His Age in a Square Year

A man was exactly $X$ years old in the year $X^2$. How old was he in $1981$?

Solution

We need a perfect square that could be a year within one lifetime of $1981$. The squares near that era are $$43^2 = 1849, \qquad 44^2 = 1936, \qquad 45^2 = 2025.$$ Of these, $1936$ is the one that lets a person also be alive in $1981$: a man who was $44$ in $1936$ was born in $1936 - 44 = 1892$, and in $1981$ he was $1981 - 1892 = 89$ years old.

The trick is that perfect squares are sparse, only one every couple of decades in this range, so the single condition “$X$ years old in year $X^2$” fixes the birth year almost completely. The same puzzle posed today turns on the next square: someone who is $45$ in $2025$ was born in $1980$.