+ First square both sides twice with a little adjustment. +
+ +\[\begin{align}& 506+(506+x)^{1\over2}=x^2 \\& (506+x)^{1\over2}=x^2-506 \\& +506+x=x^4-1012x^2+506^2 \\& x^4-1012x^2-x+505\cdot506=0 \end{align}\] + ++ Then we could tediously factor this polynomial to get + \((x-23)(x+22)(x^2+x-505)=0\) which has the solution \(x=23\). We would also + have to check for extraneous solutions and find that none of the others are + solutions. +
+ ++ Alternatively, we can observe that the left side is the function + \(f(x)=\sqrt{506+x}\) iterated to \(f(f(x))=x\). Since \(f(x)\) is + monotonic, if we choose \(x\) to be a non fixed point, it will not return to + \(x\) under iterations of \(f\). To see why, +
+ +\[f(x)<x \Rightarrow f(f(x))<f(x)<x \] + ++ Since \(f(x)\) is monotonically increasing and \(y<z\) implies + \(f(y)<f(z)\). A similar result can be shown for \(f(x)>x\). Thus, + we must have the fixed point \(f(x)=x\) which is easier to solve for than + the quartic equation above resulting from an algebra mess. +
+ +\[\sqrt{506+x}=x \Rightarrow 506+x=x^2 \Rightarrow (x+22)(x-23)=0\] + ++ We find \(x=-22\) is an extraneous solution since a square root cannot be + negative so the answer is \(x=23\). +
+