二次根式的和的有理化

hbghlyj

计算范数 例子：Let $\alpha:=\sqrt[3]{7}$ and let $K:=\mathbb{Q}[\sqrt[3]{7}]$. Consider a generic algebraic integer $a+b\alpha+c\alpha^2$, with $a,b,c\in\mathbb{Z}$. I want to find $N(a+b\alpha+c\alpha^2)$, where $N$ denotes the norm of $K$ over $\mathbb{Q}$.


Let $\beta=a+b\alpha+c\alpha^2$. We want to find the matrix of the multiplication map $M_\beta$. Well,
$$\beta\cdot 1=\beta=a+b\alpha+c\alpha^2$$
$$\beta\alpha=a\alpha+b\alpha^2+c\alpha^3=7c+a\alpha+b\alpha^2$$
$$\beta \alpha^2=a\alpha^2+b\alpha^3+c\alpha^4=7b+7c\alpha+a$$
So,
$$M_\beta=\begin{pmatrix} a & b & c\\ 7c & a & b\\ 7 b & 7c & a\end{pmatrix}$$
and thus
$$N_{K/\mathbb{Q}}(\beta)=\det M_\beta=a^3-21abc+7b^3+49b^3$$

---

In practice, if you wanted to find $N_{L/K}(x)$ for some element $x\in L$, you can use particular properties about $x$. For example, $N_{L/K}(x)=(-1)^n a_0^{\frac{n}{d}}$ where $a_0$ is the constant coefficient of the minimal polynomial of $x$ over $K$, and $d=[K(x):K]$. Or, more generally to try and reduce your calculation to a smaller subextension (the last formula is a teased out version of this, where you restrict to the subextension $K(x)/K$).

hbghlyj

hbghlyj 发表于 2024-3-27 21:05
For example, $N_{L/K}(x)=(-1)^n a_0^{\frac{n}{d}}$

例子：$\sqrt{2}$在$\Bbb Q[\sqrt{i}]$上的极小多项式为$x^2-2$，故$d=2$.
$n=[\Bbb Q[\sqrt{i}]:\Bbb Q]=4$.
所以$\sqrt{2}$在$\Bbb Q[\sqrt{i}]$上的范数$=(-1)^22^{\frac42}=4$
AlgebraicNumberNorm[Sqrt[2], Extension -> Sqrt[I]]$\;=4$

hbghlyj

如何计算$(\sqrt{3}+\sqrt{-5})(\sqrt{3}-\sqrt{-5})(-\sqrt{3}+\sqrt{-5})(-\sqrt{3}-\sqrt{-5})$的值？

我们计算$x^2-3$和$x^2+5$的结式：$\DeclareMathOperator{\res}{res}$
\begin{align*}
\res(x^2-3,x^2+5)&=\res\big(x^2-3,x^2+5-(x^2-3)\big)\\&=\res(x^2-3,8)\\&=\begin{vmatrix}8\\&8\end{vmatrix}=8^2
\end{align*}
用Mathematica验证：
Compute the smallest field that includes $\sqrt3,\sqrt5$, i.e. $\Bbb Q[\sqrt3,\sqrt5]$:

1. e = ToNumberField[{Sqrt[3], Sqrt[-5]}, All][[1, 1]]

Copy the Code

Compute the norm in the field $\Bbb Q[\sqrt3,\sqrt5]$:

1. AlgebraicNumberNorm[Sqrt[3] + Sqrt[-5], Extension -> e]

Copy the Code

64

hbghlyj

hbghlyj 发表于 2024-3-27 20:45
它们的差$\cos\frac{2i\pi}{7}-(-\sqrt[5]{5})$的范数就等于这两个多项式的结式
Resultant[(8x^3+4x^2-4x-1)/8,x^5+5,x]

