. Find all polynomials $P(x)$ with real coefficients such that

$$
(x-2010) P(x+67)=x P(x)
$$

for every integer $x$.
Taking $x=0$ in the given equality leads to $-2010 P(67)=0$, implying $P(67)=0$. Whenever $i$ is an integer such that $1 \leq i<30$ and $P(i \cdot 67)=0$, taking $x=i \cdot 67$ leads to $(i \cdot 67-2010) P((i+1) \cdot 67)=0$; as $i \cdot 67<2010$ for $i<30$, this implies $P((i+1) \cdot 67)=0$. Thus, by induction, $P(i \cdot 67)=0$ for all $i=1,2, \ldots, 30$. Hence

$$
P(x) \equiv(x-67)(x-2 \cdot 67) \ldots(x-30 \cdot 67) Q(x)
$$

where $Q(x)$ is another polynomial.

Substituting this expression for $P$ in the original equality, one obtains

$$
(x-2010) \cdot x(x-67) \ldots(x-29 \cdot 67) Q(x+67)=x(x-67)(x-2 \cdot 67) \ldots(x-30 \cdot 67) Q(x)
$$

which is equivalent to

$$
x(x-67)(x-2 \cdot 67) \ldots(x-30 \cdot 67)(Q(x+67)-Q(x))=0 .
$$

By conditions of the problem, this holds for every integer $x$. Hence there are infinitely many roots of polynomial $Q(x+67)-Q(x)$, implying that $Q(x+67)-Q(x) \equiv 0$. Let $c=Q(0)$; then $Q(i \cdot 67)=c$ for every integer $i$ by easy induction. Thus polynomial $Q(x)-c$ has infinitely many roots whence $Q(x) \equiv c$.

Consequently, $P(x)=c(x-67)(x-2 \cdot 67) \ldots(x-30 \cdot 67)$ for some real number $c$. As equation (1) shows, all such polynomials fit.