证明层层长根式等

isee

证明：$$\sqrt 5+\sqrt{22+2\sqrt{5}}=\sqrt{11+2\sqrt{29}}+\sqrt{16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}}}.$$

kuing

昨天玩的是三次方程，这回是四次了吧？

hejoseph

设
\[
\left(a+b\sqrt{11-2\sqrt{29}}\right)^2=16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}}，
\]
令
\[
\left\{
\begin{aligned}
a^2+\left(11-2\sqrt{29}\right)b^2&=16-2\sqrt{29}，\\
ab&=\sqrt{5}，
\end{aligned}
\right.
\]
解这个方程组得一组较简单的解
\[
\left\{
\begin{aligned}
a&=\sqrt 5，\\
b&=1，
\end{aligned}
\right.
\]
所以
\[
\sqrt{11+2\sqrt{29}}+\sqrt{16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}}}=\sqrt{11+2\sqrt{29}}+\sqrt{5}+\sqrt{11-2\sqrt{29}}。
\]
另外
\[
\left(\sqrt{11+2\sqrt{29}}+\sqrt{11-2\sqrt{29}}\right)^2=22+2\sqrt{5}，
\]
所以
\[
\sqrt{11+2\sqrt{29}}+\sqrt{16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}}}=\sqrt 5+\sqrt{22+2\sqrt{5}}。
\]

isee

回复 2# kuing


    转成方程当然可以啦。不过，何版已经给出了一种简解（之一）。

    又被秒了。

青青子衿

证明：\[\sqrt 5+\sqrt{22+2\sqrt{5}}=\sqrt{11+2\sqrt{29}}+\sqrt{16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}}}.\] ...
isee 发表于 2017-9-21 16:36

山克斯恒等式
美国数学月刊 1988年 6卷 534页
Shanks' identity is elementary.
ams.org/mathscinet-getitem?mr=90c%3a11005
liinwww.ira.uka.de/cgi-bin/bibshow?e=Nbui0bnfsnbuinpouimz/fyqboe ... ibtex&mode=intra

isee

山克斯恒等式
青青子衿 发表于 2017-9-21 20:17

一语中地

hbghlyj

shanks-a.pdf
shanks-b.pdf
Consider the algebraic numbers\begin{gathered}
A=\sqrt{5}+\sqrt{22+2 \sqrt{5}} \\
B=\sqrt{11+2 \sqrt{29}}+\sqrt{16-2 \sqrt{29}+2 \sqrt{55-10 \sqrt{29}}}
\end{gathered}To 25 decimals they both equal $7.3811759408956579709872669$.
Either this is an incredible coincidence or
$$\leqalignno{A=B&&(1)}
$$
is an incredible identity, since $A$ and $B$ do not appear to lie in the same algebraic field. But they do. One has
$$
\leqalignno{A=B=4 X-1,&&(2)}
$$
where $X$ is the largest root of
$$
\leqalignno{x^4-x^3-3 x^2+x+1=0 .&&(3)}
$$
The astonishing appearance of (1) stems from a peculiarity of (3). The Galois group of this quartic is the octic group (the symmetries of a square), and its resolvent cubic is therefore reducible:
$$
\leqalignno{z^{3}-8 z-7=(z+1)\left(z^{2}-z-7\right)=0&&(4)}
$$
The common discriminant of (3) and (4) equals $725=5^{2} \cdot 29$. While the quartic field $\Bbb Q(X)$ contains $\Bbb Q(\sqrt{5})$ as a subfield it does not contain $\Bbb Q(\sqrt{29})$. Yet $X$ can be computed from any root of (4). The rational root $z=-1$ gives $X=(A+1) / 4$ while $z=(1+\sqrt{29}) / 2$ gives $X=(B+1) / 4$.

It is clear that we can construct any number of such incredible identities from other quartics having an octic group. For example
$$\tag1\label1
x^{4}-x^{3}-5 x^{2}-x+1=0
$$
has the discriminant $4205=29^{2} \cdot 5$, and so the two expressions involve $\sqrt{5}$ and $\sqrt{29}$ once again. But this time $\Bbb Q(\sqrt{29})$ is in $\Bbb Q(X)$ and $\Bbb Q(\sqrt{5})$ is not.

---

计算文中给出的“Galois群为$D_4$的任意四次方程都能给出这种等式”的例子：
用Ferrari法（或Descates法）需要解resolvent cubic，可以用resolvent cubic的任意一个根来解。
而resolvent cubic有一个有理根（剩下两个根不属于$\Bbb Q$）时，用那个有理根解出来结果比用剩下两根简单的多，解\eqref{1}就得到两种“方程的根的表达式”

1. In[]:= Factor[x^4-x^3-5x^2-x+1,Extension->Sqrt[29]]
2. Out[]= -(1/4) (-2+(1+Sqrt[29]) x-2 x^2) (2+(-1+Sqrt[29]) x+2 x^2)

复制代码

$x^{4}-x^{3}-5 x^{2}-x+1=\left(x^2-\frac{\sqrt{29}+1}{2} x+1\right)\left(x^2+\frac{\sqrt{29}-1}{2} x+1\right)$

