if $x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}$ then x =? I have got a new question from my friend and it made me nervous:
$$\text{ If }\,x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}, \text{ then }\,x = \;?$$
A lot of thank you to all comments.

Note that $x$ is positive, so we can write $x=y^2-2$, for some real number $y>\sqrt{2}$ and the equation becomes
$$y^2-2=\sqrt{y^2-1}+y+\sqrt{y^2+1}$$
we obtain thus 
$\sqrt{y^2+1}=y^2-y-2-\sqrt{y^2-1}$, hence
$$y^2+1=(y^2-y-2)^2+(y^2-1)-2(y^2-y-2)\sqrt{y^2-1}.$$
This yields
$$2(y^2-y-2)\sqrt{y^2-1}=(y^2-y-2)^2-2$$
so 
$4(y^2-y-2)^2(y^2-1)=((y^2-y-2)^2-2)^2$. Hence $y$ is a positive real root of the polynomial
$$\begin{array}{rcl}
P(y)&=&((y^2-y-2)^2-2)^2-4(y^2-y-2)^2(y^2-1)\\
&=&y^8-4y^7-6y^6+28y^5+13y^4-56y^3-24y^2+32y+20\end{array}$$
There is only one real root $y>\sqrt{2}$ of this polynomial.
