\section*{Problem 6 - 091036}

For a quadratic function \(y=a x^{2}+b x+c(a \neq 0)\), consider the table

\[
\begin{array}{ccccc}
\mathrm{x} & 1 & 2 & 3 & 4 \\
\mathrm{y} & 1 & 2 & n_{1} & n_{2}
\end{array}
\]

formed.

Determine all real coefficients \(a, b, c\) for which \(n_{1}\) and \(n_{2}\) are single-digit natural numbers!
}

From the functional equation and the table, for \(x=1\) and \(x=2\) we get
(1) \(a+b+c=1\)
(2) \(4 a+2 b+c=2\)

Subtracting (1) from (2) or 2 times (1) from (2), we obtain (3) \(3 a+b=1\) and (4) \(2 a-c=0\), and from these \(b=1-3 a\) and \(c=2 a\). For the quadratic function, we thus get

\[
\text { (5) } \quad y=a x^{2}+(1-3 a) x+2 a
\]

Substituting \(x=3\) and \(x=4\) yields
(6) \(n_{1}=9 a+3(1-3 a)+2 a=2 a+3\)
(7) \(n_{2}=16 a+4(1-3 a)+2 a=6 a+4\)

Since \(n_{1}\) should be in \(\{0,1, \ldots, 9\}\), we get from (6) and (7) the following table

\[
\begin{array}{lcccccccccc}
n_{1} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\mathrm{a} & -\frac{3}{2} & -1 & -\frac{1}{2} & 0 & \frac{1}{2} & 1 & \frac{3}{2} & 2 & \frac{5}{2} & 3 \\
n_{2} & -5 & -2 & 1 & 4 & 7 & 10 & 13 & 16 & 19 & 22
\end{array}
\]

Since \(n_{2}\) should also be in \(\{0,1, \ldots, 9\}\) and \(a\) should not be 0, the only solutions for \(a\) are \(a=-\frac{1}{2}\) and \(a=\frac{1}{2}\).

For the functional equation, we then get from (5):

\[
y=-\frac{1}{2} x^{2}+\frac{5}{2} x-1 \quad \text { or } \quad y=\frac{1}{2} x^{2}-\frac{1}{2} x+1
\]