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对称无理方程化为对称多项式方程

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青青子衿 发表于 2019-2-25 20:25 |阅读模式
本帖最后由 青青子衿 于 2022-12-3 23:11 编辑 \begin{align*}
\color{red}{
\begin{align*}
&&p&=\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\\  
&\Rightarrow&\left[\left(p^2-\sigma_1\right)^2-4\sigma_2\right]^2&=64p^2\sigma_3\\
\end{align*}}
\end{align*}
\[\left[\left(p^2-x_1-x_2-x_3\right)^2-4\left(x_1x_2+x_2x_3+x_3x_1\right)\right]^2=64p^2x_1x_2x_3\]
更多元的情况怎么处理呢?
这个式子有什么意义呢?请看下面这个问题:
已知\(\,x^3 - 5 x^2 + 6 x - 1=0\,\)的三个正实根
\begin{align*}
x_1&=\frac{5}{3}+\frac{2\sqrt{7}}{3}\cos\left(\frac{1}{3}\arctan\left(3\sqrt{3}\right)\right)\\
x_2&=\frac{5}{3}-\frac{\sqrt{7}}{3}\cos\left(\frac{1}{3}\arctan\left(3\sqrt{3}\right)\right)+\frac{\sqrt{21}}{3}\sin\left(\frac{1}{3}\arctan\left(3\sqrt{3}\right)\right)\\
x_3&=\frac{5}{3}-\frac{\sqrt{7}}{3}\cos\left(\frac{1}{3}\arctan\left(3\sqrt{3}\right)\right)-\frac{\sqrt{21}}{3}\sin\left(\frac{1}{3}\arctan\left(3\sqrt{3}\right)\right)
\end{align*}
求\(\,p=\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\,\)的极小多项式.
  1. Solve[x^3 - 5 x^2 + 6 x - 1 == 0, x] // ComplexExpand
复制代码
kuing.cjhb.site/forum.php?mod=viewthread&tid=3754

\begin{align*}
S&=\begin{vmatrix}
1 & a_1 & a_2 & a_3 \\
\sqrt{x_1}& a_0 & a_3 x_1 & a_2 \\
\sqrt{x_2}& a_3 x_2 & a_0 & a_1 \\
\sqrt{x_1x_2}& a_2 x_2 & a_1 x_1 & a_0 \\
\end{vmatrix}\\
\\
T&=\begin{vmatrix}
a_0 & a_1 & a_2 & a_3 \\
a_1 x_1 & a_0 & a_3 x_1 & a_2 \\
a_2 x_2 & a_3 x_2 & a_0 & a_1 \\
a_3 x_1x_2 & a_2 x_2 & a_1 x_1 & a_0 \\
\end{vmatrix}\\
\\
\\
\dfrac{S}{T}&=\dfrac{1}{a_{0}+a_{1}\sqrt{x_1}+a_{2}\sqrt{x_2}+a_{3}\sqrt{x_{1}x_{2}}}
\end{align*}

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kuing 发表于 2019-2-25 23:03
表示 `\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}` 以及 `\sqrt{x_1x_2}+\sqrt{x_2x_3}+\sqrt{x_3x_1}` 我以前在 kuing.cjhb.site/forum.php?mod=viewthread&tid=3066 (1# 尾段 \section{一些可能没意义的等式} 部分)里用过一次。
表示 `\sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3}` 我以前在 kuing.cjhb.site/forum.php?mod=redirect&goto=findpost&ptid=5138&pid=25073 (4#)里用过一次。

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kuing 发表于 2019-2-26 02:26
来试一下四元二次根式。

照旧用 2# 链接中的方法,记
\begin{align*}
a_1 &= x^2 + y^2 + z^2 + w^2,\\
a_2 &= x^2 y^2 + y^2 z^2 + z^2 w^2 + w^2 x^2 + y^2 w^2 + x^2 z^2,\\
a_3 &= x^2 y^2 z^2 + y^2 z^2 w^2 + z^2 w^2 x^2 + w^2 x^2 y^2,\\
a_4 &= x y z w,\\
p &= x + y + z + w,\\
q &= x y + y z + z w + w x + y w + x z,\\
r &= x y z + y z w + z w x + w x y,
\end{align*}
则有
\[\led
p^2 - 2 q - a_1 &= 0,\\
q^2 - a_2 + 2 a_4 - 2 p r &= 0,\\
r^2 - a_3 - 2 a_4 q &= 0,
\endled\]
这样就可以消元了,最终得
\begin{align*}
(a_1^2 - 4 a_2 + 8 a_4)^2 - 4 (a_1^3 - 4 a_1 a_2 + 16 a_3 - 8 a_1 a_4) p^2 +
2 (3 a_1^2 - 4 a_2 - 24 a_4) p^4 - 4 a_1 p^6 + p^8 &= 0,\\
a_2^2 - 4 a_1 a_3 - 4 a_2 a_4 + 4 a_4^2 - 8 (a_3 + a_1 a_4) q -
2 (a_2 + 6 a_4) q^2 + q^4 &= 0,\\
(a_3^2 - 4 a_2 a_4^2 + 8 a_4^3)^2 -
4 (a_3^3 - 4 a_2 a_3 a_4^2 - 8 a_3 a_4^3 + 16 a_1 a_4^4) r^2 +
2 (3 a_3^2 - 4 a_2 a_4^2 - 24 a_4^3) r^4 - 4 a_3 r^6 + r^8 &= 0.
\end{align*}

