Ramanujan-type trigonometric identities with cube roots, generalizing $\sqrt[3]{\cos(2\pi/7)}+\sqrt[3]{\cos(4\pi/7)}+\sqrt[3]{\cos(8\pi/7)}$

Question

Ramanujan found the following trigonometric identity \begin{equation} \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} \end{equation} (see e.g. Ramanujan — For Lowbrows, (3.7) and around, for details and an analogue for 9 instead of 7).

Are there analogous identities for all primes $p$ of the form $3k+1$ instead of 7?

Let me try to explain what I mean. As I've learned from S. Markelov, for $p=13$ \begin{multline} \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{3\sqrt[3]{13}-7}2} \end{multline} and there are close analogues for all $p$ of the form $n^2+n+1$. For example, for $p=43=6^2+6+1$ three groups of numerators are {2, 4, 8, 16, 22, 32, 42}, {6, 10, 12, 20, 24, 38, 40}, {14, 18, 26, 28, 30, 34, 36} and the sum is $\sqrt[3]{\frac{3\sqrt[3]{86}-13}2}$.

So it looks like there is some pattern reminding of... quadratic Gauss sums, perhaps.

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For any $p=3k+1$ one can partition $\mathbb F_p^\times$ into three groups, corresponding to $\mathbb F_p^\times/\mathbb F_p^{\times3}\cong\mathbb Z/3$ — and this is precisely the partitions from the last paragraph. This explains what LHS should look like. And indeed, at least for $p=19$ there is an identity \begin{multline} \sqrt[3]{\cos\bigl(\tfrac{\pi}{19}\bigr)+\cos\bigl(\tfrac{7\pi}{19}\bigr)+\cos\bigl(\tfrac{11\pi}{19}\bigr)}+ \sqrt[3]{\cos\bigl(\tfrac{3\pi}{19}\bigr)+\cos\bigl(\tfrac{5\pi}{19}\bigr)+\cos\bigl(\tfrac{17\pi}{19}\bigr)}+\\ \sqrt[3]{\cos\bigl(\tfrac{9\pi}{19}\bigr)+\cos\bigl(\tfrac{13\pi}{19}\bigr)+\cos\bigl(\tfrac{15\pi}{19}\bigr)}=\\ \sqrt[3]{\tfrac12-3\sqrt[3]7+\tfrac32\sqrt[3]{3\sqrt[3]{49}+18\sqrt[3]7-25}+\tfrac32\sqrt[3]{3\sqrt[3]{49}+18\sqrt[3]7-44}} \end{multline} which is closely related to the fact that $2 \left( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} \right)$ is a root of the equation $\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$.

So,

more precisely: is it true, that for any $p=3k+1$ the sum of 3 cubic roots of sum of cosines (described above) can be expressed in real radicals?

what can be said about RHS in this case (say, about the number of nested radicals)?

Answer 1

$\color{brown}{Update:}$

The question is: Is it true, that for any $p=3k+1$ the sum of 3 cubic roots of sum of cosines can be expressed in real radicals?

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The answer is Yes. In general, given the roots $x_i$ of the general cubic equation,

$$x^3+ax^2+bx+c=0\tag0$$

then sums involving the cube roots of the $x_i$ can be given in the simple form,

$$(u+x_1)^{1/3}+(u+x_2)^{1/3}+(u+x_3)^{1/3} = \big(w\pm3\,\sqrt{d}\,^{1/3}\big)^{1/3}$$

where $u,w$ are the constants,

$$u = \frac{ab-9c\pm \sqrt{d}}{2(a^2-3b)}$$

$$w = -\frac{(2a^3-9ab+27c)+9\pm \sqrt{d}}{2(a^2-3b)}$$

and $d$ is the negated discriminant $D$,

$$-D = d = \frac{4(a^2-3b)^3-(2a^3-9ab+27c)^2}{27}$$

However, the fact $\pm\sqrt{d}$ gives two results implies Ramanujan's identity has a hidden partner.

