4
$\begingroup$

I was trying to generalize,

$$\sqrt[3]{\sum_{k=1}^5\cos\big(\tfrac{2^k\cdot\,2\pi}{31}\big)}+\sqrt[3]{\sum_{k=1}^5\cos\big(\tfrac{2^k\cdot\,6\pi}{31}\big)}+\sqrt[3]{\sum_{k=1}^5\cos\big(\tfrac{2^k\cdot\,10\pi}{31}\big)} = -\sqrt[3]{\tfrac{-11+3\,\sqrt[3]{62}}{2}} \tag1$$

which is a special case of an identity of Ramanujan's. It led me to a family of primes $p=x^2+27y^2$ (A014752). Let $\beta=2\pi/p$ and define,

$$p=x^2+27y^2=6m+1$$

$$x_1=2\sum_{k=1}^{m}\cos\big(2^k\times\beta\big)$$ $$x_2=2\sum_{k=1}^{m}\cos\big(2^k\times3\beta\big)$$ $$x_3=2\sum_{k=1}^{m}\cos\big(2^k\times m\beta\big)$$ and, $$a = -(x_1+x_2+x_3),\quad b=x_1x_2+x_1x_3+x_2x_3,\quad c=-x_1x_2x_3$$

I noticed some $a,b,c$ were just plain integers. In general, they were algebraic integers at most of a degree $n \leq 9$. The complete list for $p<1000$,

$$\begin{array}{|l|l|} \hline n&\quad\quad\quad p\\ \hline 1& 31, 43, 109, 157, 223, 229, 277, 283, 691, 733, 739, 811\\ 2& 433, 457\\ 3& 307, 439, 499, 643, 727, 919, 997\\ 4& 601\\ 6& (\text{for}\; p>1000?)\\ 9& 127, 397\\ \hline \end{array}$$

Questions:

  1. All these primes are also $4^m \equiv 1\pmod p$ (A016108). However, what distinguishes the primes $p$ in the first row from the others such that their $a,b,c$ are just plain integers? (It seems $p\equiv 1\pmod{24}$ have $n=2,4$.)
  2. Is there a $p$ with $n=6$? (I'm having trouble with $p=1297$.)

P.S. This is related to a MSE post of mine.

$\endgroup$
  • $\begingroup$ Isn't it the case that for any prime $p$ whatsoever, that $a,b,c$ are algebraic integers, since $x_1,x_2,x_3$ are algebraic integers? $\endgroup$ – Pace Nielsen Dec 20 '14 at 20:17
  • $\begingroup$ @PaceNielsen: Yes, I didn't phrase it properly. I meant deg $n \leq 9$. I'll edit the post. $\endgroup$ – Tito Piezas III Dec 20 '14 at 20:18
  • $\begingroup$ P.S. I've tested all primes $< 2000$ and P. Kosinar has tested all $< 6000$. If $a,b,c$ just has deg $n=1$, then I conjecture that a necessary (but not sufficient) condition is that $p=x^2+27y^2$. Perhaps someone can prove (or disprove) it. $\endgroup$ – Tito Piezas III Dec 20 '14 at 20:43
  • $\begingroup$ Something is wrong with the three displayed equations: all start with $x_1$, and the multipliers of $\beta$ are $1,3,m$ where you presumably want representatives of the multiplicative group mod cubes (in the case that $2$ generates the cubes). $\endgroup$ – Noam D. Elkies Oct 27 '16 at 2:44
  • $\begingroup$ @NoamD.Elkies: Oops, typo fixed. $\endgroup$ – Tito Piezas III Oct 27 '16 at 2:52
14
$\begingroup$

Let $\zeta = e^{2\pi i/p}$ be a primitive $p$th root of unity. Then $2 \cos (2\pi k/p) = \zeta^k + \zeta^{-k}$. The Galois group of $\mathbb Q(\zeta)$ is isomorphic to $(\mathbb Z/p \mathbb Z)^\times$ and acts transitively on the powers $\zeta^k$ with $1 \le k \le p-1$. What you want is that the Galois action fixes the set $\{x_1, x_2, x_3\}$. Now $x_j = \sum_{k \in A_j} \zeta^k$ where $A_1 = \{\pm 2^j : 0 \le k < m\} \subset \mathbb Z/p \mathbb Z$ (as a multiset) and similarly for $A_2$ and $A_3$. So you want the natural action of $(\mathbb Z/p \mathbb Z)^\times$ on $\mathbb Z/p \mathbb Z$ by multiplication to just permute these three sets. This is only possible when $A_1$ consists of all cubes in $(\mathbb Z/p \mathbb Z)^\times$. In any case, 2 is a cube (since $p = x^2 + 27 y^2$; this is a famous result due to Gauss). However, not every cube is necessarily $\pm$ a power of 2. For example, for $p = 127$, the powers of 2 give you only 7 residue classes, so $A_1$ has 14 distinct elements, but $(p-1)/3 = 42$ is larger.

The precise condition should be that

  1. the order of 2 in the group $(\mathbb Z/p \mathbb Z)^\times$ must be either $2m$ or else $m$ and $m$ must be odd (otherwise $2^{m/2} = -1$ and the positive and negative powers of 2 in $A_1$ coincide), and
  2. $3$ is not a cubic residue mod $p$ (this is the condition for $A_1$, $A_2$ and $A_3$ to be pairwise disjoint).

Note that this is consistent with your table. This answers Question 1.

To give an answer to Question 2, here is how you can determine the degree $n$ of your algebraic integers $a$, $b$, $c$. You consider the set $\{A_1, A_2, A_3\}$ and count how many images it has under the natural action of $(\mathbb Z/p \mathbb Z)^\times$. This is $n$.

For $p = 1297$, I get $n = 648$, so it is not surprising that you had trouble. (But then, 1297 is not of the form $x^2 + 27y^2$!) And $p = 3889$ has $n = 6$.

$\endgroup$
  • $\begingroup$ Thanks! Two quick questions. 1) For $a,b,c$ to be plain integers, it is indeed a necessary (but not sufficient) condition that $p=x^2+27y^2$, or is there an exception? 2) What is the degree $n$ for $p=1297$? $\endgroup$ – Tito Piezas III Dec 20 '14 at 20:54
  • $\begingroup$ The final statement in my answer is not yet entirely correct; I'll edit it soon. $\endgroup$ – Michael Stoll Dec 20 '14 at 20:59
  • $\begingroup$ The edits are made. -- It is indeed necessary that $p$ has this form. This is equivalent to $p \equiv 1 \bmod 3$ and $2$ is a cube mod $p$. Since $2 \in A_1$, $2$ must be a cube, and for the set of cubes to be one third of everything, $p$ must be 1 mod 3. $\endgroup$ – Michael Stoll Dec 20 '14 at 21:21
  • 1
    $\begingroup$ Sorry about $p=1297$. I mis-typed something. And I verified that $p=3889=3\cdot6^4+1$ does have $n=6$. I figured there had to be a $n=6$. :) $\endgroup$ – Tito Piezas III Dec 20 '14 at 21:32
  • 1
    $\begingroup$ It is not clear to me what your question is. $x_1$, $x_2$, $x_3$ and $a$, $b$, $c$ are defined as they are, and their degree is what it is. In any case, you might want to look up "Gaussian periods", since this is what your $x_1$, $x_2$, $x_3$ are when $n = 1$ (and in general, they are multiples or linear combinations of Gaussian periods). $\endgroup$ – Michael Stoll Dec 21 '14 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.