Let $f(x)=4 x-x^{2}$. Find different numbers $a$ and $b$ such that

$$
f(a)=b \quad \text { and } \quad f(b)=a \text {. }
$$
A

$$
\left.\begin{array}{l}
4 a-a^{2}=b \\
4 b-b^{2}=a
\end{array}\right\}
$$

system of equations needs to be solved.

Subtracting the second equation from the first:

$$
\begin{gathered}
4 a-4 b-a^{2}+b^{2}=b-a \\
4(a-b)-(a-b)(a+b)+(a-b)=0 \\
(a-b) \cdot[5-(a+b)]=0 \\
a-b \neq 0, \quad \text { therefore } 5-(a+b)=0
\end{gathered}
$$

Thus, we arrive at the following equation:

$$
a^{2}-5 a+5=0
$$

The solution to the problem is the following two pairs of numbers:

$$
\begin{array}{ll}
a_{1}=\frac{5+\sqrt{5}}{2} & b_{1}=\frac{5-\sqrt{5}}{2} \\
a_{2}=\frac{5-\sqrt{5}}{2} & b_{2}=\frac{5+\sqrt{5}}{2}
\end{array}
$$

$$
\left(a_{2}=b_{1}, b_{2}=a_{1} .\right)
$$

Borosnyói Adorján (Miskolc, Kós Károly Ép. Ip. Szki., II. o. t.)