If $a, a q, a q^{2}, a q^{3}$ is the sought geometric sequence, then the numbers $a, a q+6, a q^{2}+3$ and $a q^{3}-36$ form an arithmetic sequence, so

$$
2(a q+6)=a+a q^{2}+3
$$

or

$$
a\left(q^{2}-2 q+1\right)=9
$$

and

$$
2\left(a q^{2}+3\right)=(a q+6)+\left(a q^{3}-36\right)
$$

or

$$
a q\left(q^{2}-2 q+1\right)=36
$$

thus

$$
q=4 \text { and } a=1
$$

which is why the sought numbers are $1,4,16$ and 64.

(Marcel Riesz, Györ.)

The problem was also solved by: Bartók I., Demjén E., Deutsch E., Deutsch I., Eckstein J., Enyedi B., Glück J., Haar A., Harsányi Z., Hirschfeld Gy., Jánosy Gy., Kertész G., Kiss J., Kürti I., Liebner A., Neidenbach E., Pám M., Pichler S., Pivnyik I., Popoviciu M., Preisich G., Ragály B., Rássy P., Riesz K., Sárközy E., Schlesinger O., Schusler Gy., Schwemmer I., Szántó H., Szücs A., Tóth B., Weisz P.
Four numbers form a geometric sequence. If we increase the second number by 6, the third number by 3, and decrease the fourth number by 96, we get an arithmetic sequence. Let's find these four numbers.