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Let prime $p$ and given $\zeta_p = e^{2\pi i/p}$. It is well-known that the minimal polynomial of $x = \zeta_p + \zeta_p^{p-1}$ has a constant term either $\pm 1$ and, for certain $p$, the sum of four terms $x = \zeta_p +\zeta_p^{a}+\zeta_p^{p-a}+\zeta_p^{p-1}$ have as well. However, we also have,

$$F(x) = x^5 + x^4 - 28x^3 + 37x^2 + 25x + 1=0,\;\; x = \sum_{k=1}^{14} (\zeta_{71})^{23^k}$$

$$F(x) = x^6 + x^5 - 15x^4 - 28x^3 + 15x^2 + 38x - 1=0,\;\; x = \sum_{k=1}^{6} (\zeta_{37})^{11^k}$$

$$F(x) = x^7 + x^6 - 48x^5 + 37x^4 + 312x^3 - 12x^2 - 49x - 1=0,\;\; x = \sum_{k=1}^{16} (\zeta_{113})^{35^k}$$

$$F(x) = x^{11} + x^{10} - 40x^9 - 19x^8 + \dots - 1=0,\;\; x = \sum_{k=1}^{8} (\zeta_{89})^{12^k}$$

Question (edited):

What is the constraint on $p$ such that there is a minimal polynomial $F(x)$ with,

  1. root $x = \sum_{k=1}^{h} (\zeta_{p})^{a^k}$
  2. $4<h<p-1$
  3. degree $\frac{p-1}{h}$
  4. and constant term $\pm1$?

There are an infinite number of $p$ that satisfy conditions (1)-(3). But if we add the fourth, the four examples are the only ones in Kluener's Database of Number Fields. Are there others? (I've searched small $p$ and there are none.)

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Do you mean to require that (as in all your examples) the powers of $\zeta_p$ constitute a subgroup of $({\bf Z}/p{\bf Z})^*$, so that the degree of $F$ is the quotient of $p-1$ by the number of terms? Without this condition there are plenty of other examples: just multiply $s$ factors $\zeta_p^a + \zeta_p^{p-a}$ such that all $2^s$ terms in the expansion are distinct. –  Noam D. Elkies Oct 26 '13 at 20:10
    
Noam Elkies: Yes, I'll edit the question. –  Tito Piezas III Oct 26 '13 at 20:31
    
There's a fifth sporadic example at $(p,h) = (73,8)$. –  Noam D. Elkies Oct 27 '13 at 4:09
    
Ah, $-1 - 8 x + 150 x^2 - 427 x^3 - 34 x^4 + 278 x^5 - 11 x^6 - 32 x^7 + x^8 + x^9=0$. The prime $p=73$ does not appear in Kluener's database for 9T1. –  Tito Piezas III Oct 27 '13 at 5:00
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There are no further examples with $p < 1000$ (by exhaustive gp computation of all $1145$ candidate $(p,h)$ pairs). –  Noam D. Elkies Oct 27 '13 at 5:36

1 Answer 1

up vote 3 down vote accepted

This is not a complete answer, but a reformulation of your question in a way that removes the algebraic number theory, which translates the question into the realm of additive combinatorics. I assume that you also require the order of $a$ in $(\mathbf{Z}/p\mathbf{Z})^*$ to be $h$, as in your examples and Noam's comment. In this case, your conditions are equivalent to the following, where I write $C$ for the subgroup of $(\mathbf{Z}/p\mathbf{Z})^*$ generated by $a$:

  1. $C$ is a subgroup of $(\mathbf{Z}/p\mathbf{Z})^*$ with $4<\#C<p-1$
  2. writing $D_1,D_2,\dots,D_r$ for the distinct cosets of $C$ in $(\mathbf{Z}/p\mathbf{Z})^*$, the number of representations of $1$ as a sum $d_1+d_2+\dots+d_r$ with $d_i\in D_i$ differs by $1$ from the number of representations of $0$ as such a sum.

The reason for this is that the constant term of your minimal polynomial is (up to multiplication by $\pm 1$) the norm of your element $x$ from $\mathbf{Q}(\zeta_p)$ to $\mathbf{Q}$. This norm is the product of the conjugates of $x$, and we can write down these conjugates. Let $C$ be the subgroup of $(\mathbf{Z}/p\mathbf{Z})^*$ generated by your element $a$, and let $b_1,\dots,b_r$ (with $r:=(p-1)/h$) be representatives of the distinct cosets of $C$ in $(\mathbf{Z}/p\mathbf{Z})^*$. Then the conjugates of $x$ are the elements $$ \sum_{k=1}^h \zeta_p^{b_i a^k}, $$ and your question asks when the product of these $(p-1)/h$ numbers is $\pm 1$. Equivalently, $$ \pm 1 = \prod_{i=1}^r \sum_{c\in C} \zeta_p^{b_i c} = \sum_{c_1,\dots,c_r\in C} \zeta_p^{\sum_{i=1}^r b_i c_i}. $$ This expression is a $\mathbf{Z}$-linear dependence on the $p$-th roots of unity, so it must have the form $n\cdot 1 + n\cdot\zeta_p+\dots+n\cdot\zeta_p^{p-1}=0$ for some integer $n$. Thus, the collection of sums $\sum_{i=1}^r b_i c_i$ with $c_i\in C$ must consist of $n$ copies of each nonzero element of $\mathbf{Z}/p\mathbf{Z}$, together with either $n-1$ or $n+1$ copies of zero. Finally, any two nonzero elements $u,v$ of $\mathbf{Z}/p\mathbf{Z}$ always have equal numbers of representations as $\sum_{i=1}^r b_i c_i$ with $c_i\in C$, since we get a bijection between the two sets of representations by multiplying all representations of $u$ by $v/u$. Thus your condition is the same as asserting that the numbers of representations of $1$ and $0$ differ by $1$, which is the claimed reformulation.

In view of this reformulation, you might want to add a tag to your question in order to alert the additive combinatorialists, since it seems related to sum-product phenomena.

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One consequence of my reformulation is that if your condition holds then $h^{(p-1)/h}\equiv\pm 1\pmod{p}$. –  Michael Zieve Oct 27 '13 at 0:41
    
This is very close to considerations in my thesis, which I tried to pass off as Combinatorics, publishing A Combinatorial Problem in Finite Fields I and II. But I didn't fool anybody, everyone knew it was actually Number Theory. Unfortunately, I don't think there's anything in my work that contributes much to the question at hand. –  Gerry Myerson Oct 27 '13 at 5:58

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