Can we find a sequence of solutions to the equation x^2

Can we find a sequence of solutions to the equation $x^2 - 2y^2 = 1?$

Question

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

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 $x_{n+1} = 3x_n + 4y_n, \quad y_{n+1} = ax_n + by_n, \quad \text{for }n \ge 1\text{}.$ Find positive values for $a$ , $b$ such that $(x_{n+1})^2 - 2(y_{n+1})^2 = (x_n)^2 - 2(y_n)^2.$

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

(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?

Suggestion

Can you combine what we know to form an equation involving $a$ and $b$ ?

How can we use what we have learnt from (i) and (ii)?

Solution

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

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 $b = 3$ .

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.

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 $\left(\frac{x_n}{y_n}\right)^2 - 2 = \frac{1}{(y_n)^2} \approx 0 \quad \text{for large }n\text{},$ 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.