Solution to $\sqrt{\sqrt{x + 5} + 5} = x$ There are natural numbers $a$, $b$, and $c$ such that the solution to the equation
\begin{equation*}
\sqrt{\sqrt{x + 5} + 5} = x
\end{equation*}
is $\displaystyle{\frac{a + \sqrt{b}}{c}}$. Evaluate $a + b + c$.
I am not sure where I saw this problem. My guess is that it was from a high school math competition. The solution to the equation is $\frac{1 + \sqrt{21}}{2}$. This suggests use of the quadratic formula. 
The solution set to the given equation is a subset of the solution set to
\begin{equation*}
x^{2} - 5 = \sqrt{x + 5} ,
\end{equation*}
\begin{equation*}
x^{4} - 10x^{2} + 25 = x + 5
\end{equation*}
\begin{equation*}
x^{4} - 10x^{2} - x + 20 = 0 .
\end{equation*}
Using the quartic equation (or Wolfram), the solutions to this equation are computed to be
\begin{equation*}
\frac{1 \pm \sqrt{21}}{2} , \qquad  \frac{-1 \pm \sqrt{17}}{2} .
\end{equation*}

*

*You can get the four apparent solutions from $x^{4} - 10x^{2} - x + 20 = (x^2 - x - 5) (x^2 + x - 4)$ and then solving two quadratic equations 

*Squaring can produce spurious results: 


*

*for example if $x=\frac{-1 - \sqrt{17}}{2}$ then $\sqrt{\sqrt{x + 5} + 5} = -x$ rather than $+x$

*but you lost that distinction when you went to ${\sqrt{x + 5} + 5} = x^2$

*similarly $x=\frac{-1 + \sqrt{17}}{2}$ or $x=\frac{1 - \sqrt{21}}{2}$ then $\sqrt{x+5} = -(x^2-5)$ rather than $+(x^2-5)$ but you lost that distinction with the second squaring
When you use transformations which are not $1-1$ then you should check any apparent answers in the original question to see whether they are spurious or not 
