* Find all two-digit numbers such that the sum of the number and the number obtained by reversing its digits is a perfect square.
Let the required two-digit number be $ab$, then $10a+b+10b+a=m^{2}$ (where $m$ is a positive integer). That is, $11(a+b)=m^{2}$. Therefore, $(a, b)=(2,9),(3,8),(4,7),(5,6),(6,5),(7,4)$, $(8,3),(9,2)$. Thus, the numbers we are looking for are: $29,38,47,56,65,74,83,92$.