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Algebra Difficulty 7.7 National olympiad, round 2 Prove it Romania

Consider the sequence of rational numbers defined by x1=4/3x_1 = 4/3, and xn+1=xn2/(xn2−xn+1)x_{n+1} = x_n^2/(x_n^2 - x_n + 1), n≥1n \ge 1. Show that the numerator of the lowest term expression of each sum ∑k=1nxk\sum_{k=1}^n x_k is a perfect square.

Solution

It is easily seen that the xnx_n are all rational numbers greater than 11. Rewrite the recurrence formula in the form xn=1/(xn+1−1)−1/(xn−1)x_n = 1/(x_{n+1} - 1) - 1/(x_n - 1), n≥1n \ge 1, to get
∑k=1nxk=1xn+1−1−1x1−1=xn2−xn+1xn−1−3=(xn−2)2xn−1. \sum_{k=1}^{n} x_k = \frac{1}{x_{n+1} - 1} - \frac{1}{x_1 - 1} = \frac{x_n^2 - x_n + 1}{x_n - 1} - 3 = \frac{(x_n - 2)^2}{x_n - 1}.
Finally, express the positive rational number xn−1x_n - 1 in lowest terms, xn−1=a/bx_n - 1 = a/b, to deduce that (a−b)2/(ab)(a-b)^2/(ab) expresses ∑k=1nxk\sum_{k=1}^n x_k in lowest terms. The conclusion follows.

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