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## Recurrent Sequence implies Constant

Prove that the sequence which satisfies the relation $a_{n+1}=\sqrt{6-2a_n^2},\ \forall n\geq 1$ is constant.
Laurentiu Panaitopol, Romania
Solution: We can translate the recurrence relation in the following way:
$a_{n+1}^2+2a_n^2=6,\ (0)$. Therefore $a_{n+2}^2+2a_{n+1}^2=6=a_{n+1}^2+2a_n^2$, which implies $a_{n+2}^2-a_{n+1}^2=(-2)(a_{n+1}^2-a_n^2),\ (1)$.
From $(0)$ it follows that $(a_n)$ is bounded, and from $(1)$ it follows that $y_n=a_{n+1}^2-a_n^2$ is unbounded if it is not zero. This would contradict the fact that $(a_n)$ is bounded. Therefore $(y_n)$ is bounded and $(a_n^2)$ is constant. Because $a_n \geq 0,\ \forall n \geq 2$ we get that $(a_n)$ is constant.