更多元多次应该都总可以这样做,只是会越来越繁……

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 楼主| 青青子衿 发表于 2019-3-15 19:18
本帖最后由 青青子衿 于 2021-11-27 22:58 编辑
来试一下四元二次根式。
...
kuing 发表于 2019-2-26 02:26

\begin{gather*}
&\left(\left(\left(p^2-\sigma_1\right)^2-4\sigma_2\right)^2-64p^2\sigma_3+64\sigma_4\right)^2\\
{\large=}&256\sigma_4\left(3p^4-2p^2\sigma_1-{\sigma_1}^2+4\sigma_2\right)^2
\end{gather*}
\begin{cases}
\sigma_1=x+y+z+w\\
\sigma_2=xy+xz+xw+yz+yw+zw\\
\sigma_3=xyz+yzw+zwx+wxy\\
\sigma_4=xyzw\\
\end{cases}
\begin{align*}
p=\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{w}
\end{align*}
math.stackexchange.com/questions/3127229/

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 楼主| 青青子衿 发表于 2019-3-16 16:26
本帖最后由 青青子衿 于 2019-11-7 23:47 编辑
表示 $\sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3}$ 我以前在   ...
kuing 发表于 2019-2-25 23:03
第二个:

\begin{align*}
a&=x^3+y^3+z^3,\\
b&=x^3y^3+y^3z^3+z^3x^3,\\
c&=xyz,\\
t&=x+y+z,\\
u&=xy+yz+zx,
\end{align*}
则不难验证以下两个等式成立
\begin{align*}
t^3-a-3tu+3c&=0,\\
u^3-b-3ctu+3c^2&=0,
\end{align*}
消去 $u$,得到
\[t^9-3(a+6c)t^6+3(a^2-9b+3ac+9c^2)t^3-(a-3c)^3=0,\quad(*)\]
kuing 发表于 2018-1-25 01:39

{:time:}
\begin{cases}
\sigma_1=x+y+z\\  
\sigma_2=xy+xz+yz\\  
\sigma_3=xyz\\
\end{cases}
\begin{gather*}
p=\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\\
\Downarrow\\
27{\color{purple}{k^3}}+27\left(p^3-\sigma_1\right){\color{brown}{k^2}}-9\left(p^3-\sigma_1\right)\left(2p^3+\sigma_1\right){\color{brown}{k}}+\left(p^3-\sigma_1\right)^3-27p^3\sigma_2=0
\end{gather*}
\[ \sigma_3=k^3 \]

\begin{gather*}
k=\sqrt[3]{xyz}\\
\Downarrow\\
U\cdot\,\!{\color{brown}{k^2}}+V\cdot\,\!{\color{brown}{k}}+W=0
\end{gather*}

\begin{cases}
U=\phantom{+}27\left(p^3-\sigma_1\right)\\
V={\color{red}{-}}9\left(p^3-\sigma_1\right)\left(2p^3+\sigma_1\right)\\
W=\phantom{+}\left(p^3-\sigma_1\right)^3-27p^3\sigma_2+27\sigma_3
\end{cases}
\[ U^3{\sigma_3}^2+\left(V^3-3UVW\right)\sigma_3+W^3=0 \]

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 楼主| 青青子衿 发表于 2021-8-20 23:51
本帖最后由 青青子衿 于 2021-8-22 23:48 编辑 \begin{align*}
\color{black}{\prod\limits_{\substack{
\xi_1,\,\xi_2,\,\xi_3\,\\
\in\left\{ 1,\omega,\omega^2\right\}}
}\left(t-\xi _1\sqrt[3]{a}-\xi _2\sqrt[3]{b}-\xi _3\sqrt[3]{c} \right)=t^{27}+\sum\limits_{k=0}^{8}\alpha_{3k}t^{3k}}

\end{align*}

\begin{align*}
\color{black}{
\begin{aligned}
\alpha_0&= -\left[(a + b + c)^3-27 a b c\right]^3\\
\alpha_3&=9\left( a+b+c \right) ^8-81\left( ab+ac+bc \right) \left( a+b+c \right) ^6\\
&\qquad+3888abc\left( a+b+c \right) ^5-15309abc\left( ab+ac+bc \right) \left( a+b+c \right) ^3\\
&\qquad\qquad+65610a^2b^2c^2\left( a+b+c \right) ^2-59049a^2b^2c^2\left( ab+ac+bc \right)\\
\alpha_6&=-36\left( a+b+c \right) ^7+486\left( ab+ac+bc \right)\left( a+b+c \right) ^5\\
&\qquad-972abc\left( a+b+c \right) ^4-2187\left( ab+ac+bc \right) ^2\left( a+b+c \right) ^3\\
&\qquad\qquad-13122abc\left( ab+ac+bc \right) \left( a+b+c \right) ^2\\
&\qquad\qquad\qquad-124659a^2b^2c^2\left( a+b+c \right)\\
&\qquad\qquad\qquad\qquad+59049abc\left( ab+ac+bc \right) ^2\\
\alpha_9&=84\left( a+b+c \right) ^6-1215\left( ab+ac+bc \right) \left( a+b+c \right) ^4\\
&\qquad-16929abc\left( a+b+c \right) ^3+6561\left( ab+ac+bc \right) ^2\left( a+b+c \right) ^2\\
&\qquad\qquad+72171abc\left( ab+ac+bc \right) \left( a+b+c \right)\\
&\qquad\qquad\qquad+19683\left( ab+ac+bc \right) ^3+61236a^2b^2c^2\\
\alpha_{12}&=-126\left( a+b+c \right) ^5+1620\left( ab+ac+bc \right) \left( a+b+c \right) ^3\\
&\qquad+16524abc\left( a+b+c \right) ^2-6561\left( ab+ac+bc \right) ^2\left( a+b+c \right)\\
&\qquad\qquad-43740abc\left( ab+ac+bc \right)\\
\alpha_{15}&=126\left( a+b+c \right) ^4-1215\left( ab+ac+bc \right) \left( a+b+c \right) ^2\\
&\qquad+1701abc\left( a+b+c \right) +2187\left( ab+ac+bc \right) ^2\\
\alpha_{18}&=-84\left( a+b+c \right) ^3+486\left( ab+ac+bc \right) \left( a+b+c \right) -4293abc\\
\alpha_{21}&=36 (a+b+c)^2-81 (a b+a c+b c)\\
\alpha_{24}&=-9 (a+b+c)
\end{aligned}}
\end{align*}

