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This post can be generalized to,

$$\begin{align} \sqrt{ 2+ \sqrt{ 2 + \sqrt{ 2-x}}}=x&,\quad\quad x = -2\cos\left(\frac{8\pi}{9}\right)=1.8793\dots\quad\quad\quad \\ \\ \sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x&,\quad\quad x = 2\sum_{n=1}^3\cos\left(\frac{2\pi\, s_1(n)}{19}\right)=2.5070\dots\quad\\ \\ \sqrt{ 8+ \sqrt{ 8 + \sqrt{ 8-x}}}=x&,\quad\quad x = -1-2\sum_{n=1}^6\cos\left(\frac{2\pi\, s_2(n)}{37}\right)=3.3447\dots\\ \\ \sqrt{ 14+ \sqrt{ 14 + \sqrt{14-x}}}=x&,\quad\quad x =\, \color{red}? \pm 2\sum_{n=1}^{\color{red}{7\,?}}\cos\left(\frac{2\pi\, s_3(n)}{\color{red}{63\,?}}\right)=\dots\quad\quad\quad\quad\\ \\ \sqrt{ 22+ \sqrt{ 22 + \sqrt{22-x}}}=x&,\quad\quad x = -1+2\sum_{n=1}^{16}\cos\left(\frac{2\pi\, s_4(n)}{97}\right)=5.2065\dots\\ \\ \sqrt{ 32+ \sqrt{ 32 + \sqrt{ 32-x}}}=x&,\quad\quad x = -2-2\sum_{n=1}^{23}\cos\left(\frac{2\pi\, s_5(n)}{139}\right)=6.1716\dots \end{align}$$

and so on, where the sequences $s_k(n)$ are,

$$\begin{aligned} s_1(n) &= 2, 3, 5.\\ s_2(n) &= 2, 9, 12, 15, 16, 17.\\ s_3(n) &=\, \color{red}?\\ s_4(n) &= 4, 6, 9, 10, 11, 14, 15, 17, 21, 23, 25, 26, 32, 35, 39, 48. \end{aligned}$$

etc. Note that by repeated squaring, we get,

$$((x^2 - a)^2 - a)^2 - a + x = 0$$

which has two cubic factors when $a=k^2+k+2$, with the $x_i$ above as sums of $\cos(z)$, and a root of the same cubic family,

$$x^3 + k x^2 - (k^2 + 2k + 3)x - ((k + 1)^3 - k^2)=0$$

This has a negative discriminant, $D =-(4k^2+6k+9)^2$, implying all roots are real.

$$\begin{array}{|c|c|c|} k&a&\sqrt{D}\\ 0&2&9\,i\\ 1&4&19\,i\\ 2&8&37\,i\\ 3&14&\color{red}{63}\,i\\ 4&22&97\,i\\ 5&32&139\,i\\ \end{array}$$

Questions:

  1. Can anybody find the sequence $s_3(n)$? (I used Mathematica's PowerMod, but it only works for primes.)
  2. Why do the others have a constant (namely $-1,-1,-2$) added to the sum of cosines? Can we predict its value, or can it be found only by trial and error?
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    $\begingroup$ What is the pattern between the $s_k(n)$'s? $\endgroup$ – apnorton Dec 13 '14 at 5:03
  • $\begingroup$ @anorton: You can find it using Mathematica's PowerMod function, though one to do a little tinkering to get the right sequence. $\endgroup$ – Tito Piezas III Dec 13 '14 at 5:06
  • $\begingroup$ Sorry, miscomputation. $\endgroup$ – Takahiro Waki May 23 '16 at 2:05
  • $\begingroup$ You might enjoy this pair about $x^3 + x^2 - 10 x - 8$ math.stackexchange.com/questions/1980175/… AND math.stackexchange.com/questions/1980693/… $\endgroup$ – Will Jagy Oct 23 '16 at 16:16
  • $\begingroup$ $x^3 + x^2 - 10 x - 8,$ discriminant is $3844 = 62^2,$ so the Galois group is $\mathbb Z_3$ $\endgroup$ – Will Jagy Oct 23 '16 at 16:39
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(In this answer i switched $x$ with $-x$)

The polynomial $((x^2-14)^2-14)^2-x-14$ factors into the two cubic with cyclic galois groups (I already did the work in Exploring 3-cycle points for quadratic iterations)

$(x^3 + 4x^2 - 11x - 43)(x^3 - 3x^2- 18x + 55)$, whose discriminants are $49^2$ and $63^2$.

