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

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-017.jpg?height=197&width=203&top_left_y=518&top_left_x=1669)
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 know that $a^{2}=b^{2}+2001$. Therefore, $2001=a^{2}-b^{2}=(a+b)(a-b)$, and since $a$ and $b$ are integers, $a+b$ and $a-b$ are divisors of 2001. But, $2001=3 \times 23 \times 29$, so we have 4 possible ways to factor $2001$ into two factors, namely,

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

Since $(a+b)+(a-b)=2a$, it follows that

(a) $a+b=2001$ and $a-b=1$, in which case $a=\frac{2001+1}{2}=1001$;

(b) $a+b=667$ and $a-b=3$, in which case $a=\frac{667+3}{2}=335$;

(c) $a+b=87$ and $a-b=23$, in which case $a=\frac{87+23}{2}=55$;

(d) $a+b=69$ and $a-b=29$, in which case $a=\frac{69+29}{2}=49$.

Thus, the possible values for the larger side are 1001, 335, 55, and 49 cm.