For example, given the minimal polynomial for $x=2\cos\big(\frac{2\pi}7\big)$, or $x^3 + x^2 - 2x - 1 = 0$. Since $\sqrt{d} = \sqrt{49} = \mp7$, then the negative case gives $(u=0,w =5)$ while the positive case gives $(u=1,w =-4)$, yielding the pair,

$$R_1=\sqrt[3]{0+2\cos\bigl(\tfrac{2\pi}{7}\bigr)}+\sqrt[3]{0+2\cos\bigl(\tfrac{4\pi}{7}\bigr)}+\sqrt[3]{0+2\cos\bigl(\tfrac{8\pi}{7}\bigr)} = \sqrt[3]{+5\color{blue}{-}3\,\sqrt[3]{7}}$$

$$R_2=\sqrt[3]{1+2\cos\bigl(\tfrac{2\pi}{7}\bigr)}+\sqrt[3]{1+2\cos\bigl(\tfrac{4\pi}{7}\bigr)}+\sqrt[3]{1+2\cos\bigl(\tfrac{8\pi}{7}\bigr)} = \sqrt[3]{-4\color{blue}{+}3\,\sqrt[3]{7}}$$

Some remarks. If the cubic has all real roots:

1. ... then discriminant $-D=+d$, and $(u,w)$ are also real.
2. ... with a cyclic group, then $d$ is a square, so $(u,w)$ in fact are rational.
3. ... then $a^2-3b=0$ never occurs, since this implies $-D =d = -(a^3-27c)^2$, so only one root $x_1$ is real.

These pairs are for other prime $p=6m+1$, covering my old addendum for $p=19$.

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$\color{brown}{Old\; answer:}$

The general phenomenon can be explained by a beautiful cubic identity found by Ramanujan, of course. {Grigori M, ironically, you pointed out this identity in an old question of mine.}

I. Given the three roots $x_i$ of,

$$x^3-ax^2+bx-1 = 0\tag1$$

then,

$$x_1^{1/3}+x_2^{1/3}+x_3^{1/3} = (a+6+3t)^{1/3}\tag2$$

where $t$ is a root of,

$$t^3-3(a+b+3)t-\big(ab+6(a+b)+9\big) = 0\tag3$$

The form of the $p=19$ suggests it is just a particular case of this identity. Note that,

$$y=2\Bigl(\cos\bigl(\tfrac{\pi}{19}\bigr)+\cos\bigl(\tfrac{7\pi}{19}\bigr)+\cos\bigl(\tfrac{11\pi}{19}\bigr)\Bigr)$$

is a root of,

$$y^3+y^2-6y-7=0$$

One can then scale variables and make the constant term $c=-1$. I did so, and got the exact form as the one above.

II. Note that if $a+b+3=0$, then $(3)$ takes a particularly simple form. After some manipulation, I found that given,

$$x^3 + x^2 - (3 n^2 + n)x + n^3=0\tag4$$

then,

$$F_P = x_1^{1/3}+x_2^{1/3}+x_3^{1/3} = \sqrt[3]{-(6n+1)+3\sqrt[3]{nP}}\tag5$$

where $P = 9n^2+3n+1$. For $n = -1, 1, -2, 2$, we get the same cubics satisfied by the sums of cosines for $P=7,13,31,43,\dots$ thus explaining the simple sums,

$$F_7 =-\sqrt[3]{-5+3\sqrt[3]{7}} $$

$$F_{13} = \sqrt[3]{-7+3\sqrt[3]{13}} $$

and so on, though I cannot prove that the roots of $(4)$ are always sums of cosines when $P$ is prime. Maybe I'll ask it as a separate question.

Answer 2

$\color{blue}{\text{Edited 9 Jun 2017:}}$

Let $p$ be a prime number congruent to 1 modulo 6, and $g$ a primitive root modulo $p$, i.e. a generator of the group $ \mathbb F_{p} ^ {\times} \cong C_{p-1} $. By definition, put $$ S_p(g) = \sum_{k=0}^{\frac{p-4}3}\cos \left(\frac{2\pi}{p}g^{3k}\right) , S'_p(g) =\sum_{k=0}^{\frac{p-4}3}\cos \left(\frac{2\pi}{p}g^{3k+1}\right) \text{, } S''_p(g) = \sum_{k=0}^{\frac{p-4}3}\cos \left(\frac{2\pi}{p}g^{3k+2}\right) . $$

In this PDF we prove the following result.

