$a_n=\sqrt{n}+\sqrt{n+1}$ by induction For $n=1$ it is true For $n=k+1$: $(a_{k+1}-\sqrt{k}-\sqrt{k+1})(a_{k+1}+\sqrt{k}-\sqrt{k+1})=2$ $(a_{k+1}-\sqrt{k+1})^2=k+2 \to a_{k+1}=\sqrt{k+1}+\sqrt{k+2}$ 2) $ \lfloor\sqrt{n}+\sqrt{n+1}\rfloor=2022$ $2021<\sqrt{n}+\sqrt{n+1} \leq 2022$ $\sqrt{n}<1011$ and $2\sqrt{n+1}>2021$ so $n<1011^2 \to n \leq 1011^2-1$ and $n> \frac{2021^2}{4}-1=\frac{2020*2022}{4}-\frac{3}{4} \to n \geq 1010*1011$ $n \in [1010*1011,1010*1012]$

This is surely the easiest problem on the contest. Note that $a_2 = \sqrt{2} + \sqrt{3}$ by easy computation, have a quick guess and the induction follows. After obtaining the general formula, (2) is killed immediately. I think this problem should be categorized into Grade 7 level, instead of Grade 10.