<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] Side comment:  $\frac{\sqrt{m-1} + \sqrt{m+1}}2 \ge \sqrt{\sqrt{m^2 - 1}}>\sqrt m$ by $AM-GM$ theorem.
So $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} \approx 3\sqrt{x+1}$ but slightly larger.
$x \le \sqrt{x} + \sqrt{x+1} + \sqrt{x+2}< 3\sqrt{x+1} < x+1$ implies $x^2 < 9x + 9 < x^2 + 2x +1$ (the first inequality is definite, the second is aproximate).
$x^2 <9x+9$ (which is definitely true) implies $x^2 - 9x -9 = (x -\frac {9+\sqrt{81 + 4*9}}2)(x -\frac {9-\sqrt{81 + 4*9}}2) < 0$ which implies $x < \frac {9+\sqrt{81 + 4*9}}2= \frac {9+3\sqrt{13}}2\approx 9.9$ but is *definitely less than $10$.  $9$ is a definite upper limit of $x$
$9x + 9 < x^2 + 2x + 1$ (which is only approximate) implies $x^2 - 7x -8 = (x-8)(x+1) > 0$ implies $x > 8$ (or $x < 1$ which would not be possible.)  
So $9$ is a solution and $8$ is possible if $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} < 9 \le 3\sqrt{x+1}$ ... which, ... actually is the case for $x= 8$. (As $3\sqrt{8+1} = 9$ exactly.)
So solutions are $8$ and $9$.
Hmmm.... I haven't actually shown it fails for $x < 8$ but for that to hold the difference between $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2}$ and $3\sqrt{x+1}$ must be larger than $1$ and that's .... well, clearly not true for $x \approx 8$.  We can just check that $7$ fails.
I'm also assuming the rate of which $x$ increases will "outstrip" $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2}$.  Calculus verifies this. 
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