<s>[INST] Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$ 
I tried to solve this equation.
   First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2} \\\to x \in \mathbb{N}$$ so we can check $x=1,2,3,4,5,6,7,8,9,\ldots$ by a MATLAB program.  I checked the natural numbers to find solution. I found $x=8,9$ worked here.
Now my question is about somehow an analytical solving of the equation, or another idea. Can any one help me? Thanks in advance.  
 [/INST] You can use inequalities to simplify your problem. 
Since $\lfloor x \rfloor \le x$. Therefore we've 
\begin{align}
x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\
&\le \sqrt x+\sqrt {x+1}+\sqrt{x+2} \\
&\le 3\sqrt {x+2}\\
\end{align}
$$\implies x^2 \le 9(x+2) \; ; x \in \mathbb Z$$
This gives us the range $x \in [-1,10] \tag1$.
Also $\lfloor x \rfloor \ge x-1$. Therefore we've
\begin{align}
x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\
&\ge  \sqrt x+\sqrt {x+1}+\sqrt{x+2} \color{red}{-1}\\
&\ge 3\sqrt {x}-1\\
\end{align}
$$\implies x+1 \ge 3\sqrt x$$
$$\implies x^2-7x+1\ge 0 \, ; x \in \mathbb Z$$
This gives us $x \in (-\infty, 0]\cup [7,\infty)\tag2$ 
Taking intersection of $(1)$ and $(2)$, and taking care of domain, I.e. $x\ge 0$, we get
$$\color{blue}{x \in \{0,7,8,9,10\}}$$
Now you can check for $x=0,7,8,9,10$, which is quite easy now.
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