and the cubic,


with its constant expressed in terms of $(1)$ as,

$$N = \frac{1}{27}(1-3p\pm2pu)\tag3$$

and the sign $\pm u$ chosen appropriately. (The discriminant is $D=-108p^2v^2$, so all real roots.) Starting with Ramanujan's general cubic identity, they are a special case, yielding the simple,


for some rational $a,b,c,d$.

Question: Is it true that if $p=u^2+27v^2=6m+1$ is prime, then a root of $(2)$ is always a sum of the $p$th root of unity of form,

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

for some integer $k$?

Example: We have $p=127=10^2+27\cdot1^2=6\cdot21+1$, so

$$x^3+x^2-(2\times21)x+80=0,\quad\quad x =\sum_{n=1}^{2\times21}\,\exp\Bigl(\frac{2\pi\, i\, (5^n)}{127}\Bigr)$$

with all three roots $x_i$ yielding,

$$(-2+x_1)^{1/3}+(-2+x_2)^{1/3}+(-2+x_3)^{1/3}=-\big(19-3\,\sqrt[3]{254}\big)^{1/3} = -0.097378\dots$$

P.S. This generalizes Question 3 of this post, and is also related to this post.

  • $\begingroup$ cf. statement in the upd. of math.stackexchange.com/q/102736 $\endgroup$ – Grigory M Jan 3 '15 at 23:00
  • $\begingroup$ (and math.stackexchange.com/q/31485 is, of course, related) $\endgroup$ – Grigory M Jan 3 '15 at 23:01
  • $\begingroup$ and your $k$ is always an element of order $(p-1)/3$ in $\mathbb Z/p^\times$, I guess $\endgroup$ – Grigory M Jan 3 '15 at 23:03
  • $\begingroup$ (Looks like we already have all pieces of the puzzle and they start to fit together... — but I haven't yet solved it...) $\endgroup$ – Grigory M Jan 3 '15 at 23:27
  • $\begingroup$ @GrigoryM: Thanks for the links! Yes, getting close. Given the cubic eqn $(2)$ and the Ramanujan-type $(4)$, I'm trying to find a closed-form for the rationals $a,b,c,d$. $\endgroup$ – Tito Piezas III Jan 3 '15 at 23:35

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