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kuing 发表于 2021-8-24 17:06

\begin{align*}
p  &=a+b+c+d,\\
q  &=ab+ac+ad+bc+bd+cd,\\
r  &=abc+bcd+cda+dab,\\
s  &=abcd,\\
A&=a^3+b^3+c^3+d^3,\\
B&=a^3b^3+a^3c^3+a^3d^3+b^3c^3+b^3d^3+c^3d^3,\\
C&=a^3b^3c^3+b^3c^3d^3+c^3d^3a^3+d^3a^3b^3,
\end{align*}则
\begin{align*}
A&=p^3-3pq+3r,\\
B&=q^3-3pqr+3r^2+3p^2s-3qs,\\
C&=r^3-3qrs+3ps^2,
\end{align*}消去 `q`, `r` 得到
\[a_0+a_1p+a_2p^2+\cdots+a_{27}p^{27}=0,\quad(*)\]其中
\begin{align*}
a_0   &= (A^3 - 27 C)^3 + 2187 A^2 (2 A^3 - 9 A B + 27 C) s^3, \\
a_1   &= -243 s^2 (10 A^6-54 A^4 B+432 A^3 C-729 A B C+729 C^2+729 A^2 s^3), \\
a_2   &= 81 A s (4 A^6-27 A^4 B+513 A^3 C-1458 A B C+2916 C^2+2916 A^2 s^3), \\
\cdots&\cdots\text{(略去一大堆)} \\
a_{24}&= 9 A, \\
a_{25}&= 0, \\
a_{26}&= 0, \\
a_{27}&= -1,
\end{align*}设 `\omega` 为三次单位根,由 `\omega^3=1` 可知:
将 `p=a+b+c+d` 改成以下的式子,同样满足式 (*)(这些式子使 `A`, `B`, `C`, `s` 均不变)
`\omega a+\omega b+\omega c+d`、`\omega a+\omega b+c+\omega d`、……(三个 `\omega`,4 条);
`\omega a+\omega^2b+c+d`、`\omega^2a+\omega b+c+d`、……(一个 `\omega` 一个 `\omega^2`,12 条);
`\omega^2a+\omega^2b+\omega^2c+d`、`\omega^2a+\omega^2b+c+\omega^2d`、……(三个 `\omega^2`,4 条);
`\omega a+\omega b+\omega^2c+\omega^2d`、`\omega a+\omega^2b+\omega c+\omega^2d`、……(两个 `\omega` 两个 `\omega^2`,6 条);
以上总共 26 条,连同原本的 `p`,也就是说:
关于 `x` 的方程 `\sum_{k=0}^{27}a_kx^k=0` 的 27 个根就是以上所列,所以
\begin{align*}
a_0+a_1x+a_2x^2+\cdots+a_{27}x^{27}={}&(a+b+c+d-x)\\
&\times(\omega a+\omega b+\omega c+d-x)\cdots\\
&\times(\omega a+\omega^2b+c+d-x)\cdots\\
&\times(\omega^2a+\omega^2b+\omega^2c+d-x)\cdots\\
&\times(\omega a+\omega b+\omega^2c+\omega^2d-x)\cdots,
\end{align*}令 `x=0`,得
\begin{align*}
a_0={}&(a+b+c+d)\\
&\times(\omega a+\omega b+\omega c+d)\cdots\\
&\times(\omega a+\omega^2b+c+d)\cdots\\
&\times(\omega^2a+\omega^2b+\omega^2c+d)\cdots\\
&\times(\omega a+\omega b+\omega^2c+\omega^2d)\cdots,
\end{align*}变个形,把 `a_0` 代回去,即
\[
\frac1{a+b+c+d}=\frac{\begin{aligned}
&(\omega a+\omega b+\omega c+d)\cdots\\
&\times(\omega a+\omega^2b+c+d)\cdots\\
&\times(\omega^2a+\omega^2b+\omega^2c+d)\cdots\\
&\times(\omega a+\omega b+\omega^2c+\omega^2d)\cdots
\end{aligned}}{(A^3 - 27 C)^3 + 2187 A^2 (2 A^3 - 9 A B + 27 C) (abcd)^3},
\quad(**)
\]接下来继续化简上式的分子,由
\begin{align*}
(\omega a+\omega b+\omega c+d)(\omega^2a+\omega^2b+\omega^2c+d)
&=\omega^3(a+b+c)^2+(\omega+\omega^2)(a+b+c)d+d^2\\
&=(a+b+c)^2-(a+b+c)d+d^2,\\
(\omega a+\omega^2b+c+d)(\omega^2a+\omega b+c+d)
&=\omega^3(a^2+b^2)+(\omega^2+\omega^4)ab+(\omega+\omega^2)(a+b)(c+d)+(c+d)^2\\
&=a^2+b^2-ab-(a+b)(c+d)+(c+d)^2,\\
(\omega a+\omega b+\omega^2c+\omega^2d)(\omega^2a+\omega^2b+\omega c+\omega d)
&=\omega^3(a+b)^2+(\omega^2+\omega^4)(a+b)(c+d)+\omega^3(c+d)^2\\
&=(a+b)^2-(a+b)(c+d)+(c+d)^2,
\end{align*}可知式 (**) 的分子化为
\begin{align*}
&\bigl((a+b+c)^2-(a+b+c)d+d^2\bigr)\\
&\times\bigl((a+b+d)^2-(a+b+d)c+c^2\bigr)\\
&\times\bigl((a+c+d)^2-(a+c+d)b+b^2\bigr)\\
&\times\bigl((b+c+d)^2-(b+c+d)a+a^2\bigr)\\
&\times\bigl(a^2+b^2-ab-(a+b)(c+d)+(c+d)^2\bigr)\\
&\times\bigl(a^2+c^2-ac-(a+c)(b+d)+(b+d)^2\bigr)\\
&\times\bigl(a^2+d^2-ad-(a+d)(b+c)+(b+c)^2\bigr)\\
&\times\bigl(b^2+c^2-bc-(b+c)(a+d)+(a+d)^2\bigr)\\
&\times\bigl(b^2+d^2-bd-(b+d)(a+c)+(a+c)^2\bigr)\\
&\times\bigl(c^2+d^2-cd-(c+d)(a+b)+(a+b)^2\bigr)\\
&\times\bigl((a+b)^2-(a+b)(c+d)+(c+d)^2\bigr)\\
&\times\bigl((a+c)^2-(a+c)(b+d)+(b+d)^2\bigr)\\
&\times\bigl((a+d)^2-(a+d)(b+c)+(b+c)^2\bigr).
\end{align*}

