In a two-digit number, the sum of the squares of the digits is as much as the number increased by the product of the digits. If we subtract 36 from the number, the result has the digits in reverse order. Which is the number?
If $10 x+y$ is the sought number, then according to the problem

$$
x^{2}+y^{2}=10 x+x y, \quad \text { or } \quad(x-y)^{2}=10 x+y-x y
$$

and

$$
10 x+y+36=10 y+x, \quad \text { or } \quad y=x+4
$$

Substituting the value of $y$ into the first equation:

$$
\begin{gathered}
16=20 x+x+4-x(x+4) \\
x^{2}-7 x+12=0
\end{gathered}
$$

from which

$$
\begin{array}{ll}
x_{1}=4, & x_{2}=3 \\
y_{1}=8, & y_{2}=7
\end{array}
$$

Therefore, 48 and 37 are the sought numbers.

Number of solutions: 13.