4.069 Find a three-digit number whose digits form a geometric progression. If 792 is subtracted from this number, the result is a number written with the same digits but in reverse order. If 4 is subtracted from the digit representing the hundreds, and the other digits of the desired number are left unchanged, the resulting number will have digits that form an arithmetic progression.
Solution. Let the desired number have the form:

$100 x+10 y+z \Rightarrow 100 x+10 y+z=792+100 z+10 y+x \Rightarrow x=8+z \Rightarrow\left[\begin{array}{l}z=0, x=8, \\ z=1, x=9 .\end{array}\right.$

Since the digits $x, y, z$ form a geometric progression, then $z \neq 0 \Rightarrow z=1$. The possible cases are:

1) $x z=y^{2} \Rightarrow y=3$
2) $x y=z^{2} \Rightarrow \varnothing$;
3) $y z=x^{2} \Rightarrow \varnothing$.

By verification, we find that the number 931 satisfies the other conditions of the problem.

Answer: 931.