手算结式：（重复使用$\res(A,B)=\res(A-CB,B)$，$A,B$为首一多项式，且$\deg(A-CB)<\deg(A).$）
\begin{align*}
\res(\frac{8x^3+4x^2-4x-1}8 &,x^5+5)\\
=\res(\frac{8x^3+4x^2-4x-1}8&,x^5+5-(x^2-\frac12x+\frac34)\cdot\frac{8x^3+4x^2-4x-1}8)\\
=\res(\frac{8x^3+4x^2-4x-1}8&,\frac{-16 x^2 + 10 x + 163}{32})\\
=-\frac18\res(\frac{8x^3+4x^2-4x-1}8&,\frac{16 x^2 - 10 x - 163}{16})\\
=-\frac18\res(\small\frac{8x^3+4x^2-4x-1}8-(x+\frac98)\cdot\frac{16 x^2 - 10 x - 163}{16}&,\frac{16 x^2 - 10 x - 163}{16})\\
=-\frac18\res(\frac{1330 x+1451}{128}&,\frac{16 x^2 - 10 x - 163}{16})\\
=-\frac18\cdot(\frac{1330}{128})^2\res(x+\frac{1451}{1330}&,\frac{16 x^2 - 10 x - 163}{16})\\
=-\frac18\cdot(\frac{1330}{128})^2\res(x+\frac{1451}{1330}&,\frac{16 x^2 - 10 x - 163}{16}-(x -\frac{9129}{5320})\cdot(x+\frac{1451}{1330}))\\
=-\frac18\cdot(\frac{1330}{128})^2\res(x+\frac{1451}{1330}&,-\frac{3677281}{442225})\\
=-\frac18\cdot(\frac{1330}{128})^2\cdot-\frac{3677281}{442225}&\\
=\frac{3677281}{32768}&\\=112\frac{7265}{32768}&
\end{align*}这次终于对了，注意每次相除之前须把多项式化为首一的

isee

对于这种\[\frac{5}{{1 + 3\sqrt[3]{2} + \sqrt[3]{4}}},\]分母有理化，绝大多数情况下，用\[(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc,\] 其中， 令 $a=1,b=3\sqrt[3]{2},c=\sqrt[3]{4}$ 就可快速完成.

hbghlyj

在1#给出了2到5元的。今天算出了6元的

hbghlyj

一元：$x+p=0$的根$x_1$的平方根为$\sqrt{-p}$，极小多项式为$$(x-\sqrt{-p})(x+\sqrt{-p})=x^2+p$$

1. In[]:= Collect[PolynomialRemainder[Times@@(p+x^2/.p->p+c/.{{x->x-Sqrt[c]},{x->x+Sqrt[c]}})//Expand,c^2+p  c+q,c],x]
2. Out[]= p^2-4 q+2 p x^2+x^4

Copy the Code

二元：$x^2+px+q=0$的根$x_1,x_2$的平方根之和的极小多项式为$$\prod_{\mu_1,\mu_2\in\{\pm1\}}(x-(\mu_1\sqrt{x_1}+\mu_2\sqrt{x_2}))=(x^2+p)^2-4 q$$
$\sqrt{x_1}+\sqrt{x_2}$的有理化就是常数项的平方根：$$\prod_{\mu_2\in\{\pm1\}}(\sqrt{x_1}+\mu_2\sqrt{x_2})=(\sqrt{x_1}+\sqrt{x_2})(\sqrt{x_1}-\sqrt{x_2})=\sqrt{p^2-4 q}$$

1. In[]:= Collect[PolynomialRemainder[Times@@(p^2-4 q+2 p x^2+x^4/.q->q+p c/.p->p+c/.{{x->x-Sqrt[c]},{x->x+Sqrt[c]}})//Expand,c^3+p c^2+q c+r,c],x]
2. Out[]= p^4-8 p^2 q+16 q^2+(4 p^3-16 p q+64 r) x^2+(6 p^2-8 q) x^4+4 p x^6+x^8

Copy the Code

三元：$x^3+px^2+qx+r=0$的根$x_1,x_2,x_3$的平方根之和的极小多项式为$$\prod_{\mu_1,\mu_2,\mu_3\in\{\pm1\}}(x-(\mu_1\sqrt{x_1}+\mu_2\sqrt{x_2}+\mu_3\sqrt{x_3}))=\left(\left(x^2+p\right)^2-4 q\right)^2+64rx^2$$
常数项是一个平方，$\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}$的有理化就是常数项的平方根：$$\prod_{\mu_2,\mu_3\in\{\pm1\}}(\sqrt{x_1}+\mu_2\sqrt{x_2}+\mu_3\sqrt{x_3})=p^2-4q$$
验证：

