1. Determine all polynomials $P$ such that for all real numbers $x$,

$$
(P(x))^{2}+P(-x)=P\left(x^{2}\right)+P(x) .
$$

(P. Calábek)
Solution. If the polynomial $P$ is constant, i.e., $P(x)=a$, then the number $a$ satisfies the condition $a^{2}+a=a+a$ according to the problem, so $a=0$ or $a=1$. Both polynomials $P(x)=0$ and $P(x)=1$ are solutions to the problem.

If the degree $n$ of the polynomial $P$ is positive, then $P(x)=a x^{n}+Q(x)$, where $a$ is a number different from zero and $Q$ is a polynomial of degree at most $n-1$. By comparing the coefficients of the highest power $x^{2 n}$ in the equality

$$
\left(a x^{n}+Q(x)\right)^{2}+a(-x)^{n}+Q(-x)=a x^{2 n}+Q\left(x^{2}\right)+a x^{n}+Q(x)
$$

we get the condition $a^{2}=a$, from which it follows that $a=1$ (recall that $a \neq 0$). The equality (1) after substituting the value $a=1$ can be rewritten in the equivalent form

$$
2 x^{n} Q(x)+(Q(x))^{2}-Q\left(x^{2}\right)=\left[1-(-1)^{n}\right] x^{n}+Q(x)-Q(-x) .
$$

Suppose the polynomial $Q$ has a positive degree $k(k<n)$. Then on the left side of (2) stands a polynomial of degree at least $n+k$, which is a contradiction with the fact that on the right side of (2) is a polynomial of degree at most $n$. Therefore, $Q$ is a constant polynomial, i.e., $Q(x)=b$ for some number $b$. After substituting into (2), we get the condition $2 b x^{n}+b^{2}-b=\left[1-(-1)^{n}\right] x^{n}$, which is satisfied for every $x$ if and only if $2 b=1-(-1)^{n}$ and at the same time $b^{2}-b=0$. For even $n$, only $b=0$ (so $P(x)=x^{n}$) comes out, and for odd $n$, $b=1$ (so $P(x)=x^{n}+1$) comes out.

Answer: The sought polynomials are the constants 0 and 1, the monomials $x^{2}, x^{4}, x^{6}, \ldots$ and the binomials $x+1, x^{3}+1, x^{5}+1, \ldots$