\section*{Problem 5 - 181035}

Determine whether there exist real numbers \(a, b, c, d\) with the following property:

If \(f\) is the function defined for all real numbers \(x\) by the equation \(f(x)=a x^{3}+b x^{2}+c x+d\), then \(f(0)=10 ; f(1)=12 ; f(2)=4\) and \(f(3)=1\).

If such numbers \(a, b, c, d\) exist, determine all such numbers!
}

For all real \(a, b, c, d\), the required conditions are satisfied if and only if the following equations (1) to (4) hold:

\[
\begin{aligned}
& d=10 \quad(1) \quad ; \quad a+b+c+d=12 \\
& 8 a+4 b+2 a+d-1 \\
& \text { (3) } \quad ; \quad 27 a+9 b+3 c+d=1
\end{aligned}
\]

I) If (1) to (4) are satisfied for real \(a, b, c, d\), then by substituting (1) into (2), (3), and (4), we get:

\[
\begin{array}{rll}
a+b+c=2 & \text { so } & c=2-a-b \\
8 a+4 b+2 c=-6 & \text { so } & 4 a+2 b+c=-3 \\
27 a+9 b+3 c=-9 & \text { so } & 9 a+3 b+c=-3
\end{array}
\]

Substituting (5) into (6) and (7), we obtain

\[
3 a+b=-5 \quad \text { so } \quad b=-5-3 a \quad \text { (8) } \quad \text { and } \quad 8 a+2 b=-5
\]

By substituting (8) into (9), we get \(2 a=5\), from which \(a=\frac{5}{2} \quad\) (10) follows. From (10) and (8), we get \(b=-\frac{25}{2} \quad(11)\).

From this and (10) and (5), we get

\[
x=2-\frac{5}{2}+\frac{25}{2}=12
\]

Therefore, only the numbers mentioned in (10), (11), (12), (1) can satisfy equations (1) to (4). II) As can be shown by substituting these numbers into (1) to (4), they satisfy these equations. The sought numbers are thus \(a=\frac{5}{2}, b=-\frac{25}{2}, c=12, d=10\).