Both cubic have all real roots, so we can assume that the roots lie in $\Bbb Z[2\cos (2\pi/49)]$ and $\Bbb Z[2\cos (2\pi/63)]$ (I use the fact that the ring of integers of $\Bbb Q(\zeta_n)$ is $\Bbb Z[\zeta_n]$).

For the first cubic, $(\Bbb Z/49 \Bbb Z)^*/\{\pm 1\}$ has only one quotient isomorphic to $C_3$, which corresponds to $\Bbb Z[2\cos(2\pi/7)]$. But if you try to form the sum of cosines given by the subgroup you end up with $0$ so this is a case where the roots are NOT of the conjectured form.

For the second cubic, $(\Bbb Z/63 \Bbb Z)^*/\{\pm 1\}$ is isomorphic to $C_2 \times C_3 \times C_3$, so it has several quotients isomorphic to $C_3$.

To identify which quotient corresponds to the cyclic cubic extensions, we can pick a small prime for each equivalence class and see when those factors split (have a root) or stay irreducible. Doing so we get the subgroup $H = \{\pm 1, \pm 5, \pm 8, \pm 11, \pm 23, \pm 25\}$.

Then by looking at the coset of $\pm4$, we get $2(\cos(8\pi/63)+\cos(38\pi/63)+\cos(40\pi/63)+\cos(52\pi/63)+\cos(58\pi/63)+\cos(62\pi/63)) = -5.25884526118409\ldots$

If we call $A,B,C$ the quantities $\sum_{k \in H} 2 \cos(2ka\pi/63)$ for $a=1,2,4$ respectively, returning to your $x$, we have $x = -1-C$


In the general case, if the root is to be of the form $n + m A + p B$, then by summing it with its conjugates we get that $3n + (m+p).(A+B+C)$ is the coefficient of $-x^2$ in the cubic, which is $\frac {1 \pm (2k+1)}2$ (depending on the factor)

Since $A+B+C = \mu(4k^2+6k+9)$, if we could guess what $m$ and $p$ are going to be when this is nonzero, this would determine $n$ (and also possibly which factor is the right one). Clearly, if $|m|=1$ and $p=0$, $n$ is going to be around $\pm k/3$


Suppose $\delta = 4k^2+6k+9$ is a prime $p$.

Then $k \neq 0 \pmod 3$ and $p \equiv 1 \pmod 3$. Moreover, the modulus of the splitting field of $X^3 - kX^2 -(k^2+2k+3)X + (k^3+2k^2+3k+1)$ (whose discriminant is $-p^2$) is of the form $(p^m)$. Since its Galois group is cyclic of order $3$, its modulus is $(p)$ (higher powers don't introduce any new $C_3$ factor), and the corresponding subgroup $H$ is the group of cubes modulo $p$.

Let $x$ be a root of that polynomial. Since the discriminant is so small and $\Bbb Q$ doesn't have any extension with discriminant $-1$ we can deduce that the ring of integers of $\Bbb Q[x]$ is $R = \Bbb Z[x] = \langle 1,x,x^2 \rangle$. If $\sigma$ is the reciprocity symbol at $2$, then $\sigma(x) = x^2-(k^2+k+2)$ and so $R = \langle 1,x,\sigma(x) \rangle$.

Since $x+\sigma(x)+\sigma^2(x) = k$, by letting $d = (k \pm 1) /3$ we have $(x-d)+\sigma(x-d)+\sigma^2(x-d) = k-(k \pm 1) = -\pm 1$, and so we get $R = \langle 1,(x-d),\sigma(x-d)\rangle = \langle x-d, \sigma(x-d),\sigma^2(x-d)\rangle$, and we have found an integral normal basis for $R$.

