Solve $\lfloor \sqrt x \rfloor = \lfloor x/2 \rfloor$ for real $x$ I'm trying to solve $$\lfloor \sqrt x \rfloor = \left\lfloor \frac{x}{2} \right\rfloor$$ 
for real $x$.
Obviously this can't be true for any negative reals, since the root isn't defined for such. 
My approach is the following: Let $x=:n+r$, $n \in \mathbb{N}_0, 0\leq r < 1$.
For the left hand side $\lfloor \sqrt {n+r} \rfloor = \lfloor \sqrt {\lfloor n+r\rfloor} \rfloor = \lfloor \sqrt n \rfloor$ holds (without further proof).
$$\left\lfloor \frac{x}{2} \right\rfloor = \left\lfloor \frac{n+r}{2} \right\rfloor = \left\{ \begin{array}{l l} \frac{n}{2} & \quad \text{for n even} \\ \frac{n-1}{2} & \quad \text{for n odd} \end{array}\right.$$
Now I don't really know if that'd lead me in the right direction, but I'll write my thoughts down anyways.
Let $\sqrt n =: n'+r'$, $n \in \mathbb{N}_0, 0\leq r < 1$.
Therefore $\lfloor n'+r' \rfloor = n'$. And $n = (n'+r')^2 = n'^2 +2n'r' +r'^2$
For which $n',r'$ holds $$ n' < n'^2 + 2n'r' + r'^2.$$
Well and now I'm stuck and don't know how to proceed.
I'd appreciate any help.

Clearly $x$ can’t be negative, and $\sqrt{x}$ and $\frac{x}2$ can’t be too far apart. They’re equal for $x=4$, and after that $\frac{x}2$ exceeds $\sqrt{x}$ by more and more as $x$ increases. Thus, we should focus initially on relatively small non-negative values of $x$. Points where one or the other floor increases are perfect squares and even integers, so we look at intervals that have such integers as endpoints.
For $x\in[0,1)$, we clearly have
$$\lfloor\sqrt{x}\rfloor=0=\left\lfloor\frac{x}2\right\rfloor\;.$$
For $x\in[1,2)$ we have $\lfloor\sqrt{x}\rfloor=1$, but $\left\lfloor\frac{x}2\right\rfloor=0$. So far, then, we have solutions $[0,1)\cup\{2\}$.
For $x\in[2,4)$ we have $$\lfloor\sqrt{x}\rfloor=1=\left\lfloor\frac{x}2\right\rfloor\;,$$
and for $x\in[4,6)$ we have
$$\lfloor\sqrt{x}\rfloor=2=\left\lfloor\frac{x}2\right\rfloor\;,$$
so the solution set has expanded to $[0,1)\cup[2,6)$. I claim that for $x\ge 6$, however, the separation between $\sqrt{x}$ and $\frac{x}2$ is too large, and we always have
$$\lfloor\sqrt{x}\rfloor<\left\lfloor\frac{x}2\right\rfloor\;.$$
First, for $x\in[6,9)$ we have $\lfloor\sqrt{x}\rfloor=2$, but $\left\lfloor\frac{x}2\right\rfloor\ge 3$. Now suppose that $n\in\Bbb Z^+$, $n\ge 3$, and $n^2\le x<(n+1)^2$. Then $\lfloor\sqrt{x}\rfloor=n$, and
$$\left\lfloor\frac{x}2\right\rfloor\ge\left\lfloor\frac{n^2}2\right\rfloor\;.$$
We’re done if we can show that $\left\lfloor\frac{n^2}2\right\rfloor\ge n+1$ for $n\ge 3$. And this is an easy proof by induction. The inequality certainly holds when $n=3$. Suppose that it holds for some $n\ge 3$. Then
$$\left\lfloor\frac{(n+1)^2}2\right\rfloor=\left\lfloor\frac{n^2+2n+1}2\right\rfloor=\left\lfloor\frac{n^2+1}2\right\rfloor+n\ge\left\lfloor\frac{n^2}2\right\rfloor+n\ge 2n+1>n+2\;,$$
and the induction step is complete.
