Denote $\pmb{X}_n=(x_1,x_2,\dots,x_n)$. Consider the 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, 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 $$\sum_{\mu}c_{\mu,2n}=(-1)^{\binom{n}2}?$$

  • $\begingroup$ The proofs given so far are great. Now I'd like to point out that $f_n(\boldsymbol{X}_n)$ is the generating function for simple graphs by degree sequence. For example, with $n=4$, the coefficient of $x_1x_2^3x_3^2x_4^2$ is the number of graphs with degree sequence $1,3,2,2$. It would be fun to prove the two claims by an argument based on graph manipulation, but I don't know how to do it. $\endgroup$ Mar 4, 2019 at 1:11
  • $\begingroup$ That is a very interesting idea. $\endgroup$ Mar 4, 2019 at 1:25
  • $\begingroup$ @BrendanMcKay I see how it is a generating function of out-degrees of tournaments. Is it what you mean? $\endgroup$ Mar 4, 2019 at 22:22
  • $\begingroup$ @FedorPetrov, ooops you are correct. My error is all the more embarrassing given that I'm working on a paper about oriented graphs (with coauthors) that uses a generalisation of this. Thanks. $\endgroup$ Mar 4, 2019 at 23:54

3 Answers 3


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 $(-1)^{[k/2]}=(-1)^{k(k-1)/2}$.

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    $\begingroup$ An alternative to your "standard trick" is to observe that $w_i + w_j = \dfrac{w_i^2 - w_j^2}{w_i - w_j}$. This yields $\prod\limits_{i<j} \left(w_i+w_j\right) = \dfrac{\prod\limits_{i<j}\left(w_i^2 - w_j^2\right)}{\prod\limits_{i<j}\left(w_i-w_j\right)}$. But the $n$ numbers $-w_1^2, -w_2^2, \ldots, -w_n^2$ are just a permutation of the $n$ numbers $w_1, w_2, \ldots, w_n$, and thus $\dfrac{\prod\limits_{i<j}\left(w_i^2 - w_j^2\right)}{\prod\limits_{i<j}\left(w_i-w_j\right)}$ equals a power of $-1$ times the sign of this permutation. Both are easy to compute. $\endgroup$ Mar 2, 2019 at 21:26
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    $\begingroup$ Yes, this is another standard trick:) Actually possibly the shortest proof is to combine them: the absolute values equals 1 since the differences $w_i^2-w_j^2$ and $w_i-w_j$ are the same up to sign, and the sign may be obtained by looking at the argument. $\endgroup$ Mar 2, 2019 at 21:38

Here is another method using more of the theory of symmetric functions. By Enumerative Combinatorics, vol. 2, Exercise 7.30, we have $f_n(\boldsymbol{X}_n)= s_{(n-1,n-2,\dots,1)}(\boldsymbol{X}_n)$ (a Schur function). By the dual Jacobi-Trudi identity, $$ s_{(n-1,n-2,\dots,1)} = \det[e_{n-2i+j}]_{i,j=1}^{n-1},\ \ (*) $$ where $e_0=1$ and $e_k=0$ for $k<0$. Since $e_k(\boldsymbol{X}_n)=0$ for $k>n$, it follows that $\sum_\mu c_{\mu,n}$ is obtained by substituting $e_1=e_2=\cdots=e_n=1$ and $e_{n+1}=e_{n+2}=\cdots=0$ into the right-hand side of (*). When $n$ is odd, the two middle rows of the determinant are equal (in fact, they are all 1's), so the determinant is 0. If $n=2m$ then subtract row $m-1$ from row $m$, then row $m-2$ from row $m-1$, up to row 1 from row 2. Also subtract row $m+2$ from row $m+1$, row $m+3$ from $m+2$, etc. The resulting matrix $A$ can be transformed into a triangular matrix $B$ with 1's on the diagonal by row and column permutations. The permutation indexing the 1's in $A$ that become the diagonal elements of $B$ is $1,3,5,\dots,n-1,2,4,6,\dots,n-2$, which has ${m\choose 2}$ inversions, and the proof follows.

  • $\begingroup$ This is a really cool alternative proof. Thanks, Richard! $\endgroup$ Mar 3, 2019 at 22:51
  • $\begingroup$ It looks interesting that again we get that the answer is the sign of the permutation $x\to 2x+1$ of residues modulo $n$, before establishing the explicit value $(-1)^{m\choose 2}$. $\endgroup$ Mar 4, 2019 at 8:56

The polynomial in question is an instance of the Boolean product polynomials, which might give some extra insight. For example, I believe Lascoux have studied the Schur expansion of that exact expression.

  • $\begingroup$ Thanks for the pointers. $\endgroup$ Mar 4, 2019 at 22:50

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