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kuing 发表于 2021-8-24 23:21
回复 7# kuing

从便于计算的角度来说,前四个
\[\prod_{cyc}\bigl((a+b+c)^2-(a+b+c)d+d^2\bigr)=A^2p^2+3Ap(pr-3s)+9(pr-3s)^2,\]后三个
\begin{align*}
&\bigl((a+b)^2-(a+b)(c+d)+(c+d)^2\bigr)\\
&\times\bigl((a+c)^2-(a+c)(b+d)+(b+d)^2\bigr)\\
&\times\bigl((a+d)^2-(a+d)(b+c)+(b+c)^2\bigr)\\
={}&A^2+3Ar+9r^2-9p^2s,
\end{align*}然鹅中间六个没化出简单的式子……
A^4 - 27 A^2 B + 216 A C - 3 A^3 p q - 648 C p q + 9 A^2 p^2 q^2 -
3 A^3 r - 81 A B r + 810 C r - 81 A^2 p q r + 27 A p^2 q^2 r +
81 A^2 r^2 - 243 B r^2 - 243 A p q r^2 + 81 p^2 q^2 r^2 +
99 A^2 p^2 s - 243 B p^2 s - 162 A^2 q s + 729 B q s +
81 A p q^2 s + 216 A p^2 r s - 81 A q r s - 2187 p q^2 r s +
1377 p^2 r^2 s + 1701 q r^2 s + 1701 p^2 q s^2 + 729 q^2 s^2 -
4374 p r s^2 + 729 s^3

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hbghlyj 发表于 2025-1-17 16:07
本帖最后由 hbghlyj 于 2025-1-18 13:29 编辑

n = $1$: Let the root of $x+p=0$ be $x_1$.
The minimal polynomial of $\sqrt{x_1}$ is $$(x-\sqrt{-p})(x+\sqrt{-p})=x^2+p$$

n = $2$: Let the roots of $x^2+px+q=0$ be $x_1,x_2$.
The minimal polynomial of $\sqrt{x_1}+\sqrt{x_2}$ is $$\prod_{\mu_1,\mu_2\in{\pm1}}(x-(\mu_1\sqrt{x_1}+\mu_2\sqrt{x_2}))=(x^2+p)^2-4 q$$

n = $3$: Let the roots of $x^3+px^2+qx+r=0$ be $x_1,\dots,x_3$.
The minimal polynomial of $\sqrt{x_1}+\dots+\sqrt{x_3}$ is $$\prod_{\mu_1,\dots,\mu_3\in{\pm1}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_3\sqrt{x_3}))\=\left(\left(x^2+p\right)^2-4 q\right)^2+64rx^2$$

n = $4$: Let the roots of $x^4+px^3+qx^2+rx+s=0$ be $x_1,\dots,x_4$.
The minimal polynomial of $\sqrt{x_1}+\dots+\sqrt{x_4}$ is (from this post) $$\prod_{\mu_1,\dots,\mu_4\in{\pm1}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_4\sqrt{x_4}))\=(((x^2 + p)^2 - 4 q)^2 + 64r x^2 + 64 s)^2\- 256 s (3 x^4+2 p x^2-p^2+4 q)^2$$

