Let's determine $a, b, c$, and $d$ such that the fraction

$$
\frac{a x^{2}-b x+c}{x^{2}-d x-a}
$$

is maximal at $x=2$ and minimal at $x=5$, and that the maximum value of the fraction is 1, and the minimum value is 2.5.
(1) It can also be written as:

$$
(a-y) x^{2}-(b-d y) x+c+a y=0
$$

If the maximum value of the fraction is 1, then from (2) it becomes:

$$
(a-1) x^{2}-(b-d) x+c+a=0
$$

But the fraction only takes its maximum value when $x=2$, which is why both roots of the equation (3) are equal to two, and thus

$$
\frac{b-d}{a-1}=2+2=4 \quad \text { and } \quad \frac{c+a}{a-1}=2+2=4
$$

which equations can also be written as:

$$
4 a-b+d-4=0
$$

$$
3 a-c-4=0 \text {. }
$$

The minimum value of the fraction is 2.5, so from 2 it becomes:

$$
(2 a-5) x^{2}-(2 b-5 d) x+2 c+5 a=0
$$

and since the minimum value occurs at $x=5$, therefore

$$
\frac{2 b-5 d}{2 a-5}=10 \quad \text { and } \quad \frac{2 c+5 a}{2 a-5}=25
$$

or

$$
20 a-2 b+5 d-50=0
$$

$$
45 a-2 c-125=0
$$

The values of $a, b, c$ and $d$ are given by (4), (5), (6) and (7):

$$
a=3, b=10, c=5, d=2
$$

(Pál Sárközy, Pannonhalma.)

The problem was also solved by: N. Ehrenfeld, I. Erdélyi, M. Fekete, R. Krémusz, S. Pichler, B. Tóth.