Solve an equation with four nested radicals I have to solve the equation $\sqrt{x-{\sqrt{x-{\sqrt{x-{\sqrt{x-5}}}}}}}=5$.
Repeated squaring of both sides of the equation makes it very complex.
Is there any substitution or something similar which can simplify the problem ?
Does the problem have a solution if "generalized" as $\sqrt{x-{\sqrt{x-{\sqrt{x-{\sqrt{x-n}}}}}}}=n$ ?
 A: Let we put together the suggestion given in the comments. We are going to tackle the general case, under the assumption $n\geq 1$. Clearly, $x=\color{red}{n(n+1)}$ is a solution of the given equation, since it is a solution of $\sqrt{x-n}=n$. So it is enough to prove it is the only solution. For simplicity, let
$$ f_1(x)=\sqrt{x-n},\qquad f_2(x)=\sqrt{x-f_1(x)},$$ 
$$f_3(x)=\sqrt{x-f_2(x)},\quad f_4(x)=\sqrt{x-f_3(x)}$$
and $I=(n,+\infty)$. $f_1(x)$ and $f_3(x)$ are increasing and positive functions on $I$, due to the fact that
$$ \sqrt{y-n}-\sqrt{x-n} = \frac{y-x}{\sqrt{y-n}+\sqrt{x-n}} $$
has the same sign of $y-x$. That also implies that $f_3(x)$ is quite close to $\sqrt{x}$.
In particular, we cannot state that $f_4(x)$ is increasing on $I$ (as a matter of facts, it is not) but we may state that $f_4(x)$ is increasing on $J=(n+1,+\infty)$. Since every solution of $f_4(x)=n$ has to be greater than $n+1$, it follows that $x=n(n+1)$ is the only solution, as wanted.
A: For the moment, forget the question and consider this equation. 
$$
\sqrt{x-p}=p
$$
It is easy to see that :
$$
x-p=p^2
\\
\implies x=p(p+1)
$$
Now consider :
$$
x-\sqrt{x-p}=x-p=k
\\
x-\sqrt{x-\sqrt{x-p}}=x-\sqrt{k}=x-\sqrt{x-p}=x-p=p^2
$$
And so on. 
It is easy to see that the solution of $x$ always remains $p(p+1)$. 
