Review question

# Can we find a sequence of solutions to the equation $x^2 - 2y^2 = 1?$ Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8033

## Question

1. Find a pair of positive integers, $x_1$ and $y_1$, that solve the equation $(x_1)^2 - 2(y_1)^2 = 1.$

2. Given integers $a$, $b$, we define two sequences $x_1$, $x_2$, $x_3$, $\dots$ and $y_1$, $y_2$, $y_3$, $\dots$ by setting $x_{n+1} = 3x_n + 4y_n, \quad y_{n+1} = ax_n + by_n, \quad \text{for n \ge 1}.$

Find positive values for $a$, $b$ such that $(x_{n+1})^2 - 2(y_{n+1})^2 = (x_n)^2 - 2(y_n)^2.$

3. Find a pair of integers $X$, $Y$ which satisfy $X^2 - 2Y^2 = 1$ such that $X > Y > 50$.

4. (Using the values of $a$ and $b$ found in the second part) what is the approximate value of $x_n/y_n$ as $n$ increases?