n = $5$: Let the roots of $x^5+px^4+qx^3+rx^2+sx+t=0$ be $x_1,\dots,x_5$.
The minimal polynomial of $\sqrt{x_1}+\dots+\sqrt{x_5}$ is $$ \prod_{\mu_1,\dots,\mu_5\in{\pm1}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_5\sqrt{x_5}))=f_5(x)^2+t\ g_5(x)^2 $$ where $$ f_5(x)=((x^2+p)^2-4 q)^2+64 r x^2+64 s
-256s(3 x^4+2 p x^2-p^2+4 q)^2
-2048t(5 x^6+3 p x^4+7 p^2 x^2-20 q x^2+p^3-4 p q+8 r) $$ and $$ g_5(x)=128 x \Biggl(3 x^{10}+9 p x^8+\left(6 p^2+8 q\right) x^6+\left(-6 p^3+40 q p-96 r\right) x^4
+\left(-9 p^4+56 q p^2-80 q^2-192 s\right) x^2
-3 p^5-48 p q^2+24 p^3 q-32 p^2 r+128 q r-64 p s+256 t\Biggr)$$

n = $6$: Let the roots of $x^6+px^5+qx^4+rx^3+sx^2+tx+u=0$ be $x_1,\dots,x_6$.
The minimal polynomial of $\sqrt{x_1}+\dots+\sqrt{x_6}$ is $$ \prod_{\mu_1,\dots,\mu_6\in{\pm1}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_6\sqrt{x_6}))=f_6(x)^2-u\ g_6(x)^2 $$ The degree of $f_6(x)$ is $32$.
The degree of $f_6(x)-\left(f_5(x)^2+t\ g_5(x)^2\right)$ is $20$:

8192 u (209 x^20+954 p x^18+2049 p^2 x^16-792 q x^16+3928 p^3 x^14-8384 p q x^14+13824 r x^14+7650 p^4 x^12+32480 q^2 x^12-32480 p^2 q x^12+9600 p r x^12+51072 s x^12+10620 p^5 x^10+95040 p q^2 x^10-61632 p^3 q x^10+23808 p^2 r x^10-109056 q r x^10+106752 p s x^10-82944 t x^10+8714 p^6 x^8-13312 q^3 x^8+116896 p^2 q^2 x^8+30720 r^2 x^8-62032 p^4 q x^8+71296 p^3 r x^8-260608 p q r x^8-99712 p^2 s x^8+518144 q s x^8-201728 p t x^8+854016 u x^8+3864 p^7 x^6-45056 p q^3 x^6+78720 p^3 q^2 x^6+24576 p r^2 x^6-32320 p^5 q x^6+41984 p^4 r x^6+77824 q^2 r x^6-168960 p^2 q r x^6-68096 p^3 s x^6+270336 p q s x^6-49152 r s x^6-231424 p^2 t x^6+393216 q t x^6+466944 p u x^6+813 p^8 x^4+55552 q^4 x^4-48128 p^2 q^3 x^4+26656 p^4 q^2 x^4-24576 p^2 r^2 x^4+245760 q r^2 x^4+724992 s^2 x^4-7776 p^6 q x^4+640 p^5 r x^4-153600 p q^2 r x^4+35840 p^3 q r x^4-58240 p^4 s x^4-206848 q^2 s x^4+294912 p^2 q s x^4-417792 p r s x^4-2048 p^3 t x^4-32768 p q t x^4+49152 r t x^4-331776 p^2 u x^4+1081344 q u x^4+90 p^9 x^2+512 p q^4 x^2-6144 p^3 q^3 x^2+131072 r^3 x^2+4416 p^5 q^2 x^2+32768 p^3 r^2 x^2-131072 p q r^2 x^2+139264 p s^2 x^2-1088 p^7 q x^2+2304 p^6 r x^2+24576 q^3 r x^2+24576 p^2 q^2 r x^2-16896 p^4 q r x^2-14080 p^5 s x^2-102400 p q^2 s x^2+81920 p^3 q s x^2-65536 p^2 r s x^2+32768 q r s x^2+35840 p^4 t x^2+409600 q^2 t x^2-229376 p^2 q t x^2-98304 p r t x^2-327680 s t x^2+172032 p^3 u x^2-720896 p q u x^2+1572864 r u x^2+21 p^10-2048 q^5+7424 p^2 q^4-6144 p^4 q^3+2144 p^6 q^2+2048 p^4 r^2+32768 q^2 r^2-16384 p^2 q r^2+20480 p^2 s^2-32768 q s^2+131072 t^2-344 p^8 q+384 p^7 r-24576 p q^3 r+18432 p^3 q^2 r-4608 p^5 q r+384 p^6 s+16384 q^3 s-2048 p^2 q^2 s-2048 p^4 q s+8192 p^3 r s-32768 p q r s-9216 p^5 t-114688 p q^2 t+65536 p^3 q t-81920 p^2 r t+262144 q r t-65536 p s t+18432 p^4 u+131072 q^2 u-98304 p^2 q u)

The leading term of $g_6$ is $3840 x^{26}$:

