A little more on $\sqrt[3]{\cos\bigl(\tfrac{2\pi}7\bigr)}+\sqrt[3]{\cos\bigl(\tfrac{4\pi}7\bigr)}+\sqrt[3]{\cos\bigl(\tfrac{8\pi}7\bigr)}$

Question

Using a special case of an identity by Ramanujan, we find that given the roots $x_i$ of

$$x^3 + x^2 - (3 n^2 + n)x + n^3=0\tag1$$

which, since its discriminant is negative, always has three real roots, then,

$$F_p = x_1^{1/3}+x_2^{1/3}+x_3^{1/3} = \sqrt[3]{-(6n+1)+3\sqrt[3]{np}}\tag2$$

where $p = 9n^2+3n+1$.

Question:

Is it true that if $p$ is prime, then a root of $(1)$ is always a sum of the $p$th root of unity of form,

$$x = \sum_{m=1}^{(p-1)/3}\,\exp\Bigl(\frac{2\pi\, i\, k^m}{p}\Bigr)\tag3$$

for some integer $k$? We have,

$$\begin{array}{|c|c|c|} n&p&k\\ -1&7&6\\ 1&13&5\\ -2&31&15\\ 2&43&2\\ 3&73&7\\ \end{array}$$

and so on.

(P.S. If indeed true, this implies that $x$ is also a sum of cosines and would generalize,

$$F_7=\sqrt[3]{\cos\bigl(\tfrac{2\pi}7\bigr)}+\sqrt[3]{\cos\bigl(\tfrac{4\pi}7\bigr)}+\sqrt[3]{\cos\bigl(\tfrac{8\pi}7\bigr)}=-\sqrt[3]{\tfrac{-5+3\sqrt[3]7}2}$$

$$F_{13} = \sqrt[3]{\cos\bigl(\tfrac{2\pi}{13}\bigr)+\cos\bigl(\tfrac{10\pi}{13}\bigr)}+ \sqrt[3]{\cos\bigl(\tfrac{4\pi}{13}\bigr)+\cos\bigl(\tfrac{6\pi}{13}\bigr)}+ \sqrt[3]{\cos\bigl(\tfrac{8\pi}{13}\bigr)+\cos\bigl(\tfrac{12\pi}{13}\bigr)}=\\ \sqrt[3]{\tfrac{-7+3\sqrt[3]{13}}2}$$

which are just the cases $n=-1,\,1$, with the second one discussed by GrigoryM in the related question in the link above.)

Answer 1

Some computation show that the roots are in an abelian cubic extension of $\Bbb Q \subset K$, and in particular the Galois group is generated by $\sigma : x \mapsto - \frac {x^2}n - x (1+\frac 1n) + 2$.

Let $y = \sigma(x)$ and $z = \sigma(y) = -1-x-y$

The volume of a fundamental domain of the lattice $L = \langle 1,x,x^2 \rangle$ is given by $\sqrt{|\Delta|} = np$ (we embed $K$ in $\Bbb R^3$ via the $3$ real embeddings of $K$).

If we let $R = \langle x,y,z \rangle$ we have $R = \langle 1,x,y \rangle = \langle 1,x,(x^2+x)/n \rangle$. Obviously, the ring of integers $\mathcal O_K$ of $K$ contains $R$, and $L$ has index $n$ in $R$, so the volume of a fundamental domain for $R$ is $p$, and so that of $\mathcal O_K$ is a divisor of $p$.

Since $\Bbb Q$ has no unramified extension, $R$ and $\mathcal O_K$ must coincide. Then $(x,y,z)$ is an integral normal basis for $\mathcal O_K$.

$K$ has to be the class field of modulus $p$ corresponding to the subgroup $H = (\Bbb Z/p\Bbb Z)^{*3}$ of $(\Bbb Z/p\Bbb Z)^*$ (because it is only ramified over $p$ and there is essentially only one subgroup of index $3$ in all the $\Bbb Z/p^n\Bbb Z$).

Therefore $K = \Bbb Q(\zeta_p)^H$, and $\mathcal O_K = \Bbb Z[\zeta_p]^H$. An obvious integral normal basis for $\mathcal O_K$ is given by $(\sum_{k \in bH} \zeta_p^k)$ . Then using the same argument as in my other answer, we can deduce that the two normal integral basis are, up to sign, the same.

Finally, $x+y+z = -1 = \mu(p) = \sum_{k=1}^{p-1} \zeta_p^k$, which shows that that sign is $+$.