9

While trying to find a closed-form solution for particular cubics as sums of cosines (related to this question), I came across this family with all roots real. Given a prime $p=6m+1$. Define, $$F(x) = x^3+x^2-2mx+N = \Big(x-2\sum_{k=1}^{m}\cos a\Big) \Big(x-2\sum_{k=1}^{m}\cos b\Big) \Big(x-2\sum_{k=1}^{m}\cos c\Big)=0$$ where, $$a=2^k\times\beta,\;\;b=2^k\times3\beta,\;\;c=2^k\times m\beta,$$

and $\beta = \displaystyle\frac{2\pi}{p}$. I noticed that for certain primes, then $N$ is an integer. The complete list for small $p$,

$$\begin{array}{|c|c|} \hline p&N\\ \hline 31& -8\\ 43& 8\\ 109& -4\\ 157& 64\\ 223& -256\\ 229& -212\\ 277& 236\\ 283& 304\\ \hline \end{array}$$

Questions:

  1. What is the complete list of such primes for a low bound, say $p<3000$? (My old version of Mathematica conks out at $p>2000$.)
  2. What do these primes $p$ have in common that make them distinct from other primes? (Other than that their $N$ is an integer.)
  3. The coefficients of the cubic $F(x)=0$ are simple polynomials in $m$, except the constant term. Can $N$ be expressed as a polynomial in $m$?

P.S. I've checked the OEIS and it's not there, but the list I have for $p<2000$ suggests that a necessary (but not sufficient) condition is that

$$p = x^2+27y^2,\quad\text{and}\quad 2^{2m} = 1\;\text{mod}\;p$$

(A014752) and (A016108), though it would be great if someone can prove (or disprove) that if $N$ is an integer, then these must hold.

Community
  • 1
Tito Piezas III
  • 54,565
  • "but not sufficient"... Do you know any prime $6n+1$ for which $2$ is a cubic residue that is not on your list? – Will Jagy Dec 15 '14 at 19:53
  • ah: 127 is not in your list – Will Jagy Dec 15 '14 at 19:56
  • The complete list for $p<2000$ is, $$p=31, 43, 109, 157, 223, 229, 277, 283, 691, 733, 739, 811, 1051, 1069, 1327, 1423, 1459, 1471, 1579, 1627, 1699, 1723, 1789, 1831, 1999.$$ They all obey $(1)$, though one can always turn up that doesn't. – Tito Piezas III Dec 15 '14 at 20:02
  • All your primes are also expressible as $9 x^2 + 6 xy + 28 y^2. $ The set of (reduced) positive binary forms that represent, say, all three $31,43,109$ is finite and probably quite small. Worth seeing if one gets $157$ but misses $127$ – Will Jagy Dec 15 '14 at 20:22
  • Whoever figures this one out oughta send it in to Sloane's OEIS. – Robert Soupe Dec 16 '14 at 01:49
  • hmm i've just read this, but have you tried checking if those are the primes where $2$ generates the subgroup $H$ of $(\Bbb Z/p\Bbb Z)^*$ of index $3$ and where $3 \notin H$ ($3$ is not a cubic reidue) ? – mercio Dec 16 '14 at 18:59
  • @mercio: No. But I did notice that for prime $n=6m+1=x^2+27y^2$, if we define the sequence T(c)=Table[Mod[c*2^n, 6m+1], {n, 0, m-1}]], then the elements of $T(2), T(6), T(2m)$ are distinct and disjoint apparently only for those $p$ in the list. – Tito Piezas III Dec 16 '14 at 19:14
  • If my calculation is correct, the subsequent terms with $2000\leq p\leq 6000$ are: $2347$, $2749$, $2767$, $2791$, $2917$, $3163$, $3229$, $3259$, $3271$, $3463$, $3823$, $4027$, $4423$, $4519$, $4549$, $4567$, $4597$, $4651$, $4663$, $4759$, $4909$, $5101$, $5167$, $5413$, $5437$, $5503$, $5653$. They all seem to satisfy the condition (1). – Peter Košinár Dec 20 '14 at 17:18
  • @PeterKošinár: Thanks! OEIS has a list $L$ of primes up to c. 17700 that satisfy (1). I checked yours and they are all in $L$. – Tito Piezas III Dec 20 '14 at 17:45
  • @mercio $p\in {31, 223, 1327, 1423, \ldots}$ do not seem to satisfy your condition, so it is not necessary -- but it seems to be sufficient based on my test so far. Perhaps it needs to be relaxed a bit? – Peter Košinár Dec 20 '14 at 17:49
  • 1
    @PeterKošinár: A more detailed version with related questions can be found in this MO post. – Tito Piezas III Dec 20 '14 at 20:16
  • "My old version of Mathematica conks out at $p>2000$". I assume you meant $p < 2000$? – alexwlchan Dec 31 '14 at 23:52
  • @alexwlchan: I can take the calculation up to $p=2000$, but raising it to $p>2000$, and the software seems unstable. – Tito Piezas III Jan 01 '15 at 00:07
  • @TitoPiezasIII: Ah, I misread it. My bad, cheers. – alexwlchan Jan 01 '15 at 00:08
  • @alexwlchan: And happy new year. My apologies about the 5th power calculation. The paper did say it was an open problem. :( – Tito Piezas III Jan 01 '15 at 00:09

1 Answers1

2

Question 1 was answered by Peter Kosinar in the comments, and a general version of Question 2 was answered by Michael Stoll in this MO post.

The general case of Question 3 is in this post.

P.S. One nice thing about these cubics is that, starting with Ramanujan's general cubic identity, they are a special case, yielding the simple,

$$(a+b\,x_1)^{1/3}+(a+b\,x_2)^{1/3}+(a+b\,x_3)^{1/3}=\big(c+\sqrt[3]{dp}\big)^{1/3}$$

for some rational $a,b,c,d$. For example, using $p=109$, so $x^3 + x^2 - 36x - 4=0$, then,

$$(2+x_1)^{1/3}+(2+x_2)^{1/3}+(2+x_3)^{1/3}=\big({-19}+\sqrt[3]{4\cdot109}\big)^{1/3}=1.553389\dots$$

Community
  • 1
Tito Piezas III
  • 54,565