Four numbers form a geometric progression. The difference between the first and second number is 6, the difference between the third and fourth number is 5; the sum of the squares of the four numbers is 793. Which are these numbers?
Let the four sought numbers be $a+6, a, b+5$ and $b$. Then

$$
a+6: a=b+5: b
$$

from which

$$
b=\frac{5}{6} a
$$

Furthermore,

$$
(a+6)^{2}+a^{2}+(b+5)^{2}+b^{2}=(a+6)^{2}+a^{2}+\frac{25}{36}(a+6)^{2}+\frac{25}{36} a^{2}=793
$$

or

$$
a^{2}+6 a-216=0
$$

from which

$$
a_{1}=12, \quad a_{2}=-18
$$

Thus, the sought numbers are:

$18,12,15,10$

and

$$
-12,-18,-10,-15
$$

(Andor Ungár, Zenta.)