256 (15 x^26+135 p x^24+498 p^2 x^22+24 q x^22+858 p^3 x^20+744 p q x^20-1440 r x^20+165 p^4 x^18-2544 q^2 x^18+5544 p^2 q x^18-10368 p r x^18+17088 s x^18-2475 p^5 x^16-21168 p q^2 x^16+20376 p^3 q x^16-33120 p^2 r x^16+31104 q r x^16+40128 p s x^16+157440 t x^16-5940 p^6 x^14+17664 q^3 x^14-76224 p^2 q^2 x^14-147456 r^2 x^14+44784 p^4 q x^14-62464 p^3 r x^14+159744 p q r x^14-25856 p^2 s x^14+160768 q s x^14+432128 p t x^14+69632 u x^14-7524 p^7 x^12+96000 p q^3 x^12-154560 p^3 q^2 x^12-331776 p r^2 x^12+63504 p^5 q x^12-78400 p^4 r x^12-168960 q^2 r x^12+345600 p^2 q r x^12-139520 p^3 s x^12+388096 p q s x^12+233472 r s x^12+803840 p^2 t x^12-1069056 q t x^12+339968 p u x^12-6039 p^8 x^10-40704 q^4 x^10+211200 p^2 q^3 x^10-191520 p^4 q^2 x^10-184320 p^2 r^2 x^10+147456 q r^2 x^10+1953792 s^2 x^10+59472 p^6 q x^10-69888 p^5 r x^10-528384 p q^2 r x^10+419840 p^3 q r x^10-101760 p^4 s x^10+1017856 q^2 s x^10+25600 p^2 q s x^10-835584 p r s x^10+1271808 p^3 t x^10-3334144 p q t x^10+1818624 r t x^10-3502080 p^2 u x^10+10747904 q u x^10-3135 p^9 x^8-114432 p q^4 x^8+234240 p^3 q^3 x^8+393216 r^3 x^8-145824 p^5 q^2 x^8-40960 p^3 r^2 x^8+98304 p q r^2 x^8+2117632 p s^2 x^8+36144 p^7 q x^8-45248 p^6 r x^8+225280 q^3 r x^8-648192 p^2 q^2 r x^8+328960 p^4 q r x^8+23168 p^5 s x^8+1779712 p q^2 s x^8-558080 p^3 q s x^8-307200 p^2 r s x^8-2375680 q r s x^8+961024 p^4 t x^8+991232 q^2 t x^8-3223552 p^2 q t x^8+1654784 p r t x^8+1998848 s t x^8-4706304 p^3 u x^8+14942208 p q u x^8-9306112 r u x^8-990 p^10 x^6+6144 q^5 x^6-87552 p^2 q^4 x^6+126720 p^4 q^3 x^6-63168 p^6 q^2 x^6-122880 p^4 r^2 x^6-393216 q^2 r^2 x^6+589824 p^2 q r^2 x^6+401408 p^2 s^2 x^6+1146880 q s^2 x^6-13107200 t^2 x^6+13176 p^8 q x^6-20480 p^7 r x^6+262144 p q^3 r x^6-458752 p^3 q^2 r x^6+180224 p^5 q r x^6+36608 p^6 s x^6+737280 q^3 s x^6+643072 p^2 q^2 s x^6+1572864 r^2 s x^6-353280 p^4 q s x^6+655360 p^3 r s x^6-3670016 p q r s x^6+108544 p^5 t x^6-1933312 p q^2 t x^6+245760 p^3 q t x^6-819200 p^2 r t x^6+4849664 q r t x^6+1179648 p s t x^6-2248704 p^4 u x^6+2949120 q^2 u x^6+8257536 p^2 q u x^6-22806528 p r u x^6+56360960 s u x^6-150 p^11 x^4-55296 p q^5 x^4+16896 p^3 q^4 x^4+17664 p^5 q^3 x^4-393216 p^2 r^3 x^4+1572864 q r^3 x^4-10944 p^7 q^2 x^4-86016 p^5 r^2 x^4-1376256 p q^2 r^2 x^4+688128 p^3 q r^2 x^4+172032 p^3 s^2 x^4-622592 p q s^2 x^4+3538944 r s^2 x^4-4718592 p t^2 x^4+2184 p^9 q x^4-5664 p^8 r x^4+188416 q^4 r x^4+221184 p^2 q^3 r x^4-236544 p^4 q^2 r x^4+65024 p^6 q r x^4-5376 p^7 s x^4+901120 p q^3 s x^4-536576 p^3 q^2 s x^4-786432 p r^2 s x^4+99328 p^5 q s x^4-184320 p^4 r s x^4-2686976 q^2 r s x^4+1409024 p^2 q r s x^4-113664 p^6 t x^4+3080192 q^3 t x^4-3751936 p^2 q^2 t x^4-4718592 r^2 t x^4+1200128 p^4 q t x^4-491520 p^3 r t x^4+3276800 p q r t x^4-589824 p^2 s t x^4+2359296 q s t x^4-1437696 p^5 u x^4-12386304 p q^2 u x^4+9043968 p^3 q u x^4-14155776 p^2 r u x^4+26738688 q r u x^4+11796480 p s u x^4+17825792 t u x^4+3 p^12 x^2+61440 q^6 x^2-79872 p^2 q^5 x^2+42240 p^4 q^4 x^2-11520 p^6 q^3 x^2+2359296 s^3 x^2+1680 p^8 q^2 x^2-4096 p^6 r^2 x^2+262144 q^3 r^2 x^2-196608 p^2 q^2 r^2 x^2+49152 p^4 q r^2 x^2-61440 p^4 s^2 x^2-196608 q^2 s^2 x^2+294912 p^2 q s^2 x^2-524288 p r s^2 x^2-3670016 p^2 t^2 x^2+10485760 q t^2 x^2+33554432 u^2 x^2-120 p^10 q x^2-640 p^9 r x^2-163840 p q^4 r x^2+163840 p^3 q^3 r x^2-61440 p^5 q^2 r x^2+10240 p^7 q r x^2-8000 p^8 s x^2-344064 q^4 s x^2+770048 p^2 q^3 s x^2-448512 p^4 q^2 s x^2-262144 p^2 r^2 s x^2+1048576 q r^2 s x^2+101376 p^6 q s x^2-81920 p^5 r s x^2-1310720 p q^2 r s x^2+655360 p^3 q r s x^2-10240 p^7 t x^2-131072 p q^3 t x^2-98304 p^3 q^2 t x^2+73728 p^5 q t x^2+49152 p^4 r t x^2+1835008 q^2 r t x^2-655360 p^2 q r t x^2+131072 p^3 s t x^2+524288 p q s t x^2-7340032 r s t x^2-86016 p^6 u x^2+4718592 q^3 u x^2-1376256 p^2 q^2 u x^2+25165824 r^2 u x^2+393216 p^4 q u x^2+3932160 p^3 r u x^2-19922944 p q r u x^2+2359296 p^2 s u x^2-2097152 q s u x^2-6291456 p t u x^2+3 p^13+12288 p q^6-18432 p^3 q^5+11520 p^5 q^4-3840 p^7 q^3+262144 p s^3+720 p^9 q^2+4096 p^5 s^2+65536 p q^2 s^2-32768 p^3 q s^2+131072 p^2 r s^2-524288 q r s^2+524288 p^3 t^2-2097152 p q t^2+4194304 r t^2-72 p^11 q+32 p^10 r-32768 q^5 r+40960 p^2 q^4 r-20480 p^4 q^3 r+5120 p^6 q^2 r-640 p^8 q r-320 p^9 s-81920 p q^4 s+81920 p^3 q^3 s-30720 p^5 q^2 s+5120 p^7 q s-4096 p^6 r s+262144 q^3 r s-196608 p^2 q^2 r s+49152 p^4 q r s-6400 p^8 t-65536 q^4 t+458752 p^2 q^3 t-319488 p^4 q^2 t-524288 p^2 r^2 t+2097152 q r^2 t-1048576 s^2 t+77824 p^6 q t-114688 p^5 r t-1835008 p q^2 r t+917504 p^3 q r t-98304 p^4 s t+524288 q^2 s t+262144 p^2 q s t-1048576 p r s t+36864 p^7 u-1572864 p q^3 u+1376256 p^3 q^2 u-393216 p^5 q u+393216 p^4 r u+4194304 q^2 r u-2621440 p^2 q r u+786432 p^3 s u-2097152 p q s u-3145728 p^2 t u+8388608 q t u)


