Prove that $M = \mathbb Z^+$ 
Let $M$ be a nonempty subset of $\mathbb Z^+$ such that for every element $x$ in $M,$ the numbers $4x$ and $\lfloor \sqrt x \rfloor$ also belong to $M.$ Prove that $M = \mathbb Z^+$.

Suppose $a \in M$. Then so are $4ak$ and $\lfloor \sqrt{4ak} \rfloor$ for every positive integer multiple of $k$. Also we could take a multiple of $4$, then do the square root and floor it etc. in many different combinations. How do we prove that $M$ is all of $\mathbb{Z}^+$?
 A: Starting at any $m \in M$ and repeatedly applying $f(x)=[\sqrt{x}]$ (I use $[u]$ for floor of $u$) one eventually gets $1 \in M,$ then repeatedly multiplying by $4$ we have $4^k \in M,$ then applying $f$ to these we have $2^k \in M$ for all $k,$ i.e. $M$ contains the powers of $2.$
In what follows we use "integer intervals" like $[r,s)$ to mean the intersection of the corresponding intervals of reals with the integers. In particular for any $x,$ $[x,x+1)$ just means the singleton $\{x\},$ and for example $[1,5)$ consists of $\{1,2,3,4\}.$
We will be interested in the inverse image under $f$ of one of these integer intervals $[a,b),$ and claim that inverse image is $[a^2,b^2).$ This is easy to check. In turn, the inverse image of that inverse image of $[a,b)$ will be $[a^4,b^4).$ The idea of the rest of the proof that one can get a number $m$ is to show that a high enough inverse image of the interval $[m,m+1)$ will contain at least one power of $2,$ and since we know these are in $M,$ we can get to $m$ from some high enough power of $2$ on applying $f$ enough times.
So for fixed $m$ and  $k \ge 1$ let $I_k$ be the interval
$$I_k=[m^{2^k},\ (m+1)^{2^k}).$$
Any integer in $I_k$ will, after $k$ applications of $f,$ give $m.$ So we just need to show that for large enough $k$ there is a power of $2$ in $I_k.$ To avoid the problem at the right endpoint that we don't want the only such power of 2 to be that endpoint, we proceed to arrange that in fact there be at least two powers of $2$ in $I_k.$
We use the notation $\rm{lg}\  x$ for the log base $2,$ and note that $2^r \in I_k$ is equivalent to $r$ lying somewhere in the interval
$$J_k=[2^k \rm{lg} (m),\  2^k \rm{lg} (m+1)).$$
So if we take $k$ large enough that $J_k$ has length greater than $2,$ then there will be two values of $r$ in interval $J_k,$ and so there will be two different values of $k$ such that $2^k$ lies in interval $I_k$ as desired. Now the length of interval $J_k$ is $2^k\ \rm{lg}\frac{m+1}{m},$ so we can easily make $J_k$ have length greater than $2.$
Added later: As @user84413 points out in a comment, we only need the length of $J_k$ to be at least 1. It is a half-open interval of reals, and any such interval of length 1 or more will have an integer in it, which can be the $r$ from which $2^r$ lies in $I_k.$ This cuts down the size if computing an example.
A: Use the fact that $\lfloor \sqrt{x} \rfloor \in M$ for every $x \in M$ to show that $1 \in M$, hence $M$ contains every multiple of $4$. To finish, use the fact that there is always a multiple of $4$ between any two consecutive squares. 
