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Show that (b_n)^2 = 3(a_n)^ 2 + 1

Source: Canada National Mathematical Olympiad 1988 - Problem 4

October 3, 2011
number theory

Problem Statement

Let xn+1=4xn−xn−1x_{n + 1} = 4x_n - x_{n - 1}, x0=0x_0 = 0, x1=1x_1 = 1, and yn+1=4yn−yn−1y_{n + 1} = 4y_n - y_{n - 1}, y0=1y_0 = 1, y1=2y_1 = 2. Show that for all n≥0n \ge 0 that yn2=3xn2+1y_n^2 = 3x_n^2 + 1.
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