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

(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.

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 $(\mathbb{Z}/63\mathbb{Z})^\times \cong \mathrm{C}_6 \times \mathrm{C}_6$.