1. In[]:= Sum[(-1)^i x^(2^3-i)Factor[SymmetricReduction[Expand[SymmetricPolynomial[i,Total/@Tuples[{{Sqrt[a],-Sqrt[a]},{Sqrt[b],-Sqrt[b]},{Sqrt[c],-Sqrt[c]}}]]],{a,b,c},{-p,q,-r}][[1]]],{i,0,2^3,2}]
2. Out[]= (p^2-4 q)^2+4 (p^3-4 p q+16 r) x^2+2 (3 p^2-4 q) x^4+4 p x^6+x^8

Copy the Code

hbghlyj

四元：

1. In[]:= Collect[PolynomialRemainder[Times@@((p^2-4 q)^2+4 (p^3-4 p q+16 r) x^2+2 (3 p^2-4 q) x^4+4 p x^6+x^8/.r->r+q c/.q->q+p c/.p->p+c/.{{x->x-Sqrt[c]},{x->x+Sqrt[c]}})//Expand,c^4+p c^3+q c^2+r c+s,c],x]
2. Out[]= p^8-16 p^6 q+96 p^4 q^2-256 p^2 q^3+256 q^4-128 p^4 s+1024 p^2 q s-2048 q^2 s+4096 s^2+(8 p^7-96 p^5 q+384 p^3 q^2-512 p q^3+128 p^4 r-1024 p^2 q r+2048 q^2 r+1536 p^3 s-6144 p q s+8192 r s) x^2+(28 p^6-240 p^4 q+576 p^2 q^2-256 q^3+512 p^3 r-2048 p q r+4096 r^2+1280 p^2 s-7168 q s) x^4+(56 p^5-320 p^3 q+384 p q^2+768 p^2 r-1024 q r-2560 p s) x^6+(70 p^4-240 p^2 q+96 q^2+512 p r-2176 s) x^8+(56 p^3-96 p q+128 r) x^10+(28 p^2-16 q) x^12+8 p x^14+x^16

Copy the Code

$x^4+px^3+qx^2+rx+s=0$的根$x_1,x_2,x_3,x_4$的平方根之和的极小多项式为$$\prod_{\mu_1,\dots,\mu_4\in\{\pm1\}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_4\sqrt{x_4}))=(((x^2 + p)^2 - 4 q)^2 + 64r x^2 + 64 s)^2 -
256 s (-3 x^4-2 p x^2+p^2-4 q)^2$$
$\sqrt{x_1}+\dots+\sqrt{x_3}$的有理化就是常数项的平方根：$$\prod_{\mu_2,\dots,\mu_4\in\{\pm1\}}(\sqrt{x_1}+\mu_2\sqrt{x_2}+\dots+\mu_4\sqrt{x_4})=(p^2-4 q)^2-64 s$$
验证：

1. In[]:= Sum[(-1)^i x^(2^4-i)Factor[SymmetricReduction[Expand[SymmetricPolynomial[i,Total/@Tuples[{{Sqrt[a],-Sqrt[a]},{Sqrt[b],-Sqrt[b]},{Sqrt[c],-Sqrt[c]},{Sqrt[d],-Sqrt[d]}}]]],{a,b,c,d},{-p,q,-r,s}][[1]]],{i,0,2^4,2}]
2. Out[]= (p^4-8 p^2 q+16 q^2-64 s)^2+8 (p^7-12 p^5 q+48 p^3 q^2-64 p q^3+16 p^4 r-128 p^2 q r+256 q^2 r+192 p^3 s-768 p q s+1024 r s) x^2+4 (7 p^6-60 p^4 q+144 p^2 q^2-64 q^3+128 p^3 r-512 p q r+1024 r^2+320 p^2 s-1792 q s) x^4+8 (7 p^5-40 p^3 q+48 p q^2+96 p^2 r-128 q r-320 p s) x^6+2 (35 p^4-120 p^2 q+48 q^2+256 p r-1088 s) x^8+8 (7 p^3-12 p q+16 r) x^10+4 (7 p^2-4 q) x^12+8 p x^14+x^16

Copy the Code

hbghlyj

五元：

1. Collect[PolynomialRemainder[Times@@(%/.s->s+r c/.r->r+q c/.q->q+p c/.p->p+c/.{{x->x-Sqrt[c]},{x->x+Sqrt[c]}})//Expand,c^5+p c^4+q c^3+r c^2+s c+t,c],x]

