- 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 \(x_1 = 3\), \(y_1 = 2\).

- 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 \(b = 3\).

- 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.

- (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.