I want the necessary and sufficient condition for the fraction

$$
\frac{a_{1} x^{2}+b_{1} x+c_{1}}{a_{2} x^{2}+b_{2} x+c_{2}}
$$

to be a constant value for all values of $x$.
First solution. Let

$$
\frac{a_{1} x^{2}+b_{1} x+c_{1}}{a_{2} x^{2}+b_{2} x+c_{2}}=k
$$

then

$$
\left(a_{1}-a_{2} k\right) x^{2}+\left(b_{1}-b_{2} k\right) x+\left(c_{1}-c_{2} k\right)=0
$$

This equation can only hold for all values of $x$ if

$$
a_{1}-a_{2} k=0, b_{1}-b_{2} k=0, c_{1}-c_{2} k=0
$$

from which

$$
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=k
$$

This condition is not only necessary but also sufficient, because if we substitute the values of $a_{1}$, $b_{1}$, and $c_{1}$ from (2) into (1), we get:

$$
\frac{a_{2} k x^{2}+b_{2} k x+c_{1} k}{a_{2} x^{2}+b_{2} x+c_{2}}=k
$$

(Lajos Velics, Kassa.)

Second solution. Since the given fraction is equal to $k$ for all values of $x$, substituting 1, (-1), and 0 for $x$ should also yield $k$. Thus, after these substitutions, (1) becomes:

$$
\begin{gathered}
a_{1}+b_{1}+c_{1}=a_{2} k+b_{2} k+c_{2} k \\
a_{1}-b_{1}+c_{1}=a_{2} k-b_{2} k+c_{2} k \\
c_{1}=c_{2} k
\end{gathered}
$$

From these equations, we again obtain that

$$
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}
$$

(Zoltán Gádor, Losoncz.)