The measures in centimeters of the sides of each of the two squares are integers. If the smaller square had $2001 \mathrm{~cm}^{2}$ more in area, the two squares would be equal. How much can the side of the larger square measure?

![](https://cdn.mathpix.com/cropped/2024_05_01_9ca8b777b16e5f5c5786g-2.jpg?height=394&width=377&top_left_y=974&top_left_x=1342)
If $a$ is the side length of the larger square and $b$ is the side length of the smaller square, then from the statement we have

$$
a^{2}=b^{2}+2001
$$

Thus:

$$
2001=a^{2}-b^{2}=(a+b)(a-b)
$$

Since $a$ and $b$ are integers, it follows that $a+b$ and $a-b$ are divisors of 2001. But, $2001=3 \times 23 \times 29$, so

$$
(a+b)(a-b)=2001 \times 1=667 \times 3=87 \times 23=69 \times 29
$$

Consequently, there are 4 possible ways to factor 2001 into two factors:

- if $a+b=2001$ and $a-b=1 \Longrightarrow a=\frac{(a+b)+(a-b)}{2}=1001$;
- if $a+b=667$ and $a-b=3 \Longrightarrow a=\frac{667+3}{2}=335$;
- if $a+b=87$ and $a-b=23 \Longrightarrow a=\frac{87+23}{2}=55$;
- if $a+b=69$ and $a-b=29 \Longrightarrow a=\frac{69+29}{2}=49$.

Therefore, the possible values for the larger side are: $1001, 335, 55$, and 49.