In order to solve $\lfloor \sqrt{x}\rfloor =\sqrt{\lfloor x\rfloor }$ , how to express the set defined by : $z^2 \leq x\lt z^2+1 (\space z\in Z$ ) My original goal is to solve for $x$ in $\mathbb R$ :
$$\lfloor \sqrt{x}\rfloor =\sqrt{\lfloor x\rfloor }$$
(1) Since $\lfloor \sqrt{x}\rfloor \in \mathbb Z$ (by definiton), it must be the case that : $\sqrt{\lfloor x\rfloor }  \in \mathbb Z$
( 2) In general, if $\sqrt A =z , \text {with}\space \space  z \in \mathbb Z$, $A$ can be written as $A= z^2$. Here, $A= \lfloor x\rfloor$, so $\lfloor x\rfloor = z^2, \space z\in \mathbb Z$.
(3) In general : $ \lfloor x\rfloor = F_{\in \mathbb Z } \iff  F\leq x \lt F+1$. Here, $F= z^2$ , so  $\lfloor x\rfloor = z^2 \iff z^2 \leq x \lt z^2 +1$, with, as before, $z\in \mathbb Z$.
(4) Finally $x$ must be a positive or null real number , unless $\lfloor x\rfloor \ngeq 0 $, and $\sqrt {\lfloor x\rfloor}$ would not be defined.
Consequently, the solution set $S$ of the original equation is :
$$S= \{x| x\geq 0 \land z^2 \leq x \lt z^2 +1 \space \space\text {for some } \space  z\in \mathbb Z\}$$
My questions are :

(1) is the expression of the solution set correct?
(2) what is this solution set equal to? Is it possible to find a more explicit definition of $S$?

I can see that $[0,1)$ is a subset of $S$, because every $x$ in this interval satisfies
$ 0^2 \leq x \lt 0^2 +1$.
Same thing for $[1, 2)$ since every $x$ in this interval satisfies $ 1^2 \leq x \lt 1^2 +1$.
Same thing for $[4, 5)$, with here $z=2$.
But what if $x\in [2, 3)$?
Certainly if $x\in [2,3)$ there is a $z\in \mathbb Z$ such that $z^2 \leq x$, and the closest one is $1$ . But, no $x$  in this interval satisfies $1^2\leq x \lt 1^2 +1$.
However, Wolfram Alpha suggests that the solution set  of the equation  $ \lfloor \sqrt{x}\rfloor -\sqrt{\lfloor x\rfloor} = 0 $ is the positive part of the $X-$ axis, that is $\mathbb R^{+}$.

 A: Edit: at first, I did not notice that your concern was How to express the solution set.
It simply is the union of intervals $S =\displaystyle \bigcup_{k = 0,1,2,\ldots} \left[ k^2, k^2+1 \right)$.
The solution I originally wrote is very similar to yours.

A first remark is that any solution must be nonnegative, since otherwise, $\sqrt{x}$ has no sense.
Let us show that the set of solutions is $\bigcup_{k\in \{0,1,\ldots,\}} [k^2,k^2+1)$.
First, assume that $x\geqslant 0$ satisfies $\lfloor \sqrt{x} \rfloor = \sqrt{\lfloor x \rfloor}$.
Let $k = \lfloor \sqrt{x} \rfloor$.
Then $\lfloor x \rfloor = \sqrt{\lfloor x \rfloor}^2 = k^2$.
Hence, by definition
$$
k^2 \leqslant x < k^2+1.
$$
Conversly, let $x \geqslant 0$ be such that its integer part is a square, that is there exists $k\in \{0,1\ldots,\}$ such that $k^2 \leqslant x < k^2+1$.
Since the square-root function is strictly increasing on $\Bbb R_+$, we have $k \leqslant \sqrt{x} < \sqrt{k^2+1}$.

*

*If $k=0$, there is nothing else to prove.

*Assume that $k>0$.
Recall that for all $h\geqslant 0$, we have $\sqrt{1+h} \leqslant 1+ \frac{1}{2}h$.
It follows that
$\sqrt{k^2+1} = k\sqrt{1+\frac{1}{k^2}} \leqslant k + \frac{1}{2k} \leqslant k + 1$.
Hence,
$$
k \leqslant \sqrt{x} < k+1,
$$
and therefore, $ \sqrt{\lfloor x \rfloor} = k = \lfloor\sqrt{x}\rfloor$.

