TITLE: Find the floor value of a finite continued surd
QUESTION [6 upvotes]: Given $x=20062007$, and let
$$A=\sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}.$$
Find the greatest integer not exceeding $A$.

REPLY [7 votes]: $$10x+1<\sqrt{100x^2+39x+\sqrt{3}}<10x+2$$
$$4x+1<\sqrt{16x^2+10x+1}<\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}<\sqrt{16x^2+10x+2}<4x+2$$
$$2x+1=\sqrt{4x^2+4x+1}<\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}<\sqrt{4x^2+4x+2}<2x+2$$
$$x+1=\sqrt{x^2+2x+1}<A<\sqrt{x^2+2x+2}<x+2$$
Thus $\lfloor A \rfloor=x+1=20062008$.