, 2006

Let $P(x)$ be a polynomial satisfying the condition of the problem. We have:
$$P(x^2) + x(3P(x) + P(-x)) = (P(x))^2 + 2x^2, \quad x \in \mathbb{R}, \quad (1)$$

It is clear that $\deg P > 0$. Consider the cases:
i) $\deg P = 1$.
$P(x)$ has the form $P(x) = ax + b$, $a \neq 0$. Substituting it into (1) gives:
$$(a^2 - 3a + 2)x^2 + 2b(a - 2b)x + b^2 - b \equiv 0 \quad (2)$$
It is evident that:
$$(2) \Leftrightarrow (a = 1, b = 0), (a = 2, b = 0), (a = 2, b = 1).$$
So we get the polynomials: $P(x) = x$, $P(x) = 2x$, and $P(x) = 2x + 1$.

Put $P(x) = ax^n + S(x)$, $a \neq 0$, (3).
Where $S(x)$ is a polynomial of degree $\deg S(x) = k < n$.
Substituting (3) into (1) gives:
$$(a^2 - a)x^{2n} + (S(x))^2 - S(x^2) + 2ax^n \cdot S(x) \equiv \\ \equiv (3 + (-1)^n)ax^{n+1} + (3S(x) + S(-x))x - 2x^2. \quad (4)$$
Since the degree of left hand side of (4) is equal to $(n + 1)$ and since $n + 1 < 2n$, (4) implies that $a^2 - a = 0$ and so $a = 1$. Therefore, from (4) we obtain:
$$2x^n S(x) + (S(x))^2 - S(x^2) \equiv (3 + (-1)^n)x^{n+1} + (3S(x) + S(-x))x - 2x^2 \quad (5)$$

Since degree of the left hand side of (5) is equal to $(n + k)$ and the degree of right hand side of (5) is equal to $(n+1)$, (5) implies that $k = 1$.
Moreover, by substituting $x = 0$ into (5), we obtain $(S(0))^2 - S(0) = 0$, i.e $S(0) = 0$ or $S(0) = 1$. So $S(x) = px$ or $S(x) = px + 1$.

a) In case $S(x) = px$, by substituting it into (5), we get:
$$(3 + (-1)^n - 2p)x^{n+1} - (p^2 - 3p + 2)x^2 \equiv 0$$
Which is equivalent to:
$$\begin{cases} 3 + (-1)^n - 2p = 0 \\ p^2 - 3p + 2 = 0 \end{cases}$$
$$\Leftrightarrow p = 1, n \equiv 1 \pmod{2}$$
$$\text{or } p = 2, n \equiv 0 \pmod{2}.$$
So we obtain the polynomials:
$$P(x) = x^{2n+1} + x \text{ and } P(x) = x^{2n} + 2x.$$
Direct verifications shows that these polynomials satisfy (1).

b) In case $S(x) = px + 1$, by substituting it in to (5) we get:
$$(3 + (-1)^n - 2p)x^{n+1} - 2x^n - (p^2 - 3p + 2)x^2 - 2(p-2)x \equiv 0$$
which implies that $2 \equiv 0$. This contradiction show that in this case, there is no $S(x)$ satisfying (5) and so there is no $P(x)$ satisfying (1).

All in all, the polynomials $P(x)$ satisfying the condition of the problem are:
$$P(x) = x, \quad P(x) = x^{2n} + 2x, \quad n \in \mathbb{N}$$
$$P(x) = x^{2n+1} + x, \quad n \in \mathbb{N}.$$