Copy the Code

这里的%是(上楼的)四元的极小多项式.
$x^5+px^4+qx^3+rx^2+sx+t=0$的根$x_1,x_2,x_3,x_4,x_5$的平方根之和的极小多项式为一个$2^5=32$次多项式
\begin{gather*}\prod_{\mu_1,\dots,\mu_5\in\{\pm1\}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_5\sqrt{x_5}))=\left(\left(\left(\left(x^2+p\right)^2-4 q\right)^2+64 r x^2+64 s\right)^2-256 s \left(-3 x^4-2 p x^2+p^2-4 q\right)^2-10240 t x^6-6144 p t x^4-73728 p^2 t x^2-1572864 q t x^2+2048 \left(29 p^2+788 q\right) t x^2+4718592 p q t+14155776 r t-2048 \left(p^3+2300 q p+6920 r\right) t\right)^2\\
+16384 t x^2 \left(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\right)^2\end{gather*}

1. ((((x^2+p)^2-4q)^2+64r x^2+64s)^2-256 s (p^2-4 q-2 p x^2-3 x^4)^2+63488 t x^6-153600 p t x^4+2048 (29 p^2 t+788 q t) x^2-2048 (p^3 t+2300 p q t+6920 r t)+(4718592 p q t+14155776 r t-73728 p^2 t x^2-1572864 q t x^2+147456 p t x^4-73728 t x^6))^2+16384 t x^2 (3 p^5-24 p^3 q+48 p q^2+32 p^2 r-128 q r+64 p s-256 t+9 p^4 x^2-56 p^2 q x^2+80 q^2 x^2+192 s x^2+6 p^3 x^4-40 p q x^4+96 r x^4-6 p^2 x^6-8 q x^6-9 p x^8-3 x^10)^2

Copy the Code

常数项是一个平方，$\sqrt{x_1}+\dots+\sqrt{x_5}$的有理化就是常数项的平方根：
$$\prod_{\mu_2,\dots,\mu_5\in\{\pm1\}}(\sqrt{x_1}+\mu_2\sqrt{x_2}+\dots+\mu_5\sqrt{x_5})=\left(\left(p^2-4 q\right)^2-64 s\right)^2-2048 t \left(p(p^2-4q)+8 r\right)$$

hbghlyj

六元：

1. Collect[PolynomialRemainder[Times@@(%/.t->t+s c/.s->s+r c/.r->r+q c/.q->q+p c/.p->p+c/.{{x->x-Sqrt[c]},{x->x+Sqrt[c]}})//Expand,c^6+p c^5+q c^4+r c^3+s c^2+t c+u,c],x]

Copy the Code

这里的%是(上楼的)五元的极小多项式.
输出很长，$x^6+px^5+qx^4+rx^3+sx^2+tx+u=0$的根$x_1,x_2,x_3,x_4,x_5,x_6$的平方根之和的极小多项式为一个$2^6=64$次的多项式。
配方得（办法是把$u$替换成平方，然后Factor平方差）
$$\prod_{\mu_1,\dots,\mu_6\in\{\pm1\}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_6\sqrt{x_6}))=f^2-ug^2$$其中，