Theorem 1. Suppose $p$ equals $9m^2+3m+1$ for some $m\in \mathbb Z$. Then for any primitive root $g$ we have $$\sqrt[3]{S_p(g)} + \sqrt[3]{S'_p(g)} + \sqrt[3]{S''_p(g)} =\sqrt[3]{ 3\sqrt[3]{mp}-(6m+1) } . $$

Warning: I am not sure about my English. If you find some mistakes please write me.

Answer 3

I tried to find such rationals $a,b,c,d$ so that, $$(a+b\,x_1)^{1/3} + (a+b\,x_2)^{1/3} + (a+b\,x_3)^{1/3} = (c+\sqrt[3]{dp})^{1‌​/3}$$ So, we have identities for all primes $p=1\ mod\ 3$ such that $p<100$ $$\sqrt[3]{1+\sum_{k=0}^{5}\cos(\frac{2\pi }{19}\cdot2^{3k}) }+\sqrt[3]{1+\sum_{k=0}^{5}\cos(\frac{2\pi }{19}\cdot2^{3k+1}) }+\sqrt[3]{1+\sum_{k=0}^{5}\cos(\frac{2\pi }{19}\cdot2^{3k+2}) }=\sqrt[3]{8-3\sqrt[3]{19}}\qquad\\ \sqrt[3]{-1+\sum_{k=0}^{11}\cos(\frac{2\pi }{37}\cdot2^{3k}) }+\sqrt[3]{-1+\sum_{k=0}^{11}\cos(\frac{2\pi }{37}\cdot2^{3k+1}) }+\sqrt[3]{-1+\sum_{k=0}^{11}\cos(\frac{2\pi }{37}\cdot2^{3k+2}) }=\sqrt[3]{-10+3\sqrt[3]{37}}\qquad\\ \sqrt[3]{-1+\sum_{k=0}^{19}\cos(\frac{2\pi }{61}\cdot2^{3k}) }+\sqrt[3]{-1+\sum_{k=0}^{19}\cos(\frac{2\pi }{61}\cdot2^{3k+1}) }+\sqrt[3]{-1+\sum_{k=0}^{19}\cos(\frac{2\pi }{61}\cdot2^{3k+2}) }=\sqrt[3]{14-3\sqrt[3]{3\cdot 61}}\qquad\\$$

$$\sqrt[3]{1+\sum_{k=0}^{21}\cos(\frac{2\pi }{67}\cdot2^{3k}) }+\sqrt[3]{1+\sum_{k=0}^{21}\cos(\frac{2\pi }{67}\cdot2^{3k+1}) }+\sqrt[3]{1+\sum_{k=0}^{21}\cos(\frac{2\pi }{67}\cdot2^{3k+2})}=\sqrt[3]{-16+3\sqrt[3]{3\cdot 67}}\qquad\\ \sqrt[3]{-2+\sum_{k=0}^{25}\cos(\frac{2\pi }{79}\cdot3^{3k}) }+\sqrt[3]{-2+\sum_{k=0}^{25}\cos(\frac{2\pi }{79}\cdot3^{3k+1})}+\sqrt[3]{-2+\sum_{k=0}^{25}\cos(\frac{2\pi }{79}\cdot3^{3k+2})}=\sqrt[3]{-13+3\sqrt[3]{79}}\qquad\\ \sqrt[3]{3+\sum_{k=0}^{31}\cos(\frac{2\pi }{97}\cdot5^{3k}) }+\sqrt[3]{3+\sum_{k=0}^{31}\cos(\frac{2\pi }{97}\cdot5^{3k+1})}+\sqrt[3]{3+\sum_{k=0}^{31}\cos(\frac{2\pi }{97}\cdot5^{3k+2})}=\sqrt[3]{14-3\sqrt[3]{97}} $$

