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

## Solution

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

By inspection or by trying a few small values, we can find the solution $x_1 = 3$, $y_1 = 2$.

1. 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 $(x_{n+1})^2 - 2(y_{n+1})^2 = (x_n)^2 - 2(y_n)^2.$

Substituting the expressions for $x_n$ and $y_n$ into this equation, we get $(3x_n + 4y_n)^2 - 2(ax_n + by_n)^2 = (x_n)^2 - 2(y_n)^2$ that is, $(8 - 2a^2)(x_n)^2 + (24 - 4ab)x_ny_n + (18-2b^2)(y_n)^2 = 0.$

This equation will certainly hold if all three coefficients are zero. In this case, we have

$2a^2 = 8, \quad 4ab = 24, \quad 2b^2 = 18.$

These equations DO solve, to give $a = \pm 2, b = \pm 3$. Since we also require that $a$ and $b$ are positive, we have $a = 2$ and $b = 3$.

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

We start with our previously found small solution, $x_1 = 3$ and $y_1 = 2$.

Now we can use the sequences defined above with $a=2, b=3$ to generate new solutions. We have $x_1 = 3, \quad y_1 = 2;$ $x_2 = 3 \times 3 + 4 \times 2 = 17, \quad y_2 = 2 \times 3 + 3 \times 2 = 12;$ $x_3 = 3 \times 17 + 4 \times 12 = 99, \quad y_3 = 2 \times 17 + 3 \times 12 = 70.$ So $X = 99$ and $Y = 70$ are such a pair.

1. (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?

Let’s suppose that $\dfrac{x_n}{y_n}$ tends to a limit $L$ as $n$ tends to infinity.

Then for large $n,$ $L \approx \dfrac{x_{n+1}}{y_{n+1}} = \dfrac{3x_{n}+4y_n}{2x_n+3y_{n}}$ $=\dfrac{3\dfrac{x_{n}}{y_{n}}+4}{2\dfrac{x_{n}}{y_{n}}+3}$ $\approx \dfrac{3L+4}{2L+3}.$

Thus $L \approx \dfrac{3L+4}{2L+3}$, which becomes equality when we allow $n$ to tend to infinity. Thus by solving the equation, $L = \sqrt{2}.$

Alternatively, we could argue like this:

For all the pairs $(x_n,y_n)$, we have $(x_n)^2 - 2(y_n)^2 = 1.$

Also, because of the way $x_n$ and $y_n$ are constructed, both form increasing sequences.

So if we divide through by $(y_n)^2$, we see that $\left(\frac{x_n}{y_n}\right)^2 - 2 = \frac{1}{(y_n)^2} \approx 0 \quad \text{for large n},$ so $x_n/y_n \approx \sqrt{2}$, since $x_n$ and $y_n$ are both positive.

This question may remind you of Staircase sequences.