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Consider $a_{1} = \alpha^{N-n}, a_{2} = \alpha^{N-n+1}, a_{3} = \alpha^{N-n+2}, a_{4} = \alpha^{N-n+3}, \cdots, a_{n} = \alpha^{N-1}$ where $\alpha$ is a complex $N$th root of unity where $N = 2 + (n-1)2^{n+1}$.

Let:

$P_{1}(\alpha) = a_{1}a_{2}a_{3} \cdots a_{n}$

$P_{2}(\alpha) = (a_{1} + a_{2})(a_{1} + a_{3})\cdots(a_{n-1} + a_{n})$

$P_{3}(\alpha) = (a_{1} + a_{2} + a_{3})(a_{1} + a_{2} + a_{4})\cdots(a_{n-2} + a_{n-1} + a_{n})$

$\vdots$

$P_{n}(\alpha) = (a_{1} + a_{2} + a_{3} + \cdots + a_{n})$

Is there an elementary expression for $\Pi(\alpha) := P_{1}(\alpha)P_{2}(\alpha) \cdots P_{n}(\alpha)$?

Or given $\Pi(\alpha)$, can one say anything about $\Pi(\alpha^{r})$ where $2 \le r \le N-1$?

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    $\begingroup$ Do you have some reason to think there is an elementary expression? Does it come out nicely for small values of $n$? Is there any particular reason to be interested in this problem only in the case where $N=2+(n-1)2^{n+1}$? Is there any particular reason to be interested in this problem at all? Have you done any work on it? Do you have any partial results? Shouldn't it be tagged number theory, rather than rings-and-algebras? $\endgroup$
    – Gerry Myerson
    Commented Nov 5, 2010 at 0:38
  • $\begingroup$ $N = 2 + (n-1)2^{n+1}$ comes from degree of product of all degree $n$ polynomials with $0$/$1$ coefficients. I think the product of $P_{i}$ can be thought of product of Monomial_symmetric_polynomials. en.wikipedia.org/wiki/… My belief that there may be relation bw diff prod comes from the fact that the symm poly are related to power symm poly. Also angle of prod seem to have simple relation. I felt the magnitude might also have nice relation. I am only an engineer. There might be connections to alg geom & num theory that I dont see. $\endgroup$
    – Turbo
    Commented Nov 5, 2010 at 0:55
  • $\begingroup$ Also product of any two of the Monomial symmetric polynomial seem to have some relation. See 1.4 of this: mathcircle.berkeley.edu/BMC3/SymPol.pdf $\endgroup$
    – Turbo
    Commented Nov 5, 2010 at 1:01
  • $\begingroup$ One more thing I am almost sure is $\frac{1}{N}\displaystyle \sum_{r=0}^{N-1}\Pi(\alpha^{r}) = 1$, $\frac{1}{N}\displaystyle \sum_{r=0}^{N-1}\Pi(\alpha^{r})\alpha^{r(N-t)} = 0$ $\forall t \in [1,n]$ with $n \ge 1$. $\endgroup$
    – Turbo
    Commented Nov 5, 2010 at 1:45
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    $\begingroup$ Thanks for the clarifications. It sounds like you have done some computations. Why not let people know what you have found, to save us the effort of repeating what you've already done? $\endgroup$
    – Gerry Myerson
    Commented Nov 5, 2010 at 4:41

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