\section*{Problem 6A - 161246A}

All polynomials

\[
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}
\]

with real coefficients \(a_{0}, a_{1}, \ldots, a_{n}\) are to be specified, which have the following property:

For all real numbers \(x\), \(x \cdot f(x-1)=(x-2) \cdot f(x)\).
}

For \(x=2\) it follows: \(2 \cdot f(1)=0\), i.e., \(x=1\) is a root of \(f(x)\).

For \(x=1\) it follows: \(1 \cdot f(0)=0\), i.e., \(x=0\) is a root of \(f(x)\).

Thus, \(f(x)\) can be represented as

\[
f(x)=x \cdot(x-1) \cdot g(x)
\]

with a polynomial \(g(x)\). Substituting this ansatz into the functional equation, we get:

\[
x \cdot[(x-1)(x-2) g(x-1)]=(x-2) \cdot[x \cdot(x-1) g(x)]
\]

From this, for all \(x \neq 0,1,2\), the relation \(g(x-1)=g(x)\) follows, i.e., \(g(x)\) is a periodic function with period 1. Since \(g(x)\) is a polynomial, it follows that:

\[
g(x)=c=\text{const.}
\]

Thus, all polynomials of the required type can be represented in the form

\[
f(x)=c \cdot x \cdot(x-1)
\]

with real \(c\). Substituting the found representation into the functional equation, we get the identity

\[
x \cdot c \cdot(x-1)(x-2)=(x-2) \cdot c \cdot x \cdot(x-1)
\]
so that indeed every polynomial of the form (1) satisfies the conditions of the problem.

Adapted from [5]