18
$\begingroup$

This is a somewhat more explicit version of a question I have recently asked.

Let $p$ be an odd prime, and write $\zeta:=\exp(2\pi i/p)$ (any other primitive $p$th root of unity will do as well). For integer $n\in[0,p]$, the products $$ {\mathcal P}_p(n) := \prod_{\substack{A\subseteq{\mathbb F}_p \\ |A|=n}}\ \sum_{a\in A} \zeta^a $$ are rational integers; can they be found "explicitly"?

It is immediately seen that ${\mathcal P}_p(0)=0$ and ${\mathcal P}_p(1)=1$, and I can show that ${\mathcal P}_p(2)=\left(\frac2p\right)$ (it is not difficult to see that ${\mathcal P}_p(2)=\pm 1$, but to determine the sign is trickier). Also, we have ${\mathcal P}_p(p-n)=(-1)^{\binom{p}{n}}{\mathcal P}_p(n)$, so that ${\mathcal P}_p(p)=0$, ${\mathcal P}_p(p-1)=-1$, and ${\mathcal P}_p(p-2)=\pm 1$.

$\quad$ What are the values of ${\mathcal P}_p(n)$ for $n\in[3,p-3]$?

Some numerical data (thanks to Talmon Silver for the programming):

$\quad {\mathcal P}_5(3)=-1$
$\quad {\mathcal P}_7(3)=-2^7$
$\quad {\mathcal P}_{11}(3)=23^{11}$
$\quad {\mathcal P}_{13}(3)=159^{13}$
$\quad {\mathcal P}_{17}(3)=-24617^{17}$
$\quad {\mathcal P}_{19}(3)=-611009^{19}$
$\quad {\mathcal P}_{23}(3)=1265401351^{23}$

$\quad$ If finding the individual values ${\mathcal P}_p(n)$ is difficult, can we at least find explicitly the product $$ {\mathcal P}_p(1){\mathcal P}_p(2)\dotsb{\mathcal P}_p(p-2){\mathcal P}_p(p-1) = \prod_{\varnothing\ne A\subsetneq{\mathbb F}_p} \sum_{a\in A} \zeta^a \ ?$$

Denoting this product by ${\mathcal P}_p$,

$\quad {\mathcal P}_3=-1$
$\quad {\mathcal P}_5=-1$
$\quad {\mathcal P}_7=-2^{14}$
$\quad {\mathcal P}_{11}=-(3\cdot 23^4 \cdot 67\cdot 89)^{22}$
$\quad {\mathcal P}_{13}=-(3^{12}\cdot 5\cdot 53^6 \cdot 79^4\cdot 131^2 \cdot 157^2 \cdot 313\cdot 547\cdot 599\cdot 911)^{26}$

The problem can be restated in a purely combinatorial way, as hinted to in Ofir's comment below. Write $N:=\binom pn$, let $A_1,\dotsc,A_N$ be all the $n$-element subsets of ${\mathbb F}_p$, and for $z\in{\mathbb F}_p$ denote by $r_n(z)$ the number of representations $z=a_1+\dotsb+ a_N$ with $a_1\in A_1,\dotsc, a_N\in A_N$. We have then ${\mathcal P}_p(n)=\sum_{z\in{\mathbb F}_p}r_n(z)\zeta^z$, and from the fact that ${\mathcal P}_p(n)$ is an integer, it follows that $r_n(z)$ are actually equal to each other for all $z\in{\mathbb F}_p\setminus\{0\}$; as a result, we have, say, ${\mathcal P}_p(n)=r_n(0)-r_n(1)$. On the other hand, $$ r_n(0)+(p-1)r_n(1) = \sum_{z\in{\mathbb F}_p} r_n(z) = |A_1|\dotsb|A_N| = n^{\binom pn}. $$ This yields $$ {\mathcal P}_p(n) = \frac1{p-1} \left( p\,r_n(0)-n^{\binom pn}\right). $$ Thus, the problem boils down to finding $r_n(0)$, the number of all zero-sum $N$-tuples $(a_1,\dotsc,a_N)$ with the components $a_i$ representing each of the $n$-element subsets of ${\mathbb F}_p$.

