Let $k$ be a positive integer. Find all polynomials with real coefficients which satisfy the equation $P(P(x))=\left(P(x)\right)^k$.

Question

Let $k$ be a positive integer. Find all polynomials with real coefficients which satisfy the equation $$P(P(x))=\left(P(x)\right)^k.$$

I simply don't even know how to think about this problem.

I've tried simple stuff just to get my head on the problem.

For example for $P(x)=x^n$ I have $P(P(x))=(P(x))^n$, and I think that any polynomial $P(x)=x^n+x^{n-1} +\cdots +c$ can't be a solution as I would have $P(x)=P(x)q_1(x) +R $.

After that I simply stare at the problem.

Can you guys give some help ?

Note: I would like to understand how to tackle these kind of problems, so I would be really grateful if you would explain the thinking process behind the solution. (This is optional, so feel free to give an answer as you prefer.)

Thanks in advance.

Answer 1

First we note that only the constant polynomials $P \equiv 0$ and $P \equiv 1$, as well als $P \equiv -1$ for odd $k$, satisfy the equation. (This holds true for $k>1$; for $k=1$ any constant polynomial will do.)

Now if $P$ is not constant, then the range $Y = \{ P(x) : x \in \mathbb{R} \}$ of $P$ is an infinite set. For every $y \in Y$ we have $P(y) = y^k$, which means that the polynomial $Q(x) = P(x) - x^k$ is zero on $Y$. Since $Y$ is infinite, this implies that $Q$ is the zero polynomial. In conclusion, $P(x) = x^k$.

Answer 2

Note, first, that if $P$ is constant, then $P = P^k$, so $P$ can be any real solution of the equation $u^k - u = 0$. From now on, we shall assume that $P$ is not constant.

Let $P = cx^n + Q$ with $n \ge 1$, $c \ne 0$ and $\deg Q < n$. The equality $P(P(x)) = (P(x))^k$ means $c (cx^n + Q)^n + Q = (cx^n + Q)^k$. Developing this and equating the terms of highest degree, we get that $c^{n+1} x^{n^2} = c^k x^{nk}$, so $n = k$, which implies $c^{k+1} = c^k$. Since $c \ne 0$ we get $c = 1$.

Rewriting the conclusion above gives $(x^k + Q)^k + Q = (x^k + Q)^k$, i.e. $Q=0$, so $P = x^k$.

Therefore, the only solutions are the roots of the equation $u^k - u = 0$ and $x^k$.

Answer 3

Let $\deg(P(x))=n$,then $\deg(P(P(x)))=n^2,\deg(P(x))^k=nk$,so $n=k$ or $n=0$,

if $n=0$, it easy to find $P(x)=1$

if $n=k$, then let $$P(x)=a_{k}x^k+a_{k-1}x^{k-1}+\cdots+a_{0}$$ then $$[x^{k^2}](P(P(x))=(a_{k})^{k+1},[x^{k^2}](P(x))^k=(a_{k})^k$$ so we have $$a_{k}=1$$ so $$P(x)=x^k+a_{k-1}x^{k-1}+\cdots+a_{0}$$ so we have $$P(P(x))=(x^k+a_{k-1}x^{k-1}+\cdots+a_{0})^k+a_{k-1}(x^k+a_{k-1}x^{k-1}+\cdots+a_{0})^{k-1}+\cdots+a_{0}$$ and $$(P(x))^k=(x^k+a_{k-1}x^{k-1}+\cdots+a_{0})^k$$ so have $$a_{k-1}=a_{k-2}=\cdots=a_{0}=0$$

Answer 4

Suppose P(x) is a polynomial of order n.

Thus:

$P(x)=\sum_{r=o}^{n}a_{r}x^{r}$.

And so we have:

$P(P(x))=P(\sum_{r=o}^{n}a_{r}x^{r})= \sum_{r=o}^{n}a_{r}(\sum_{r=o}^{n}a_{r}x^{r})^{r}$.

From this we can deduce that $P(P(x))$ is of order $n^{2}$.

Now $[P(x)]^{k} =[\sum_{r=o}^{n}a_{r}x^{r}]^{k} $, hence $[P(x)]^{k}$ is of the order $nk$.

Since these polynomials are the same, they must be of the same order, hence, $nk=n^{2}$ and rearranging gives $0=n(n-k)$.

Assuming that the polynomial is non constant, (i.e isn't of $0th$ order) we must therefore conclude that $n=k$.

Note that:

$P(P(x))=a_{n}(a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0})^{n}+\sum_{r=o}^{n-1}a_{r}(\sum_{r=o}^{n}a_{r}x^{r})^{r}$

And that: $[P(x)]^{n} =(a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0})^{n}=(a_nx^{n})^{n} +Q(x)+(a_{0})^{n}$ where $Q(x)$ is the rest of the polynomial that isn't really relevant to this argument.

Comparing the coefficients of $x^{n}$ leads us to conclude that $(a_{n})^{n+1} =(a_{n})^{n}$, thus $a_{n}=1$.

Thus we have $P(P(x))=(x^{n}+a_{n-1}x^{n-1}+...+a_{0})^{n} +\sum_{r=o}^{n-1}a_{r}(\sum_{r=o}^{n}a_{r}x^{r})^{r}=[P(x)]^{n}=(x^{n}+a_{n-1}x^{n-1}+...+a_{0})^{n}$.

Thus:$\sum_{r=o}^{n-1}a_{r}(\sum_{r=o}^{n}a_{r}x^{r})^{r}=0$

Hence $a_{r}$=0 for all $r=0,1,2,3,...,n-1$.

Answer 5

Let $\deg p = n$. Then we have $n^2 = nk$, thus $n = 0$ or $n = k$.

If $n = 0$, let $p (x) = c$, where $c$ is a constant. Then we have $c = c^k$, which implies $p (x) \equiv 0$ or $p (x) \equiv \pm 1$.

Assume now $n = k$. Then, $$p (x) = c x^k + q (x),$$ where $\deg q < k$. By the original equation, we have $$c (c x^k + q (x))^k + q (x) = (c x^k + q (x)) ^k$$ and $c^{k + 1} = c^k$, so $c = 1$. Then, $p (x) = x^k + q (x)$. Therefore, $$(x^k + q(x))^k + q (x) = (x^k + q (x))^k,$$ by which we have $q (x) \equiv 0$. Hence, $p (x) = x^k$.

Edit: Now that I saw the other answers, I think all of our solutions are somewhat the same.

Answer 6

I want to add a different way to see that the non constant solutions to the equation must be polynomials with no non zero roots.

Let $P(x)=(x-a_1)\cdot...\cdot(x-a_n)^1$. Then the roots of $[P(x)]^k$ are the same as the roots of $P(x)$ and must also be the same as the roots of $P(P(x))$. Now $P(P(x))=\prod_{i=1}^{n}(P(x)-a_i)$. Suppose that $a_i \neq 0$ for some $i$. Then by the Fundamental Theorem of Algebra $(P(x)-a_i)$ must have at least one root which, by definition of $P(P(x))$, is also a root of $P(P(x))$. But $(P(x)-a_i) \neq 0$ for all $a_i \in {\{a_j\}}_{j=1}^{n}$. So $P(P(x))$ has some root that differs from every $a_i$ which is a contradiction. Thus $a_i=0$ for all $i$.

$^1$ We can write this because $P(x)$ of degree $n$ can be factored over $\mathbb{C}$.