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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 $\sqrt{x-{\sqrt{x-{\sqrt{x-{\sqrt{x-n}}}}}}}=n$ ?

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    Try to prove that $x=30$ is the only solution. $x=n(n+1)$ in the general case. – 2017-01-13
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    the conditions for $n$ need to be carefully considered. – 2017-01-13
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    @SteveX: What do you mean? For any $n>0$, the function $\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-n}}}}$ is well defined and increasing on $x\geq n$. And obviously $x=n(n+1)$ is a solution. – 2017-01-13
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    @JackD'Aurizio I was speaking to the OP since he/she didn't specify the range of $n$ in the question. – 2017-01-13
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    @JackD'Aurizio how does one come up with the solution $n(n+1) $? – 2017-01-13
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    @RSerrao: an inspired guess. Inspired by the fact that $x=n(n+1)$ is a solution of $\sqrt{x-\sqrt{x-n}}=n$. – 2017-01-13
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    @JackD'Aurizio More simply, the solution to $\sqrt{x-n}=n$. – 2017-01-13
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    @JackD'Aurizio The function is decreasing on $x<5.002$ – 2017-01-13
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    @SimpleArt: you are right, the naive argument has to be fixed. Luckily, it is not too difficult to prove that every solution has to be $\geq(n+1)$, then prove that the given function is increasing on the interval $(n\color{red}{+1},+\infty)$ – 2017-01-13
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    @JackD'Aurizio XD I was graphing, and I noticed that if there are an even amount of radicals, it is decreasing on a very small interval. If there is an odd amount of radicals, it is always increasing. I imagine these are related. – 2017-01-13
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    @JackD'Aurizio And you might want to fix the restriction $n\ge0$ to something like $n\ge0.32$... – 2017-01-13
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    @SimpleArt: you are right again. Can you check if my fixed answer is ok, now? – 2017-01-13
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    Question specifically for $n=5$ is already here: [How to solve this equation:$\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}}=5$](http://math.stackexchange.com/questions/216179/how-to-solve-this-equation-sqrtx-sqrtx-sqrtx-sqrtx-5-5) – 2017-01-15

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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.

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    Looks pretty good man. Nice trick to show intervals it was increasing. (Such pain trying to show it was increasing D-: – 2017-01-13
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    @SimpleArt: the original idea was more elegant, but it did not really work, so one needs to get creative to save his own creatures :D – 2017-01-13
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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 $p(p+1)$.

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    But how to show that is the only solution? (hint, it's not always the only solution) – 2017-01-13
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    @SimpleArt Does it have something to do with $x \ge p^2$ ? – 2017-01-13
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    $x\ge p^2$ is not necessarily the case... imagine $p=0.3$ and $x=0.39$ and $x=0.79$. Both solutions, so it is not unique. – 2017-01-13