2. Find all polynomials $p$ of odd degree $z$ with real coefficients, for which

$$
p(p(x)) \leq (p(x))^{3}
$$

holds for all $x \in \mathbb{R}$ and which have the coefficient of $x^{2}$ equal to 0.
III/2. Let's write $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$, where $a_{n} \neq 0$. The leading term of the polynomial $p(p(x))$ is $a_{n}\left(a_{n} x^{n}\right)^{n}=a_{n}^{n+1} x^{n^{2}}$, and the leading term of the polynomial $p(x)^{3}$ is $\left(a_{n} x^{n}\right)^{3}=a_{n}^{3} x^{3 n}$, both of which are of odd degree. Since the polynomial $p(x)^{3}-p(p(x))$ is non-negative everywhere, it must be of even degree, so the mentioned leading terms must cancel each other out. Therefore, $a_{n}^{n+1} x^{n^{2}}=a_{n}^{3} x^{3 n}$ or $n^{2}=3 n$ and $a_{n}^{n+1}=a_{n}^{3}$. It follows that $n=3$ and $a_{n}=1$, since $a_{n} \neq 0$, i.e., $p$ is a polynomial of degree 3 with a leading coefficient of 1. Since the coefficient of the polynomial $p$ at $x^{2}$ is 0, the polynomial $p$ is of the form $p(x)=x^{3}+a x+b$. When we consider this in the given inequality and simplify the inequality, we get $a x^{3}+a^{2} x+a b+b \leq 0$ for every $x \in \mathbb{R}$. This is only possible if $a=0$, since a polynomial of odd degree always takes positive values. The inequality simplifies to $b \leq 0$ in this case. The polynomials that satisfy the conditions of the problem are exactly the polynomials of the form $p(x)=x^{3}+b$, where $b \leq 0$.

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=51&width=1633&top_left_y=1551&top_left_x=220)

The degree of the polynomial $p^{3}(x)-p(p(x))$ must be even .............................. 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=57&width=1633&top_left_y=1645&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=57&width=1633&top_left_y=1693&top_left_x=220)

Substituting the assumption $p(x)=x^{3}+a x+b$ into the inequality .............................. 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=49&width=1631&top_left_y=1803&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=60&width=1631&top_left_y=1849&top_left_x=224)

## Addendum:

We can also determine that the constant term is less than or equal to 0 in another way: since $p$ is of odd degree, it has a real root $x_{0}$. Then $a_{0}=p(0)=p\left(p\left(x_{0}\right)\right) \leq p\left(x_{0}\right)^{3}=0 . \mathbf{1}$ point

In addition, points can also be awarded for:

Guessing at least one correct solution

1 point

Excluding a polynomial of degree 1 in the analysis

1 point

Correctly writing the general inequality $p(p(x)) \leq p(x)^{3}$