If you take any natural number greater than three, take the square root and round it off, from the result you take the square root and round it off again, and so on. Show that ,at some point, this process leads to $2$ or $3$
I tried finding patterns in different examples, but I can't see any consistency:
$\lfloor(\sqrt{1234})\rfloor=35 ⇒ \lfloor\sqrt{35}\rfloor=5 ⇒ \lfloor\sqrt{5}\rfloor=2$
$\lfloor\sqrt{53123}\rfloor=230 ⇒ \lfloor \sqrt{230}\rfloor=15 ⇒ \lfloor\sqrt{15}\rfloor=3$
...
I also tried proving it by Induction, but I don't know that I even have to show..
If $k$ is that natural number define $a_1=k$ and $a_{n+1}=\lfloor\sqrt{a_n}\rfloor$. $$\lim_{n\to\infty}a_n=1$$ Also, $a_n$ is integer for every $n$. Then, the sequence is eventually constant.
Let $r$ be the least natural number such that $a_r=1$ (well-ordered property for that). Then $a_{r-1}$ is $2$ or $3$.