Given $f(n) = n + \left \lfloor{\sqrt n}\right \rfloor$
Prove that for any natural number $n$, the sequence $n, f(n), f(f(n)),.......$ must always contain at least one perfect square.
Now obviously if we take $n$ to be a perfect square, we are done since the sequence contains $n$. So I started by looking at the behaviour of the function at $n = 2$, and within $2$ iterations we reach $4$. Then I looked at $n = 5$, and that reaches $9$ in $2$ iterations. However, $6$ is missed in the sequence as $f(5) = 7$ and this function is strictly increasing. So I looked at the sequence when $n = 6$, which reaches $16$ in $4$ iterations.
Trying to generalise this is a bit difficult, however, as I tried to let $k^2 < a < (k+1)^2$ for some natural number $a$.
Clearly, $f(a) = a + k$, however I am stuck here. Is this approach worth taking or is there something completely different required?