1. In[]:= Factor[x^4-x^3-5x^2-x+1,Extension->1/4 (-7+Sqrt[5]+Sqrt[2 (19-7 Sqrt[5])])]
2. Out[]= -(1/64) (14-2 Sqrt[5]+2 Sqrt[2 (19-7 Sqrt[5])]+(4+3 Sqrt[2 (19-7 Sqrt[5])]+Sqrt[10 (19-7 Sqrt[5])]) x-8 x^2) (-14+2 Sqrt[5]+2 Sqrt[2 (19-7 Sqrt[5])]+(-4+3 Sqrt[2 (19-7 Sqrt[5])]+Sqrt[10 (19-7 Sqrt[5])]) x+8 x^2)

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\begin{gather*}x^{4}-x^{3}-5 x^{2}-x+1=
\\
\left(x^2-\frac{(3+\sqrt{5})\sqrt{38-14 \sqrt{5}}+4}{8}x+\frac{\sqrt{5}-\sqrt{38-14 \sqrt{5}}-7}{4}\right)
\\
\left(x^2+\frac{(3+\sqrt{5})\sqrt{38-14 \sqrt{5}}-4}{8}x+\frac{\sqrt{5}+\sqrt{38-14 \sqrt{5}}-7}{4}\right)\end{gather*}
这两种“方程的根的表达式”相等。令最大的根相等，见这帖

hbghlyj

Galois groups of irreducible quartic polynomials
The resolvent cubic of an irreducible quartic polynomial $P(x)$ can be used to determine its Galois group $G$; that is, the Galois group of the splitting field of $P(x)$. Let $m$ be the degree over $k$ of the splitting field of the resolvent cubic (it can be either $R_4(y)$ or $R_5(y)$; they have the same splitting field). Then the group $G$ is a subgroup of the symmetric group $S_4$. More precisely:^[4]
    If $m = 1$ (that is, if the resolvent cubic factors into linear factors in $k$), then $G$ is the group $\{e, (12)(34), (13)(24), (14)(23)\}$.
    If $m = 2$ (that is, if the resolvent cubic has one and, up to multiplicity, only one root in $k$), then, in order to determine $G$, one can determine whether or not $P(x)$ is still irreducible after adjoining to the field $k$ the roots of the resolvent cubic. If not, then $G$ is a cyclic group of order 4; more precisely, it is one of the three cyclic subgroups of $S_4$ generated by any of its six 4-cycles. If it is still irreducible, then $G$ is one of the three subgroups of $S_4$ of order 8, each of which is isomorphic to the dihedral group of order 8.
    If $m = 3$, then $G$ is the alternating group $A_4$.
    If $m = 6$, then $G$ is the whole group $S_4$.

    References   
[4]  Kaplansky, Irving (1972), "Fields: Cubic and quartic equations", Fields and Rings, Chicago Lectures in Mathematics (2nd ed.), University of Chicago Press
maths.ed.ac.uk/~v1ranick/papers/kapfield.pdf

hbghlyj

假设$a,b,D\inQ$，$D\not\inQ^2$，

• $D\times(a^2-b^2D)\inQ^2$.
  contained in cyclic extensions of degree 4
  可以写成正/余弦：
  $\sqrt{2-\sqrt2}=2\sin \left(\frac{\pi }{8}\right)$
  
  1. In[]:= ResourceFunction["GaloisGroupProperties"][MinimalPolynomial[Sqrt[2-Sqrt[2]],x],x]
  2. Out[]= CyclicGroup[4]

  复制代码
  
  $\sqrt{5+\sqrt5}=2 \cos \left(\frac{\pi }{4}+\frac{\pi }{10}\right)+2 \cos \left(\frac{\pi }{4}-\frac{\pi }{10}\right)$
  
  1. In[]:= ResourceFunction["GaloisGroupProperties"][MinimalPolynomial[Sqrt[5+Sqrt[5]],x],x]
  2. Out[]= CyclicGroup[4]

  复制代码
  
  $\sqrt{5-\sqrt5}=2 \cos \left(\frac{\pi }{4}+\frac{3\pi }{10}\right)+2 \cos \left(\frac{\pi }{4}-\frac{3\pi }{10}\right)$
• $a^2-b^2D\inQ^2$.
  可以写成两个二次根式：$\sqrt{2-\sqrt{3}}=\sqrt{\frac32}-\sqrt{\frac12}$
  $$\left\{\begin{array}{l}
  \delta=\sqrt{a^2-b} \\
  a>0
  \end{array} \Longrightarrow \sqrt{a \pm \sqrt{b}}=\sqrt{\frac{a+\delta}{2}} \pm \sqrt{\frac{a-\delta}{2}} .\right.$$
  
  1. In[]:= ResourceFunction["GaloisGroupProperties"][MinimalPolynomial[Sqrt[2-Sqrt[3]],x],x]
  2. Out[]= AbelianGroup[{2,2}]

  复制代码
  
  也可以写成正/余弦：
  $\sqrt{2-\sqrt{3}}=2\sin\left(\frac{\pi }{12}\right)$
• 以上两种都不满足。无法化简。$\sqrt{2-\sqrt{5}}$
  
  1. In[]:= ResourceFunction["GaloisGroupProperties"][MinimalPolynomial[Sqrt[2-Sqrt[5]],x],x]
  2. Out[]= DihedralGroup[4]

  复制代码
  
  $D_4$不是交换群，根据这帖不能用有理度角的sin/cos和有理数写出