$\color{blue}{\text{Added June 27:}}$

$$\sqrt[3]{\frac{74}{43}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k}) }+\sqrt[3]{\frac{74}{43}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k+1}) }+\sqrt[3]{\frac{74}{43}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k+2}) }=\sqrt[3]{\frac{392}{43}}$$ $$\sqrt[3]{\frac{5105}{11349}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k}) }+\sqrt[3]{\frac{5105}{11349}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k+1}) }+\sqrt[3]{\frac{5105}{11349}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k+2}) }=\sqrt[3]{\frac{21}{1261}}$$ $$\sqrt[3]{-\frac{2306997866696}{1047656140569}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k}) }+\sqrt[3]{-\frac{2306997866696}{1047656140569}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k+1}) }+\sqrt[3]{-\frac{2306997866696}{1047656140569}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k+2}) }=-\sqrt[3]{\frac{118149192000}{1908298981}}$$ $$\sqrt[3]{\frac{9658771264742899051}{5361029308457632889}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k}) }+\sqrt[3]{\frac{9658771264742899051}{5361029308457632889}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k+1}) }+\sqrt[3]{\frac{9658771264742899051}{5361029308457632889}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k+2}) }=\sqrt[3]{\frac{60079542508626277}{3881990809889669}}$$ $$\sqrt[3]{\frac{1512950552654226074845360221799154}{22816093930106873589073278570630913}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k}) }+\sqrt[3]{\frac{1512950552654226074845360221799154}{22816093930106873589073278570630913}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k+1}) }+\sqrt[3]{\frac{1512950552654226074845360221799154}{22816093930106873589073278570630913}+\sum_{k=0}^{1}\cos(\frac{2\pi }{7}\cdot3^{3k+2}) }=-\sqrt[3]{\frac{9567191523845032860388297048}{16721690052047565734068821733}}$$ and so on.

Answer 4

Suppose $p=3k+1$. Then by a theorem of Gauss $4p=A^2+27B^2$ and cubic sums like $\displaystyle\sum_{t\in\mathbb Z/p}\cos\left(\frac{2\pi t^3}p\right)$ are roots of the equation $x^3-3px-Ap=0$. And if we have three roots $x_i$ of a cubic equation, Ramanujan's lemma gives the value of $\sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3}$. That's precisely what's going on, I believe.

Answer 5

(Addendum to my answer.) I just realized that given the $p=19$ identity cited by the OP,

$$\sqrt[3]{\cos\bigl(\tfrac{\pi}{19}\bigr)+\cos\bigl(\tfrac{7\pi}{19}\bigr)+\cos\bigl(\tfrac{11\pi}{19}\bigr)}+\\ \sqrt[3]{\cos\bigl(\tfrac{3\pi}{19}\bigr)+\cos\bigl(\tfrac{5\pi}{19}\bigr)+\cos\bigl(\tfrac{17\pi}{19}\bigr)}+\\ \sqrt[3]{\cos\bigl(\tfrac{9\pi}{19}\bigr)+\cos\bigl(\tfrac{13\pi}{19}\bigr)+\cos\bigl(\tfrac{15\pi}{19}\bigr)}=\\ \sqrt[3]{\tfrac12-3\sqrt[3]7+\tfrac32\sqrt[3]{3\sqrt[3]{49}+18\sqrt[3]7-25}+\tfrac32\sqrt[3]{3\sqrt[3]{49}+18\sqrt[3]7-44}}=0.815731\dots$$

a small tweak can give a simpler, but different one,

$$\sqrt[3]{ -\tfrac{1}{2}+\cos\bigl(\tfrac{\pi}{19}\bigr)+\cos\bigl(\tfrac{7\pi}{19}\bigr)+\cos\bigl(\tfrac{11\pi}{19}\bigr)}+\\ \sqrt[3]{-\tfrac{1}{2}+\cos\bigl(\tfrac{3\pi}{19}\bigr)+\cos\bigl(\tfrac{5\pi}{19}\bigr)+\cos\bigl(\tfrac{17\pi}{19}\bigr)}+\\ \sqrt[3]{-\tfrac{1}{2}+\cos\bigl(\tfrac{9\pi}{19}\bigr)+\cos\bigl(\tfrac{13\pi}{19}\bigr)+\cos\bigl(\tfrac{15\pi}{19}\bigr)}=\\ \sqrt[3]{-4+\tfrac{3}{2}\sqrt[3]{19}}=0.1375504\dots$$

It's because there is only a linear transformation between the cubic involved in,

$$\sqrt{ a+ \sqrt{ a + \sqrt{ a-x}}}=x$$

and the cubic for,

$$y_1^{1/3}+y_2^{1/3}+y_3^{1/3} = \sqrt[3]{-(6n+1)+3\sqrt[3]{nP}}$$

with $P = 9n^2+3n+1$, and we use $n=\frac{1}{2}$.

See these comments.