[Algebra]
Problem: Dan was born in a year between 1300 and 1400. Steve was born in a year between 1400 and 1500. Each was born on April 6 in a year that is a perfect square. Each lived for 110 years. In what year while they were both alive were their ages both perfect squares on April 7?

Solution: First, we determine the perfect squares between 1300 and 1400 and between 1400 and 1500.

Since $\sqrt{1300} \approx 36.06$, then the first perfect square larger than 1300 is $37^{2}=1369$.

The next perfect squares are $38^{2}=1444$ and $39^{2}=1521$.

Since Dan was born between 1300 and 1400 in a year that was a perfect square, then Dan was born in 1369.

Since Steve was born between 1400 and 1500 in a year that was a perfect square, then Steve was born in 1444.

Suppose that on April 7 in some year, Dan was $m^{2}$ years old and Steve was $n^{2}$ years old for some positive integers $m$ and $n$. Thus, Dan was $m^{2}$ years old in the year $1369+m^{2}$ and Steve was $n^{2}$ years old in the year $1444+n^{2}$.

Since these represent the same years, then $1369+m^{2}=1444+n^{2}$, or $m^{2}-n^{2}=1444-$ $1369=75$.

In other words, we want to find two perfect squares less than 110 (since their ages are less than 110) whose difference is 75.

The perfect squares less than 110 are $1,4,9,16,25,36,49,64,81,100$.

The two that differ by 75 are 100 and 25 .

Thus, $m^{2}=100$ and $n^{2}=25$.

This means that the year in which the age of each of Dan and Steve was a perfect square was the year $1369+100=1469$.

Final answer: 1469