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I came across this equation as a special case of this question and I wanted to understand the space of polynomial solutions. This special case cannot be treated in a similar away as the other cases, since it does have non-trivial solutions; in particular, it is clear that $P(0)=0$ and therefore $P(-1)=0$ or $P(-1)=-1$. Both branches yield at least one solution, namely

$$P(x)=x(x+1),~P(x)=x$$

I believe these are the only solutions, but I haven't been able to reach a convincing argument.

For example, by assuming that $P(0)=P(-1)=0$, we can infer the decomposition

$$P(x)=x^r(x+1)^sQ(x), \quad Q(0), Q(-1)\neq 0$$

By doing some case work after substituting this into the equation, one can show that $r=s$, but all my other attempts at milking the equation have failed.

EDIT: Per the comments below, it is clear there are more solutions to this equation than the ones stated above. Is it possible to construct a classification of all possible solutions?

Mohsen Shahriari
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K. Grammatikos
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    With $f(x) = x^2+x$, you have $P(f(x)) = f(P(x))$. So you're asking for when do two polynomials commute, which is well understood. For example, $f^2(x) = x^4 + 2x^3 + 2x^2 + x = x(x+1)(x^2+x+1)$ is another solution. Does this answer your question? If so, you can close it as a duplicate. – Calvin Lin Feb 18 '25 at 22:41
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    I would appreciate a complete characterization of the solutions to this equation, which is not trivial to infer from the linked question. Indeed, the solutions seem to be combinations of iterates of $x+1$ with $x^2+x$, but how do I construct all of them? – K. Grammatikos Feb 19 '25 at 14:45

1 Answers1

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This article analyzes the general question of when two polynomials commute under composition, in an elementary way. Consider the following statement in the article, in which $ F $ is assumed to be an algebraically closed field of characteristic $ 0 $ (like the field of complex numbers):

Corollary 2.10. Fix an integer $ k \ge 1 $ and let $ f ( x ) $ be a polynomial in $ F [ x ] $ of degree $ n > 1 $. Then there are at most $ n - 1 $ polynomials of degree $ k $ that commute with $ f $.

In the case of the problem at hand, taking $ f ( x ) = x ^ 2 + x $, the Corollary above implies that for any $ k > 1 $, there is at most one polynomial $ P ( x ) $ of degree $ k $ such that $ P ( x ^ 2 + x ) = P ( x ) ^ 2 + P ( x ) $.

On the other hand, we have the following classic result proved by Ritt and Julia, and quoted in the mentioned article:

Theorem 1.2. Let $ P ( x ) $ and $ Q ( x ) $ be nonlinear, nonconstant polynomials with coefficients in the complex numbers that commute under composition. Then one of the following holds:

  1. $ P $ and $ Q $ are similar, via the same map $ \lambda $, to Chebyshev polynomials;
  2. $ P $ and $ Q $ are similar, via the same map $ \lambda $, to monomials;
  3. There exist positive integers $ m $ and $ n $ such that $ P ^ m = Q ^ n $.

Here, $ P ^ m $ and $ Q ^ n $ denote, respectively, the $ m $-th iterate of $ P $ and the $ n $-th iterate of $ Q $. Also, the notion of similarity is defined as follows: $ P $ and $ Q $ are similar via $ \lambda $ whenever $ \lambda ( x ) = a x + b \in \mathbb C [ x ] $ with $ a \ne 0 $ and $ Q = \lambda ^ { - 1 } \circ P \circ \lambda $. Note that similar polynomials must have the same degree.

Now, first note that $ f ( x ) = x ^ 2 + x $ is neither similar to the quadratic monomial $ Q ( x ) = x ^ 2 $ nor to the quadratic Chebyshev polynomial $ T _ 2 ( x ) = 2 x ^ 2 - 1 $. Also, note that if $ P ( x ) $ is such that $ P ^ m = f ^ n $ for positive integers $ m $ and $ n $, then we must have $ ( \deg P ) ^ m = ( \deg f ) ^ n $, and therefore $ \deg P $ must be a power of $ 2 $. As $ f ^ n $ is a polynomial of degree $ 2 ^ n $ that commutes with $ f $, these are the only polynomials with degree no less than $ 2 $ that commute with $ f $ under composition. Verifying that the constant zero function and the identity function are the only polynomials with degree less than two that commute with $ f $ under composition, the list becomes complete.

Mohsen Shahriari
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