9.5. The sides of a rectangle are expressed in integers. What should their lengths be so that the perimeter of the rectangle is numerically equal to its area?
Solution: Denoting the sides of the rectangle by $x$ and $y$, we form the equation

$$
2 x+2 y=x y, \text{ from which } x=\frac{2 y}{y-2}
$$

Since $x$ and $y$ must be positive, the number $y-2$ must also be positive, i.e., $y$ must be greater than 2.

Notice now that

$$
x=\frac{2 y}{y-2}=\frac{2(y-2)+4}{y-2}=2+\frac{4}{y-2}
$$

Since $x$ must be an integer, the expression $\frac{4}{y-2}$ must also be an integer. However, for $y>2$, this is only possible if $y$ equals 3, 4, or 6. The corresponding values of $x$ will be $6, 4, 3$.

Thus, the desired figure is either a rectangle with sides 3 and 6, or a square with side 4.

Answer: rectangle with sides 3 and 6; square with side 4.