1. f=(((((x^2+p)^2-4q)^2+64r x^2+64s)^2-256 s (p^2-4 q-2 p x^2-3 x^4)^2+63488 t x^6-153600 p t x^4+2048 (29 p^2 t+788 q t) x^2-2048 (p^3 t+2300 p q t+6920 r t)+(4718592 p q t+14155776 r t-73728 p^2 t x^2-1572864 q t x^2+147456 p t x^4-73728 t x^6))^2+16384 t x^2 (3 p^5-24 p^3 q+48 p q^2+32 p^2 r-128 q r+64 p s-256 t+9 p^4 x^2-56 p^2 q x^2+80 q^2 x^2+192 s x^2+6 p^3 x^4-40 p q x^4+96 r x^4-6 p^2 x^6-8 q x^6-9 p x^8-3 x^10)^2)+8192 u (21 p^10-344 p^8 q+2144 p^6 q^2-6144 p^4 q^3+7424 p^2 q^4-2048 q^5+384 p^7 r-4608 p^5 q r+18432 p^3 q^2 r-24576 p q^3 r+2048 p^4 r^2-16384 p^2 q r^2+32768 q^2 r^2+384 p^6 s-2048 p^4 q s-2048 p^2 q^2 s+16384 q^3 s+8192 p^3 r s-32768 p q r s+20480 p^2 s^2-32768 q s^2-9216 p^5 t+65536 p^3 q t-114688 p q^2 t-81920 p^2 r t+262144 q r t-65536 p s t+131072 t^2+18432 p^4 u-98304 p^2 q u+131072 q^2 u+90 p^9 x^2-1088 p^7 q x^2+4416 p^5 q^2 x^2-6144 p^3 q^3 x^2+512 p q^4 x^2+2304 p^6 r x^2-16896 p^4 q r x^2+24576 p^2 q^2 r x^2+24576 q^3 r x^2+32768 p^3 r^2 x^2-131072 p q r^2 x^2+131072 r^3 x^2-14080 p^5 s x^2+81920 p^3 q s x^2-102400 p q^2 s x^2-65536 p^2 r s x^2+32768 q r s x^2+139264 p s^2 x^2+35840 p^4 t x^2-229376 p^2 q t x^2+409600 q^2 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+813 p^8 x^4-7776 p^6 q x^4+26656 p^4 q^2 x^4-48128 p^2 q^3 x^4+55552 q^4 x^4+640 p^5 r x^4+35840 p^3 q r x^4-153600 p q^2 r x^4-24576 p^2 r^2 x^4+245760 q r^2 x^4-58240 p^4 s x^4+294912 p^2 q s x^4-206848 q^2 s x^4-417792 p r s x^4+724992 s^2 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+3864 p^7 x^6-32320 p^5 q x^6+78720 p^3 q^2 x^6-45056 p q^3 x^6+41984 p^4 r x^6-168960 p^2 q r x^6+77824 q^2 r x^6+24576 p r^2 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+8714 p^6 x^8-62032 p^4 q x^8+116896 p^2 q^2 x^8-13312 q^3 x^8+71296 p^3 r x^8-260608 p q r x^8+30720 r^2 x^8-99712 p^2 s x^8+518144 q s x^8-201728 p t x^8+854016 u x^8+10620 p^5 x^10-61632 p^3 q x^10+95040 p q^2 x^10+23808 p^2 r x^10-109056 q r x^10+106752 p s x^10-82944 t x^10+7650 p^4 x^12-32480 p^2 q x^12+32480 q^2 x^12+9600 p r x^12+51072 s x^12+3928 p^3 x^14-8384 p q x^14+13824 r x^14+2049 p^2 x^16-792 q x^16+954 p x^18+209 x^20);
2. g=256 (3 p^13-72 p^11 q+720 p^9 q^2-3840 p^7 q^3+11520 p^5 q^4-18432 p^3 q^5+12288 p q^6+32 p^10 r-640 p^8 q r+5120 p^6 q^2 r-20480 p^4 q^3 r+40960 p^2 q^4 r-32768 q^5 r-320 p^9 s+5120 p^7 q s-30720 p^5 q^2 s+81920 p^3 q^3 s-81920 p q^4 s-4096 p^6 r s+49152 p^4 q r s-196608 p^2 q^2 r s+262144 q^3 r s+4096 p^5 s^2-32768 p^3 q s^2+65536 p q^2 s^2+131072 p^2 r s^2-524288 q r s^2+262144 p s^3-6400 p^8 t+77824 p^6 q t-319488 p^4 q^2 t+458752 p^2 q^3 t-65536 q^4 t-114688 p^5 r t+917504 p^3 q r t-1835008 p q^2 r t-524288 p^2 r^2 t+2097152 q r^2 t-98304 p^4 s t+262144 p^2 q s t+524288 q^2 s t-1048576 p r s t-1048576 s^2 t+524288 p^3 t^2-2097152 p q t^2+4194304 r t^2+36864 p^7 u-393216 p^5 q u+1376256 p^3 q^2 u-1572864 p q^3 u+393216 p^4 r u-2621440 p^2 q r u+4194304 q^2 r u+786432 p^3 s u-2097152 p q s u-3145728 p^2 t u+8388608 q t u+3 p^12 x^2-120 p^10 q x^2+1680 p^8 q^2 x^2-11520 p^6 q^3 x^2+42240 p^4 q^4 x^2-79872 p^2 q^5 x^2+61440 q^6 x^2-640 p^9 r x^2+10240 p^7 q r x^2-61440 p^5 q^2 r x^2+163840 p^3 q^3 r x^2-163840 p q^4 r x^2-4096 p^6 r^2 x^2+49152 p^4 q r^2 x^2-196608 p^2 q^2 r^2 x^2+262144 q^3 r^2 x^2-8000 p^8 s x^2+101376 p^6 q s x^2-448512 p^4 q^2 s x^2+770048 p^2 q^3 s x^2-344064 q^4 s x^2-81920 p^5 r s x^2+655360 p^3 