Denote $\pmb{X}\_n=(x\_1,x\_2,\dots,x\_n)$. Consider the [symmetric polynomial](https://en.wikipedia.org/wiki/Symmetric_polynomial)
$$f\_n(\pmb X\_n)=\prod\_{1\leq i<j\leq n}(x\_i+x\_j).$$
Expand these in terms of [elementary symmetric polynomials](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial), say
$$f\_n(\pmb{X}\_n)=\sum\_{\mu}c\_{\mu,n}\cdot e\_{\mu}(\pmb{X}\_n).$$


For example,
\begin{align\*} f\_3&=-e\_{(3)}+e\_{(2,1)} \\
f\_5&=-e\_{(5,5)}+2e\_{(5,4,1)}+e\_{(5,3,2)}-e\_{(5,2,2,1)}-e\_{(4,4,1,1)}-e\_{(4,3,3)}+e\_{(4,3,2,1)}.
\end{align\*}



> 
> **QUESTION 1.** Is it true that, for integers $n \geq 1$, we have
>  $$\sum\_{\mu}c\_{\mu,2n+1}=0?$$
> 
> 
> 


**POSTSCRIPT.** Fedor's reply (to Question 1) shown below suggests to me to ask:



> 
> **QUESTION 2.** Is it true that, for integers $n \geq 1$, we have
>  $$\sum\_{\mu}c\_{\mu,2n}=(-1)^{\binom{n}2}?$$
> 
> 
> 



The vanishing of sum of coefficients: symmetric polynomials
Choose $n$ numbers $x\_1,\dots,x\_n$ for which all elementary symmetric polynomials are equal to 1 and substitute them to our $f\_n$. We should get zero value for odd $n$. Well, what are these numbers? The roots of $x^{n}-x^{n-1}+x^{n-2}-\ldots-1=(x^{n+1}-1)/(x+1)$. This polynomial indeed has two roots with sum equal to 0 when $n$ is odd.


If $n=2k$ is even, we substitute the roots $w\_1,\dots,w\_n$ of the polynomial $f(x)=x^{2k}-x^{2k-1}+\ldots+1=(x^{2k+1}+1)/(x+1)=(x-w\_1)\dots (x-w\_n)$. Then your claim reads as $$A:=\prod\_{1\leqslant i<j\leqslant n} (w\_i+w\_j)=(-1)^{\binom{k}2}.$$ This is done by the standard trick (and is well known itself). At first, 
$$
|A|^2=\prod\_{i=1}^n \prod\_{j\ne i,1\leqslant j\leqslant n}|w\_i+w\_j|=2^{-n}\prod\_{i=1}^n \prod\_{j=1}^n|(-w\_i)-w\_j|=2^{-n}\prod\_{i=1}^n |f(w\_i)|=\\=2^{-n}\prod\_{i=1}^n\left|\frac{(-w\_i)^{2k+1}+1}{-w\_i+1}\right|=1,
$$
since $1+(-w\_i)^{2k+1}=2$ for all $i=1,2,\dots,n$ and $\prod\_{i=1}^n (1-w\_i)=f(1)=1$.


At second, we need to find the argument of the complex number $A$. This may be done for example as follows: all pairs $w\_i+w\_j$ for which $w\_i$ and $w\_j$ are not complex conjugate are partitioned onto complex conjugate pairs. In each pair the product is positive reals. If $w\_i$ and $w\_j$ are complex conjugate, the sum $w\_i+w\_j$ is a real number whose sign is the sign of the real part of $w\_i$. Therefore $A$ is the real number whose sign equals $(1)^{m/2}$, where $m$ is the number of $w$'s in the left half-plane. It is easy to see that $m/2=[k/2]$ and that $(-1)^{[k/2]}=(-1)^{k(k-1)/2}$.