For $n>1$, let the roots of $x^n+\sum_{k=0}^{n-1}a_{n-k}x^k=0$ be $x_1,\dots,x_n$.

A procedure to obtain the minimal polynomial of $\sqrt{x_1}+\dots+\sqrt{x_n}$ is squaring:

n = $1$: $x=\sqrt{x_1}$

$$x^2-(\sqrt{x_1})^2=x^2+p$$

n = $2$: $x=\sqrt{x_1}+\sqrt{x_2}$

$$x^2-(\sqrt{x_1}+\sqrt{x_2})^2=x^2+p-2\sqrt{x_1x_2}$$

$$(x^2+p)^2-(2\sqrt{x_1x_2})^2=(x^2+p)^2-4q$$

n = $3$: $x=\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}$

$$x^2-(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3})^2=x^2+p-2(\sqrt{x_1x_2}+\sqrt{x_1 x_3}+\sqrt{x_2 x_3})$$

$$(x^2+p)^2-(2(\sqrt{x_1x_2}+\sqrt{x_1 x_3}+\sqrt{x_2 x_3}))^2=(x^2+p)^2-4q-8\sqrt{x_1x_2x_3}(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3})\=(x^2+p)^2-4q-8\sqrt{x_1x_2x_3}x$$

$$((x^2+p)^2-4q)^2-(8\sqrt{x_1x_2x_3}x)^2=((x^2+p)^2-4q)^2+r(8x)^2$$

n = $4$: $x=\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}$

$$x^2-(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4})^2=x^2+p-2(\sqrt{x_1 x_2}+\sqrt{x_1 x_3}+\sqrt{x_2 x_3}+\sqrt{x_1 x_4}+\sqrt{x_2 x_4}+\sqrt{x_3 x_4})$$

$$(x^2+p)^2-(2(\sqrt{x_1 x_2}+\sqrt{x_1 x_3}+\sqrt{x_2 x_3}+\sqrt{x_1 x_4}+\sqrt{x_2 x_4}+\sqrt{x_3 x_4}))^2 =(x^2+p)^2-4q-8(\sqrt{x_1^2 x_2 x_3}+\sqrt{x_1 x_2^2 x_3}+\sqrt{x_1 x_2 x_3^2}+\sqrt{x_1^2 x_2 x_4}+\sqrt{x_1 x_2^2 x_4}+\sqrt{x_1 x_2 x_4^2}+\sqrt{x_1^2 x_3 x_4}+\sqrt{x_1 x_3^2 x_4}+\sqrt{x_1 x_3 x_4^2}+\sqrt{x_2^2 x_3 x_4}+\sqrt{x_2 x_3^2 x_4}+\sqrt{x_2 x_3 x_4^2}+3\sqrt{x_1x_2 x_3 x_4})$$

