
Problem 4. Find all polynomials $P(x)$ with real coeficients such that

$$
P(x) \cdot P(x+1)=P\left(x^{2}\right)
$$

for any real $x$.


Solution: We show first that for any natural number $n$ there exists at most one polynomial $P(x)$ of degree $n$ such that $P(x) \cdot P(x+1)=$
$P\left(x^{2}\right)$. Indeed, if

$$
P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}
$$

comparing the coefficients in front of the same degrees of $x$ in $P(x) . P(x+1)$ and $P\left(x^{2}\right)$, we get $a_{n}=1$ and $a_{n-1}=-\frac{n}{2}$. It is easily seen that each of the subsequent coefficients is a solution of an equation of first degree of the form $2 x+b=0$, where $b$ is a function of already chosen coefficients. Therefore, if a polynomial of degree $n$ with the desired property exists it is unique one.

We prove now that a polynomial $P(x)$ of odd degree such that $P(x) \cdot P(x+1)=P\left(x^{2}\right)$ does not exist. Let $P(x)=x^{m}(x-1)^{n} \cdot Q(x)$ ( $m$ or $n$ could be zero), where $Q(x)$ is a polynomial such that $Q(0) \neq$ 0 and $Q(1) \neq 0$. A substitution gives

$$
(x+1)^{m} \cdot x^{n} \cdot Q(x) \cdot Q(x+1)=x^{m} \cdot(x+1)^{n} \cdot Q\left(x^{2}\right)
$$

If $m \neq n$ the substitution $x=0$ leeds to a contradiction. Therefore $m=n$ and so $Q(x) \cdot Q(x+1)=Q\left(x^{2}\right)$, where $Q(x)$ is of odd degree. Therefore there exists a real number $x_{0} \neq 0,1$, such that $Q\left(x_{0}\right)=0$. It is clear that $x_{0}=-1$ since otherwise the substitution $x=-1$ shows that $Q(1)=0$ which is impossible. A substitution $x=x_{0}$ gives that $Q\left(x_{0}^{2}\right)=0$, thus $Q\left(x_{0}^{4}\right)=0$ and so on. Since $x_{0} \neq \pm 1,0$ it follows that there are infinitely many distinct terms in the sequence $x_{0}, x_{0}^{2}, \ldots, x_{0}^{2^{k}}$, which is impossible.

Direct verification shows that for any even natural number $n=$ $2 k$, the polynomial $P(x)=x^{k}(x-1)^{k}$ is of degree $n$ and is a solution of the problem.

Therefore all polynomials such that $P(x) \cdot P(x+1)=P\left(x^{2}\right)$ are $P(x)=x^{k}(x-1)^{k}$.
