Henry Ernest Dudeney/Puzzles and Curious Problems/46 - A Square Family/Solution

by : $46$

 * A Square Family

Solution

 * $2, 5, 8, 11, 14, 17, 20, 23, 26$.

Proof
Let $a$ years be the age of the youngest child.

Let $d$ years be the interval between consecutive children.

Let $f$ be the age of the father.

We have:

We have that $f^2$ is a multiple of $3$ and so because it is square a multiple of $9$.

So $9 a^2 + 72 a d + 204 d^2$ is also a multiple of $9$ and so $204 d^2$ is a multiple of $9$.

That is, $d^2$ is a multiple of $9$ and so $d$ is a multiple of $3$.

Because of the nature of the question, it can be assumed that $d = 3$, as otherwise there is the unfeasible gap of $48$ years between eldest and youngest child.

So we have:

We now need to find two squares which differ by $60$.

By the Odd Number Theorem we have:

Thus if $14 = a + 12$ then $a = 2$ and we have:


 * $48^2 = 2^2 + 5^2 + 8^2 + 11^2 + 14^2 + 17^2 + 20^2 + 23^2 + 26^2$

and the solution is complete.