$$((x^2+p)^2-4q)^2-(8(\sqrt{x_1^2 x_2 x_3}+\sqrt{x_1 x_2^2 x_3}+\sqrt{x_1 x_2 x_3^2}+\sqrt{x_1^2 x_2 x_4}+\sqrt{x_1 x_2^2 x_4}+\sqrt{x_1 x_2 x_4^2}+\sqrt{x_1^2 x_3 x_4}+\sqrt{x_1 x_3^2 x_4}+\sqrt{x_1 x_3 x_4^2}+\sqrt{x_2^2 x_3 x_4}+\sqrt{x_2 x_3^2 x_4}+\sqrt{x_2 x_3 x_4^2}+3\sqrt{x_1x_2 x_3 x_4}))^2=((x^2 + p)^2 - 4 q)^2 + 64r x^2 + 64 s-128\sqrt{s}\left(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\right)\left(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_4}\right) \left(\sqrt{x_1}+\sqrt{x_3}+\sqrt{x_4}\right) \left(\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}\right)$$

$$(((x^2 + p)^2 - 4 q)^2 + 64r x^2 + 64 s)^2-16^2s\left(8\left(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\right)\left(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_4}\right) \left(\sqrt{x_1}+\sqrt{x_3}+\sqrt{x_4}\right) \left(\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}\right)\right)^2
=(((x^2 + p)^2 - 4 q)^2 + 64r x^2 + 64 s)^2- 256 s (3 x^4+2 p x^2-p^2+4 q)^2$$ the last line used: $$3 x^4 + 2 p x^2 - p^2 + 4 q=8 \left(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\right) \left(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_4}\right) \left(\sqrt{x_1}+\sqrt{x_3}+\sqrt{x_4}\right) \left(\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}\right)$$


By the last step of the procedure, the minimal polynomial of $\sqrt{x_1}+\dots+\sqrt{x_n}$ is of the form $$f_n(x)^2-(-1)^na_ng_n(x)^2$$ for some polynomials $f_n(x),g_n(x)$ and $f_n(x)=f_n(-x),g_n(x)=g_n(-x)$.

Note that $(-1)^na_n=x_1\dots x_n$.

Let $$P_0=\prod_{\substack{\mu_1,\dots,\mu_n\in{\pm1}\\mu_1\dots\mu_n=1}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_n\sqrt{x_n}))
P_1=\prod_{\substack{\mu_1,\dots,\mu_n\in{\pm1}\\mu_1\dots\mu_n=-1}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_n\sqrt{x_n}))$$ Let $$P_0=f_n(x)-\sqrt{x_1\dots x_n}g_n(x)
P_1=f_n(x)+\sqrt{x_1\dots x_n}g_n(x)$$ then $$f_n(x)=\frac{P_0+P_1}2
g_n(x)=\frac{P_1-P_0}{2\sqrt{x_1\dots x_n}}$$

Note that $P_0,P_1$ are products of $2^{n-1}$ polynomials with leading term $x$,
so $P_0,P_1$ are polynomials with leading term $x^{2^{n-1}}$,
so $f_n(x)$ is a polynomial with leading term $x^{2^{n-1}}$.

Note that $f_n(x),g_n(x)$ are unchanged under each automorphism $\phi_i:\sqrt{x_i}\mapsto-\sqrt{x_i}$,
so $f_n(x),g_n(x)$ contain no square roots.

Also, the leading term of $g_n(x)$ is $2^{n-1}(n-1)!x^{2^{n-1}-n}$. This is proved in this answer.


Setting $x_n=0$ in the expression of $P_0$:

$$P_0=\prod_{\substack{\mu_1,\dots,\mu_n\in{\pm1}
\mu_1\dots\mu_n=1}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_{n-1}\sqrt{x_{n-1}}))
=\prod_{\substack{\mu_1,\dots,\mu_{n-1}\in{\pm1}}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_{n-1}\sqrt{x_{n-1}}))$$ it becomes the minimal polynomial for $n-1$.

Similarly, setting $x_n=0$ in the expression of $P_1$, it becomes the minimal polynomial for $n-1$.

So, setting $x_n=0$ in the expression of $f_n$, it becomes the minimal polynomial for $n-1$.

Consider the difference of $f_n(x)$ with the minimal polynomial for $n-1$ $$\tag1\label1 f_n(x)-\left(f_{n-1}(x)^2-(-1)^{n-1}a_{n-1}g_{n-1}(x)^2\right)$$ As a polynomial in $x_n$, \eqref{1} has a root $x_n=0$, so it is divisible by $x_n$.

By symmetry, \eqref{1} is divisible by $x_1\dots x_n$,
so \eqref{1} is divisible by $a_n$.


For $n=2,3$, \eqref{1} is identically zero.

My question is: Is the following true?

For $n\ge4$, the degree of \eqref{1} is $2(\deg g_n)-(\deg f_n)$.

I verified it for $n=4,5,6$.

For example, $n=5$ in the above,
the leading term of $f_5(x)$ is $x^{16}$,
the leading term of $g_5(x)$ is $384x^{11}$,
the difference $f_5(x)-\left(f_4(x)^2-s\ g_4(x)^2\right)=-2048t(5x^6+\cdots)$ has degree $6=2\times11-16$.

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hbghlyj 发表于 2025-1-17 16:08
kuing 发表于 2021-8-24 15:21
更多元多次应该都总可以这样做


math.stackexchange.com/questions/4898529/patterns-in-the-minimal ... -sum-of-square-roots
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怎么说来,MSE上的网友把递推式得到了?😮  发表于 2025-1-18 21:22

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