q r s x^2-1310720 p q^2 r s x^2-262144 p^2 r^2 s x^2+1048576 q r^2 s x^2-61440 p^4 s^2 x^2+294912 p^2 q s^2 x^2-196608 q^2 s^2 x^2-524288 p r s^2 x^2+2359296 s^3 x^2-10240 p^7 t x^2+73728 p^5 q t x^2-98304 p^3 q^2 t x^2-131072 p q^3 t x^2+49152 p^4 r t x^2-655360 p^2 q r t x^2+1835008 q^2 r t x^2+131072 p^3 s t x^2+524288 p q s t x^2-7340032 r s t x^2-3670016 p^2 t^2 x^2+10485760 q t^2 x^2-86016 p^6 u x^2+393216 p^4 q u x^2-1376256 p^2 q^2 u x^2+4718592 q^3 u x^2+3932160 p^3 r u x^2-19922944 p q r u x^2+25165824 r^2 u x^2+2359296 p^2 s u x^2-2097152 q s u x^2-6291456 p t u x^2+33554432 u^2 x^2-150 p^11 x^4+2184 p^9 q x^4-10944 p^7 q^2 x^4+17664 p^5 q^3 x^4+16896 p^3 q^4 x^4-55296 p q^5 x^4-5664 p^8 r x^4+65024 p^6 q r x^4-236544 p^4 q^2 r x^4+221184 p^2 q^3 r x^4+188416 q^4 r x^4-86016 p^5 r^2 x^4+688128 p^3 q r^2 x^4-1376256 p q^2 r^2 x^4-393216 p^2 r^3 x^4+1572864 q r^3 x^4-5376 p^7 s x^4+99328 p^5 q s x^4-536576 p^3 q^2 s x^4+901120 p q^3 s x^4-184320 p^4 r s x^4+1409024 p^2 q r s x^4-2686976 q^2 r s x^4-786432 p r^2 s x^4+172032 p^3 s^2 x^4-622592 p q s^2 x^4+3538944 r s^2 x^4-113664 p^6 t x^4+1200128 p^4 q t x^4-3751936 p^2 q^2 t x^4+3080192 q^3 t x^4-491520 p^3 r t x^4+3276800 p q r t x^4-4718592 r^2 t x^4-589824 p^2 s t x^4+2359296 q s t x^4-4718592 p t^2 x^4-1437696 p^5 u x^4+9043968 p^3 q u x^4-12386304 p q^2 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^4-990 p^10 x^6+13176 p^8 q x^6-63168 p^6 q^2 x^6+126720 p^4 q^3 x^6-87552 p^2 q^4 x^6+6144 q^5 x^6-20480 p^7 r x^6+180224 p^5 q r x^6-458752 p^3 q^2 r x^6+262144 p q^3 r x^6-122880 p^4 r^2 x^6+589824 p^2 q r^2 x^6-393216 q^2 r^2 x^6+36608 p^6 s x^6-353280 p^4 q s x^6+643072 p^2 q^2 s x^6+737280 q^3 s x^6+655360 p^3 r s x^6-3670016 p q r s x^6+1572864 r^2 s x^6+401408 p^2 s^2 x^6+1146880 q s^2 x^6+108544 p^5 t x^6+245760 p^3 q t x^6-1933312 p q^2 t x^6-819200 p^2 r t x^6+4849664 q r t x^6+1179648 p s t x^6-13107200 t^2 x^6-2248704 p^4 u x^6+8257536 p^2 q u x^6+2949120 q^2 u x^6-22806528 p r u x^6+56360960 s u x^6-3135 p^9 x^8+36144 p^7 q x^8-145824 p^5 q^2 x^8+234240 p^3 q^3 x^8-114432 p q^4 x^8-45248 p^6 r x^8+328960 p^4 q r x^8-648192 p^2 q^2 r x^8+225280 q^3 r x^8-40960 p^3 r^2 x^8+98304 p q r^2 x^8+393216 r^3 x^8+23168 p^5 s x^8-558080 p^3 q s x^8+1779712 p q^2 s x^8-307200 p^2 r s x^8-2375680 q r s x^8+2117632 p s^2 x^8+961024 p^4 t x^8-3223552 p^2 q t x^8+991232 q^2 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^8-6039 p^8 x^10+59472 p^6 q x^10-191520 p^4 q^2 x^10+211200 p^2 q^3 x^10-40704 q^4 x^10-69888 p^5 r x^10+419840 p^3 q r x^10-528384 p q^2 r x^10-184320 p^2 r^2 x^10+147456 q r^2 x^10-101760 p^4 s x^10+25600 p^2 q s x^10+1017856 q^2 s x^10-835584 p r s x^10+1953792 s^2 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^10-7524 p^7 x^12+63504 p^5 q x^12-154560 p^3 q^2 x^12+96000 p q^3 x^12-78400 p^4 r x^12+345600 p^2 q r x^12-168960 q^2 r x^12-331776 p r^2 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^12-5940 p^6 x^14+44784 p^4 q x^14-76224 p^2 q^2 x^14+17664 q^3 x^14-62464 p^3 r x^14+159744 p q r x^14-147456 r^2 x^14-25856 p^2 s x^14+160768 q s x^14+432128 p t x^14+69632 u x^14-2475 p^5 x^16+20376 p^3 q x^16-21168 p q^2 x^16-33120 p^2 r x^16+31104 q r x^16+40128 p s x^16+157440 t x^16+165 p^4 x^18+5544 p^2 q x^18-2544 q^2 x^18-10368 p r x^18+17088 s x^18+858 p^3 x^20+744 p q x^20-1440 r x^20+498 p^2 x^22+24 q x^22+135 p x^24+15 x^26);

