#### SpecialSequence IMO

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Let $\{a_n\}$ be a sequence defined as $a_n=\lfloor{n\sqrt{2}}\rfloor$ where $\lfloor{x}\rfloor$ indicates the largest integer not exceeding $x$. Show that this sequence has infinitely many square numbers.

If a sequence $\{a_n\}$ satisfies $a_1=1$ and $a_{n+1}=\frac{1}{16}\big(1+4a_n+\sqrt{1+24a_n}\big)$, find the general term of $a_n$.

Find all numbers $n \ge 3$ for which there exists real numbers $a_1, a_2, ..., a_{n+2}$ satisfying $a_{n+1} = a_1, a_{n+2} = a_2$ and$a_{i}a_{i+1} + 1 = a_{i+2}$for $i = 1, 2, ..., n.$