4. Find a two-digit number, the digits of which are different and the square of which is equal to the cube of the sum of its digits.
Answer: 27.

Solution: Let's write the condition as $\overline{A B}^{2}=(A+B)^{3}$. Note that the sum of the digits ( $A+B$ ) does not exceed 17, i.e., $\overline{A B}^{2} \leq 17^{3}$. Moreover, this number $\overline{A B}^{2}=n^{6}$, where $n-$
is some natural number that does not exceed 4. However, 1 and 2 do not work, since their cubes are single-digit numbers. Only 3 and 4 remain, and direct verification shows that $27^{2}=(2+7)^{2}=729$.