There's a problem statement:
For each $m \in \mathbb{N}$, we construct a sequence $m_0$, $m_1$, $m_2,\dots$ denoted $S_m$, recursively via $m_0=m$ and
$$m_{i+1} = m_i + \left\lfloor \sqrt{m_i} \right\rfloor$$
for all $i \ge 0$. Here, $\lfloor x \rfloor$ is the floor of $x$, the greatest integer less than or equal to $x$. Hence, we have $\left\lfloor \sqrt{10} \right\rfloor=3$ and $\left\lfloor \sqrt{29} \right\rfloor=5$.
Show that for each positive integer $m$, the sequence $S_m$ contains the square of some integer.
I'm pretty certain that this can be proved with induction. I am just not quite sure what to induct on. Examining examples shows that $S_m$ always results in a perfect square eventually, though I'm not sure how to prove it.
Suppose our $m$ is less than some square, say, $k^2$. Then let's go up until we get over $k^2$. (If we hit it, we just stop right there.) Let's denote the first term above $k^2$ as $k^2+l$. Since the last step was less than $k$, it follows that $l<k$. It can be said that we missed this square by $l$.
Fine. Let's make two more steps. The first would get us to $k^2+k+l$, which is still less than the next square, so the $\lfloor \sqrt{m_i} \rfloor$ is still $k$. The second would get us to $k^2+2k+l = (k+1)^2+l-1$.
See that? We missed the next square by $l-1$. Well, let's do two more steps and miss the next square after that by $l-2$, and so on, until we hit some square spot on. That would be it.
Q.e.d.