Let $n = 2m$ be an even integer and let $F_n(X)$ be the polynomial $$F_n(X):=\sum_{k=0}^n\binom{n}{k}(-1)^kX^{k(n-k)}.$$ I observed (but cannot prove) that the polynomial $F_n$ is always divisible by $(1-X)^m$, and the quotient $F_n(X)/(1-X)^m$ always has positive coefficients.
Similarly, with $m, n$ as above, let $G_n(X)$ be the polynomial $$G_n(X):=\sum_{k=0}^n\binom{n}{k}(-1)^k X^{(k+1)(n-k)}.$$ Since all the exponents of $X$ are even, the polynomial $G_n$ is essentially a polynomial of $X^2$.
Now it turns out (still without proof) that $G_n$ is always divisible by $(1-X^2)^m$, and the quotient $G(X)/(1-X^2)^m$ always have positive coefficients.
Are these polynomials well known? Could someone give a proof of any of the assertions?
My ideas
I spent some time on this, trying to attack by establishing recurrence relations among them, as well as looking at the generating functions (i.e. $\sum F_nT^n$).
I also noted that they are both special values of the polynomial $$H_n(X,Y):=\sum_{k=0}^n\binom{n}{k}(-1)^kX^{k(n-k)}Y^k.$$
However, all my usual tools eventually failed, because of the strange exponent $k(n-k)$ (which is morally $k^2$). I only have seen this kind of exponent in Jacobi triple product before, but that is also not quite relevent…
To give some examples:
\begin{eqnarray*}
F_2(X) &=& 2(1-X)\\
F_4(X) &=& 2(1-X)^2(1 + 2X + 3X^2)\\
F_6(X) &=& 2(1-X)^3(1 + 3X + 6X^2 + 10X^3 + 15X^4 + 15X^5 + 10x^6)\\
F_8(X) &=& 2(1-X)^4(1 + 4X + 10X^2 + 20X^3 + 35X^4 + 56X^5 + 84X^6 + 112X^7 + 133X^8 + 140X^9+126X^{10}+84X^{11}+35X^{12})\\
\\
G_2(X) &=& (1-X^2)\\
G_4(X) &=& (1-X^2)^2(1+2X^2)\\
G_6(X) &=& (1-X^2)^3(1+3X^2+6X^4+5X^6)\\
G_8(X) &=& (1-X^2)^4(1+4X^2+10X^4+20X^6+28X^8+28X^{10}+14X^{12})
\end{eqnarray*}
The examples are calculated using sagecell.
UPDATE
Using the method of Joe Silverman, I am able to prove the following: $$ F_{2m}(X)=2(1-X)^m(a_{m,0} + a_{m,1}X + a_{m,2}X^2 + \cdots),$$ where the coefficients $a_{m,k}$ are given by: $$a_{m,k} = \frac{1}{2}\sum_{j = 0}^{\infty}(-1)^j\binom{2m}{j}\binom{m-1+k-2mj+j^2}{m-1}.$$
Here the convention is that $\binom{z}{y} = 0$ for all integers $z < y$.
A remark: since the degree of $F_{2m}$ is equal to $m^2$, we must have $a_{m,k} = 0$ for all $k > m^2-m$. This, however, is not obvious from the above explicit formula…
For $k \leq m^2-m$, the above formula can also be written as $$a_{m,k} = \sum_{j = 0}^{m}(-1)^j\binom{2m}{j}\binom{m-1+k-2mj+j^2}{m-1}.$$
Now the problem is to show that all the $a_{m,k}$ are positive. According to the comment of Zach Teitler, we have the following guess:
For fixed $m$, the function $k\mapsto a_{m,k}$ is the Hilbert function of an ideal of $k[\mathbb{P}^{m-1}]$ generated by $2m$ general degree $2m-1$ forms.
The positivity of all $a_{m,k}$ would follow from this, since the Hilbert functions, being dimensions, are positive.
Best Answer
It is clear that $F_n(1)=0$. On the other hand, the derivative can be expressed in terms of $F_{n-2}$. Thus $$ \begin{aligned} F_n'(X) &=\sum_{k=0}^n\binom{n}{k}(-1)^k(k(n-k))X^{k(n-k)-1} \\ &= \sum_{k=1}^{n-1} \frac{n!}{(k-1)!(n-k-1)!} (-1)^k X^{k(n-k)-1} \\ &= n(n-1) \sum_{k=1}^{n-1} \frac{(n-2)!}{(k-1)!(n-k-1)!} (-1)^k X^{k(n-k)-1} \\ &= n(n-1) \sum_{k=1}^{n-1} \binom{n-2}{k-1}(-1)^k X^{k(n-k)-1} \\ &= n(n-1) \sum_{k=0}^{n-2} \binom{n-2}{k}(-1)^k X^{(k+1)(n-k-1)-1} \\ &= n(n-1) \sum_{k=0}^{n-2} \binom{n-2}{k}(-1)^k X^{k(n-2-k)+n-2} \\ &= n(n-1)X^{n-2}F_{n-2}(X). \\ \end{aligned} $$ Repeating, one find that the $\ell$'th derivative of $F_n(X)$ is a $\mathbb Z[X]$ linear combination of $F_{n-2}(X),\ldots,F_{n-2\ell}(X)$. So it vanishes at $X=1$ until you get to $\ell=n/2$. Possibly one can get the positivity of the coefficients from this recursion, too, but I haven't tried.