Something strange about $\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$ and its friends 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:


*

*Can anybody find the sequence $s_3(n)$? (I used Mathematica's PowerMod, but it only works for primes.)

*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? 


(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.
