Let $n$ be a positive integer and $S\_n$ be the symmetric group on $\{1,2,\ldots,n\}$. Let $\mathcal{A}\_n$ be the anti-symmetrization operator on $\mathbb{Z}[x\_1,x\_2,\ldots,x\_n]$ such that for any $f(x\_1,x\_2,\ldots,x\_n)\in \mathbb{Z}[x\_1,x\_2,\ldots,x\_n]$,
$$\mathcal{A}\_n(f)=\sum\_{w\in S\_n}\varepsilon(w)f(x\_{w(1)},x\_{w(2)},\ldots,x\_{w(n)}),$$
where $\varepsilon(w)=(-1)^{l(w)}$ is the sign of $w$.


**My question:** 


For any integer $n>1$ with $n(n-1)\equiv 0 \pmod 4$, is it true that we can always choose $\displaystyle \frac{n(n-1)}{4}$ different polynomials
$$x\_{i\_k}+x\_{j\_k},\ 1\leq k\leq \frac{n(n-1)}{4}$$
from the $\displaystyle \frac{n(n-1)}{2}$ polynomials
$$x\_i+x\_j,\ 1\leq i<j\leq n,$$
such that
$$\mathcal{A}\_n\left(\prod\_{1\leq k\leq \frac{n(n-1)}{4}}(x\_{i\_k}+x\_{j\_k})^2\right)\neq0\ ?$$


For example,
\begin{align\*}
n=4,\ \ &\mathcal{A}\_4\left((x\_1+x\_2)^2(x\_2+x\_3)^2(x\_3+x\_4)^2\right)\neq0;\\
n=5,\ \ &\mathcal{A}\_5\left((x\_1+x\_2)^2(x\_2+x\_3)^2(x\_3+x\_4)^2(x\_4+x\_5)^2(x\_5+x\_1)^2\right)\neq0.
\end{align\*}



The anti-symmetrization of a kind of polynomials in $\mathbb{Z}[x_1,x_2,\ldots,x_n]$
Hope that now it works, but please check.


For $n=4k$ or $n=4k+1$ we consider the following graph $H\_n$ on the set $V\_n$, $|V\_n|=n$ with $n(n-1)/4$ edges: $V\_n=\{a\_i,b\_i,c\_i,d\_i,1\leqslant i\leqslant k\}$ for $n=4k$ and $V\_n=V\_{n-1}\cup \{e\}$ for $n=4k+1$; the edges of $H\_n$ are the edges of $H\_{n-4}$ plus $a\_kb\_k,b\_kc\_k,c\_kd\_k$ and all edges from $b\_k,c\_k$ to $V\_{n-4}$.


The graph $H\_n$ has $2^k$ automorphisms generated by simultaneous changes $a\_i\leftrightarrow d\_i$, $b\_i\leftrightarrow c\_i$. It is important that these automorphisms are even permutations of $V\_n$. Note that $H\_n$ has unique directed Hamiltonian path up to automorphisms.


For each $u\in V\_n$ take a variable $x\_u$ and consider the polynomial $F=\prod\_{(u,v)\notin H\_n} (x\_u+x\_v-n)(x\_u+x\_v-(n-1))$. It has degree $n(n-1)/2$. I claim that its antisymmetrization is non-zero. It coincides with antisymmetrization of the polynomial $G=\prod\_{(u,v)\notin H\_n} (x\_u+x\_v)^2$, since the difference of two polynomials has degree less than $n(n-1)/2$, therefore its antisymmetrization equals to 0 (there are no other antisymmetric polynomials of such small degree). 


For proving this we let the variables $\{x\_v\}$ take the values $0,1,\dots,n-1$. Note that when $y\_1,\dots,y\_n$ is a permutation of $0,1,\dots,n-1$, and $F(y\_1,\dots,y\_n)\ne 0$, we get $y\_i+y\_j\ne n$ and $y\_i+y\_j\ne n-1$ for $(i,j)\notin H\_n$, thus all pairs $(i,j)$ for which $y\_i+y\_j\in \{n,n-1\}$ belong to $H\_n$. Such pairs form a Hamiltonian path in $H\_n$. But all Hamiltonian paths of $H\_n$ are obtained one from another by an even automorphism of $H\_n$, thus the corresponding summands are equal and non-zero.


