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Consider the equation $P\circ P'=P'\circ P$ for polynomials (real polynomials of one variable).

For any degree of $P$ it has a solution: $P_n(x)=\frac{1}{n^{n-1}}x^n$.

For degree 2 it gives $\frac{1}{2}x^2$, while all solutions are $\frac{1}{2}x^2+c$.

For degree 3 it gives $\frac{1}{9}x^3$, while there is one more solution $\frac{1}{9}x^3-x^2+9$

For $n=4$ and $n=5$ I calculated $P\circ P'-P'\circ P$ in Mathematica and by elliminating the highest powers of this difference I found that all coefficients without the leading one must be equal 0, hence $\frac{1}{n^{n-1}}x^n$ is the only solution.

Is it true that $P\circ P'=P'\circ P$ has only one solution for $\deg P>3$ ?

I googled "polynomials commuting with their derivatives" but without finding the answer.

larry01
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    This answer references the result that "polynomials that commute under composition must be, up to a linear homeomorphism, either both powers of x, both iterates of the same polynomial, or both Chebychev polynomials". The first case can be shown to reduce here to $\frac{1}{n^{n-1}}x^n$. If it's proved that for $\deg P \ge 4$ it is not possible for $P,P'$ to be either linear mappings of Chebyshev polynomials, or iterates of the same polynomial, then uniqueness would follow. – dxiv May 23 '21 at 19:34

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As dxiv points out in the comments, given this result by Ritt it suffices to restrict the analysis to polynomials of a specific form. In particular, if a pair of polynomials $(P, Q)$ of respective degrees $n, m>1$ with $\gcd(n,m)=1$ satisfy $P \circ Q = Q \circ P$, they must be conjugate to $(E_n, E_m)$ or to $(T_n, T_m)$ under a linear homeomorphism, where $E_n(x)=x^n$ and $T_n(x)$ is the $n$-th Chebyshev polynomial of the first kind (the condition on the degrees precludes $P$ and $Q$ being iterates of the same polynomial).

Say $P$ has degree $n>2$ (the case $n\le 2$ is easy to solve); then its derivative $P'$ has degree $n-1$, and $\gcd(n,n-1)=1$, so we are in the previous case. Let $X$ denote either $E$ or $T$; by the previous paragraph we can find a solution whenever $X'_n$ is equal to $\alpha X_{n-1} + \beta$ for some $\alpha\neq 0$ and $\beta$. If this holds, $P(x)=\alpha X_n\left( \frac{x-\beta}{\alpha} \right) + \beta$ can be checked to satisfy $P \circ P' = P' \circ P$.

This happens in two cases:

  • For any $n$, $E'_n = n E_{n-1}$. Hence, the polynomial $n E_n(\frac{x}{n}) = \frac{1}{n^{n-1}}x^n$ commutes with its derivative.

  • For $n=3$, $T'_3 = 6 T_2 + 3$. Hence, the polynomial $6 T_3 \left(\frac{x}{6}-\frac12 \right) +3 = \frac19 x^3-x^2+9$ again satisfies the desired condition.

We can rule out $T_n$ for $n>3$ using the identity $T'_n=n U_{n-1}$, where $U_n$ denotes the $n$-th Chebyshev polynomial of the second kind. The coefficients of $T_n$ and $U_n$ are known explicitly:

$$T_n(x) = 2^{n-1} x^n - n 2^{n-3} x^{n-2} + \ldots$$

$$U_n(x) = 2^n x^n - (n-1) 2^{n-2} x^{n-2} + \ldots$$

By expanding the equation

$$\alpha T_{n-1}(x)+\beta= n U_{n-1}(x)$$

and looking at the coefficients of the two highest powers of $x$, one obtains the two inconsistent equations $\alpha = 2(n+1)$ and $\alpha = 2(n^2-1)/n$.

For completeness, we list the solutions of degree $n\le 2$:

  • The polynomial $\frac12 x^2 + c$ is a solution for any $c$.
  • The polynomial $ax + a - a^2$ is a solution for any $a$ (this case is missing from the question. For $a=0$ this gives the constant solution $P(x)=0$).
pregunton
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