\section*{Problem 1 - 151241}

Determine whether there exists a polynomial \(P(x)\) of degree three such that \(P(0)=74, P(1)=19\), \(P(2)=65\) and \(P(3)=92\).

If such a polynomial exists, find \(P(4)\) and \(P(5)\).
A sequence \(P(0), P(1), P(2), \ldots\) is described by a polynomial of degree \(k\) with \(k>0, k \in \mathbb{N}\) if and only if its difference sequence \(P^{\prime}(0):=P(1)-P(0), P^{\prime}(1):=P(2)-P(1), \ldots\) is described by a polynomial of degree \((k-1)\).

Applied to the numerical values of the problem, one obtains:

\[
P^{\prime}(0)=-55, \quad P^{\prime}(1)=46, \quad P^{\prime}(2)=27, \quad P^{\prime \prime}(0)=101, \quad P^{\prime \prime}(1)=-19
\]

thus, no polynomial of second degree \(P^{\prime \prime \prime}(0)=-120\) suffices.

If one now continues \(P^{\prime \prime \prime}\) with the value of \(P^{\prime \prime \prime}(0)\), one obtains a cubic polynomial for \(P\) with the required initial values.

In particular, one obtains

\[
\begin{aligned}
& P^{\prime \prime \prime}(1)=-120, \quad P^{\prime \prime \prime}(2)=-120 ; \quad P^{\prime \prime}(2)=-139, \quad P^{\prime \prime}(3)=-259 \\
& P^{\prime}(3)=-112, \quad P^{\prime}(4)=-371 ; \quad P(4)=-20 \quad \text { and } \quad P(5)=-391
\end{aligned}
\]

Remark: The function \(P^{\prime}\) is the "discrete derivative" of the function \(P\).