Copy the Code

常数项是一个平方，$\sqrt{x_1}+\dots+\sqrt{x_6}$的有理化就是常数项的平方根：
$$\prod_{\mu_2,\dots,\mu_6\in\{\pm1\}}(\sqrt{x_1}+\mu_2\sqrt{x_2}+\dots+\mu_6\sqrt{x_6})=\left(\left(\left(p^2-4 q\right)^2-64 s\right)^2-2048 t \left(p(p^2-4q)+8 r\right)\right)^2-8192 u \left(15 p^{10}-232 p^8 q+384 p^7 r+1312 p^6 q^2+1152 p^6 s-4608 p^5 q r+3072 p^5 t-3072 p^4 q^3-10240 p^4 q s+2048 p^4 r^2-18432 p^4 u+18432 p^3 q^2 r-16384 p^3 q t+8192 p^3 r s+1792 p^2 q^4+26624 p^2 q^2 s-16384 p^2 q r^2+98304 p^2 q u+16384 p^2 r t-4096 p^2 s^2-24576 p q^3 r+16384 p q^2 t-32768 p q r s-65536 p s t+2048 q^5-16384 q^3 s+32768 q^2 r^2-131072 q^2 u+32768 q s^2+131072 t^2\right)$$
七元：？？
有什么规律

hbghlyj

hbghlyj 发表于 2024-4-13 13:57
有什么规律

只发现一个小规律：
n元极小多项式配成平方后，剩下部分关于x是平方式，
且关于x的最高次项系数：$-4,64,-2304,147456,\dots$
$(-4)^{n-1}((n-1)!)^2$
A002454 Central factorial numbers: a(n) = 4^n (n!)^2.

n个二次根式的和的有理化 是否存在“移项平方”的有限步骤
把这个问题发到MSE试试看：
math.stackexchange.com/questions/4898529/patterns-in-the-minimal ... -sum-of-square-roots