1. Find a polynomial $P(x)$ for which

$$
x \cdot P(x-1)=(x-1990) \cdot P(x)
$$
Solution. The polynomial $P(x)(x-1990)$ is divisible by $x$, so $P(x)$ must be divisible by $x$, i.e., there exists a polynomial $P_{1}(x)$ such that

$$
P(x)=x P_{1}(x)
$$

Substituting this into the given equation, we get

$$
\begin{aligned}
& x(x-1) P_{1}(x-1)=(x-1990) x P_{1}(x) \\
& (x-1) P_{1}(x-1)=(x-1990) P_{1}(x)
\end{aligned}
$$

Similarly, there will exist a polynomial $P_{2}(x)$ such that

$$
P_{1}(x)=(x-1) P_{2}(x)
$$

and if this is substituted, we get

$$
\begin{aligned}
& (x-1)(x-2) P_{2}(x-1)=(x-1990)(x-1) P_{2}(x) \\
& (x-2) P_{2}(x-1)=(x-1990) P_{2}(x)
\end{aligned}
$$

If this process is continued, we obtain polynomials $P_{3}(x), P_{4}(x), \ldots, P_{1990}(x)$, and for this last polynomial, we will have

$$
\begin{aligned}
& (x-1990) P_{1990}(x-1)=(x-1990) P_{1990}(x) \\
& P_{1990}(x-1)=P_{1990}(x)
\end{aligned}
$$

From this, it follows that the polynomial $P_{1990}(x)-P_{1990}(0)$ takes the value 0 for every integer, which is possible only if $P_{1990}(x)-P_{1990}(0)$ is the zero polynomial. Let

$$
P_{1990}(0)=C,
$$

then

$$
\begin{aligned}
P(x) & =x P_{1}(x)=x(x-1) P_{2}(x)=\ldots . .=x(x-1)(x-2) \ldots(x-1989) P_{1990}(x)= \\
& =C x(x-1)(x-2) \ldots(x-1989)
\end{aligned}
$$