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


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,


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

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

  • $\begingroup$ Given your recent and continued interest in this type of problems, could you perhaps shed some light on how to solve this question ? $\endgroup$ – Lucian Dec 23 '14 at 4:00
  • $\begingroup$ @Lucian: I gave it a shot. It was the case $n=-1$ of the identity above. (P.S. There was a small typo in the original question.) $\endgroup$ – Tito Piezas III Dec 23 '14 at 4:38
  • $\begingroup$ Noam Elkies has finally found a quintic that uses $\cos \pi/11$ (analogous to Ramanujan's cubic above). See this MO post. $\endgroup$ – Tito Piezas III Aug 2 '16 at 19:25

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 $+$.

  • $\begingroup$ the fun thing is that we have already 2 families of cubic that gives those sum of cosines. I haven't checked if the primes represented overlap or not. Also I'm beginning to wonder if the discriminants of ALL the abelian cubics extensions of Q are that small (exactly the smallest conductor for that extension, from what i've seen) $\endgroup$ – mercio Dec 30 '14 at 18:41
  • $\begingroup$ I just noticed their dicriminants do overlap. This family (let's call it $F_1$) has discriminant $p=9n^2+3n+1$, while the other family (call it $F_2$) has $D=4k^2+6k+9$. However, they were asked from different directions: $F_1$ has nice forms for $x_1^{1/3}+x_2^{1/3}+ x_2^{1/3}$, while $F_2$ is involved in $\sqrt{ a+ \sqrt{ a + \sqrt{ a-x}}}=x$. $\endgroup$ – Tito Piezas III Dec 30 '14 at 19:10
  • $\begingroup$ Ah, the two cubics are the same! The one from the other post is, $$x^3 +kx^2 −(k^2 +2k+3)x−((k+1)^3 −k^2 )=0$$ The transformations, $$x=−(1+2y)/(2n),\quad k=1/(2n)$$ transforms it to, $$y^3 +y^2 −(3n^2 +n)y+n^3 =0$$ which is the cubic in this post. $\endgroup$ – Tito Piezas III Dec 31 '14 at 4:12
  • $\begingroup$ the cubics may be the same over $\Bbb Q$, but not over $\Bbb Z$. Except for $7$ and $13$ that are represented by both, the values represented are disjoint. (so I didn't exactly do twice the same thing). And also, if $x$ generates a cyclic cubic extension there always is an affine transformation $f$ such that the Galois action on $f(x)$ is of the form $y \mapsto y^2+c$ for some rational $c$. So it's not surprising. $\endgroup$ – mercio Dec 31 '14 at 16:01
  • $\begingroup$ I found a general formula. Kindly see this updated answer. $\endgroup$ – Tito Piezas III Jan 4 '15 at 5:37

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