Discriminant of numerator of inverse logarithmic derivative operator iteration Let $T:\mathbb Q(x)\to \mathbb Q(x)$ be the operator of inverse logarithmic derivative, i.e. $$Tf=\frac{f}{f'}.$$ Define $$p_n(x)=T^n\left(x-\frac{x^2}{2}\right).$$ Let $f_n(x) \in \mathbb Z[x]$ be the numerator of $p_n(x)$, taken with positive leading term. The sequence $D_n$ of discriminants of our polynomials starts with
$$\begin{aligned}
D_0&=4,\\ D_1&=4,\\ D_2&=4,\\ D_3&=-256=2^8,\\ D_4&=2^{32}5^2,\\ D_5&=2^{128}5^6331^2, \\D_6&=2^{512}5^{20}331^646599695357^2.\\
\end{aligned}
$$
Is it true that we always have $D_n=\pm a_n^2$ for some integer $a_n$? What else can be said about the arithmetic properties of $a_n$? For example, is it true that for all $n \geq 2$ we have $\nu_2(D_n)=2^{2n-3}$?
 A: First, let's make the substitution $x\to 1-x$, so the function $x-\frac12x^2$
becomes $\frac12(1-x^2)$. The key here is that it is an even function, and the
discriminant of even and odd polynomials are essentially squares. More precisely,
$$\begin{aligned}
  \text{Disc}\bigl( A(x^2) \bigr) &= \pm\text{Resultant}\bigl(A(x^2),2xA'(x^2)\bigr)\\
  &= \pm 2^{\deg(A)} \cdot A(0)\cdot \text{Disc}\bigl( A(x) \bigr)^2
\end{aligned}
$$
and
$$\begin{aligned}
  \text{Disc}\bigl( xA(x^2) \bigr) &= \pm\text{Resultant}\bigl(xA(x^2),A(x^2)+2x^2A'(x^2)\bigr)\\
   &= \pm A(0)\text{Resultant}\bigl(A(x^2),2x^2A'(x^2)\bigr)\\
   &= \pm 2^{\deg(A)} \cdot A(0)^3\cdot \text{Disc}\bigl( A(x) \bigr)^2.\\
\end{aligned}
$$
Next, define three operators on the ring of rational functions:
$$ T(f)=\frac{f}{f'},\quad D(f)=\frac{f'}{f},\quad I(f)=\frac{1}{f}. $$
The question asks about iterates of $T$, but note that $T=I\circ D$ and
$D\circ I=-D$, so
$$ T^n(f) = (I\circ D)^n(f) = I\circ(D\circ I)^{n-1}\circ D(f) = (-1)^{n-1}I\circ D^n. $$
So up to sign, the numerator and denominator of $T^n(f)$ are the reverse of those for $D^n(f)$,
so it's enough to look at the logarithmic derivative $D$. Finally, we note that
if we start with an even function, then $Df$ is odd, and likewise if we start with an odd function,
then $Df$ is odd. Hence if $f$ is an even (or odd) rational function, then for all $n\ge1$ we
have $D^nf=A_n/B_n$ with one of $A_n$ and $B_n$ odd and the other even.
From above, the absolute values of the discriminants of $A_n$ and $B_n$ are squares,
up to the power of $2$ and the $A(0)$ term. But it's easy to check that for the particular
function $\frac12(1-x^2)$, the power of $2$ will be even and the $A(0)$ term will be $\pm1$.
