22.6. (New-York, 75). Find all polynomials $P(x)$ satisfying the condition $P(0)=0$ and the identity

$$
P(x)=\frac{1}{2}(P(x+1)+P(x-1)), \quad x \in \mathbf{R}
$$
22.6. Any polynomial of the form $P(x)=a x$, where $a$ is a constant, satisfies the conditions of the problem. We will prove by induction on $n \in \mathbf{Z}^{+}$ that for each desired polynomial $P(x)$, the equalities $P(n)=n P(1)$ hold. For $n=0$ and $n=1$, these equalities are true. Suppose they have already been proven for the numbers $n-1$ and $n$, where $n \in \mathbf{N}$. Then $P(n+1)=2 P(n)-P(n-1)=$ $=(n+1) P(1)$, which means the equality is also valid for the number $n+1$. Since the polynomial $P(x)-P(1) x$ has infinitely many roots $x=0,1,2, \ldots$, it is equal to zero. Therefore, the desired polynomials have the form $P(x)=a x$.