Henry Ernest Dudeney/Puzzles and Curious Problems/46 - A Square Family/Solution
Puzzles and Curious Problems by Henry Ernest Dudeney: $46$
- A Square Family
- A man had nine children, all born at regular intervals,
- and the sum of the squares of their ages was equal to the square of his own.
- What were the ages of each?
- Every age was an exact number of years.
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:
\(\ds f^2\) | \(=\) | \(\ds a^2 + \paren {a + d}^2 + \paren {a + 2 d}^2 + \cdots + \paren {a + 8 d}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^8 \paren {a + d i}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^8 \paren {a^2 + 2 a d i + \paren {d i}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^8 a^2 + \sum_{i \mathop = 0}^8 2 a d i + \sum_{i \mathop = 0}^8 d^2 i^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a^2 \sum_{i \mathop = 0}^8 1 + 2 a d \sum_{i \mathop = 1}^8 i + d^2 \sum_{i \mathop = 1}^8 i^2\) | as two of the summations vanish at $i = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 9 a^2 + 2 a d \dfrac {8 \paren {8 + 1} } 2 + d^2 \dfrac {8 \paren {8 + 1} \paren {2 \times 8 + 1} } 6\) | Closed Form for Triangular Numbers, Sum of Sequence of Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds 9 a^2 + 72 a d + 204 d^2\) | evaluating |
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:
\(\ds \paren {3 b}^2\) | \(=\) | \(\ds 9 a^2 + 216 a + 1836\) | for some integer $b$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds b^2\) | \(=\) | \(\ds a^2 + 24 a + 204\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a + 12}^2 + 60\) |
We now need to find two squares which differ by $60$.
By the Odd Number Theorem we have:
\(\ds 14^2 + \paren {2 \times 14 + 1}\) | \(=\) | \(\ds 15^2\) | ||||||||||||
\(\ds 15^2 + \paren {2 \times 15 + 1}\) | \(=\) | \(\ds 16^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 16^2 - 14^2\) | \(=\) | \(\ds 60\) |
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.
$\blacksquare$
Sources
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $46$. -- A Square Family
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $41$. A Square Family