On the polynomial $\sum_{k=0}^n\binom{n}{k}(-1)^kX^{k(n-k)}$ 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.

Here is a proof for the positivity of coefficients. Suppose that 
$$F_{2m}\,(x)=2(1-x)^m\sum_ka_{m,k}\,x^k,$$
where $a_{m,k}=0$ if $k<0$ or $k>m^2-m$.
We shall show that $a_{m,k}>0$ for all other $k$.
First of all, a negative sign was missed in Joe Silverman's deduction from the third last step. The correct formula is 
$$F_n'(x)=-n(n-1)x^{n-2}F_{n-2}\,(x).$$
Using this, the derivative of $F_{2m}(x)$ is 
\begin{align}
F_{2m}'\,(x)
&=-2m(2m-1)x^{2m-2}F_{2m-2}\,(x)\\
&=-2(1-x)^{m-1}\sum_{k}2m(2m-1)a_{m-1,\,k-2m+2}\ \ x^{k}.
\end{align}
The derivative can be computed alternatively as
\begin{align}
F_{2m}'(x)
&=-2m(1-x)^{m-1}\sum_{k}a_{m,k}x^k+2(1-x)^m\sum_{k}a_{m,k}\ k\,x^{k-1}\\
&=2(1-x)^{m-1}\sum_{k}(\,(k+1)a_{m,\,k+1}-(m+k)a_{m,k}\,)\ x^k.
\end{align}
Comparing the coefficient of $x^k$ in the sums of the above two formulas we find
$$a_{m,k}=\frac{2m(2m-1)a_{m-1,\,k-2m+2}\,+(k+1)a_{m,\,k+1}}{m+k},\quad \text{for $m\ge1$}.$$
The initial value to use this recurrence for each $m$ is the leading coefficient of the sum part in $F_{2m}(x)$, that is, 
$$a_{m,\,m^2-m}=\frac{1}{2}\binom{2m}{m}=\binom{2m-1}{m},\quad\text{for $m\ge1$}.$$
which is a positive integer. The desired positivity follows from the recurrence immediately by induction on $m$.
For instance, by the initial value formula, we have $a_{1,0}=1>0$. That is all we need to show for $m=1$. For $m=2$, we have $a_{2,2}=3$. Using the recurrence formula for $m=2$, we obtain 
$$
a_{2,1}=\frac{12a_{1,-1}+2a_{2,2}}{3}=2\quad\text{and}\quad
a_{2,0}=\frac{12a_{1,-2}+a_{2,1}}{2}=1.$$
For $m=3$, the initial value formula gives $a_{3,6}=10$; the recurrence formula allows us to compute $a_{3,k}$ for $k=5,4,3,2,1,0$ one by one:
$$a_{3,5}=\frac{30a_{2,1}+6a_{3,6}}{8}=15,\quad a_{3,4}=\dots.$$
