Three integers form a geometric progression. If we add 8 to the second number, an arithmetic progression is formed. If we add 64 to the third term of this arithmetic progression, we again get a geometric progression. What are the original numbers?
Let the numbers be

$$
a, a q, a q^{2}
$$

According to the main task, we have

$$
\begin{gathered}
2(a q+8)=a+a q^{2} \\
(a q+8)^{2}=a\left(a q^{2}+64\right)
\end{gathered}
$$

From (1),

$$
a\left(1-2 q+q^{2}\right)=16
$$

From (2),

$$
a(q-4)=-4
$$

Divide (3) by (4), then we get

$$
16(q-4)=-4\left(1-2 q+q^{2}\right)
$$

or

$$
q^{2}+2 q-15=0
$$

from which

$$
q=3, \quad q^{\prime}=-5
$$

and

$$
a_{1}-4, \quad a_{1}^{\prime}=\frac{4}{9}
$$

Therefore, the original integers are

$4,12,36$.