12

(This extends this post.) Define the function,

$$\sqrt[3]{G(t)} = \sqrt[3]{t+x_1}+\sqrt[3]{t+x_2}+\sqrt[3]{t+x_3}\tag1$$

where the $x_i$ are roots of the cubic,

$$x^3+ax^2+bx+c=0\tag2$$

While $G(t)$ generally is a root of a $9$th deg equation, davidoff303 found that, using $\cos(\frac{\pi\,k}{7})$ and special rational $t$, then $G(t)$ can in fact be rational. For example,

$$\sqrt[3]{\tfrac{74}{43}+2\cos\big(\tfrac{2\pi }{7}\big)}+\sqrt[3]{\tfrac{74}{43}+2\cos\big(\tfrac{4\pi }{7}\big)}+\sqrt[3]{\tfrac{74}{43}+2\cos\big(\tfrac{8\pi }{7}\big)}=2\,\sqrt[3]{\tfrac{49}{43}}$$

This post generalizes to $\cos(\frac{\pi\,k}{19})$, $\cos(\frac{\pi\,k}{37})$, etc. The trick is to use rational $t,w$ such that they obey,

$$t^3-at^2+bt-c=w^3\tag3$$

If $(3)$ holds, then there is the nice relation, $$Y(Y+d) = X^3\tag4$$

where $X,Y,$ and discriminant $d$ are,

$$27d^2 = 4 (a^2 - 3 b)^3 - (2 a^3 - 9 a b + 27 c)^2\tag5$$

$$X = b - 2 a t + 3 t^2 + (-a + 3 t) w + 3 w^2\tag6$$

$$Y =\tfrac{1}{2}(-a b - 9 c - d) + (a^2 t + 6 b t - 9 a t^2 + 9 t^3) + 3 (b - 2 a t + 3 t^2) w - 3 (a - 3 t) w^2\tag7$$

However, the product $Y(Y+d)$ by itself does not guarantee that $G(t)$ is rational.

Questions:

  1. If all three rational Diophantine eqns below, $$t^3-at^2+bt-c=w^3\\ Y = u^3\\ Y+d = v^3$$ are satisfied with $Y$ defined in $(7)$, then is $G(t)$ also rational?
  2. Let $(2)$ be the minimal cubic eqn of sums of $\cos(\frac{\pi\,k}{d})$ with $d=6n+1$. What is the necessary criteria for $d$ such that all three equations of Question 1 hold? (I observed only $d=7,19,37$ within the search bounds I used.)
Community
  • 1
Tito Piezas III
  • 54,565
  • I just realized that the third condition is just $u^3+d =v^3$ so $d$ must be expressible as the sum of two rational cubes. Not all prime $d=6n+1$ qualify such as $d=73, 109, 181, 199, 307,\dots$ – Tito Piezas III Jun 30 '16 at 01:30
  • However, the formula for $d$ in $(5)$ does not necessarily yield a prime. For example, the minimal cubic eqn for sums of $\cos{\frac{\pi,k}{31}}$ is $x^3+x^2-10x-8=0$ and the formula yields $d=2\times31$ so this complicates things. – Tito Piezas III Jun 30 '16 at 01:57
  • Are you assuming that $d$ is rational and not a square root of a rational ? if you work over $\Bbb Q(d)$ ($d$ is rational iff the initial cubic has cyclic or trivial galois group), the elliptic curve $(3)$ is equivalent to $y^2 = x^3 + d^2/4$, and its unramified $3$-cover that can parametrize $G(t)$ is equivalent to $y^2 = x^3 - 27d^2/4$. Those must have the same rank (over $\Bbb Q(d)$), so there is a nonzero density of points in $(3)$ that have a preimage over $\Bbb Q(d)$ in the cover. Also, only the original curve has $3$-torsion, so this density is $3^{-k}$ where $1 \le k \le rank(E)+1$. – mercio Jul 01 '16 at 15:23
  • if $\Bbb Q(d) = \Bbb Q(\sqrt{-3})$ then both curves have $3$-torsion over $\Bbb Q(\sqrt{-3})$ so maybe there is a chance that $k=0$ in that case. Though again one of them might have more torsion since we're in $\Bbb Q(\sqrt{-3})$ and not $\Bbb Q$. – mercio Jul 01 '16 at 15:35
  • A generalization of Ramanujan's cubic identity to quintics has finally been found by Noam Elkies using $\cos \pi/11$. See this MO post. – Tito Piezas III Aug 02 '16 at 19:28

0 Answers0