Problem 4. A rectangle and a square have equal areas. The numerical values of their dimensions are natural numbers from the first ten. If the width of the rectangle is $2 \mathrm{~cm}$, determine the dimensions of the rectangle and the square.
Solution. Let $P_{1}$ be the area of a square with side $a$. Let $P_{2}$ be the area of a rectangle with length $b$. Then, from the conditions of the problem, we have $P_{1}=a \cdot a=2 \cdot b=P_{2}$, where $a, b$ are numbers from the first ten and $a \neq 2$. From $a \cdot a=2 \cdot b$, we get that the number $2 \mid a$, i.e., $a \in\{4,6,8\}$. For $a=4$, we get $b=8$, and then the square has a side of $4 \, \text{cm}$, and the rectangle has sides of $2 \, \text{cm}$ and $8 \, \text{cm}$. The obtained solution is unique.

Indeed, for $a=6$ we get $b=18$, but then 18 is not a number from the first ten. The same applies to $a=8 \, \text{cm}$.