On the other hand, since we have an integral normal basis for $\Bbb Z[\zeta]$ we have another integral normal basis for $R$, which is $\langle \sum_{n \in H} \zeta^{bn} \mid b=1,2,4\rangle$

But integral normal basis are very rare. If you have one $\langle a,b,c \rangle$ and another one with $a' = xa+yb+zc$ then $b' = xb+yc+za, c' = xc+ya+zb$, and so the index of $\langle a',b',c' \rangle$ in $\langle a,b,c \rangle$ is given by the determinant which can be factored into $(x+y+z)(x+\zeta_3y+\zeta_3^2z)(x+\zeta_3^2y+\zeta_3z)$. Since this has to equal $\pm 1$ and since there are only $6$ units in $\Bbb Z[\zeta_3]$, we can quickly deduce that $a' \in \{\pm a, \pm b, \pm c\}$.

This proves that $x = d \pm \sum_{n \in H} \zeta^{bn}$ with $b \in \{1,2,4\}$.


This can still be carried out when $\delta$ is squarefree and composite if we assume that the modulus isn't a strict divisor of $\delta$.

When $\delta$ is not squarefree and the sum of cosines is nonzero, the latter half can be adapted to work (basis of the form $\langle 1,x,\sigma(x)\rangle$ with $x+\sigma(x)+\sigma^2(x) = 0$ are also very few), but I'm not sure how to show that $\langle 1,x,x^2\rangle$ is an integral basis for the ring of integers.

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    $\begingroup$ Ah, so, $$\sqrt{ 14+ \sqrt{ 14 + \sqrt{14-x}}}=x,\quad\quad x =-1-2\sum_{n=1}^6\cos\left(\frac{2\pi\, s_3(n)}{63}\right)=4.25884\dots$$ where $s_3(n) = 4, 19, 20, 26, 29, 31$. So good to finally see this elusive creature. (I had assumed it would involve 7 cosines.) $\endgroup$ – Tito Piezas III Dec 14 '14 at 17:32
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    $\begingroup$ I still don't get why those expressions are so simple. You need only one coset in all of those examples and that's really surprising. And the coefficient in front of those sums is always $\pm 1$, I would like to know the reason behind all that $\endgroup$ – mercio Dec 14 '14 at 18:05
  • $\begingroup$ I just made another question also about sums of cosines. $\endgroup$ – Tito Piezas III Dec 14 '14 at 20:04
  • $\begingroup$ Alright so i checked for $|k| \le 20$ and we always have $|x| + |round(k/3)| = |2 \sum_{h \in aH} \cos(2\pi h/(4k^2+6k+9))|$ except when the sums of cosines are all $0$ (which can only happen when $4k^2+6k+9$ isn't squarefree, I'm not sure about the reverse implication), when $k = -22,-21,-10,-4,-3,8,17$ $\endgroup$ – mercio Dec 16 '14 at 15:12
  • $\begingroup$ oh lol, barring possible overflow errors, $k=22$ doesn't work $\endgroup$ – mercio Dec 16 '14 at 15:14
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Explicit spliting: $$(x^{3}-3x^{2}-18x+55)(x^{3}+4x^{2}-11x-43)=(x-1-\sum_{t=0}^{1}\sum_{k=0}^{5}\cos(\frac{2\pi }{63}\cdot2^{3t}\cdot5^{k}))\cdot(x-1-\sum_{t=0}^{1}\sum_{k=0}^{5}\cos(\frac{2\pi }{63}\cdot2^{3t+1}\cdot5^{k}))\cdot(x-1-\sum_{t=0}^{1}\sum_{k=0}^{5}\cos(\frac{2\pi }{63}\cdot2^{3t+2}\cdot5^{k}))\cdot (x+6\cos\frac{3\pi }{7}+2\cos\frac{\pi }{7})\cdot(x+3-4\cos\frac{3\pi }{7}-6\cos\frac{\pi }{7})\cdot(x+1-2\cos\frac{3\pi }{7}+4\cos\frac{\pi }{7})$$ {2,5} is a generating set of $(\mathbb{Z}/63\mathbb{Z})^\times \cong \mathrm{C}_6 \times \mathrm{C}_6$.

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