# How to find all polynomials that follows the equation $P(P'(x)) = P'(P(x))$

I want to find all polynomials that has the following property

$$P(P'(x)) = P'(P(x))$$

where P(x) is a polynomial.

Can you please tell me a way to solve this problem?

• I don't think it can be monic. Look at the leading term of the composition. – open problem Jan 24 at 3:15
• $P(x)=x$ works, as does $P(x)=0$ – Rhys Hughes Jan 24 at 3:17
• Ah yes exceptions for 1st and 0th order. I should have been more clear. – open problem Jan 24 at 3:20

Let $$ax^{n}$$ be the leading coefficient, then the leading coefficient of both sides of the equation are equal giving:

$$a(anx^{n-1})^{n} = an(ax^{n})^{n-1}$$

So $$a^{n+1}n^{n} = a^{n}n$$

So $$a= \frac{1}{n^{n-1}}$$

So one such family is $$\frac{x^{n}}{n^{n-1}}$$

Now consider the constant coefficient K and the linear coefficient L:

$$P'(K) = P(L)$$

So if $$P(x) = Lx+K$$ is linear then, LK=K

So linear functions of the form

$$P(x) = x+k$$

Work.

$$P(x) = ax^{2}+Lx+K$$

Has $$a=\frac{1}{2}$$ and $$\frac{L^{2}}{2}+L^{2} + K = K + L$$

So K is free, and Either L=0 or $$L = \frac{2}{3}$$.

$$L = \frac{2}{3}$$ doesn't work when plugging back into $$P'(P(x)) = P(P'(x))$$

So we get $$\frac{x^{2}}{2}+k$$.

At third order and higher P and P' are both non-linear which forces K to be a solution of a polynomial of order higher than 2 and so K can only take on so many values. It is likely that the first family given will describe all of the remaining polynomials. We will demonstrate for the third order polynomials.

$$\frac{x^{3}}{9} + bx^{2} + Lx + K$$

Looking at 5th order terms:

$$\frac{b}{27} = \frac{2b}{81}$$

Which forces $$b=0$$.

Then looking at second order terms:

$$\frac{L^{2}}{3} = 0$$

which forces $$L=0$$.

$$\frac{K^{2}}{3} + 2bK + L = \frac{L^{3}}{9} + bL^{2} + L^{2} + K$$

So $$\frac{K^{2}}{3} - K = 0$$.

So K=3 or K=0.

$$\frac{1}{3}(\frac{x^{3}}{9} + K )^2 = \frac{1}{9}(\frac{x^{2}}{3})^3 + K$$

Looking at the 3rd order term forces K=0.

So $$\frac{x^{3}}{9}$$ is the only 3rd order solution.