{w=u_k}-\nabla_w\Phi(v_k,w)_{w=v_k}\Vert}\right\}
\end{align*}

青青子衿

\begin{align*}
&\quad\>\>(54 a^2 b - 108 a b^2 + 70 b^3)^3 \\
&\qquad\qquad+ (27 a^3 - 81 a^2 b + 117 ab^2 - 35 b^3)^3 \\
&\qquad\qquad\qquad+ (81 a^2 b - 27 a^3 - 117 a b^2 + 91 b^3)^3 \\
&= 6 (36 a^2 b - 72 ab^2 + 56 b^3)^3
\end{align*}

\begin{align*}
&\quad\>\>\left(\dfrac{54 a^2 b - 108 a b^2 + 70 b^3}{36 a^2 b - 72 ab^2 + 56 b^3}\right)^3 \\
&\qquad\quad\>\>\>+\left(\dfrac{27 a^3 - 81 a^2 b + 117 ab^2 - 35 b^3}{36 a^2 b - 72 ab^2 + 56 b^3}\right)^3 \\
&\qquad\qquad\qquad+ \left(\dfrac{81 a^2 b - 27 a^3 - 117 a b^2 + 91 b^3}{36 a^2 b - 72 ab^2 + 56 b^3}\right)^3 \\
&= 6
\end{align*}

青青子衿

青青子衿 发表于 2023-3-15 10:38
\begin{gather*}
a^3+b^3+c^3=d^3\\
\alpha^3+\beta^3+\gamma^3=\delta^3\\
A=a\alpha^2+b\beta^2+c\gamma^2-d\delta^2\\
B=\alpha\,\!a^2+\beta\,\!b^2+\gamma\,\!c^2-\delta\,\!d^2\\
\\
(aA-\alpha\,\!B)^3 + (bA-\beta\,\!B)^3 + (cA-\gamma\,\!B)^3= (dA-\delta\,\!B)^3\\
\end{gather*}

\begin{align*}
&\qquad\,\,a^3+b^3+c^3=kd^3\\
&\qquad\alpha^3+\beta^3+\gamma^3=k\delta^3\\
\\
&\left(a-\frac{a^2\alpha +b^2 \beta +c^2 \gamma -k d^2 \delta}{a \alpha ^2+b \beta ^2+c \gamma ^2-k d \delta ^2}\alpha\right)^3\\

&\quad+\left(b-\frac{a^2\alpha +b^2 \beta +c^2 \gamma -k d^2 \delta}{a \alpha ^2+b \beta ^2+c \gamma ^2-k d \delta ^2}\beta\right)^3\\
&\qquad\>\>\>+\left(c-\frac{a^2\alpha +b^2 \beta +c^2 \gamma -k d^2 \delta}{a \alpha ^2+b \beta ^2+c \gamma ^2-k d \delta ^2}\gamma\right)^3\\

&=k \left(d-\frac{a^2\alpha +b^2 \beta +c^2 \gamma -k d^2 \delta}{a \alpha ^2+b \beta ^2+c \gamma ^2-k d \delta ^2}\delta\right)^3
\end{align*}