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.
 A: $$x=\sqrt{x+1}+\sqrt{x+2}+\sqrt{x+3}$$
$$x-\sqrt{x+1}=\sqrt{x+2}+\sqrt{x+3}$$
$$x^2+x+1-2x\sqrt{x+1}=2x+5+2\sqrt{x^2+5x+6}$$
$$x^2-x-4=2(\sqrt{x^3+x^2}+\sqrt{x^2+5x+6})$$
$$x^4-2x^3-7x^2+8x+16=4(x^3+2x^2+5x+6+2\sqrt{x^5+6x^4+11x^3+6x^2})$$
$$x^4-6x^3-15x^2-12x-8=8\sqrt{x^5+6x^4+11x^3+6x^2}$$
$$x^8-12x^7+6x^6+156x^5+353x^4+456x^3+384x^2+192x+64=64x^5+384x^4+704x^3+384x^2$$
$$x^8-12x^7+6x^6+92x^5-31x^4-248x^3+192x+64=0$$
This equation has $6$ real roots, but only the one I mentioned in my comment fulfills the original equation. Therefore, it is the only real solution.
A: Since $x$ is the sum of three square roots, $x$ is positive. Thus there exists a positive real $y$ such that $x=y^2-2$. 
The equation then becomes
$$y^2-2=\sqrt{y^2-1}+\sqrt{y^2}+\sqrt{y^2+1}$$
$$y^2-y-2=\sqrt{y^2-1}+\sqrt{y^2+1}$$
Squaring gives
$$y^4-2y^3-3y^2+4y+4=2y^2 + 2\sqrt{y^4-1}$$
$$y^4-2y^3-5y^2+4y+4=\sqrt{4y^4-4}$$
Squaring again
$$y^8-4 y^7-6 y^6+28 y^5+17 y^4-56 y^3-24 y^2+32 y+16=4y^4-4$$
$$y^8-4 y^7-6 y^6+28 y^5+13 y^4-56 y^3-24 y^2+32 y+20=0$$
This octic has six real roots, of which five are extraneous. The approximate result is $$y \approx 3.56021$$
Using $x=y^2-2$ this gives $$x \approx 10.67507$$
A: 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.