Improve this question
$\endgroup$
15
  • 3
    $\begingroup$ Two observations: 1. This is, in a way, the "dual" of the coefficients of $x^p-1 = \prod_{a=0}^{p-1} (x- \zeta^a)$, which are (up to sign) $\sum_{ A\subseteq \mathbb{F}_{p}, |A|=n} \prod_{a \in A} \zeta ^a$. 2. The case $n=1$ is related to $N(\zeta)=1$, the case $n=2$ to $N(1+\zeta) = 1$ (for $p \neq 2$). The case $n=-3$ is related to $N((1+\zeta)^p+1)$, which is computable mod $p$ as 1. $\endgroup$
    – Ofir Gorodetsky
    Commented Jan 23, 2015 at 15:01
  • 2
    $\begingroup$ Up to sign, they appear to be powers of primes, at least for p, n small. I notice $2^7$, $3^8$, $11^{10}$, and $19^9$ cropping up. In particular, $\mathcal{P}_p(n)$ appears to be zero or of the form $\pm q^p$ for $q$ apparently prime (or unity). $\endgroup$
    – Steve Huntsman
    Commented Jan 23, 2015 at 16:13
  • 1
    $\begingroup$ I can calculate ${\mathcal P}_p(n) \mod p$ (it's usually 1). Let $A_{1}, \cdots, A_{\binom{p}{n}}$ be the subsets of $\mathbb{F}_{p}$ of size $n$. If we let $X_i$ denote number of solutions to $a_1 + \cdots + a_{\binom{p}{n}} = i \mod p$ for $a_j \in A_j$, we see that ${\mathcal P}_p(n) = \sum_{i=0}^{p-1} X_i \zeta^i$. Since this must be an integer, $X_i$ is constant for $i \neq 0$, and we find ${\mathcal P}_p(n) = X_0 - X_1$. On the other hand, $n^{\binom{p}{n}} = \sum X_i = X_0 + (p-1)X_1 \equiv X_0 - X_1 \mod p$, hence ${\mathcal P}_p(n)\equiv n^{\binom{p}{n}} \mod p$. $\endgroup$
    – Ofir Gorodetsky
    Commented Jan 23, 2015 at 18:10
  • 4
    $\begingroup$ @SteveHuntsman I can explain the $p$'th power phenomena. Let $X=\sum_{a \in A} \zeta^a$ be one of the sums appearing in the product. Let $X_{\Delta} = \sum_{a \in A} \zeta^{a+\Delta} = \zeta^{\Delta} X$ be another sum appearing in the product and corresponding to $A+\Delta$. The sums $\{ X+\Delta\}_{\Delta=0}^{p-1}$ are distinct (assuming $0<n<p$), and their product is a $p$'th power of an element of $\mathbb{Z}[\zeta]$, as $\prod_{\Delta=0}^{p-1} X_{\Delta} = X_{\Delta}^{p} \zeta^{\binom{p}{2}} = X_{\Delta}^{p}$. $\endgroup$
    – Ofir Gorodetsky
    Commented Jan 23, 2015 at 18:41
  • 1
    $\begingroup$ (cont.) Hence, $\mathcal{P}_{p}(n)^{p-1} = N(\mathcal{P}_{p}(n)) = N(a^p) = N(a)^p$ for some $a \in \mathbb{Z}[\zeta]$. As $N(a)$ is an integer, we find $\mathcal{P}_{p}(n)$ is a $p$'th power. $\endgroup$
    – Ofir Gorodetsky
    Commented Jan 23, 2015 at 18:44

1 Answer 1

Reset to default
3
$\begingroup$

Only a partial answer:

I can prove that ${\cal P}_p(3)=\epsilon_p v_p^p$, with $\varepsilon_p^2=1$, for every integer $p\ge 3$ (not necessarily prime). The proof is too technical to present here. The values of $v_p$ are given by $$ v_p^6=\frac{9W_p}{(2^p-(-1)^p)^2(1-(-2)^p)}, $$ where $$ W_p=\prod_\zeta\{(1+\zeta)^p-(-1)^p\}; $$ except for the sign these $W_p$ are the numbers of sequence A096964 in OEIS (Wendt's determinant).

Also $W_p=\det(M+(1-(-1)^p) I)$ where $M$ is the circulant matrix with first line $\binom{p}{k}$ for $0\le k\le p-1$.

The sequence $\varepsilon_p$ appear to have a simple periodic structure, but this I have not proved.

Improve this answer
$\endgroup$
5
  • $\begingroup$ @Seva My formulas are valid for any natural number $p\ge3$. $\endgroup$
    – juan
    Commented Jan 29, 2015 at 22:04
  • $\begingroup$ Since your $v_p$ seems to be defined only up to a 6th root of unity, why do you also need to include $\epsilon_p$ before it? $\endgroup$
    – LSpice
    Commented Oct 8, 2021 at 15:57
  • $\begingroup$ @Seva If my computation are correct, $v_p^6$ is a positive real number for $p$ odd. So I take its positive $6$-root and then only remain a sign. $\endgroup$
    – juan
    Commented Oct 8, 2021 at 17:45
  • $\begingroup$ @Seva I will send you my computation if I find your email. $\endgroup$
    – juan
    Commented Oct 8, 2021 at 17:48
  • $\begingroup$ @Seva I tried the email address in your web page. But my message is not send. I am not sure what is the problem. Perhaps if you send a mail to me I can answer to it. $\endgroup$
    – juan
    Commented Oct 8, 2021 at 18:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .