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

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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 \[x_{n+1} = 3x_n + 4y_n, \quad y_{n+1} = ax_n + by_n, \quad \text{for $n \ge 1$}.\]

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?