Competition problem — Algebra — Olympiad Maths Prep

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Algebra Difficulty 5.2 AIME, harder Prove it Mongolia

The sequence $\{x_n\}_{n \geq 1}$ is defined as follows: $x_1 = x_2 = 1$ and $x_{n+2} = x_{n+1} + x_n + 2\sqrt{x_{n+1} x_n + 3}$ for all $n \geq 2$ . Prove that $x_n$ is integer for all $n \geq 2$ .

(proposed by Bat. Bayarjargal)

Solution

We will show that by induction method.
For $n = 1$ is trivial. Now suppose that $x_1, x_2, ..., x_n, x_{n+1}$ are integers. From the given recurrence we get $x_n - x_{n+1} - x_{n+2} = 2\sqrt{x_{n+1} \cdot x_{n+2} + 3}$ . Hence $2\sqrt{x_{n+1} \cdot x_{n+2} + 3}$ is integer. This is showing us $x_{n+3} = x_{n+2} + x_{n+1} + 2\sqrt{x_{n+1} \cdot x_{n+2} + 3}$ is integer. Proof is completed.

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