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For any odd natural number $n$, denote $t \equiv \frac{n-1}{2}$. Let $K$ be a field such that $\operatorname{char} K \neq 2$. Working over the polynomial ring $K\left[x_1,x_2,...,x_{n} \right]$, denote by $\Pi_n$ the product of all possible linear combinations of the indeterminants $x_1,x_2,...,x_{n}$, where each linear combination has $i$ coefficients which are $-1$ and its remaining $n-i$ coefficients are $1$, and $0 \le i \le t$.

Problem

Find an expression for the coefficient of the monomial $\prod_{j=1}^{n}x_{j}^{p_j}$ in the expansion of the aforementioned product $\Pi_n$.

Examples of $\Pi_n$

  • For $n=1$ obtain that $t=0$ and that $i \in \{0\}$; then $\Pi_1$ is $$x_1$$
  • For $n=3$ obtain that $t=1$ and that $i \in \{0,1\}$; then $\Pi_3$ is $$\left(x_1+x_2+x_3 \right)\left(-x_1+x_2+x_3 \right)\left(x_1-x_2+x_3 \right)\left(x_1+x_2-x_3 \right)$$
  • For $n=5$ obtain that $t=2$ and that $i \in \{0,1,2\}$; then $\Pi_5$ is $$ \begin{align} & \left(x_1+x_2+x_3+x_4+x_5 \right)\\ & \left(-x_1+x_2+x_3+x_4+x_5 \right)\left(x_1-x_2+x_3+x_4+x_5 \right)\left(x_1+x_2-x_3+x_4+x_5 \right)\\& \left(x_1+x_2+x_3-x_4+x_5 \right)\left(x_1+x_2+x_3+x_4-x_5 \right)\\ &\left(-x_1-x_2+x_3+x_4+x_5 \right)\left(-x_1+x_2-x_3+x_4+x_5 \right)\left(-x_1+x_2+x_3-x_4+x_5 \right)\\& \left(-x_1+x_2+x_3+x_4-x_5 \right)\left(x_1-x_2-x_3+x_4+x_5 \right)\left(x_1-x_2+x_3-x_4+x_5 \right)\\& \left(x_1-x_2+x_3+x_4-x_5 \right)\left(x_1+x_2-x_3-x_4+x_5 \right)\left(x_1+x_2-x_3+x_4-x_5 \right)\\& \left(x_1+x_2+x_3-x_4-x_5 \right)\\ \end{align} $$

Quick observations

  • In general, $\Pi_n$ has $\binom{n}{i}=\binom{2t+1}{i}$ factors which correspond to each $0 \le i \le t$; hence the total number of factors is $\sum_{i=0}^{\frac{n-1}{2}} \binom{n}{i}=\sum_{i=0}^{t} \binom{2t+1}{i}=2^{2t+1-1}=2^{2t}=4^{t}$.
  • Any one of the indeterminants $x_1,x_2,...,x_{n}$ has coeffient $-1$ in the linear combinations appearing in exactly $\sum_{j=0}^{\frac{n-1}{2}-1} \binom{n-1}{j}=\sum_{j=0}^{t-1} \binom{2t}{j}=2^{2t-1}-\frac{1}{2}\binom{2t}{t}$ of the factors of $\Pi_n$.
  • Any $k$ of the indeterminants $x_1,x_2,...,x_{n}$ all have coeffient $-1$ in the linear combinations appearing in exactly $\sum_{j=0}^{\frac{n-1}{2}-k} \binom{n-k}{j}=\sum_{j=0}^{t-k} \binom{2t+1-k}{j}$ of the factors of $\Pi_n$.

It seems that Inclusion-Exclusion is in order to solve this problem, as there is a matter of choice of "how many minus signs" each of the indeterminants $x_1,x_2,...,x_{n}$ "gets".

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  • 2
    $\begingroup$ Further observations: $\Pi_n^2 = \prod (\pm x_1 \pm x_2 \pm \cdots)$. The sum of the coefficients can be found by setting all the variables to $1$, giving $\prod_{i=0}^t (n-2i) \binom n i$ $\endgroup$ – Peter Taylor Oct 4 '19 at 6:45

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