49.* A two-digit number, when added to the number written with the same digits but in reverse order, gives a perfect square. Find all such numbers.
Let $N=10 a+b$. Then, according to the problem:

$$
10 a+b+10 b+a=11(a+b)=k^{2}
$$

Thus, $k^{2}$ is divisible by $11$, $k$ is divisible by $11$, and then $k^{2}$ is divisible by $121$. Therefore, $a+b=11$. So, when $a=2,3,4,5,6,7,8,9$, respectively, $b=9,8,7,6,5,4,3,2$, i.e., the sought numbers are: $29,38,47,56,65,74,83,92$.

## § 2. Formulas for Short Multiplication and Division