In a recent post of mine in this site, that's MSE 4244256, I've tried to state claims and conjectures in an attempt to relate my ideas and the called Casas-Alvero conjecture. What I'm trying is to get an equivalent formulation of the conjecture due to Casas-Alvero in terms of derivatives of polynomials. Again in this post I evoke the same strategy. Wikipedia has the article Casas-Alvero conjecture.
Throughout this post $P(x)$ is a polynomial of degree $n>1$ defined over a field $K$ of characteristic zero, and we denete its derivatives as $P^{(i)}(x)$ with $P^{(0)}(x)=P(x)$. One can to prove easily the following statement.
Claim. If $P(x)$ has the form $P(x)=a_n(x-\alpha)^n$, for $n>1$ ($a_n=\frac{P^{(n)}(0)}{n!}$) and $\alpha\in K$, then our polynomial satisfies $$\frac{\frac{d}{dx}\left(P(x)P^{(l)}(x)\right)}{P^{(l+1)}(x)}=\frac{2n-l}{n-l}\cdot P(x)\tag{1}$$ for each integer $l$ over the interval $0\leq l<n-1$.
Remark. Additionally and in relation with my cited post I add the following if can be inspiring here: if our polynomial in previous claim satisfies $(1)$ and $P(x)=a_n\left(\frac{n-l}{\frac{d}{dx}\log P^{(l)}(x)}\right)^{n}$ for same indexes, also $P'(x)=n\cdot a_n\left(\frac{n-l}{\frac{d}{dx}\log P^{(l)}(x)}\right)^{n-1}$ holds.
Conjecture 1. Let $P(x)$ be a polynomial of degree $n>1$ satisfisfying $(1)$ for each integer $0\leq l<n-1$. Then $P(x)$ has the form $\frac{P^{(n)}(0)}{n!}(x-\alpha)^n$ for some element $\alpha\in K$.
Conjecture 2. Casas-Alvero conjecture is equivalent to previous Conjecture 1.
Question. I would like to know if it is possible to prove or refute previous Conjecture 1 and Conjecture 2. Many thanks.
I don't know if these equations and conjectures are in the literature in a more or less explicit way, in such case please add a comment or the reference. Feel free to comment if the mathematical content of the conjectures can be improved to be more interesting.
References:
[1] Eduardo Casas-Alvero, Higher order polar germs, J. Algebra. 240 (1), (2001) pp. 326–337.
I prove conjecture 1, assuming only that equation (1) holds for $\ell=1$.
Since $(PP^{(\ell)})' = P'P^{(\ell)}+PP^{(\ell+1)}$, equation (1), namely $$\frac{(PP^{(\ell)})'}{P^{(\ell+1)}} = \frac{2n-\ell}{n-\ell}P$$ is equivalent to $$\frac{P'P^{(\ell)}}{P^{(\ell+1)}} = \frac{n}{n-\ell}P.$$
Assume that it holds for $\ell=1$. Then $(n-1)P'^2 = nPP''$, i.e.
$$(n-1)\frac{P'}{P} = n \frac{P''}{P'}.$$ Thus \begin{eqnarray*} \Big(\frac{(P')^n}{P^{n-1}}\Big)' &=& \frac{n(P')^{n-1}P'' \times P^{n-1} - (P')^n \times (n-1)P^{n-2}P'}{P^{2n-2}} \\ &=& \frac{(P')^{n-1} \times P^{n-2} \times \big(nPP''- (n-1)(P')^2\big)}{P^{2n-2}} \\ &=& 0. \end{eqnarray*} Therefore, $(P')^n = cP^{n-1}$ for some $c \in \mathbf{K}^*$.
Let $\alpha$ be a root of $P$ (in some extension of $\mathbf{K}$), call $m$ its multiplicity. The equality $(P')^n = cP^{n-1}$ shows that $\alpha$ is also a root of $P'$, with multiplicity $m-1$ since $P^{(k)}(\alpha) = 0$ for all $k \in [0,m-1]$ but $P^{(m)}(\alpha) \ne 0$ (here, we use the assumption that the characteristic of $\mathbf{K}$ is $0$). Hence $n(m-1)=(n-1)m$, so $m=n$.
Since $P$ has degree $n$, we derive that $P=a_n(X-\alpha)^n$ with $\alpha \in \mathbf{K}^*$. Last, $\alpha \in \mathbf{K}$ since the coefficient of $X^{n-1}$ in $P$ is $-n\alpha \in \mathbf{K}$. We are done.
Nota. I used the fact that a rational fraction in $\mathbf{K}(C)$ with null derivative is constant. Indeed, assume that $A$ and $B$ are relative prime polynomials and $(A/B)'=0$. Then $A'B-AB'=0$, so $B$ divides $AB'$. Since $A$ and $B$ are relative prime, $B$ divides $B'$. Because of the degrees, the only possibility is that $B$ is constant. In the same way, $A$ is constant. We used the null characacteristic of $\mathbf{K}$ again to ensure that the derivative of any polynomial with degree $d \ge 1$ has degree $d-1$.