Henry Ernest Dudeney/Modern Puzzles/86 - A New Street Puzzle/Solution

by : $86$

 * A New Street Puzzle

Solution
Brown lived at No. $84$ on a street of $119$ houses.

Proof
Let there be $m$ houses in the street, where we are told $20 < m < 500$.

Let the man live at no. $n$.

We have that:

Hence this problem is equivalent to finding the answers to Pell's Equation:
 * $2 x^2 - 1 = y^2$

or equivalently:
 * $y^2 - 2 x^2 = -1$

From the solution to Pell's Equation, the solutions to this are:


 * ${p_r}^2 - 2 {q_r}^2 = \paren {-1}^r$ for $r = 1, 2, 3, \ldots$

where $\dfrac {p_r} {q_r}$ are the convergent of the Continued Fraction Expansion of $\sqrt 2$.

It is the odd integer values of $r$ that we need in order to make the equal to $-1$.

From Continued Fraction Expansion of Root 2:

Hence we have $x$ and $y$, and thence $n = \dfrac {x - 1} 2$ and $m = \dfrac {y - 1} 2$ as:


 * $\begin{array} {r|r} x & y & n & m \\ \hline

1 & 1 & 0 & 0 \\ 5 & 7 & 2 & 3 \\ 29 & 41 & 14 & 20 \\ 169 & 239 & 84 & 119 \\ 985 & 1393 & 492 & 696 \\ \end{array}$

Because there are between $20$ and $500$ houses in the street, we know the man lives at no. $84$ in a $119$-house street.