Problem 3. Marco was born between 1300 and 1400, and Darko between 1400 and 1500. Both were born on April 6th, in years that are perfect squares of natural numbers. Both lived for 110 years. In which year, while they were alive, were their ages on April 7th perfect squares of natural numbers.
Solution. Since $36^{2}=1296, 37^{2}=1369, 38^{2}=1444, 39^{2}=1521$, it is clear that the only perfect square between 1300 and 1400 is 1369, and between 1400 and 1500 is 1444. Therefore, Marco was born in 1369, and Darco was born in 1444.

Assume that on April 7th of some year, Marco was $m^{2}<111$ years old, and Darco was $n^{2}<111$ years old, where $m, n \in \mathbb{N}$. Marco turned $m^{2}$ years old in the year $1369+m^{2}$, and Darco turned $n^{2}$ years old in the year $1444+n^{2}$. From the condition of the problem, it follows that in the same year, they must have ages that are perfect squares of natural numbers, so

$$
1369+m^{2}=1444+n^{2} \text{, i.e., } m^{2}-n^{2}=75
$$

The perfect squares less than 11 are $1, 4, 9, 16, 25, 36, 49, 64, 81$, and 100, and the only ones whose difference is 75 are $m^{2}=100$ and $n^{2}=25$. Therefore, in the year $1369+10^{2}=1469$, Marco was 100 years old, and Darco was 25 years old.

Note. In the set of natural numbers, the equation $m^{2}-n^{2}=75$ can be solved by using the factorization

$$
(m-n)(m+n)=1 \cdot 75=3 \cdot 25=5 \cdot 15
$$

where, since $0<m-n<m+n$, we get three systems of linear equations, whose solutions are $m=38, n=37, m=14, n=11$, and $m=10, n=5$, of which only in the solution $m=10, n=5$ the squares of the found numbers are less than 111.