More nested radicals

$\begin{aligned} \sqrt{\sqrt{x+5}+5} & = x & \small \color{#3D99F6} \text{Squaring both sides} \\ \sqrt{x+5} + 5 & = x^2 & \small \color{#3D99F6} \text{Adding both sides with }x \\ \sqrt{x+5} + x + 5 & = x + x^2 \\ \sqrt{x+5} + (\sqrt{x+5})^2 & = x + x^2 & \small \color{#3D99F6} \text{Equating powers on both sides} \\ \implies x & = \sqrt{x+5} & \small \color{#3D99F6} \text{Squaring both sides} \\ x^2 & = x + 5 \\ x^2 - x - 5 & = 0 \\ \implies x & = \frac {1\pm \sqrt{21}}2 \end{aligned}$

Let us check the two likely solutions with $\sqrt{\sqrt{x+5}+5} = x$ .

For $x=\dfrac {1-\sqrt{21}}2$ ,

$\begin{aligned} \sqrt{\sqrt{x+5}+5} & = \sqrt{\sqrt{\frac {1-\sqrt{21}}2+5}+5} = \sqrt{\sqrt{\frac {11-\sqrt{21}}2}+5} = \sqrt{\frac {\sqrt{21}-1}2+5} = \sqrt{\frac {\sqrt{21}+9}2} \ne x \end{aligned}$

Therefore, the solution is not valid.

For $x=\dfrac {1+\sqrt{21}}2$ ,

$\begin{aligned} \sqrt{\sqrt{x+5}+5} & = \sqrt{\sqrt{\frac {1+\sqrt{21}}2+5}+5} = \sqrt{\sqrt{\frac {11+\sqrt{21}}2}+5} = \sqrt{\frac {1+\sqrt{21}}2+5} = \sqrt{\frac {11+\sqrt{21}}2} = \frac {1+\sqrt {21}}2 = x \end{aligned}$

Therefore, the solution is valid.

Therefore, $a+b+c=1+21+2 = \boxed{24}$ .

Let $f(x) = \sqrt{x+5}$ . This equation can be represented as $f(f(x)) = x$ .

We can see that this function is strictly increasing, which means that in order for $f(f(x)) = x$ to be true, we must also have $f(f(x)) = f(x) = x$ .

We can now just solve $f(x) = x$ , or $\sqrt{x+5}=x$ , which results in the solutions $x = \frac{1 \pm \sqrt{21}}{2}$ .

The left side of the equation must be positive since it's a radical, so we discard the extraneous negative answer, giving us that the only solution is $x = \frac{1 + \sqrt{21}}{2}$

Let $\sqrt{x+5} = t$

Then we have two equations:

1. $\sqrt{x+5} = t$

2. $\sqrt{t+5} = x$

Both the equations are satisfied when $t = x$ . The rest follows by solving the quadratic equation.

From $\sqrt{\sqrt{x+5}+5}=x$ we can deduce that $x>\sqrt5$ . We also reason that
$\begin{aligned} \sqrt{x+5} &= x^2-5 \\ 0 &= x^4 - 10x^2 -x +20 \\ &= (x^2 -x - 5)(x^2 +x -4) \\ x&= \frac{1\pm\sqrt{21}}2,\frac{-1\pm\sqrt{17}}2 \end{aligned}$
Only $\frac{1+\sqrt{21}}2>\sqrt5$ , so the answer is $24$ .

By squaring and rearranging terms we get $\sqrt{5+x}=x^2-5$ .

[And by squaring again, you get $x^4-10x^2-x+20=0$ which, according to the main theorem of algebra, can have up to 4 (complex) solutions.]

Let $f(x) = \sqrt{5+x}$ and $g(x) = x^2-5$ . The graphs are part of of the two parabolas $x=y^2-5$ and $y=x^2-5$ which intersect at 4 points in the real plane, once in each quadrant. So, all of the 4 complex roots of the 4th degree polynomial are in fact real, but... when talking about real roots, the root sign is reserved for the nonnegative solution by definition, so the original expression rules out negative values for $x=\sqrt{y+5}$ and for $y=\sqrt{x+5}$ .

If we plug in $x=\sqrt{5+x}$ into the original expression, we end up with the identity $\sqrt{5+x}=\sqrt{5+x}$ , so $x=\sqrt{5+x}$ is a solution. Square both sides to get $x^2=5+x$ which solves to $x=\frac{1+\sqrt{21}}{2}$ (or $x=\frac{1-\sqrt{21}}{2}$ , which we discard). This solution in the first quadrant has a+b+c=1+21+2=24.